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https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 1 | AVOGADRO EXAM 2010
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
13 MAY 2010 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. Top performers are eligible for a prize. The names of the top 200 students
will be published in the September issue of Chem 13 News.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 11 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 11 Chemistry
in a semestered school
Code 3 Ontario , Grade 11 Chemistry
already completed
Code 4 Any other Ontario student
Code 5 Manitoba or Saskatchewan high school
student
Code 6 Québec high school student
Code 7 not used
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 2 |
2 / AVOGADRO EXAM © 2010 UNIVERSITY OF WATERLOO 1 Compared to an electron, a proton has
A the same charge and about the same mass
B the same charge but a much greater mass
C the opposite charge and much less mass
*D the opposite charge and a much greater mass
E no charge and a much smaller mass
2 Argon has three isotopes with relative atomic masses
of 36.0, 38.0 and 40.0. Given that the relative atomic
mass of naturally occurring argon is 39.95, which of the following statements must be correct?
A
40Ar is less abundant than 38Ar.
*B 40Ar is more abundant than either 36Ar or 38Ar.
C 38Ar is more abundant than 36Ar.
D 36Ar is more abundant than 40Ar.
E Another isotope of lesser mass must exist.
3 An incomplete equation describing the nuclear decay
of boron-9 is given below. How many neutrons or
protons are also produced?
9
5B → 8
4Be + ?
A one neutron
*B one proton
C one neutron and one proton
D two protons
E two neutrons
4 When 50.0 mL of water and 50.0 mL of ethanol are
mixed, the total volume is found to be 96.5 mL. What
is the density of this water-ethanol solution?
A 1.78 g/mL
B 0.895 g/mL
C 0.211 g/mL
D 3.45 mL
*E 0.927 g/mL 5 Which of the following has a linear geometry?
A O
3
B NO 2−
*C C
2H2
D H2S
E F 2O
6 Which of the following elements has properties that
most closely resemble those of calcium, Ca?
A sodium, Na
B potassium, K
*C magnesium, Mg
D bromine, Br
E krypton, Kr
7 What is the formula of lead(II) nitrate?
A Pb
3N2
B Pb 2N3
C Pb 2NO 3
*D Pb(NO 3)2
E PbNO 3
8 Which of the following reacts with moisture in the air
to form acid rain?
*A sulfur trioxide, SO 3
B nitrogen, N
2
C carbon dioxide, CO 2
D methane, CH 4
E ozone, O
3
9 Which of the following is an example of chemical
change?
A boiling water
B dissolving alcohol in water
C heating copper metal
D compressing a gas
*E rusting of iron
AVOGADRO EXAM 2010 - Answers
Densities, in g/mL:
Water, 1.00
Ethanol , 0.789 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 3 |
© 2010 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 3 10 What is [Na+] in a solution obtained by mixing 50.0 mL
of 0.100 mol/L NaNO 3(aq) and 25.0 mL of
0.100 mol/L Na 2CO 3(aq)?
*A 0.133 mol L−1
B 0.200 mol L−1
C 0.300 mol L
−1
D 0.167 mol L−1
E 0.125 mol L−1
11 What is the mass of 0.67 mol Na?
A 29 mg
*B 15 g
C 10 g
D 23 g
E 0.67 g
12 One litre of oxygen gas is compared to one litre of
carbon dioxide gas, both at 25 oC and 100 kPa.
Which statement is correct?
A The density of the oxygen gas is greater than that
of the carbon dioxide gas.
B On average, the kinetic energy of a carbon
dioxide molecule is greater than that of an oxygen
molecule.
C On average, a carbon dioxide molecule moves
faster than does an oxygen molecule.
D On average, the kinetic energy of carbon dioxide
molecule is less than that of an oxygen molecule.
* E The two samples contain the same number of
molecules.
13 What is the net ionic equation for the reaction of
Na
2CO 3(aq) and CaCl 2(aq)?
A Na+(aq) + Cl−(aq) → NaCl(s)
B Na 2CO 3(aq) + CaCl 2(aq)
→ 2 NaCl(aq) + CaCO 3(s)
C Ca+(aq) + CO 3−(aq) → CaCO 3(s)
*D Ca2+(aq) + CO 32−(aq) → CaCO 3(s)
E 2 Na+(aq) + CO 32−(aq) → CO 2(g) + Na 2O(s) 14 In an experiment, 16.0 g SO 2 is treated with 6.0 g O 2
and 18.0 g SO 3 is obtained. A balanced chemical
equation for the reaction is given below.
2 SO 2(g) + O 2(g) → 2 SO 3(g)
What is the percentage yield of SO 3 in this experiment?
A 25%
B 38%
C 67%
D 60%
*E 75%
15 What amount of C
8H10O2N4 contains the same
number of C atoms as 2 mol CO 2?
A 2 mol
B 8 mol
C 4 mol
*D 0.25 mol
E 0.5 mol
16 In which region of the peri odic table would you find the
elements of highest electronegativity?
A top, left
* B top, right
C near the middle
D bottom, left
E bottom, right
17 Which of the following has an odd-number of
electrons?
A NO 3−
*B NO 2
C N
2O
D NO+
E NO 2− Molar masses
(in g/mol)
SO 2 64.1
O2 32.0
SO 3 96.2
Using the molar masses given, the “correct”
answer is E. However, the molar mass given for
SO 3 is wrong. Question 14 was deleted. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 4 |
4 / AVOGADRO EXAM © 2010 UNIVERSITY OF WATERLOO 18 What is the correct electron arrangement for a
scandium (Sc) atom? (The first number in each list refers to the number of electrons in the first shell; the
second number refers to the number of electrons in the
second shell; and so on.)
* A 2, 8, 9, 2
B 2, 8, 2, 8, 1
C 2, 8, 8, 3
D 10, 10, 1
E 4, 4, 4, 4, 1
19 A 10.0 L cylinder containing neon gas with a measured
pressure of 550 kPa at 298 K is connected through a
valve to a 2.50 L cylinder containing 275 kPa of helium gas at 298 K. The valve is opened and the gases mix
with no change in temperature. What is the final total
pressure in the system?
A 277 kPa
B 326 kPa
C 413 kPa
* D 495 kPa
E 599 kPa
20 What is the H-N-H angle in the NH
3 molecule?
Choose the closest value.
A 45o
B 90o
*C 109o
D 120o
E 180o
21 Which of the following molecules has the strongest
carbon-carbon bond?
A ethanol, CH
3CH 2OH
B ethanoic acid, CH 3CO 2H
C ethane, C 2H6
D ethene, C 2H4
*E ethyne, C 2H2
22 Consider the Lewis structure below for the CH 3CCH
molecule.
What is the maximum number of atoms that can lie in
the same plane?
A three
B four
*C five
D six
E seven
23 The following ions all have the same number of
electrons.
O
2−, F−, Na+, Mg2+
In which of following lists are these ions arranged in
order of increasing radius (from smallest to largest)?
A O
2− < F− < Na+ < Mg2+
* B Mg2+ < Na+ < F− < O2−
C Na+ < Mg2+ < O2− < F−
D Mg2+ < Na+ < O2− < F−
E F− < O2− < Na+ < Mg2+
24 Which of the following is not a Bronsted-Lowry
conjugate acid-base pair?
A NH
3 and NH 2−
B OH− and O2−
*C H 3O+ and OH−
D HCl and Cl−
E NH 4+ and NH 3
CC C HH
HH |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 5 |
© 2010 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 5 25 A 0.350 g sample of acid HX requires 25.4 mL of
0.140 mol/L NaOH(aq) for complete reaction. What is the molar mass of the acid?
A 42.3 g/mol
B 68.4 g/mol
* C 98.4 g/mol
D 121 g/mol
E 84.6 g/mol
26 What is the density of carbon dioxide gas at 0.00
oC
and 101.3 kPa?
* A 1.96 g/L
B 0.0446 g/L
C 22.4 g/L
D 44.6 g/L
E 0.509 g/L
27 An element M forms an ion M
3+. The atom M and the
ion M3+ have the same
* A number of protons
B number of electrons
C radius
D ionization energy
E chemical properties
28 Methanoic acid, HCOOH, is a weak electrolyte. In a
solution prepared by dissolving 0.10 mol HCOOH in water to make 1.0 L of solution, approximately 4.1% of
the HCOOH molecules ionize. What is the pH of this
solution?
A 0.61
B 1.39
* C 2.39
D 4.10
E 6.10
29 In March of this year, the International Union of Pure
and Applied Chemistry (IUPAC) officially approved the name and atomic symbol (Cn) for element 112.
What is the official name of element 112?
*A copernicium
B californium
C cupenium
D cernium
E cuternium
30 Element 114 would be placed directly below lead
(element 82). At the present time, nuclear
scientists have managed to synthesize only a few atoms of element 114 at any one time and thus,
the physical appearance of a larger sample is not
yet known. Based on its position in the periodic table, element 114 is most likely to be a
A reddish-brown volatile liquid
B a pale yellow green gas
C a colourless crystal
* D a gray-silvery metal
E a black powdery solid
31 Sodium hydroxide, NaOH, is most likely found in
which household product?
A vinegar
B soap
C bleach
D window cleaner
* E drain cleaner
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 6 |
6 / AVOGADRO EXAM © 2010 UNIVERSITY OF WATERLOO 32 A balloon was filled with helium gas to a volume of
3.0 L on a day when the atmospheric pressure was 101 kPa and the temperature was 31
oC. The
following day, the atmospheric pressure and temperature were measured as 98.3 kPa and 33
oC,
respectively. The volume of the balloon had not changed. Which of the following statements is consistent with these data?
A Based on the changes in pressure and
temperature that occurred, the volume of the
balloon would not be expected to change.
B The balloon absorbed some air from the
atmosphere.
* C Some helium gas leaked out of the balloon.
D Helium atoms in the balloon lost energy to the
surroundings.
E Based on the changes in pressure and
temperature that occurred, the volume of the
balloon should have decreased.
33 To prepare exactly 250 mL of 0.10 mol/L HCl(aq)
starting from 1.0 L of 0. 20 mol/L HCl(aq), one should
A slowly add exactly 125 mL of 0.20 mol/L HCl(aq)
to exactly 125 mL of water.
*B slowly add exactly 125 mL of 0.20 mol/L
HCl(aq) to about 100 mL of water and then
dilute with water to a total volume of 250 mL.
C evaporate 750 mL of water from 1.0 L of 0.20
mol/L HCl(aq).
D slowly add exactly 125 mL of water to exactly
125 mL of 0.20 mol/L HCl(aq).
E add 750 mL of 0.10 mol/L NaOH to 1.0 L of
0.20 mol/L HCl(aq).
34 Which of the following dilute solutions would allow a
chemist to distinguish between dilute solutions of
NaCl(aq) and NaNO 3(aq)?
A NaOH(aq)
B HCl(aq)
C NH 3(aq)
D H 2SO 4(aq)
*E AgNO 3(aq) 35 Compared to a chlorine atom, a sodium atom has a
larger
* A radius
B mass
C number of electrons
D ionization energy
E electronegativity
36 Which of the following bonds has the greatest ionic
character?
A C−H
B O−H
C O−F
* D H−F
E C−O
37 A compound is found to be 85.62% carbon by mass
and 14.38% hydrogen. What is the simplest formula
of this compound?
A CH
* B CH
2
C CH 3
D CH 4
E C3H4
38 Mercury(II) sulfide, HgS, is practically insoluble in pure
water. Its solubility at 25 oC is probably no more than
3×10−25 g/L. Of the following quantities of pure water,
which is the smallest quantity that could be used to
make a saturated solution of HgS?
A 20,000 L
B 1000 L
C 10,000 L
* D 2000 L
E 200 L
Molar masses
(in g/mol)
Hg 200.6
S 32.07 For question 33, the intended answer was
“B”, but the volume of water was mistakenly given as 200 mL when 100 mL is what was
intended. Question 33 was deleted. Answer
“A” is not the correct answer because 125 mL of HCl(aq) and 125 mL of water may not
give exactly 250 mL of solution because
volumes are not exactly additive. See
question 4 for an extreme example. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 7 |
© 2010 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 7 39 What is the pH of a solution prepared by mixing
50.0 mL of 0.010 mol/L HCl(aq) and 50.0 mL of
0.010 mol/L Ca(OH) 2(aq)? Assume the temperature
is 25 oC.
A 2.00
B 2.30
C 7.00
*D 11.70
E 12.00
40 Consider the Lewis structure below. What is the
charge on this molecule or ion?
A −2
* B −1
C 0
D +1
E +2
FB r FKw = 1.0×10−14 at 25 oC |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2010.pdf | 8 |
8 / AVOGADRO EXAM © 2010 UNIVERSITY OF WATERLOO
DATA SHEET
AVOGADRO EXAM 2010
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 1023 mol1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K1 mol1 0oC = 273.15 K
= 8.3145 kPa L K1 mol1
= 8.3145 J K1 mol1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) 24
2bb a cx
a
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
9
8B
10
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57-71
La-Lu 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89-103
Ac-Lr 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Ds 111
Sg 112
Cn
57
La
138.9 58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
89
Ac
227. 90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2009.pdf | 1 | AVOGADRO EXAM 2009
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
21 MAY 2009 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. Top performers are eligible for a prize. The names of the top 200 students
will be published in the September issue of Chem 13 News.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 11 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 11 Chemistry
in a semestered school
Code 3 Ontario , Grade 11 Chemistry
already completed
Code 4 Any other Ontario student
Code 5 Manitoba or Saskatchewan high school
student
Code 6 Québec high school student
Code 7 not used
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2009.pdf | 2 |
2 / AVOGADRO EXAM © 2009 UNIVERSITY OF WATERLOO 1 The “lead” of a pencil is mostly
A lead, Pb
*B carbon, C
C silicon dioxide, SiO 2
D silicon, Si
E calcium carbonate, CaCO 3
2 How many protons, neutrons and electrons are there in
a single atom of
209
84Po?
A 84 protons, 84 neutrons, 209 electrons
B 84 protons, 209 neutrons, 84 electrons
C 209 protons, 125 neutrons, 209 electrons
D 125 protons, 84 neutrons, 125 electrons
*E 84 protons, 125 neutrons, 84 electrons
3 The mass of one atom of 12C is exactly 12 atomic mass
units. With the assumption that a proton and a neutron are equally massive, what is the total number of protons
and neutrons in the body of a 75-kg person? (You may
neglect the mass of an electron is negligible compared to that of a proton or neutron.)
A 2.2 × 1027
*B 4.5 × 1028
C 8.0 × 1021
D 3.8 × 1023
E 8.0 × 1024
4 Mercury, Hg( l), has a density of 13.6 g mL−1 at 25 °C.
What is the volume of 4.25 grams of Hg( l) at 25 oC?
A 0.0173 mL
B 3.20 mL
C 0.0562 mL
*D 0.313 mL
E 0.0735 mL 5 Which of the following molecules has the same number
of electrons as a water molecule?
*A HF
B BH 3
C CO
D H2S
E F 2
6 Which of the following elements is a liquid at room
temperature and atmospheric pressure?
A chlorine
B phosphorus
C sulfur
*D bromine
E iodine
7 What is the formula of the binary compound formed
between Mg and P?
A MgP
B Mg 2P
C MgP 2
D Mg 2P3
*E Mg 3P2
8 Which of the following elements has no known stable
compounds?
*A neon, Ne
B xenon, Xe
C gold, Au
D platinum, Pt
E uranium, U
9 Which of the following elements is believed to be the
most abundant in the earth’s crust?
A hydrogen
*B oxygen
C carbon
D nitrogen
E silicon AVOGADRO EXAM 2009 - Answers |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2009.pdf | 3 |
© 2009 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 3 10 Which of the following has the highest concentration at
equilibrium when one mole of HCl is dissolved in 1.0 L of water at 25
oC?
*A Cl−
B Cl+
C Cl 2
D H 2
E HCl
11 What is the symbol for the atom or ion that results from
the addition of two protons to a single atom of 42
20Ca ?
A 42 2+
22Ca
B 44 2+
22Ca
C 42
22Ti
*D 44 2+
22Ti
E 44 2+
20Ti
12 In a mixture of N 2 and O 2 gases, all the N 2 molecules
and the O 2 molecules have the same
A average speed
* B average kinetic energy
C partial pressure
D average molecular mass
E average momentum
13 When ethanol, CH 3CH 2OH, is burned in excess oxygen,
carbon dioxide and water are the only products. What
is the coefficient of O 2 when the chemical equation
representing the combustion reaction is balanced using
the smallest whole number coefficients ?
A 1
B 2
*C 3
D
7
E none of the above 14 In an experiment, 16 g of methane and 32 g of oxygen
react to produce 11 g of carbon dioxide. A balanced chemical equation for the reaction is given below.
CH
4(g) + 2 O 2(g) → CO 2(g) + 2 H 2O(g)
What is the percentage yield of carbon dioxide in this
experiment?
A 10%
B 25%
*C 50%
D 67%
E 75%
15 If an oxide of nitrogen contains 25.9% by mass of
nitrogen, what is its empirical formula?
A NO
B N 2O
C
NO 2
D N 2O4
*E N 2O5
16 What is the percentage by mass of sodium in a mixture
containing 1.00 mol NaCl and 1.00 mol NaF?
A 39.3%
* B 45.8%
C 47.1%
D 50.0%
E 54.8%
17 When the hydrides of the group 16 elements are
arranged in order of increasing boiling point, the order is
*
A H 2S H 2Se H 2Te H 2O
B H 2O H 2S H 2Se H 2Te
C H2Te H 2Se H 2S H 2O
D H 2O H 2Te H 2Se H 2S
E H 2S H 2O H 2Se H 2Te |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2009.pdf | 4 |
4 / AVOGADRO EXAM © 2009 UNIVERSITY OF WATERLOO 18 How many unpaired electrons are there in a ground
state Mn2+ ion?
A zero
B one
C two
D three
* E more than three
19 What is the pressure (in mmHg) of the gas inside the
apparatus below if P atm = 750 mmHg, Δh1 = 40 mm
and Δh2 = 30 mm?
A 710 mmHg
* B 790 mmHg
C 720 mmHg
D 780 mmHg
E 820 mmHg
20
What is the HCH bond angle in a formaldehyde (H 2CO)
molecule? Choose the closest value.
A 45o
B 90o
C 109o
*D 120o
E 180o
21 Which of the following diatomic molecules has the
strongest bond?
*A N 2
B O 2
C F 2
D
Cl 2
E
Br 2 22 Which of the following molecules or ions is planar? (The
central atom is underlined and all other atoms are bonded to it.)
A NH3
B NH4+
C SF4
D SO32−
*E SO3
23 What is the formula of iron(II) sulfate?
A Fe 2S
B FeS 2
*C FeSO 4
D FeSO 3
E
Fe 2(SO 4)3
24
The pH of lemon juice is about 2.3. What is [H+] in
lemon juice?
A 0.36 mol L−1
B 0.83 mol L−1
C 0.10 mol L-1
*D 5.0×10−3 mol L−1
E 0.071 mol L−1
25 Solid aluminum dissolves in hydrochloric acid solution
according to the following chemical equation.
2 Al( s) + 6 HCl( aq) → 2 AlCl 3(aq) + 3 H 2(g)
A reaction mixture contains 0.500 mol HCl and
0.400 mol Al. Assuming the reaction goes to completion, how many moles of the excess reactant remain?
A 0.000 mol
B 0.100 mol
C 0.167 mol
* D 0.233 mol
E 0.400 mol Patm
Δh1
mercury (Hg) Δh2 Gas |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2009.pdf | 5 |
© 2009 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 5 26 What volume does 11 kg of carbon dioxide occupy
at 0 oC and 101.3 kPa?
A 246 m3
*
B 5.6 × 103 L
C 11 L
D 0.25 L
E 0.22 m3
27 What is the ground state electron configuration of an
isolated sulfur (S) atom?
A 1s2 2s2 2p2 3s2 3p2 4s2 3d2 4p2
B 1s2 2s2 2p6 3s1 3p3 3d5
*C 1s2 2s2 2p6 3s2 3p4
D 1s2 2s2 2p6 3s2 3p6
E
1s2 2s2 2p6 3s2 3d6
28 What volume of 0.123 mol/L aqueous H 2SO 4 is needed
to neutralize 40.0 mL of 0.175 mol/L aqueous NaOH? A balanced chemical equation for the reaction is given
below.
H
2SO 4(aq) + 2 NaOH( aq) → Na 2SO 4(aq) + 2 H 2O(l)
*A 28.5 mL
B 56.9 mL
C 114 mL
D 80.0 mL
E 40.0 mL
29 Three successive elements, in order of increasing
atomic number, have these first ionization energies:
1680 2080 494 kJ/mol
Which of following sets represents the three elements?
A N O F
B O F N
C Ne Na Mg
*D F Ne Na
E Na Mg Al 30 Which of the following gases does not burn, does
not support combustion, and has no effect on lime water, Ca(OH)
2(aq)?
A hydrogen, H 2
B oxygen, O 2
C carbon monoxide, CO
* D nitrogen, N 2
E carbon dioxide, CO 2
31 Which of the following elements would you expect to be
the most similar in chemical properties to element 20?
A element 19
B element 21
C element 18
D element 4
*
E element 38
32 A weather balloon filled with helium gas, He( g), has a
volume of 2.00 ×103 m3 at ground level where the
atmospheric pressure is 1.000 atm and the temperature is 27
oC. After the balloon rises high above the earth to
a point where the atmospheric pressure is 0.400 atm, its volume increases to 4.00
×103 m3. What is the
temperature of the atmosphere at this altitude?
*A −33 oC
B −22 oC
C −73 oC
D 22 oC
E 240 oC
33 In which of these compounds is the oxidation state of O
the highest (i.e., the most positive)?
*
A F2O
B O2
C O3
D H 2O2
E H 2SO 4 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2009.pdf | 6 |
6 / AVOGADRO EXAM © 2009 UNIVERSITY OF WATERLOO 34 The molar volumes of C 2H6(g) and H 2(g), measured at
300 K and 10.0 atm, are 2.30 L and 2.51 L, respectively. Which of the following statements accounts for the
observation that the molar volume of C
2H6(g) is smaller
than that of H 2(g)?
A C 2H6 molecules are larger than H 2 molecules.
B The intermolecular attractions in C 2H6(g) are
weaker than they are in H 2(g).
*C The intermolecular attractions in C 2H6(g) are
stronger than they are in H 2(g).
D The average kinetic energy of H 2 molecules is
greater than that of C 2H6 molecules.
E The average kinetic energy of H 2 molecules is less
than that of C 2H6 molecules.
35 When aqueous sodium carbonate, Na 2CO 3, is treated
with dilute hydrochloric acid, HCl, the products are sodium chloride, water and carbon dioxide gas. What
is the
net ionic equation for this reaction?
A Na 2CO 3(aq) + 2 HCl( aq)
→ 2 NaCl( aq) + CO 2(g) + H 2O(l)
B CO 32−(aq) + 2 HCl( aq)
→ H2O(l) + CO 2(g) + 2 Cl−(aq)
*C CO 32−(aq) + 2 H+(aq) → H 2O(l) + CO 2(g)
D Na 2CO 3(s) + 2 H+(aq)
→ 2 Na+(aq) + CO 2(g) + H 2O(l)
E H+(aq) + OH−(aq) → H2O(l)
36 Which of the following is the best Lewis structure (i.e.,
the best electron dot structure) for the N 2O molecule?
A
B
C
*
D
E
37 A 2.4917-g sample of a hydrate of cobalt (II)
fluoride, ⋅x2 2CoF H O , was heated to drive off all of the
water of hydration. The remaining solid weighed
1.4290 g. What is the formula of the hydrate?
A ⋅2 2CoF H O
B ⋅2 2CoF 2H O
C ⋅2 2CoF 3H O
* D ⋅2 2CoF 4H O
E ⋅2 2CoF 5H O
38 How many isomers are there for C 4H8? Consider both
structural (i.e. constitutional) isomers and
stereoisomers.
A one
B two
C three
D four
* E more than four
39 Which of the following combinations reagents react to
form an insoluble precipitate?
A HNO 3(aq) and Ca(OH) 2(aq)
B Zn( s) and HCl( aq)
C Zn( s) and Cu(NO 3)2(aq)
D NaHCO 3(aq) and NaOH( aq)
*E Na 2CO 3(aq) and CaCl 2(aq)
40 Which of the following will occur if a 0.10 mol L−1
solution of acetic acid (CH 3COOH) is diluted to
0.010 mol L−1 at constant temperature?
A the pH will decrease
B the dissociation constant of CH 3COOH will increase
C the dissociation constant of CH 3COOH will decrease
D the hydrogen ion concentration will decrease to
0.010 mol L−1
*
E the percentage ionization of CH 3COOH will increase NN O
NN O
NN O
NN O
NNO |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2009.pdf | 7 |
© 2009 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 7
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
←
9
8B
10
→
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57
La
138.9 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89
Ac
227.0 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Uun 111
Uuu 112
Uub 113
Uut
58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) DATA SHEET
AVOGADRO EXAM 2009
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 × 1023 mol−1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K−1 mol−1 0oC = 273.15 K
= 8.3145 kPa L K−1 mol−1
= 8.3145 J K−1 mol−1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) −± −=24
2bb a cx
a |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 1 | AVOGADRO EXAM 2008
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
15 MAY 2008 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. Top performers are eligible for a prize. The names of the top 200 students
will be published in the September issue of Chem 13 News.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 11 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 11 Chemistry
in a semestered school
Code 3 Ontario , Grade 11 Chemistry
already completed
Code 4 Any other Ontario student
Code 5 Manitoba or Saskatchewan high school
student
Code 6 Québec high school student
Code 7 not used
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 2 |
2 / AVOGADRO EXAM © 2008 UNIVERSITY OF WATERLOO 1 Which of the following elements is not a metal?
*A Se
B Sn
C Sr
D Sc
E Cs
2 A colourless, odourless gas is thought to be oxygen.
Which of the following experimental results would support this conclusion?
A Burning the gas in air produces only water.
B The gas extinguishes a flame.
C The gas turns a Ca(OH)
2 solution milky.
*D A glowing piece of wood bursts into flames in the
gas.
E The gas tarnishes silver.
3 Which of the following particles is the most massive?
*A α-particle
B β-particle
C electron
D proton
E neutron
4 What volume of 5.0 mol L
−1 H2SO 4(aq) must be diluted
with water to make 1.00 L of 0.45 mol L−1 H2SO 4(aq)?
*A 0.090 L
B 0.44 L
C 0.090 mL
D 0.045 L
E 2.22 mL
5 How many neutrons are there in the nucleus of
131I?
A 44
B 53
*C 78
D 131
E 184
6 Which group of elements contains no metals or
metalloids?
A group 13
B group 14
C group 15
D group 16
*E group 17
7 Which of these chloride salts is least likely to exist?
A NaCl
B CuCl
C CaCl 2
D FeCl 3
*E MgCl
8 When a sample of atomic hydrogen gas is heated, it
emits violet, blue, green and red light. Which of the
following statements best explains this observation?
*A The energy of the electron in a hydrogen atom is
restricted to certain values.
B The energy of the electron in a hydrogen atom is
not restricted in any way.
C The electron in a hydrogen atom is restricted to one
of only four possible circular orbits.
D The distance between the electron and the nucleus
in a hydrogen atom is restricted to certain values.
E none of the above
AVOGADRO EXAM 2008 - Answers |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 3 |
© 2008 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 3 9 Which of the following is not a mixture?
A seawater
*B table sugar
C brass
D cement
E smoke
10 Radioactive 131I is used to treat thyroid cancer. An
incomplete chemical equati on for the radioactive decay
of 131I is given below.
131I → ? + −0
1e
What is the missing product in the equation above?
A 130I
B 129I
*C 131Xe
D
131Te
E 131I+
11 Which of the following has the highest concentration in
air at STP?
A He
B H 2O
C CO 2
*D N 2
E O 2
12 The average mass of a solid copper penny is 2.63 g.
What is the mass of one mole of pennies?
* A 1.58×1024 g
B 6.02×1023 g
C 6.36×1023 g
D 63.6 g
E 1.58×1023 g
13 What is the sum of the coefficients when the following
equation is balanced using the smallest whole number coefficients?
__ P
4 + __ Cl 2 → __ PCl 3
A 12
*B 11
C 6
D 5
E 3
14
How many litres of gaseous methane (CH 4) must be
burned in oxygen to produce enough H 2O and CO 2 to
fill a 3.0-L balloon? Assume that H 2O and CO 2 are the
only combustion products and that the temperature and pressure remain constant.
*A 1.0 L
B 1.5 L
C 2.0 L
D 2.5 L
E 3.0 L
15 A compound that contains only Fe and O is 69.9% Fe
by mass. What is the empirical formula of this
compound?
A FeO
B FeO 2
*C Fe 2O3
D Fe 2O
E Fe 3O4
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 4 |
4 / AVOGADRO EXAM © 2008 UNIVERSITY OF WATERLOO 16 If 17.0 grams of sodium chloride are dissolved in water
to make 0.5 L of solution, then what is the final concentration of the solution? Give your answer with
the correct number of significant figures.
*
A 0.6 mol L−1
B 0.58 mol L−1
C 0.581 mol L−1
D 0.3 mol L−1
E 0.291 mol L−1
17 What is the effect of adding a catalyst to a reaction
mixture?
A It increases the equilibrium concentrations of the
products.
B It decreases the enthalpy change of the reaction.
* C It reduces the activation energy of the reaction.
D It increases the value of the equilibrium constant for
the reaction.
E It increases the time it takes for the reaction to
reach equilibrium.
18 How many valence electrons are there in one Al3+ ion?
A 2
B 4
C 6
*
D 8
E 10
19 What volume of He( g) contains the same number of
moles of gas as 1.00 L of N 2(g) at the same
temperature and pressure?
A 7.00 L
* B 1.00 L
C 0.143 L
D 35.7 mL
E 4.00 L 20 What is the HNH bond angle in an ammonia (NH 3)
molecule? Choose the closest value.
A 90o
B 45o
C 120o
*D 109o
E 180o
21 Which of the following types of radiation has the lowest
energy per photon?
*A radio waves
B ultraviolet radiation
C infrared radiation
D
x-rays
E purple laser light
22 An incomplete Lewis structure (i.e. electron dot
structure) for the O 3 molecule is given below.
How many lone pairs of electrons are there in the
completed structure?
A two
B four
C five
*D six
E eight
23 Which of the following is not a common oxide of
nitrogen?
A NO
B NO 2
C N 2O4
D N 2O
*E NO 3
O O O |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 5 |
© 2008 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 5 24 In an experiment, 0.12 L of 0.10 mol L−1 H2SO 4(aq) and
0.20 L of 0.10 mol L−1 NaOH( aq) are combined. Which
of the following statements is true?
* A The pH of the resulting solution is less than 7.
B The pH of the resulting solution is greater than 7.
C The pH of the resulting solution is close to 7.
D The pH of the resulting solution is exactly 7.
E None of the statements above are true.
25 Solid aluminum dissolves in hydrochloric acid solution
according to the following chemical equation.
2 Al( s) + 6 HCl( aq) → 2 AlCl 3(aq) + 3 H 2(g)
How many moles of H 2 are produced if 17.5 moles of Al
are added to a solution containing 24.8 moles of HCl?
A 26.3 mol
* B 12.4 mol
C 7.30 mol
D 17.5 mol
E 24.8 mol
26 Which of the following choices does not involve a
chemical change?
A evaporation and neutralization
B neutralization and sublimation
C oxidation and sublimation
* D evaporation and sublimation
E neutralization and oxidation
27
Which of the following atoms or ions has the electron
configuration 1s2 2s2 2p6 3s1 in its ground electronic
state?
A Na−
*B Mg+
C K
D Ca+
E
Al3+
28 Which of the following is a brittle solid and an electrical
insulator at room temperature, but an excellent electrical conductor in its liquid form?
A sulphur
* B sodium chloride
C aluminum
D mercury
E carbon
29 Which of the following salts produces a basic solution
when it is dissolved in water?
A KCl
B NH 4Cl
*C K 2CO 3
D NaNO 3
E CuBr 2
30 Which of the following de scribes the pr ocess that
produces Fe( s) from Fe 2O3(s)?
A combustion
B precipitation
C hydrolysis
* D reduction
E oxidation
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 6 |
6 / AVOGADRO EXAM © 2008 UNIVERSITY OF WATERLOO 31 Which one of the following solutions will be the worst
electrical conductor at 25oC?
A 0.10 mol L−1 Na 2SO 4 (aq)
B 0.10 mol L−1 NaCl( aq)
C 0.10 mol L−1 CaSO 4(aq)
* D 0.10 mol L−1 CH 3OH(aq)
E 0.10 mol L−1 CsCl( aq)
32 Which of the following atoms is not present in large
numbers in biological molecules?
A C
* B F
C O
D N
E H
33
In which of these compounds is the oxidation state of Cl
the highest?
A HClO 2
B ClO 2
C Cl2O5
D Cl 2O
* E HClO 4
34 Which of the gases most closely resembles an ideal
gas at standard temperature and pressure?
A CO 2
B NH 3
C HI
*D H 2
E H 2O
35 Which of the following have ground state electron
configurations of the type ns2 np2 ?
A group 2 atoms
B group 4 atoms
C group 6 atoms
*D group 14 atoms
E group 16 atoms
36 Which of the species in the reaction below are
Brønsted-Lowry acids?
−
4HSO + −
3HCO U −2
4SO + H 2CO 3
A −
4HSO and −
3HCO
*
B −
4HSO and H 2CO 3
C −
3HCO and −2
4SO
D −2
4SO and H 2CO 3
*
E −
4HSO and −2
4SO
37 Which of the following is not an alkane?
*
A C 2H4
B C 3H8
C C 4H10
D
C 5H12
E C 6H14
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 7 |
© 2008 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 7 38 What happens when a solution of lithium chloride (LiCl)
and a solution of ammonium nitrate (NH 4NO 3) are
mixed?
A A precipitate forms.
B A new salt is formed.
C A gas is evolved.
D A metal is formed.
* E No reaction occurs.
39 An average person expends approximately 100 kJ to
walk 1 km. How far will the average car travel by the
time it expends the same amount of energy (i.e. 100 kJ)
as a person who walked 1 km? Use the data given below to determine the answer. Choose the closest
answer.
A 2 km
B 0.2 km
* C 0.02 km
D 20 km
E 200 km
40 How many structural isomers are there for C 5H12?
A less than three
* B three
C four
D five
E more than five
Fuel consumption
of an average car, 8 km L−1
Heat of combustion of
gasoline, 50 kJ g−1
Density of gasoline, 0.7 g mL−1 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2008.pdf | 8 |
8 / AVOGADRO EXAM © 2008 UNIVERSITY OF WATERLOO
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
←
9
8B
10
→
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57
La
138.9 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89
Ac
227.0 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Uun 111
Uuu 112
Uub 113
Uut
58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) DATA SHEET
AVOGADRO EXAM 2008
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 × 1023 mol−1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K−1 mol−1 0oC = 273.15 K
= 8.3145 kPa L K−1 mol−1
= 8.3145 J K−1 mol−1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) −± −=24
2bb a cx
a |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2007.pdf | 1 | AVOGADRO EXAM 2007
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
17 MAY 2007 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. Top performers are eligible for a prize.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 11 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 11 Chemistry
in a semestered school
Code 3 Ontario , Grade 11 Chemistry
already completed
Code 4 Any other Ontario student
Code 5 Manitoba or Saskatchewan high school
student
Code 6 Québec high school student
Code 7 not used
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
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2 / AVOGADRO EXAM © 2007 UNIVERSITY OF WATERLOO 1 Which atom has the most neutrons?
A
F18
9
*B O18
8
C C14
6
D N15
7
E B11
5
2
Which of the following pairs of atomic symbols and
elements is incorrect?
A Fe, iron
B Mg, magnesium
C Ca, calcium
*D Br, boron
E Mn, manganese
3 Which of the following particles is not a charged
particle?
A α-particle
B β-particle
C electron
D proton
*E neutron
4 The formula of a compound is X 2O. Which of the
following is X least likely to be?
*A barium (Ba)
B sodium (Na)
C cesium (Cs)
D hydrogen (H)
E copper (Cu)
5 How many protons are there in the nucleus of I127
53?
A 7
*B 53
C 74
D 127
E 180
6 Which group of elements has the greatest electron
affinity?
A group 14
B group 15
C group 16
*D group 17
E group 18
7 The difference between deuterium, H2
1, and the more
common form hydrogen is that deuterium
A does not occur naturally.
B is radioactive.
C has one more atom per molecule.
D has one more proton in the nucleus.
*E has one more neutron in the nucleus.
8 Which group of atoms and ions contain the same
number of electrons?
A F, Ne, Na
B O2-, S2-, Se2-
C Mg, Al, Si
D Ca2+, Fe3+, Zn2+
*E Cl-, Ar, K+
AVOGADRO EXAM 2007 - Answers |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2007.pdf | 3 |
© 2007 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 3 9 Which of the following is an ionic solid?
A N 2O
B HCl
*C LiCl
D CO 2
E CH 4
10
What volume of CO 2 is produced when you burn
exactly 1.0 litre of gaseous propane (C 3H8) in the
presence of excess oxygen in your backyard
barbecue? Assume H 2O and CO 2 are the only
combustion products and that the pressure and
temperature remain constant.
A 1.0
B 1.5
C 2.0
D 2.5
*E 3.0
11 Polonium-210 (210Po) is radioactive, extremely toxic,
and it decays according to the chemical equation below. What is the missing product in the equation?
210Po → + He4
2
A 214Po
B 212Tl
*C 206Pb
D 214Rn
E 210Po
12
The bubbles in boiling water are mostly
A He
*B H 2O
C CO 2
D N 2
E O 2 13 An element, X, from group 1 of the periodic table,
combines to form a stable compound with an element, Y, from group 16. The formula of that
compound is most likely to be
A X 3Y
B XY 3
C XY
*D X 2Y
E XY 2
14 After a large meal the pH of your stomach drops to
1.78. What is [H+] in your stomach after the meal?
*
A 1.66 x 10-2 mol L-1
B 0.250 mol L-1
C 1.78 mol L-1
D 1.83 x 10-3 mol L-1
E 6.03 x 10-2 mol L-1
15 The chemical formula of barium perrhenate is
Ba(ReO 4)2. What is the charge on the perrhenate ion?
A +2
B +1
C 0
*D -1
E -2
16
These three compounds have been isolated: NaCl,
Na2O, and AlCl 3. What is the formula of aluminum
oxide?
A Al 2O
* B Al 2O3
C Al 3O
D AlO
E AlO 3
? |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2007.pdf | 4 |
4 / AVOGADRO EXAM © 2007 UNIVERSITY OF WATERLOO 17 The average car in Canada uses 0.93 L of gasoline to
go 100 km. If it is assumed that gasoline is pure octane (C
8H18), with a density of 0.70 g/mL and a
molar mass of 114.2 g/mol, then how many moles of
octane are consumed by driving 100 km?
A 0.93 mol
*
B 5.7 mol
C 11 mol
D 5.7 x 10-4 mol
E 1.1 x 10-3 mol
18 How many moles of gas are present in a 15.0-L
scuba tank, if the pressure in the tank is 23.0 MPa
and the temperature is 298 K? Assume the gas behaves ideally.
A 23 mol
B 72 mol
C 44 mol
D 14.1 mol
* E 139 mol
19 Chlorine has two abundant stable isotopes, 35Cl and
37Cl, with atomic masses of 34.97 amu and 36.96 amu
respectively. What is the percent abundance of the
heavier isotope?
A 78%
* B 24%
C 64%
D 50%
E 36%
20 Which of the following is not a gas at 298 K?
A Ar
B He
*C Br 2
D H 2
E O 2 21 Which of the following types of radiation has the
highest energy per photon?
A radio waves
B ultraviolet radiation
C
infrared radiation
*D x-rays
E
purple laser light
22 The Lewis structure (i.e. electron dot) structure for the
HCN molecule is given below.
The bond angle is nearest to
A 60o
B 90o
C 105o
D 120o
*E 180o
23 What volume of 0.100 mol L-1 NaOH( aq) is required to
neutralize 0.245 L of 0.200 mol L-1 H3PO 4(aq)?
A 0.490 L
B 0.500 L
*C 1.47 L
D
2.30 L
E 1.47 mL
24 Which of the following molecules forms hydrogen
bonds amongst themselves?
A dimethyl ether (CH 3OCH 3)
B methane (CH 4)
C hydrogen sulfide (H 2S)
*D ethanol (CH 3CH 2OH)
E formaldehyde (H 2CO)
H C N
1 MPa = 1 ×103 kPa |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2007.pdf | 5 |
© 2007 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 5 25 Aluminum dissolves in acidic solution according to the
chemical equation below.
2 Al( s) + 6 HCl( aq) → 2 AlCl 3(aq) + 3 H 2(g)
How many grams of aluminum (27 g mol-1) are
required to produce 0.50 mol H 2?
A 20 g
*
B 9.0 g
C 14 g
D 27 g
E 0.24 g
26 For which of the following reactions is the change in
energy equal to the first ionization energy of oxygen?
A O-(g) + e- → O2-(g)
B O(g) + 2e- → O2-(g)
* C O(g) → O+(g) + e-
D O(g) + e- → O-(g)
E O( g) → O2+(g) + 2e-
27
How does the pH of a solution change as HCl is
added to a solution of NaOH?
* A The pH decreases and may go below 7.
B The pH will not change.
C The pH decreases until it reaches a value of 7
and then stops.
D The pH increases until it reaches a value of 7 and
then stops.
E The pH increases and may go above 7.
28
The volume of a gas, initially at 1 atm and 20oC, is
increased from 40.0 mL to 80.0 mL. If the pressure remains constant, what is the final temperature of
the gas?
A
40.080.0 K 293+
B ×80.020 C 40.0D
* C
40.080.0 K 293×
D
80.040.0 K 293×
E ×40.020 C 80.0D
29
Which drawing shows a pipet correctly filled for
delivery?
A 1
*B 2
C 3
D 4
E none of the above
30 What is the mass percentage of copper in CuCl 2 ?
A 12.1%
B 64.2%
C 91.2%
D 25.2%
*E 47.3% 1 2 3 4 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2007.pdf | 6 |
6 / AVOGADRO EXAM © 2007 UNIVERSITY OF WATERLOO 31 Which one of the following solutions will be the best
electrical conductor at 25oC?
*A 0.10 mol L−1 Na 2SO 4(aq)
B 0.10 mol L−1 NaCl( aq)
C 0.10 mol L−1 H2SO 4(aq)
D 0.10 mol L−1 HNO 3(aq)
E 0.10 mol L−1 CsCl( aq)
32 What is the coefficient of O 2 when the following
equation is balanced with the smallest whole-number
coefficients?
__Cr 2O3 + ___ KOH + ___O 2 → __K 2CrO 4 + __H 2O
A 2
* B 3
C 4
D 5
E 6
33 What is the oxidation state of N in HNO 2?
A +5
* B +3
C +1
D −1
E −3
34 If the Kelvin temperature of a sample of ideal gas doubles (e.g. from 200 K to 400 K), then the average kinetic energy of the molecules in the sample
A increases by a factor of 2
B decreases by a factor of 2
*C increases by a factor of 2
D increases by a factor of 4
E remains the same
35 The ground state electronic configuration of a certain neutral atom is [Xe] 6s
2 4f14 5d10 6p4. To which group
of the periodic table does this atom belong?
A group 1
B group 3
C group 6
D group 14
*E group 16
36 How many moles of water are there in 1.80 L of
H2O(l) at 1.00 atm and 298 K? The density of water
is 1.00 g/mL at 1.00 atm and 298 K.
A 1.00 mol
B 0.0736 mol
C 55.6 mol
*D 1.00 x 102 mol
E 13.6 mol
37 The reaction 2 Al(s) + 6 HCl(aq) → 2 AlCl 3(aq) + 3 H2(g)
is an example of
A a precipitation reaction
B an acid-base reaction
C a decomposition reaction
*D an oxidation-reduction reaction
E an isomerization reaction
38 If equal volumes of 0.10 mol L-1 solutions of NaOH
and HCl are mixed, what is the pH of the resulting
solution at 298 K?
A 1
B 13
* C 7
D 1.3
E 12.7
This question was NOT marked.
The electron configuration was
mistakenly given as:
[Xe] 6s2 5f14 6d10 6p4 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2007.pdf | 7 |
© 2007 UNIVERSITY OF WATERLOO AVOGADRO EXAM / 7 39 A calcium chloride solution was prepared by
dissolving 11.00 g CaCl 2 in water to make 500 mL of
solution. What is the correct way to report the
concentration of this solution?
*
A 0.2 mol L-1
B 0.1982 mol L-1
C 0.198 mol L-1
D 0.2000 mol L-1
E 0.20 mol L-1
40 A compound of carbon and hydrogen is found to be
85.6 % carbon, by mass, and 14.38% hydrogen. What
is the simplest formula of the compound?
A CH
*
B CH 2
C CH 3
D CH 4
E C 3H4
CaCl 2, 110.98 g mol-1
The number of significant figures in the volume is
ambiguous (i.e. we don’t know if the zeros are significant),
so we must assume the wors t: that the volume is known
only to 1 significant figure. Therefore, we report the
concentration to 1 significant figure only. If the volume had
been recorded as 0.500 L, then we could have given the
concentration as 0.198 mol L−1. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/avogadro-exam-solution-2007.pdf | 8 |
8 / AVOGADRO EXAM © 2007 UNIVERSITY OF WATERLOO
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
←
9
8B
10
→
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57
La
138.9 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89
Ac
227.0 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Uun 111
Uuu 112
Uub 113
Uut
58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) DATA SHEET
AVOGADRO EXAM 2007
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 × 1023 mol−1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K−1 mol−1 0oC = 273.15 K
= 8.3145 kPa L K−1 mol−1
= 8.3145 J K−1 mol−1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) −± −=24
2bb a cx
a |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-solution-2011.pdf | 1 |
2 /CHEM 13 NEWS EXAM © 2011 UNIVERSITY OF WATERLOO 1 At 25 oC and 100 kPa, most of the known elements are
A monatomic gases
B diatomic gases
C liquids
*D metallic solids
E non-metallic or semi -metallic solids
2 Which of the following series lists the compounds in
order of increasing boiling point? (from lowest to
highest)
A H2Te < H 2Se < H 2S < H 2O
*B H2S < H2Se < H 2Te < H 2O
C H2S < H 2O < H 2Se < H 2Te
D H2O < H 2S < H 2Se < H 2Te
E H2O < H 2Te < H 2Se < H 2S
3 In which of the following compounds does oxygen
have the highest oxidation state ?
A CsO 2
B H2O
C O2
D H2O2
*E OF 2
4 Which of the following processe s is the most
endothermic ?
A H2O(l) H2O(g)
B F(g) + e− F−(g)
C NaCl(s) 2H O NaCl(aq)
*D Na(g) Na+(g) + e−
E K+(g) + Cl−(g) KCl(s)
5 Which of the following atoms has electrons in its
outermost shell arranged in the configuration 4s2 4p3 ?
Assume each atom is in its lowest energy state .
A Rb
B Kr
*C As
D Cr
E Sb
6 The following reaction reaches equilib rium in a closed
reaction vessel at 200 oC.
CO(g) + 3 H 2(g) CH4(g) + H2O(g), Ho=−206 kJ
Which of the following actions causes the reaction to
proceed from left to right in order to restore
equilibrium?
A increasing the volume of the container , holding
temperature constant
B adding some CH 4 gas to the system, with volume
and temper ature held constant
*C adding some H 2 gas to the system, with volume
and temperature held constant
D increasing the temperature, holding the pressure
constant
E removing some CO gas from the system, with
volume and temperature held constant
7 At a ce rtain temperature, the following equilibrium
constants have been measured.
A2(s) + 2 B(g) 2 C(g) K1 = 36
D(s) + 2 E(g) C(g) K2 = 20
What is the equilibrium constant at the same
temperature for the reaction below ?
½ A 2(s) + B(g) D(s) + 2 E(g)
A 720
B 1.8
C 0.56
*D 0.30
E 0.090
CHEM 13 NEWS EXAM 20 11 - Answers |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-solution-2011.pdf | 2 |
© 2011 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 3 8 In a particular solution , [Br− ] = 0.020 mol L−1 and
[CrO 42− ] = 0.0030 mol L−1. Finely -divided solid silver
nitrate, AgNO 3, is slowly added to the solution. What
is [Br− ] when Ag 2CrO 4(s) just begins to precipitate?
*A 2.1×10−8 mol L−1
B 6.0×10−8 mol L−1
C 2.7×10−7 mol L−1
D 5.2×10−13 mol L−1
E 6.4×10−4 mol L−1
9 What is the formula of the stable compound for med
by magnesium and nitrogen?
A MgN
B Mg 2N
*C Mg 3N2
D Mg 2N3
E MgN 2
10 Which of the following ions has the smallest tendency
to be protonated when dissolved in liquid acetic acid,
CH 3COOH(l)?
A hydroxide,OH−
B fluoride, F−
C chloride, Cl−
D bromide, Br −
*E iodide, I−
11 X-ray radiation is more energetic than microwave
radiation because
A photons of X-ray radiation travel faster than those
of microwave radiation
B photons of X-ray radiation are heavier than those
of microwave radiation
*C X-ray rad iation has a higher frequency than does
microwave radiation
D X-ray radiation has a longer wavelength than
does microwave radiation
E photons of X-ray radiation travel slower than
those of microwave radiation
12 Which of the following contains only single bonds ?
A NO+
B CO
C CN−
D N22−
*E O22−
13 What is the empirical formula of a compound that is
66.64% carbon , 7.45% hydrogen and 25.91%
nitrogen by mass ?
*A C3H4N
B C3H4N2
C C3H3N
D C4H4N
E C4H3N2
14 Let DC=C represent the C=C bond dissociation energy
in ethene, H 2C=CH 2, and DC−C the C -C bond
dissociation energy in ethane, H 3C−CH 3. How do
these bond dissociation energies compare?
A DC=C equals DC−C
B DC=C is exactly equal to 2 × DC−C
C DC=C is exactly equal to ½ × DC−C
*D DC=C is greater than DC−C but less than 2 × DC−C
E DC=C is greater than 2 × DC−C
15 Which of the following bonds is most polar?
A B−O
*B B-F
C C-O
D C=O
E C−F
Ksp
Ag2CrO 4 1.9×10−12
AgBr 5.2×10−13 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-solution-2011.pdf | 3 |
4 /CHEM 13 NEWS EXAM © 2011 UNIVERSITY OF WATERLOO 16 Consider the following energy level diagram for the
reaction R → P.
Which of the following statements is false ?
*A The conversion of R to P occurs via a two-step
process.
B X and Y represent reaction intermediates .
C The conversion of R to P is endothermic.
D At equilibrium, the rate of conversion of R to P is
equal to the rate of conversion of P to R.
E The rate -limiting step is the conversion of X to Y.
17 A solution in which the bromide concentration is
2.0×10−5 mol L−1 is in equilibrium with solid AgBr and
solid AgI. What is the concentration of iodide ion ?
A 2.6×10−8 mol L−1
*B 5.8×10−9 mol L−1
C 1.5×10−16 mol L−1
D 7.5×10−12 mol L−1
E 2.9×10−4 mol L−1
18 Consider the hydrogen halides HF, HCl, H Br and HI.
Which of the statements about them is true?
A They are all strong acids.
B They are all weak acids.
C The boiling point increases with molar mass.
D The bond dissociation energy increases with
molar mass .
*E none of above
19 For the reaction below, Kc = 1.0×10−20.
2 A(g) + B(g) C(g)
In an experiment, 1.0 mol each of A, B and C are
placed in an empty 1.0 L container and then the
container is quickly sealed. When equilibrium is
established, w hich of the following will be true?
A [A] < [B] < [C]
*B [A] > [B] > [C]
C [A] = [B] = [C]
D [A] = [B] < [C]
E [A] > [B] = [C]
20 What percentage of CH 3COOH molecules are ionized
in 1.8×10−5 mol L−1 CH 3COOH(aq) ?
A 1.8%
B 4.2%
C 42%
*D 62%
E almost 100%
21 A technician rec orded the following curve during a
titration.
The curve represents the titration of a
A weak acid by adding strong base
B strong aci d by adding weak base
C strong base by adding weak acid
D strong base by adding strong acid
*E a weak base by adding strong acid Y
R X P
Reaction progress Potential
Energy
Volume of reagent added
from burette (in mL) pH
Ksp
AgBr 5.2×10−13
AgI 1.5×10−16 Ka(CH 3COOH) = 1.8×10−5 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-solution-2011.pdf | 4 |
© 2011 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 5 Use the table of standard reduction potentials
given below to answer questions 22 through 25.
22 Which of the following is the strongest oxidizing agent
under standard conditions ?
*A Ag+(aq)
B Ag(s)
C H+(aq)
D Al(s)
E Al3+(aq)
23 When Ag+(aq) reacts completely with exactly one
mole of H2(g) under stan dard conditions, how many
moles of solid Ag are produced ?
A 1 mol
*B 2 mol
C 0.5 mol
D 4 mol
E 0.25 mol
24 What is Eo for the reaction 2 H2(g) + O2(g) → 2 H2O(l)?
*A 1.23 V
B 0.43 V
C 4.06 V
D 0.43 V
E 2.06 V 25 Which of the following reagents would spontaneously
reduce Ni2+(aq) to Ni(s) under standard conditions ?
A Ag+(aq)
B Ag(s)
*C Zn(s)
D Sn(s)
E Al3+(aq)
26 Consider the ions K+, Ca2+, Cl− and S2−. In which
series are the species listed in order of decreasing
radius? (from largest to smallest)
*A S2− > Cl− > K+ > Ca2+
B K+ > Ca2+ > S2− > Cl−
C S2− > Ca2+ > Cl− > K+
D Ca2+ > K+ > Cl− > S2−
E Ca2+ > K+ > S2− > Cl−
27 A solu tion is prepared by completely dissolving a
solid mixture of NaOH and Mg(OH) 2 in water. For the
resulting solution, which of the following conditions
must be satisfied?
A [Na+ ] = [Mg2+ ] = [OH− ]
B [Na+ ] = [Mg2+ ] = 3 [OH− ]
C [Na+ ] + [Mg2+ ] = 3 [O H− ]
*D [Na+ ] + 2 [Mg2+ ] = [OH− ]
E [Na+ ] + [Mg2+ ] = [OH− ]
28 What is the minimum volume of water needed to
dissolve completely 1.0 g SrF 2?
*A 9.0 L
B 150 L
C 10.5 L
D 5.6 L
E 2.8 L
Half–Reaction Eo
Ag+(aq) + e− Ag(s) +0.80 V
O2(g) + 2 H2O(l) + 4e– 4 OH–(aq) +0.40 V
2 H+(aq) + 2e− H2(g) 0.0 V
Sn2+(aq) + 2e− Sn(s) –0.14 V
Ni2+(aq) + 2e− Ni(s) –0.25 V
Fe2+(aq) + 2e− Fe(s) –0.41 V
Zn2+(aq) + 2e− Zn(s) –0.76 V
2 H2O(l) + 2e− H2(g) + 2 OH−(aq) –0.83 V
Al3+(aq) + 3 e− Al(s) –1.66 V
Ksp(SrF 2) = 2.8×10−9
Sr, 87.62 g mol−1
F, 19.00 g mol−1 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-solution-2011.pdf | 5 |
6 /CHEM 13 NEWS EXAM © 2011 UNIVERSITY OF WATERLOO 29 What is the molecular geometry of SF 4?
A T-shaped
B tetrahedral
*C see-saw
D square planar
E square pyramidal
30 In the incomplete equation below, NH 3 acts as a
Bronsted -Lowry acid and “X” represents a Bronsted -
Lowry base. What is the conjugate base of NH 3?
NH 3 + X → ?
A X
B XH+
C NH 4+
*D NH 2−
E OH−
31 What is the general trend observed for the first
ionization energies of the elements in groups 13
through 17?
A Ionization energies tend to increase from left to
right in a period, and are approximately constant
in a g roup.
*B Ionization energies tend to increase from left to
right in a period, and decrease from top to bottom
in a group.
C Ionization energies tend to decrease from left to
right in a period, and increase from top to bottom
in a group.
D Ionization energies tend to decrease from left to
right in a period, and decrease from top to bottom
in a group.
E Ionization energies are approximately constant in
a period, and decrease from top to bottom in a
group.
32 What is the hybridization of the sulfur atom in the
SO 32− ion?
A sp
B sp2
*C sp3
D sp3d
E sp3d2
33 The phase diagram for an unidentified substance is
shown below.
Which of the following statements is true?
A Liquid can be converted to solid by increasing the
pressure at constant temperature .
B The melting temperature of the solid increases as
pressure increases.
C Solid cannot be converted into gas without first
being converted to liquid.
*D There is only one combination of temperature and
pressure for which solid, liquid and gas can
coexist.
E More than one of the statements above are true.
34 When the following equation is balanced using the
smallest whole number coefficients, what is the
coefficient of O 2?
NH 3 + O 2 → NO + H 2O
A 2
B 3
C 4
*D 5
E 6 Pressure
gas liquid
solid
Temperature |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-solution-2011.pdf | 6 |
© 2011 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 7 35 What is [CH 3COOH] at equilibrium if 0.10 moles of
CH 3COOH and 0.15 moles of NaOH are dissolved in
enough water to make 1.0 L of solution at 25 oC? For
CH 3COOH, Ka = 1.8×10−5 at 25 oC.
A 0 mol L−1
B 1.8×10−5 mol L−1
C 5.6×10−10 mol L−1
*D 1.1×10−9 mol L−1
E 1.3 × 10−3 mol L−1
36 The following diagram is sometimes used to illustrate
the structure of benzene, C 6H6.
Which of the statements concerning the structure of
benzene is false ?
*A The double bonds oscillate rapidly back and forth
between adjacent pairs of carbon atoms.
B The H -C-C angles are 120o.
C The carbon atoms form a flat hexagonal ring.
D The oxidation state of carbon is −1.
E The carbon -carbon bonds are all the s ame length.
37 A particular substance, X, decomposes such that its
concentration decreases by a factor of two every 35 s.
If the initial concentration of X was 1.0 mol L−1, what is
[X] after exactly 140 s?
A 0.33 mol L−1
B 0.13 mol L−1
C 0.25 mol L−1
*D 0.063 mol L−1
E 0.67 mol L−1
38 The bond dissociation energies for F 2 and Cl 2 are
approximately 158 and 242 kJ mol−1, respectively.
Given that the enthalpy change for the reaction below
is ΔH = −54 kJ mol−1, what is the bond dissociation
energy for the F -Cl bond?
½ F2(g) + ½ Cl 2(g) → FCl(g)
A 200 kJ mol−1
*B 254 kJ mol−1
C 146 kJ mol−1
D 454 kJ mol−1
E 346 kJ mol−1
39 Which of the following has the greatest number of
unpaired electrons in its ground electronic state?
A Al
B Cl
*C Ti2+
D Zn2+
E S2−
40 Let HA represent a weak monoprotic acid with
Ka = 1.0×10−5. In an experiment, a 50.0 mL
sample of 0.1 0 mol L−1 HA(aq) is titrated with
0.10 mol L−1 NaOH(aq) . At which point during the
titration are the equi librium concentrations of H+
and OH− equal?
A after the addition of exactly 25.0 mL of NaOH(aq)
*B after the addition of slightly less than 50.0 mL of
NaOH(aq)
C after the addition of exactly 50.0 mL of NaOH(aq)
D after the addition of more than 50.0 m L of
NaOH(aq)
E The equilibrium concentrations of H+ and OH− are
never equal .
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 1 | CHEM 13 NEWS EXAM 2010
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
20 MAY 2010 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. T op performers are eligible for a prize.
The names of the top 200 students will be published in the September issue of Chem 13 News.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 12 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 12 Chemistry
in a semestered school
Code 3 Ontario , Grade 12 Chemistry
already completed
Code 4 Any other Ontario student
Code 5 Manitoba or Saskatchewan high school
student
Code 6 Québec high school student
Code 7 Québec CEGEP student
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 2 |
2 /CHEM 13 NEWS EXAM © 2010 UNIVERSITY OF WATERLOO 1 Of the first 18 elements, how many are gases at
25 oC and 100 kPa?
A less than seven
B seven
*C eight
D nine
E more than nine
2 Which of the following substances has the highest
vapour pressure at 25 oC?
A CH
3OH
B CH 3CH 2CH 2OH
C LiF
*D H 2CO
E Li
3 Which of the following compounds has the highest
melting point?
A LiF
*B ZnO
C LiCl
D NaF
E NaCl
4 For a given substance, which of the following phase
transitions is the most exothermic?
A solid → liquid
B gas → liquid
C liquid → gas
D solid → gas
*E gas → solid
5 What is the ground state electron configuration of
selenium, Se?
A 1s
2 2s2 2p6 3s2 3p6 4s2 4p4
B 1s2 2s2 2p6 3s2 3p6 3d10 4p6
*C 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p4
D 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
E 1s2 2s2 2p6 3s2 3p6 4s2 4p4 4d10 6 The reaction below reaches equilibrium in a closed
reaction vessel.
C 6H12O6(s) 2 C 2H5OH(l) + 2 CO 2(g), ∆Ho = −72 kJ
Which of the following actions causes an increase in
the value of Kc?
(i) adding some CO 2(g)
(ii) transferring the reaction mixture to a vessel of
larger volume
(iii) increasing the temperature
A (i) only
B (ii) only
C (iii) only
D (i) and (ii)
*E none of the above
7 Given that
2 Hg
2+(aq) + 2 e− Hg 22+(aq) E o = 0.920 V
Ag+(aq) + e− Ag(s) E o = 0.799 V
what is E o for the reaction below?
2 A g+(aq) + Hg 22+(aq) 2 Ag(s) + 2 Hg2+(aq)
A 0.121 V
*B −0.121 V
C 0.678 V
D −0.678 V
E 0.339 V
8 Given that
F e2+(aq) + 2 e− Fe(s) E o = −0.40 V
2 H+(aq) + 2 e− H2(g) E o = 0.00 V
B r 2(l) + 2 e− 2 Br−(aq) E o = +1.09 V
which of the following is the strongest reducing agent
under standard conditions?
A Fe2+(aq)
B H+(aq)
C Br 2(l)
D Br−(aq)
*E H 2(g)
CHEM 13 NEWS EXAM 2010 - Answers
Ionic Radii (in pm)
Li+, 68 F−, 136
Zn2+, 74 O2−, 140
Na+, 97 Cl−, 181 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 3 |
© 2010 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 3 9 What is the coefficient of O 2 when the following
equation is balanced?
1 C 10H8(s) + x O2(g) → y CO 2(g) + z H2O(l)
A 1
B 6
C 7
*D 12
E 14
10 Which of the following will react appreciably with
water at room temperature and pressure to produce
hydrogen?
*A NaH
B NH
3
C CH 4
D HCl
E H 2S
11 Cesium forms a number of compounds with oxygen.
A particular compound is found to be 26.5% oxygen
by mass. What is the formula of this compound?
A Cs 2O
B Cs 2O2
C CsO 2
*D CsO 3
E CsO 4
12 Which of the following is the strongest acid in water?
A HBr
B HOBrO
2
C HF
D HOIO 2
*E HI
13 Let the energy of the 2s level in a hydrogen atom
be –E. What is the energy of the 3s level?
A
E23
*B E49
C E32
D E94
E E3
14 Natural oils, such as vegetable oil, are converted into
solid, edible fats by a process called
A fusion
*B hydrogenation
C crystallization
D flash freezing
E saponification
15 The value for the activation energy of the forward
reaction is represented by which letter in the diagram
below?
* A A
B B
C C
D D
E E
reactants
productsA
C
B E
D
Reaction progress Energy Molar masses
(in g/mol)
O, 16.00
Cs, 132.9 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 4 |
4 /CHEM 13 NEWS EXAM © 2010 UNIVERSITY OF WATERLOO 16 The heat of combustion of C(s) is −394 kJ/mol and
that of CO(g) is −111 kJ/mol. What is the enthalpy
change for the reaction below?
CO(g) → C(s) + ½ O 2(g)
A 505 kJ
* B 283 kJ
C 111 kJ
D −283 kJ
E −505 kJ
17 Exactly 850 mL of O
2 gas is collected over water at
30.0 oC using the setup below. Given that the
barometric pressure was 98.5 kPa and the vapour
pressure of water is 4.24 kPa at 30 oC, what volume
would the pure O 2 gas occupy at 98.5 kPa and 30 oC?
* A 813 mL
B 818 mL
C 850 mL
D 882 mL
E 888 mL
18 How are the boiling and freezing points of water
affected by the addition of a soluble salt?
A The freezing and boiling points are both lowered.
B The freezing and boiling points are both raised.
* C The freezing is lowered and the boiling point is
raised.
D The freezing is raised and the boiling point is
lowered.
E The boiling and freezing po ints are not affected.
19 The reaction below comes to equilibrium in a closed
reaction vessel of volume 2.50 L.
2 NO
2(g) 2 NO(g) + O 2(g)
At equilibrium, there are 3.0 mol NO, 4.00 mol O 2 and
22.0 mol NO 2. What is the value of Kc for the reaction
above?
*A 0.0298
B 33.6
C 1.83
D 13.4
E 0.218
20 Which of the following occurs if a 0.10 mol/L solution
of a weak acid is diluted to 0.010 mol/L at constant temperature?
A The hydrogen ion concentration decreases to
0.010 mol/L.
B The pH decreases.
C The ionization constant,
Ka, decreases.
*D The percentage ionization increases.
E all of the above
21 What is the equilibrium concentration of Ag
+ in
solution when 0.50 L of 0.10 mol/L AgNO 3(aq) and
0.50 L of 0.20 mol/L NaCl(a q) are mixed? Assume the
temperature is 25 oC.
A 0 mol/L
* B 3.6×10−9 mol/L
C 9.0×10−10 mol/L
D 1.3×10−5 mol/L
E 0.05 mol/L
22 In which ionic compound does the cation have the
same number of electrons as the anion?
A LiF
B NaCl
C CaO
*D MgF 2
E KI
O2(g)
water
For AgCl, Ksp = 1.8×10−10
at 25 oC. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 5 |
© 2010 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 5 23 How many moles of NaOH or HCl should be added to
1.0 L of 0.010 mol L−1 formic acid (HCOOH) solution
to obtain a solution with pH = 3.50? Assume no
change in volume. (Choose the closest value.)
*A 3.6×10
−3 mol NaOH
B 3.6×10−3 mol HCl
C 5.8×10−3 mol NaOH
D 5.8×10−3 mol HCl
E 3.2×10−4 mol HCl
24 For the reaction below, Kc = 6.3×104 at 25 oC.
2 NO(g) + Cl 2(g) 2 NOCl(g)
In an experiment, carried out at 25 oC, 1.0 mol NO
and 1.0 mol Cl 2 are added to an evacuated reaction
vessel of volume 1.0 L and then the vessel is quickly
sealed. What is the equilibrium concentration of NO?
A 0.50 mol/L
*B 5.6×10−3 mol/L
C 2.8×10−3 mol/L
D 1.6×10
−5 mol/L
E 7.9×10−6 mol/L
25 What is the molecular geometry of the BrF
3 molecule?
The Br atom is the central atom and all the F atoms
are bonded directly to Br.
A trigonal planar
B trigonal bipyramidal
* C T-shaped
D square planar
E trigonal pyramidal
26 When 0.012 moles of a monoprotic acid is dissolved in
water to give 1.0 L of solution at 25
oC, the final pH is
1.95. What is Ka for this acid?
A 2.9×10−1
B 1.1×10−2
* C 1.6×10
−1
D 1.3×10−4
E 1.5×10−6
27 A 1.00 mol/L solution of Cu(NO 3)2(aq) is electrolyzed
using the setup illustrated below . What is the reaction
occurring at the anode?
A Cu
2+(aq) + 2 e− → Cu(s)
B Cu(s) → Cu2+(aq) + 2 e−
C 2 H 2O(l) + 2 e− → H 2(g) + 2 OH−(aq)
*D 2 H 2O(l) → O2(g) + 4 H+(aq) + 4 e−
E Pt(s) → Pt4+(aq) + 4 e−
28 Which of the following forms of radiation has the
longest wavelength?
A infrared
B x-ray
* C microwave
D ultraviolet
E visible
Ka = 1.8×10−4 for HCOOH
Cu(s)
Cu(NO 3)2(aq) battery
Pt(s) − + |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 6 |
6 /CHEM 13 NEWS EXAM © 2010 UNIVERSITY OF WATERLOO 29 In the unbalanced chemical equation below, x, y and
z are coefficients to be determined.
1 Fe2+ + x Br2 → y Fe3+ + z Br −
When the equation is properly balanced, what is the
value of z?
*A 1
B 2
C
12
D 4
E 14
30 If the pH of a solution changed from 4.0 to 8.0, what
happened to the hydrogen ion concentration?
A It increased by a factor of two.
B It decreased by a factor of two.
C It increased by a factor of 104.
*D It decreased by a factor of 104.
E It decreased by a factor of 102.
31 Which of the following compounds displays only
covalent bonding?
A NH
4OH
B Li2O
* C HOCN
D NaNO
3
E KH
32 How many sigma ( σ) and pi ( π) bonds are there in the
allene molecule, H 2CCCH 2?
* A six σ bonds and two π bonds
B two σ bonds and six π bonds
C four σ bonds and four π bonds
D eight σ bonds and no π bonds
E two σ bonds and six π bonds 33 What is the oxidation state of each sulfur atom in the
peroxydisulfate ion, S 2O82−? In the structure below,
lone pairs are not shown.
A −2
B 0
C +4
* D +6
E +7
34 A Lewis structure for POCl
3 is shown below.
Which of the following statements is correct?
A This is most important Lewis structure for the
POCl
3 molecule.
B The phosphorus atom is sp2-hybridized.
C The Cl-P-Cl angles are 90
o.
D The oxidation state of phosphorus is +4.
* E None of the statements above are true.
35 What is the maximum number of electrons that can
have a principal quantum number of 4 within one atom?
A two
B four
C eight
D sixteen
* E thirty-two OSO
OO O SO
O
O2
PO
Cl
ClCl |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 7 |
© 2010 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 7 36 How many unpaired electrons are there in a Mn2+ ion in
its ground electronic state? The atomic number of manganese is Z = 25 .
A 0
B 2
C 3
* D 5
E 6
37 The skeletal structure below for the CH
2CHOCN
molecule is incomplete; additional bonding pairs or
lone pairs must be added. When the structure is
properly completed, how many lone pairs are there in this molecule?
A none
B one
C two
* D three
E four
38 When temperature is increased, the rate of a reaction
also increases. This obse rvation is best explained by
A an increase in the frequency of molecular
collisions
B a decrease in the activation energy,
Ea, for the
reaction
C an increase in the activation energy, Ea, for the
reaction
D a decrease in the enthalpy change, ∆H, for the
reaction
* E an increase in the fraction of molecules that have
enough energy to react
39 Which of the following would need the smallest
quantity of heat to change the temperature of 5 g by 10°C?
A I
2(s)
B H2O(l)
* C Au(s)
D He(g)
E Cu(s)
40 Let HA represent a weak monoprotic acid with
Ka = 1.0×10−5. What is the pH at the equivalence
point in the titration of 50. 0 mL of 0.20 mol/L HA(aq)
with 0.20 mol/L NaOH(aq)?
A 5.00
*B 9.00
C 7.00
D 3.00
E 11.00
Specific Heat
(in J g−1 °C−1)
I2(s) 0.158
H2O(l) 4.18
Au(s) 0.129 He(g) 5.19
Cu(s) 0.385
HC
HC
HO C N |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2010.pdf | 8 |
8 /CHEM 13 NEWS EXAM © 2010 UNIVERSITY OF WATERLOO
DATA SHEET
CHEM 13 NEWS EXAM 2010
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 1023 mol1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K1 mol1 0oC = 273.15 K
= 8.3145 kPa L K1 mol1
= 8.3145 J K1 mol1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) 24
2bb a cx
a
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
9
8B
10
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57-71
La-Lu 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89-103
Ac-Lr 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Ds 111
Sg 112
Cn
57
La
138.9 58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
89
Ac
227. 90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 1 | CHEM 13 NEWS EXAM 2009
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
14 MAY 2009 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. T op performers are eligible for a prize.
The names of the top 200 students will be published in the September issue of Chem 13 News.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 12 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 12 Chemistry
in a semestered school
Code 3 Ontario , Grade 12 Chemistry
already completed
Code 4 Any other Ontario student
Code 5 Manitoba or Saskatchewan high school
student
Code 6 Québec high school student
Code 7 Québec CEGEP student
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 2 |
2 /CHEM 13 NEWS EXAM © 2009 UNIVERSITY OF WATERLOO 1 In the third period of the elements, how do the atomic
radii of the elements vary?
A The radii increase steadily from Na to Ar.
B The radii increase from Na to Al and decrease
from Al to Ar.
C There is no regular pattern.
D The radii decrease from Na to S and increase
from S to Ar.
*E The radii decrease steadily from Na to Ar.
2 Which of the following compounds has the highest
boiling point?
3 Nitrous acid, HNO
2, is a weak acid in water. Which of
the following statements concerning NO 2− is true?
*A NO 2− is a weak base.
B NO 2− is a strong base.
C NO 2− is a weak acid.
D NO 2− is a strong acid.
E NO 2− is neither an acid nor a base.
4 How many moles of NaOH should be added to 1.0 L
of 0.10 mol L−1 HCOOH( aq) to obtain a solution
having a final pH of 4.0 at 298 K? Assume no change
in volume. (Choose the closest value.)
A 0.018 mol
B 1.8 mol
C 0.26 mol
*D 0.064 mol
E 0.0099 mol
5 Which of the following molecules do not
form
hydrogen bonds amongst themselves?
A CH 3COOH
B H2O2
*C CH 3OCH 3
D HF
E CH
3CH 3OH
6 The reaction below reaches equilibrium in a closed
reaction vessel.
4 HCl( aq) + MnO 2(s)
U Cl 2(g) + 2 H 2O(l) + Mn2+(aq) + 2 Cl−(aq), ΔH < 0
Which of the following actions increases the mass of
Cl2(g) in the equilibrium mixture?
A adding some MnO 2(s)
B increasing the temperature
C adding some MnCl 2(s)
D decreasing the volume of the reaction vessel
*E adding something that precipitates Mn
2+
Ka = 1.8×10−4 at 298 K
for HCOOH. CHEM 13 NEWS EXAM 2009 - Answers
Ka = 7.2×10−4 at
298 K for HNO 2. CC
CH3Cl CH3
Cl
CC
CH3Cl Cl
CH3
CC ClA
B
C
DCH3CC C H3
Cl
CC
CH3CH3 CH3
CH3E* |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 3 |
© 2009 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 3
7 What is
oE for the cell described above?
A 0.15 V
*B 0.91 V
C 2.43 V
D 3.49 V
E 6.53 V
8 In the cell described above, where does reduction
occur?
A at the aluminum electrode
*B at the zinc electrode
C at the voltmeter
D in the salt bridge
E in the aluminum nitrate solution
9 Considering the standard reduction potentials given in
the box on the right, which of the following is the strongest reducing agent under standard conditions?
*A Al(s)
B Al3+(aq)
C Zn( s)
D Zn2+(aq)
E impossible to determine
10 What is E
cell equal to when the cell described in the
box reaches equilibrium at 25 oC?
A −2.43 V
B +5.62 V
*C 0 V
D 1.06 V
E none of the above
11 A compound is 54.6% C, 36.2% O and 9.2% H by
mass. What is the empirical formula of the
compound?
A CH 2O
*B C 2H4O
C C 3H6O2
D C 4H4O
E C 6H6O
12 What is the pH of a 1.25 ×10−7 mol L−1 HCl( aq)?
A 6.90
*B 6.74
C 7.00
D 6.67
E less than 6.67
Use the following information and diagram
to answer questions 7-10.
A galvanic cell is constructed by placing a strip of zinc into a 1.0 mol L
−1 solution of zinc nitrate
and a strip of aluminum into a 1.0 mol L−1
solution of aluminum nitrate. The two metal strips are connected to a voltmeter by wires
and a salt bridge connects the solutions. (See
the diagram below.) The temperature is 25
oC.
The following standard reduction potentials
apply:
Al3+(aq) + 3e– U Al(s) oE= −1.67 V
Zn2+(aq) + 2e– U Zn( s) oE= −0.76 V
Zn(s) Al(s)
1.0 mol L−1
Zn(NO 3)2(aq) 1.0 mol L−1
Al(NO 3)3(aq) salt bridge Voltmete r |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 4 |
4 /CHEM 13 NEWS EXAM © 2009 UNIVERSITY OF WATERLOO 13 Which of the following statements is true?
A A single covalent bond consists of a single
delocalized electron.
B For a bond formed between a given pair of atoms,
the bond dissociation energy increases as the
bond order decreases.
C The bond dissociation energy for a C=C bond is
twice that of a C −C bond.
D A polar covalent bond results from the transfer of
one or more electrons from one atom to another.
*E none of the above
14 How many isomers are there for C 5H12?
A one
B two
*C three
D four
E more than four
15 The enthalpy change for the reaction below is
ΔH = −58 kJ (per mole of N
2O4 formed).
2 NO 2(g) + 2 I−(aq)
−⎯⎯→←⎯⎯k
k1
1 2 N 2O4(aq) + I 2(s)
If k1 and k−1 are the rate constants for the forward
and reverse reactions, respectively, and Kc is the
equilibrium constant for the reaction as written, then what effect does increasing the temperature have on
the values of k
1, k−1 and Kc?
A k1 increases, k−1 decreases, Kc increases
B k1 increases, k−1 increases, Kc increases
* C k
1 increases, k−1 increases, Kc decreases
D k1 increases, k−1 decreases, Kc decreases
E k1 decreases, k−1 decreases, Kc decreases
16 Consider the thermochemical equations below.
C
2H4(g) + 3 O 2(g) → 2 CO 2(g) + 2 H 2O(l)
ΔH° = −1411 kJ (per mol C 2H4)
2 C(s) + 3 H 2 (g) + ½ O 2(g) → C 2H5OH(l)
ΔH° = −278 kJ (per mol C 2H5OH)
C2H4(g) + H 2O(l) → C2H5OH(l)
ΔH° = −44 kJ (per mol C 2H4)
What is ΔH° for the following reaction? All the
answers below are for the combustion of one mole of
C2H5OH.
C2H5OH(l) + 3 O 2(g) → 2 CO 2(g) + 3 H 2O(l)
A −1089 kJ
B 632 kJ
C −1455 kJ
D −1733 kJ
*E −1367 kJ
17 A 10.0-L gas cylinder contains neon gas with a
measured pressure of 5.50 atm at 298 K. The 10.0-L
cylinder is then connected to an empty gas cylinder of unknown volume, and the neon gas expands to fill
both cylinders. If the final pressure is found to be
3.76 atm at 298 K, then what is the volume of the second cylinder?
A 14.6 L
B 6.52 L
C 10.0 L
* D 4.63 L
E 9.26 L
18 Which of the following correctly describes what
happens when aqueous solutions of ammonium
carbonate, (NH
4)2CO 3, and potassium bromide, KBr,
are mixed?
A Br− neutralizes NH 4+.
B K2CO 3(s) precipitates.
C HBr is formed.
D NH
4Br(s) precipitates.
* E none of the above
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 5 |
© 2009 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 5
19 For the titration descri bed above, which of the
following is true at the equivalence point?
A [HA] = [Na
+]
B [A− ] = [HA]
* C [Na+ ] = [A− ]
D [H+] = [OH−]
E [A− ] = [H+]
20 Based on the titration curve above, what was the
concentration of the original sample solution (i.e.
before the titration started)?
A 0.63 mol L
−1
*B 0.15 mol L−1
C 0.24 mol L−1
D 0.067 mol L−1
E 0.20 mol L−1
21 Based on the titration curve above, what is the
ionization constant ( Ka) for the acid, HA?
A 10−12.40
B 10−8.91
* C 10−4.89
D 10−2.86
E 10+2.86
22 How many unpaired electron s are there in the nickel
(Ni) atom in its ground state electron state?
A 5
B 4
C 6
*D 2
E 0
23 Equal volumes of 0.1 mol L−1 HCl( aq) and 0.1 mol L−1
HF(aq) are titrated in separate experiments with
0.1 mol L−1 NaOH( aq). Which of the following would be
equal for both titrations?
A the initial pH (i.e. the pH before any NaOH
is added)
B the pH when half the acid has been neutralized
(i.e. the pH at the half-neutralization point)
C the pH at the equivalence point
*D the volume of NaOH required to reach the
equivalence point
E none of the above
24
For the reaction below, Kc = 7.8 ×108. What is the
equilibrium concentration of NH 3 when 1.00 mol each
of Zn(NO 3)2 and NH 3 are dissolved in water to make
1.0 L of solution?
Z n2+(aq) + 4 NH 3(aq) U Zn(NH 3)42+(aq)
A 0 mol L−1
B 1.3×10−9 mol L−1
C 0.75 mol L−1
D 0.25 mol L
−1
*E 4.5×10−3 mol L−1
Use the information and diagram below to answer
questions 19-21.
A 40.0-mL sample of a weak monoprotic acid, HA, is
titrated with 0.20 mol L
−1 NaOH( aq). The titration curve
is shown below.
02468101214
0 1 02 03 04 05 0VNaOH (in mL)pHpH = 12.40
pH = 8.91
pH = 4.89
pH = 2.86 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 6 |
6 /CHEM 13 NEWS EXAM © 2009 UNIVERSITY OF WATERLOO 25 Which of the following molecules is polar?
A CS
2
* B N 2O
C CCl 4
D PF
5
E SO 3
26 Consider the compounds NaCl, AgCl and CO
2 in terms
of their solubilities in water. Which of these
compounds exhibits an increase in solubility if the temperature is lowered and the pressure is increased?
A NaCl only
B AgCl only
* C CO
2 only
D NaCl and AgCl
E NaCl, AgCl and CO 2
27 When a 1.00 mol L−1 solution of M2+(aq) is
electrolyzed with a current of 2.5 amperes for
0.2 hours, 0.485 g of M( s) are deposited. What is the
identity of M? (Note: 1 ampere = 1 C s−1)?
*A Cr
B Rh
C Na
D Mg
E Ag
28 Iron (III) oxide, Fe
2O3, reacts with hydrochloric acid to
produce only water and a salt . What is the formula of
the salt?
* A FeCl 3
B FeCl
2
C FeCl
D Fe2Cl3
E FeCl 6
29 The unbalanced chemical equation for the oxidation
of Zn by NO 3− is given below. The reaction occurs in
aqueous basic solution.
Zn + NO 3− → Zn(OH) 42− + NH 3
How many moles of NO 3− are required to oxidize
exactly one mole of Zn?
A 1 mol
* B ¼ mol
C 4 mol
D 8 mol
E ⅛ mol
30 Two students each made four measurements of the
mass of an object. Their results are shown in the
table below.
Student A Student B
Measurements: 51.6 g 50.1 g
50.8 g 49.6 g
52.2 g 51.0 g 5 0 . 2 g
49.4 g
Average: 51.3 g 50.0 g
If the exact mass of the object is 51.0 g, then which
of the following statements is true?
A Student A’s results are more accurate and more
precise.
B Student B’s results are more accurate and more
precise.
*C Student A’s results are more accurate but less
precise.
D Student B’s results are more accurate but less
precise.
E The two sets of results are equally precise.
31 Which of the following compounds displays the
greatest ionic character in its bonds?
A NO
2
B CO 2
* C H 2O
D NH
3
E F 2O
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 7 |
© 2009 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 7 32 What is the oxidation state of oxygen in FOCN? The
molecular structure of FOCN is shown below.
*A zero
B +2
C −2
D +1
E −1
33 Experiment shows that in the formamide molecule,
H
2NCHO, the H-N-H angle is 119o and the N-C-O
angle is 124o. Which of the following structures is an
acceptable structure for H 2NCHO and is consistent
with the experimentally-determined bond angles?
A
B
C
* D
E
34 Which of the following best describes the bonding in
the N
2 molecule?
A one σ bond and one π bond
B two σ bonds and one π bond
C two π bonds
D three π bonds
* E one σ bond and two π bonds
35 When building up the electron configuration of a
neutral atom, which orbital fills immediately after the 5s orbital?
* A 4d
B 4f
C 5p
D 6s
E 5d
36 Perovskite is a mineral containing Ca, O and Ti. The
smallest repeating unit in the structure of perovskite is shown below. (There is a single titanium atom at the centre of the cube.) By considering the total number of atoms of each type that lie inside
the cube below,
determine the formula of perovskite. What is the formula of perovskite?
A CaOTi
B CaO
6Ti
C Ca
8O6Ti
D Ca 2O3Ti
* E CaO 3Ti
FOCN
Ca O Ti HNH
CO
H
HNH
CO
H
HNH
CO
H
HNH
CO
H
HNH
CO
H |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 8 |
8 /CHEM 13 NEWS EXAM © 2009 UNIVERSITY OF WATERLOO 37 Consider the Lewis structure shown below for the
polyatomic ion, EOF 22−. The central atom, E, is an
unidentified element. Which of the following atoms
could E represent?
A oxygen (O)
* B sulfur (S)
C bromine (Br)
D nitrogen (N)
E xenon (Xe)
38 Consider the following reaction mechanism.
(CH
3)3CBr R (CH 3)3C+ + Br −
(CH 3)3C+ + N 3− → (CH 3)3CN 3
According to this mechanism, (CH 3)3C+ is
A a reaction product
* B a reaction intermediate
C an activated complex
D a catalyst
E a Lewis base
39 What is the final temperature when 100.0 mL of
water at 90.0 °C and 200.0 mL of water at 10.0 °C
are mixed? Assume no heat is lost to the surroundings. Choose the closest value.
* A 40
oC
B 50 oC
C 70
oC
D 80 oC
E 100 oC
40 Which of the following is present in the greatest
number in a dilute sulfuric acid (H
2SO 4) solution?
A H2SO 4 molecules
B HSO 4− ions
C SO 42− ions
*D H+ ions
E OH− ions
Properties of water
density = 1.0 g mL−1
specific heat = 4.18 J g−1 oC−1
heat of vaporization = 2260 J g−1 F EO
F2− |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2009.pdf | 9 |
© 2009 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 9
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
←
9
8B
10
→
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57
La
138.9 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89
Ac
227.0 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Uun 111
Uuu 112
Uub 113
Uut
58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) DATA SHEET
CHEM 13 NEWS EXAM 2009
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 × 1023 mol−1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K−1 mol−1 0oC = 273.15 K
= 8.3145 kPa L K−1 mol−1
= 8.3145 J K−1 mol−1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) −± −=24
2bb a cx
a |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2008.pdf | 1 | CHEM 13 NEWS EXAM 2008
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
8 MAY 2008 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. T op performers are eligible for a prize.
The names of the top 200 students will be published in the September issue of Chem 13 News.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 12 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 12 Chemistry
in a semestered school
Code 3 Ontario , Grade 12 Chemistry
already completed
Code 4 Any other Ontario student
Code 5 Manitoba or Saskatchewan high school
student
Code 6 Québec high school student
Code 7 Québec CEGEP student
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2008.pdf | 2 |
2 /CHEM 13 NEWS EXAM © 2008 UNIVERSITY OF WATERLOO 1 Which of the following statements about the group 17
elements is false ?
A The ionization energy decreases down the group.
*B The group contains both metals and non-metals.
C Electronegativity decreases down the group.
D The melting point increases down the group.
E The most common ion formed by these elements
is X−.
2 Which of the following has the highest melting point?
A I 2(s)
B C 60(s)
C NaCl( s)
*D LiF(s)
E Xe( s)
3 The acid ionization constant for HNO 2 is Ka = 4.5×10−4
at 298 K. What is the pH of 0.100 mol L−1 HNO 2(aq) at
298 K? (Choose the closest value.)
A 1.00
*B 2.17
C 1.67
D 3.23
E 6.53
4 A 0.100 mol L−1 solution of which of the following salts
has the highest pH at 298 K?
A NaF
B NaIO 3
*C NaCN
D NH 4F
E NH 4IO3
5 A solution is prepared by dissolving 4.50 grams of
solid NaOH in 1.00 L of 0.100 mol L
−1 HNO 2(aq) at
298 K? What is the pH of this solution? Assume that
the final volume is 1.00 L.
A 7.00
B 1.90
C 2.45
*D 12.10
E 13.05
6 If 1.00 L of 0.100 mol L
−1 HNO 2(aq) is diluted with water
to a final volume of 4.00 L, then which of the following
statements regarding the new solution is true?
A The percent ionization of the acid decreases and
the pH remains the same.
B The percent ionization of the acid increases and
the pH decreases.
*C The percent ionization of the acid increases and
the pH increases.
D The percent ionization of the acid decreases and
the pH decreases.
E The percent ionization of the acid increases
and the pH remains the same.
7 Which of the following equilibria shifts to the left when
the external pressure is increased and shifts to the
right when the temperature is increased?
A N
2(g) + O 2(g) U 2 NO( g) ΔH > 0
B 2 H 2O(g) U O2(g) + 2 H 2(g) ΔH < 0
C PCl 3(g) + Cl 2(g) U PCl 5(g) ΔH > 0
D N 2(g) + 3 H 2(g) U 2 NH 3(g) ΔH < 0
*E 2 CO 2(g) U 2 CO( g) + O 2(g) ΔH > 0
Ionization constants
(at 298 K)
HIO 3, Ka = 1.7×10−2
HF, Ka = 6.3×10−4
HCN, Ka = 6.2×10−10
NH 3, Kb = 1.8×10−5
H2O, Kw = 1.0×10−14 Ionization constants
(at 298 K)
HNO 2, Ka = 4.5×10−4
H2O, Kw = 1.0×10−14 CHEM 13 NEWS EXAM 2008 - Answers |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2008.pdf | 3 |
© 2008 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 3 Use the table of standard redu ction potentials given below
to answer questions 8 through 10.
Half–Reaction
Eo
Ag+(aq) + e− U Ag(s) +0.80 V
O2(g) + 2H 2O(l) + 4e– U 4 OH–(aq) +0.40 V
Cu2+(aq) + 2e− U Cu(s) +0.34 V
2H+(aq) + 2e− U H2(g) 0.0 V
Sn2+(aq) + 2e− U Sn(s) –0.14 V
Ni2+(aq) + 2e− U Ni(s) –0.25 V
Fe2+(aq) + 2e− U Fe(s) –0.41 V
Cr3+(aq) + 3e− U Cr(s) –0.74 V
Zn2+(aq) + 2e− U Zn(s) –0.76 V
2H2O (l) + 2e− U H2(g) + 2OH−(aq) –0.83 V
Al3+(aq) + 3e− U Al(s) –1.66 V
8 Which of the following is the best reducing agent
under standard conditions?
A Cu( s)
*B Zn( s)
C Al3+(aq)
D Fe2+(aq)
E Ag( s)
9 The metal X dissolves in HCl( aq) but does not react in
pure water, even its powdered form. It is a better
reducing agent than Ni( s). It forms an oxide with the
formula X 2O3. What is X?
A silver, Ag
B copper, Cu
C zinc, Zn
D aluminum, Al
*E chromium, Cr
10 Sacrificial anodes are attached to the hulls of ships to
protect the iron (Fe) in the hull from corrosion. Which of the following metals could be used as a sacrificial
anode for protecting the iron hull of a ship?
A nickel, Ni
*B zinc, Zn
C tin, Sn
D copper, Cu
E silver, Ag
11 The phase diagram for carbon dioxide is shown
below. The temperature and pressure at the triple point (TP) and the critical point (CP) are shown.
Which of the following account s for the fact that liquid
CO
2 is not observed when a piece of solid CO 2 (dry
ice) is placed on a lab bench at 25oC and 1 atm?
A The triple point temperature is less than the
critical point temperature.
B The critical temperature is greater than 25oC.
C The triple point temperature is less than 25oC.
D The critical pressure is greater than 1 atm.
*E The triple point pressure is greater than 1 atm.
12 When 1.50 grams of a compound containing only
carbon, hydrogen, nitrogen and oxygen is burned
completely in excess O 2, 1.72 g CO 2, 0.585 g NO
and 1.23 g H 2O are produced. What is the empirical
formula for the compound?
*A C 2H7O2N
B C 2H14O2N
C CH 7ON
D C 2H7ON 2
E CH 7O2N
13 What is the hybridization of the carbon atoms in
benzene, C 6H6?
A sp2 and sp3
B sp3 only
C sp , sp2 and sp3
*D sp2 only
E sp only
CP
TP
−57oC 31oC5.2 atm72 atm
gas liquid solid |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2008.pdf | 4 |
4 /CHEM 13 NEWS EXAM © 2008 UNIVERSITY OF WATERLOO 14 How many structural isomers are there for C 4H8?
A one
B two
C three
D four
*E more than four
15 The reaction below was studied at 40oC using the
method of initial rates. Data are given in the table
below.
−2
28SO (aq) + 2 I−(aq) → 2 −2
4SO (aq) + I 2(s)
run []−2
28SO
(in mol L-1) []−I
(in mol L-1) Initial Rate
(in mol L−1 s−1)
1 0.010 0.10 3.5×10−4
2 0.020 0.20 1.4×10−3
3 0.020 0.40 2.8×10−3
What are the correct value and units of the rate
constant, k?
A 0.35 mol L−1 s−1
B 3.5 mol L−1 s−1
* C 0.35 mol
−1 L s−1
D 0.35 mol−2 L2 s−1
E 1.8×102 mol−1 L s−1
16 For the reaction below, ΔHo= −879.6 kJ.
3 N 2O(g) + 2 NH 3(g) → 4 N 2(g) + 3 H 2O(g)
Given that ΔHo
f = −45.9 kJ mol−1 for NH 3(g) and
ΔHo
f= −241.8 kJ mol−1 for H 2O(g), what is ΔHo
f
for N 2O(g)?
A 684 kJ mol−1
B −504 kJ mol−1
C −684 kJ mol
−1
* D 82.0 kJ mol−1
E The answer cannot be determined with the
information provided.
17 The following figure shows the contents and
pressures of three vessels of gas which are joined by a connecting tube.
After the valves on the vessels are opened, the final
pressure is measured and found to be 0.675 atm.
What is the total volume of the connecting tube? All
vessels are at a constant temperature of 25ºC.
A 0.53 L
* B 0.056 L
C 0.094 L
D 0.040 L
E 0.023 L
18 At a certain temperature, the equilibrium constant for
the reaction below is
Kp = 0.100.
P4(g) U 2 P 2(g)
In an experiment, some P 4 gas was added to an
empty reaction vessel and then the vessel was
quickly sealed. The total pr essure at equilibrium was
1.00 atm. What was the initial pressure of P 4 used in
this experiment?
A 1.00 atm
B 0.730 atm
C 0.752 atm
* D 0.865 atm
E 0.667 atm
valve
He
0.75 atm
1.0 L Xe
0.45 atm
2.5 L Ar
1.20 atm
1.0 L |
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© 2008 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 5 19 In acidic aqueous solution, zinc metal is oxidized to
Zn2+. The net ionic equation for the reaction is given
below.
Zn(s) + 2 H+(aq) → Zn2+(aq) + H 2(g)
In an experiment, 5.0 grams of Zn( s) were added to
100 mL of 1.0 mol L−1 HCl( aq). Which of the following
changes to the procedure would not affect the initial
rate of the reaction?
A warming the HCl solution before adding the zinc
B using zinc powder instead of zinc granules
* C using 50 mL of 1.0 mol L
−1 HCl( aq)
D using 200 mL of 0.50 mol L−1 HCl( aq)
E using 100 mL of 1.0 mol L−1 H2SO 4(aq)
20 Which of the following groups of ions and atoms is
comprised of species having exactly the same ground
state electron configuration?
A Al
3+, O2−, Ne, Cl−
B Ca, Ti2+, Cl−, S2−
C H−, He, Li, Be2+
D Ne, Ar, Kr, Xe
*E Ca2+, Ti4+, Cl−, S2−
21 Proteins are polymers of which kind of acids?
* A amino acids
B strong acids
C binary acids
D inorganic acids
E lactic acids
22 In separate experiments, a 50.0-mL sample of each
of the two solutions listed below is titrated with 0.10 mol L
−1 NaOH( aq).
0.10 mol L−1 HCl( aq), pH = 1.0
0.10 mol L−1 HCN( aq), pH = 5.1
Which of the following statements is true?
A For both titrations, the pH at the equivalence
point is 7.00.
B It takes a greater volume of the NaOH solution to
reach the equivalence point for the titration of the
HCl solution than it does for the titration of the
HCN solution.
C For both titrations, the pH at the equivalence
point is greater than 7.00.
D HCN is a stronger acid than HCl.
*E For both titrations, it takes 50.0 mL of the NaOH
solution to reach the equivalence point.
23 At high temperatures, sodium hydrogen carbonate,
NaHCO 3, decomposes according to the chemical
equation given below.
2 NaHCO 3(s) U Na 2CO 3(s) + H 2O(g) + CO 2(g)
What is the equilibrium co nstant expression for this
reaction?
A [] [ ] [ ]
[]=cK22 2 3
3H O CO Na CO
NaHCO
B [] [ ] [ ]
[]=cK22 2 3
2
3H O CO Na CO
NaHCO
*C [] [ ]=cK22HO C O
D []
[] [ ] [ ]=cK2
3
22 2 3NaHCO
H O CO Na CO
E [] [ ]=cK
221
HO C O
|
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6 /CHEM 13 NEWS EXAM © 2008 UNIVERSITY OF WATERLOO 24 Which of the following statements is always true?
A A nonelectrolyte is ionized completely in aqueous
solution.
B Most ionic compounds of the Group 1 elements
are insoluble.
C A 1 mol L−1 solution of NH 3(aq) is a better
conductor of electric current than a 1 mol L−1
solution of HCl( aq).
*D A weak acid is partially ionized in aqueous
solution.
E Cl− will precipitate Na+ from solution.
25
Given the data below, what is the bond dissociation
energy for the H-Cl bond?
*
A 430 kJ mol−1
B 384 kJ mol−1
C 123 kJ mol−1
D 92 kJ mol−1
E 767 kJ mol−1
26 Of the following organic compounds, which is least
soluble in water at 298 K?
A methanol, CH 3OH
B ethanol, CH 3CH 2OH
* C dimethyl ether, H 3COCH 3
D ethylene glycol, HOCH 2CH 2OH
E ethanoic acid, CH 3COOH
27 The temperature-time graph is shown below for
heating H 2O at a constant rate of 1.00 kJ s−1. What
does the line segment DE represent?
A warming of ice
B fusion
C warming of liquid
*D vaporization
E condensation
28 The unbalanced chemical equation for the oxidation
of −Br by −
4MnO is given below. The reaction occurs
in aqueous acidic solution.
−Br + −
4MnO → Br2 + Mn2+
How many moles of −
4MnO are required to oxidize
exactly 1.0 mol −Br?
A 1.0 mol
* B 0.2 mol
C 5.0 mol
D 0.1 mol
E 10 mol
29 Which of the following is the best choice to measure
accurately 22.5 mL of a solution?
*A a 50 mL buret
B a 50 mL Erlenmeyer flask
C a 50 mL beaker
D a 50 mL graduated cylinder
E a 50 mL volumetric pipet H-H bond dissociation energy = 432 kJ mol−1
Cl-Cl bond dissociation energy = 244 kJ mol−1
ΔfHofor HCl = −92 kJ mol−1 AB
C D E
time temperature |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2008.pdf | 7 |
© 2008 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 7 30 Solid NH 4NO 3 is added to a solution of sodium
hydroxide, NaOH, and the solution is warmed. Which of the following gases is produced?
A nitrogen, N 2
B oxygen, O 2
C dinitrogen oxide, N 2O
D hydrogen, H 2
*E ammonia, NH 3
31 Which one of the following solutions does not conduct
electricity at 25oC?
*A 0.10 mol L−1 CH 3CH 2OH(aq)
B 0.10 mol L−1 H2SO 4(aq)
C 0.10 mol L−1 CH 3COOH( aq)
D 0.10 mol L−1 HNO 3(aq)
E 0.10 mol L−1 NH 3(aq)
32 In which of the following compounds is the oxidation
state of chlorine equal to +5?
A HCl
B ClF 3
* C HClO 3
D PCl 5
E HClO 2
33 The structure of which of the following is not a hybrid
of two or more equivalent resonance structures?
A −2
3CO
B −3
4PO
C C6H6
D O 3
* E C 2H4
34 For the reaction
2 HBr + ½ O 2 → H 2O + Br 2,
the following mechanism has been proposed. HBr + O
2 → HOOBr fast
HOOBr + HBr → 2 HOBr slow
HOBr + HBr → H 2O + Br 2 fast
What is the predicted rate law for the overall reaction?
* A Rate = k [HBr]2 [O2]
B Rate = k [HBr]2 [O2]1/2
C Rate = k [HBr] [O 2]
D Rate = k [HBr] [O 2]2
E Rate = k
21 / 2
222[H O][Br ]
[HBr] [O ]
35
Which of the following substances is the most soluble
in hexane, C 6H14(l)?
A NaCl
* B Cl2
C CH 3Cl
D HCl
E CaCl 2
36 Which oxide of nitrogen is 36.8% N by mass?
A N2O4
B NO
* C N2O3
D NO 2
E N 2O
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2008.pdf | 8 |
8 /CHEM 13 NEWS EXAM © 2008 UNIVERSITY OF WATERLOO 37 A student drew the following Lewis structures for NO,
CO, NH 3 and BH 3. Which of the following structures
is (are) correct?
( 1 ) ( 2 )
( 3 ) ( 4 )
A (1) only
B (3) and (4)
*
C (1) and (2)
D (2) and (3)
E (3) only
38 The reaction N 2(g) + 3 H 2(g) U 2 NH 3(g) is
exothermic. The reaction is allowed to reach
equilibrium in a closed vessel. Which of the following
will lead to an increase in the number of moles of ammonia in the equilibrium mixture?
(1) increasing the temperature
(2) adding a catalyst (3) increasing the external pressure
(4) adding N
2 to the reaction vessel
A (1) only
B (1) and (2) only
C (2) and (3) only
* D (3) and (4) only
E (1), (2) and (4)
39 What is the maximum mass of nickel metal that can be deposited from an aqueous solution of Ni(NO
3)2
by the passage of three moles of electrons?
A 29 g
B 39 g
C 59 g
* D 88 g
E 176 g
40 Ethanoic acid, CH 3COOH, is a weak acid in water.
What happens when 0.01 moles of HCl are added to
a 0.1 mol L−1 solution of ethanoic acid?
A The pH of the solution increases and the percent ionization of ethanoic acid increases.
B The pH of the solution decreases and the percent
ionization of ethanoic acid increases.
*C The pH of the solution decreases and the percent
ionization of ethanoic acid decreases.
D The pH of the solution increases and the percent
ionization of ethanoic acid decreases.
E The weak acid is neutralized by the strong acid
and the pH of the solution is 7.00.
NH
HHBH H
HNO CO |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2008.pdf | 9 |
© 2008 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 9
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
←
9
8B
10
→
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57
La
138.9 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89
Ac
227.0 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Uun 111
Uuu 112
Uub 113
Uut
58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) DATA SHEET
CHEM 13 NEWS EXAM 2008
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 × 1023 mol−1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K−1 mol−1 0oC = 273.15 K
= 8.3145 kPa L K−1 mol−1
= 8.3145 J K−1 mol−1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) −± −=24
2bb a cx
a |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 1 | CHEM 13 NEWS EXAM 2007
UNIVERSITY OF WATERLOO
DEPARTMENT OF CHEMISTRY
10 MAY 2007 TIME: 75 MINUTES
This exam is being written by several thousand students. Please be sure that you follow the instructions below.
We'll send you a report on your performance. Top performers are eligible for a prize.
1. Print your name here:
2. Print your school name
and city on your STUDENT
RESPONSE sheet.
3. Select, and enter on the STUDENT RESPONSE sheet, one of the following CODE numbers:
Code 1 Ontario , now studying Grade 12 Chemistry
in a nonsemestered school
Code 2 Ontario , now studying Grade 12 Chemistry
in a semestered school
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already completed
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student
Code 6 Québec high school student
Code 7 Québec CEGEP student
Code 8 Alberta or British Columbia high school
student
Code 9 New Brunswick, Newfoundland, Nova Scotia,
or Prince Edward Island high school student
Code 10 Northwest Territories, Nunavut, or Yukon
high school student
Code 11 High school student outside Canada
Code 12 Teacher
4. Print
your name (last name, first name and optional
middle initial) on the STUDENT RESPONSE sheet .
Also fill in the corresponding circles below your printed
name.
5. Carefully detach the last page. It is the datasheet.
6. Now answer the exam questions. Questions are not in
order of difficulty. Indicate your choice on the STUDENT RESPONSE sheet by marking one letter beside the question number.
• Mark only one answer for each question. • Questions are all of the same value. • There is a penalty (1/4 off) for each incorrect
answer, but no penalty if you do not answer.
7. Take care that you make firm, black pencil marks, just
filling the oval.
Be careful that any erasures are complete—make the
sheet white again.
Carefully detach the last page.
It is the Data Sheet. |
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2 / CHEM 13 NEWS EXAM © 2007 UNIVERSITY OF WATERLOO 1 In which of the following series are the atomic orbitals
given in order of increasing energy?
A 3d, 4s, 4p, 4d, 4f, 5s
B 2s, 3s, 2p, 3p, 3d, 4s
C 4s, 3d, 4p, 4d, 4f, 5s
*D 4s, 3d, 4p, 5s, 4d, 5p
E 1s, 2s, 3s, 4s, 2p, 3p
2 What is the ground state electron configuration of Ar?
*A 1s2 2s2 2p6 3s2 3p6
B 1s2 2s2 2p6
C 1s2 2s2 3s2 3p6
D 1s2 2s2 2p3 3s2 3p3
E 1s2 1p6 2s2 2p6 3s2 3p6
3 Which of the following ions, in its ground electronic
state, does not
have the same electronic
configuration as a ground state Ar atom?
A P3−
B Cl−
C K+
D Ca2+
*E Sc2+
4 Which of the following molecules is linear?
A H 2O
B O 3
C NH 3
*D HCN
E HONO
5 Which of the following molecules has polar bonds but
is nonpolar?
A N
2H4
*B CCl
4
C HNO 3
D CH 2Cl2
E F 2O
6 Why is the boiling point of iodine chloride (I-Cl)
greater than that of bromine (Br 2)?
A ICl is heavier than Br 2.
B ICl is a covalent compound and Br 2 is not.
C The I-Cl bond is stronger than the Br −Br bond.
*D ICl is a polar molecule and Br 2 is nonpolar.
E ICl is an ionic compound and Br 2 is not.
7 What is the molecular geometry of phosphorus
pentachloride, PCl
5 ?
A square pyramidal
*B trigonal bipyramidal
C pentagonal
D trigonal pyramidal
E octahedral
8 Which of the following correctly characterizes the
bonds and geometry of C
2H4?
A four σ bonds, one π bond and an H-C-C bond
angle very close to 109o
B five σ bonds, no π bonds and an H-C-C bond
angle very close to 90o
* C five σ bonds, one π bond and an H-C-C bond
angle very close to 120o
D three σ bonds, two π bonds and an H-C-C bond
angle very close to 109o
E four σ bonds, two π bonds and an H-C-C bond
angle very close to 120o
CHEM 13 NEWS EXAM 2007 - Answers |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 3 |
© 2007 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 3
9 What is the oxidation state of rhenium (Re) in ReO
4−?
A 0
B +1
C +3
D +4
*E +7
10 What is the coefficient of zinc (Zn) when the equation
above for the reaction is balanced using the smallest
whole number coefficients?
A 1
B 2
*C 7
D 16
E none of the above
11 For the reaction above, what element or ion is the
reducing agent?
A Re( s)
*B Zn( s)
C ReO
4−(aq)
D Zn2+(aq)
E H+(aq)
12 In the galvanic cell shown below, what is the reaction
that occurs at the cathode?
A H
2(g) → 2H+(aq) + 2e−
B 2H+(aq) + 2e− → H2(g)
C Cu( s) → Cu2+(aq) + 2e−
*D Cu2+(aq) + 2e− → Cu(s)
E Pt( s) + H 2(g) + 4Cl−(aq)
→ PtCl 42−(aq) + 2H+(aq) + 4e−
13 In the statements below, X refers to one of Ca, Fe, Pb,
Cu or Pt. What is the identity of X?
• X( s) reacts spontaneously in 1 mol L
−1 HCl( aq) to
give XCl 2(aq) and H 2(g).
• The reaction 3X2+(aq) + 2Al( s) → 3X(s) + 2Al3+(aq)
is spontaneous under standard conditions.
• X( s) is a better reducing agent than Co( s) under
standard conditions.
A Ca
*B Fe
C Pb
D Cu
E Pt
Use the following information to
answer questions 9-11.
In acidic solution, zinc metal reacts spontaneously with
ReO 4−. The unbalanced chemical equation for the
reaction is given below.
Zn(s) + ReO 4−(aq) + H+(aq) → Re(s) + Zn2+(aq) + H 2O(l)
Half-reaction Eo
Ca2+(aq) + 2e− → Ca(s) −2.84 V
Al3+(aq) + 3e− → Al(s) −1.66 V
Fe2+(aq) + 2e− → Fe(s) −0.44 V
Co2+(aq) + 2e− → Co(s) −0.28 V
Pb2+(aq) + 2e− → Pb(s) −0.13 V
2H+(aq) + 2e− → H2(g) 0.00 V
Cu2+(aq) + 2e− → Cu(s) 0.34 V
Pt2+(aq) + 2e− → Pt(s) 1.18 VHCl(aq)
1 mol L−1 H2(g)
1 atm
Pt(s)Cu(s)
Cu(NO 3)2(aq)
1 mol L−1 salt bridge 0.34 V
e−e− |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 4 |
4 / CHEM 13 NEWS EXAM © 2007 UNIVERSITY OF WATERLOO 14 In the laboratory, one must never dip a stirring rod
into a reagent bottle. This is because
A the bottle may tip over
B the stirring rod might break
C the rod might puncture the bottle
* D the contents of the bottle may become
contaminated
E reagent can creep up the rod and come in contact
with one’s hand
15 What is the most accurate and precise way to
measure one litre of water?
A Use a 1-L graduated cylinder.
*B Use a 1-L volumetric flask.
C Use a 100-mL volumetric flask ten times.
D Use a 100-mL pipette ten times.
E Weigh 1 kg of water using a balance that weighs
to ±1 g.
16 Examine the diagrams below carefully. Which of the
burets shown below is/are ready for use?
(1) (2) (3) (4)
A (1) only
B (2) only
C (3) only
* D (4) only
E (1), (3) and (4)
17 An aqueous solution is 5.0% ethanoic acid (HC 2H3O2)
by mass and its density is 0.96 g mL−1. What is the
molar concentration of ethanoic acid in this solution?
* A 0.80 mol L−1
B 4.8 mol L−1
C 12 mol L−1
D 0.087 mol L−1
E 16 mol L−1
18 Which reagents react to give ethyl benzoate
(C
6H5COOC 2H5) and water? The structure of ethyl
benzoate is given below.
A and
* B and CH
3CH 2OH
C and C
6H5CH 2OH
D CH 3CH 2OH and C 6H5OH
E none of the above
19 Which of the following is not
a pair of isomers?
A ethyl benzene (C 6H5-C2H5) and
dimethyl benzene, C 6H4(CH 3)2
B 1-propanol (CH 3CH 2CH 2OH) and
2-propanol (CH 3CHOHCH 3)
C ethanol (C 2H5OH) and dimethyl ether (CH 3OCH 3)
* D 2-butanone (CH 3COCH 2CH 3) and
1-butanol (CH 3CH 2CH 2CH 2OH)
E urea (NH 2CONH 2) and ammonium cyanate
(NH 4CNO)
5049482 1 0
5049482 1 0
5049482 1 0
5049482 1 0 HC 2H3O2, 60.05 g mol−1
H3CCO
OH C6H5CO
OH
C6H5CO
OH
H3CCO
OHC6H5CO
OC H 2CH3 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 5 |
© 2007 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 5 20 What is the IUPAC name for the compound below?
A 2-chloro-1,4-dimethylpentane
B 3-chloro-1,1,4-trimethylbutane
*C 4-chloro-2-methylhexane
D 3-chloro-5-methylhexane
E 3-chloroheptane
21 Which of the following compounds is a solid at room
temperature?
A H-C≡C-H
B CH 3CH 2CH 3
C CH 3CH 2CH 2OH
D C 8H18
*E C 6H5OH
22 How many different structural isomers are there for the
compound chlorobutane (C
4H9Cl)?
A two
B three
* C four
D five
E more than five
23 According to the reaction profile below, what is ΔH for
the reaction 4HBr( g) + O
2(g) → 2H 2O(g) + 2Br 2(g)?
A 276 kJ
* B −276 kJ
C 434 kJ
D −434 kJ
E 158 kJ
24 The enthalpy change for the reaction below is
ΔH = −58 kJ (per mole of N
2O4 formed).
2 N O 2(g)
−⎯⎯⎯→←⎯⎯⎯1
1k
k N 2O4(g)
If k1 and k−1 are the rate constants for the forward and
reverse reactions, respectively, and Kc is the
equilibrium constant for the reaction as written, then what is the effect of adding a catalyst on the values of
k
1, k−1 and Kc?
A k1 increases, k−1 increases, Kc increases
B k1 decreases, k−1 decreases, Kc decreases
*C k1 increases, k−1 increases, Kc remains the same
D k1 decreases, k−1 decreases, Kc remains the same
E k1 remains the same, k−1 remains the same,
Kc remains the same
158 kJ
434 kJ
Reaction progress Potential
Energy
4HBr( g) + O 2(g)
2H2O(g) + 2Br 2(g) HCH
HCCHH
CCCC
H
H ClH
HHH
HH
H H
For 20, D is not correct because you must number the C atoms so that the
substituents have the lowest numbers possible (4+2 is less than 3+5). |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 6 |
6 / CHEM 13 NEWS EXAM © 2007 UNIVERSITY OF WATERLOO 25 The reaction below reaches equilibrium in a closed
reaction vessel.
CaCO 3(s) U CaO( s) + CO 2(g), ΔH = 178 kJ
Which of the following actions cause(s) an increase in
the partial pressure of CO 2(g)?
(i) increasing the temperature (ii) adding some CaCO
3(s)
(iii) increasing the volume of the reaction vessel
*A (i) only
B (i) and (ii)
C (i), (ii) and (iii)
D (ii) only
E (i) and (iii)
26 The reaction below was studied using the method of
initial rates.
−
3BrO (aq) + 5Br−(aq) + 6H+(aq) → 3Br 2(aq) + 3H 2O(l)
The rate law for the reaction was determined to be
Rate = k [−
3BrO ] [Br − ] [H+ ]2, where Rate refers to the
rate of consumption of−
3BrO . Which of the following
statements is false ?
* A If concentrations are measured in mol L−1 and
time is measured in seconds ( s), then the units of
k are mol L−1 s−1.
B The rate of consumption of Br − is five times
greater than the rate of consumption of −
3BrO .
C The conversion of reactants into products must
involve two or more simpler reactions.
D If the concentrations of all reactants are doubled,
the rate of consumption of−
3BrO will increase by a
factor of sixteen.
E When the reaction reaches a state of dynamic
equilibrium, [−
3BrO ] stops changing.
27
Which of the following reagents could be used to
separate the metal ions in an aqueous mixture of Fe(NO
3)3 and AgNO 3?
A NH 3
B KOH
*C NaCl
D
HNO 3
E
CaCO 3
28 The reaction below was studied using the method of
initial rates.
2 HgCl 2(aq) + C 2O42−(aq) → products
The following data were recorded. ( Rate refers to the
initial rate of consumption of C 2O42−.)
ExperimentInitial [HgCl 2]
(in mol L-1) Initial [C 2O42−]
(in mol L-1) Rate
(in mol L-1 hr-1)
1 0.0836 0.202 0.260
2 0.0836 0.404 1.04
3 0.0334 0.404 0.416
What is the rate law for the reaction?
*A Rate = k [HgCl 2] [C2O42− ]2
B Rate = k [HgCl 2]2
[C2O42− ]
C Rate = k [HgCl 2] [C2O42− ]
D Rate = k [HgCl 2]2
[C2O42− ]2
E Rate = k [HgCl 2]½
[C2O42− ]
29
A concentrated solution of ethanoic acid (HC 2H3O2)
has a concentration of 17.4 mol L−1. What volume of
this solution is needed to prepare 0.25 L of
0.30 mol L−1 HC 2H3O2(aq)?
A 4.7 mL
*B 4.3 mL
C 3.0 mL
D 2.5 mL
E 2.2 mL
|
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 7 |
© 2007 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 7 30 Which of the following is a valid set of quantum
numbers for an electron in a p orbital?
A n = 1, l = 1, ml = 0, ms = ½
B n = 3, l = 1, ml = 2, ms = ½
* C n = 2, l = 1, ml = −1, ms = ½
D n = 2, l = 0, ml = 0, ms = ½
E n = 2, l = 2, ml = 0, ms = ½
31 For the reaction below, ΔHo = −518.02 kJ per mole of
H2S. What is ΔHo
f for H 2S(g)?
H 2S(g) + 3
2 O2(g) → SO 2(g) + H 2O(g)
*A −20.63 kJ mol−1
B 41.26 kJ mol−1
C 20.63 kJ mol−1
D −497.39 kJ mol−1
E −41.26 kJ mol−1
32 What is the pH of 0.10 mol L−1 HClO 2(aq)?
A 1.98
B 5.11
* C 1.55
D 2.52
E 1.00
33 Consider the reaction below.
2 SO 2(g) + O 2(g) U 2 SO 3(g)
In an experiment, 0.10 mol of O 2 and 0.10 mol of SO 3
are added to an empty 1.0-L flask and then the flask
is sealed. Which of the following must be true at
equilibrium?
A [SO 2] = [O 2] = [SO 3]
B [O2] < [SO 3]
C [O2] = 2 [SO 2]
D [O 2] = [SO 2]
* E [SO 3] < [O 2] 34 Which of the following statements concerning the
structure below is true?
A There are eight σ bonds in this structure.
B The nitrogen atom is sp-hybridized.
C The H-C-H bond angle is 90o.
D The structure above is the most important
structure for the CH 3NCO molecule.
*
E None of the statements above are true.
35 When a 10.0-g sample of a mixture of CH 4 and C 2H6
is burned excess oxygen, exactly 525 kJ of heat is
produced. What is the percentage by mass of CH 4 in
the original mixture?
CH 4(g) + 2 O2(g) → CO 2(g) + 2 H2O(l)
∆H = −890.4 kJ (per mol CH 4)
C2H6(g) + 7
2 O2(g) → 2 CO 2(g) + 3 H2O(l)
∆H = −1560.0 kJ (per mol C 2H6)
* A 17%
B 21%
C 34%
D 59%
E 87%
36 Which of the following is an acceptable Lewis
structure for the thiocyanate ion, SCN− ?
A
B
C
*D
E ΔHo
f
(in kJ mol−1)
SO 2(g) −296.83
H2O(g) −241.82
Ka = 1.1×10−2 for HClO 2 CH
H
HNC O
CH 4, 16.042 g mol−1
C2H6, 30.068 g mol−1
SCN
SCN
SCNSCN
SCNThe reaction must
go ← to establish
equilibrium. |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 8 |
8 / CHEM 13 NEWS EXAM © 2007 UNIVERSITY OF WATERLOO 37 What is the pressure (in mmHg) of the gas inside the
apparatus below if P atm = 750 mmHg, Δh1 = 20 mm
and Δh2 = 50 mm?
A 20 mmHg
B 50 mmHg
C 700 mmHg
* D 730 mmHg
E 770 mmHg
38 Consider the compounds HF, HCl, HBr and HI. Of
these compounds, which one has the highest boiling
point and which one is the strongest acid in water?
A HF has the highest boiling point and is the
strongest acid
B HI has the highest boiling point and is the
strongest acid
*
C HF has the highest boiling point and HI is the
strongest acid
D HI has the highest boiling point and HF is the
strongest acid
E HI has the highest boiling point and HCl is the
strongest acid
39 Ethanoic acid, CH 3COOH, is a weak acid in water.
Which substance, when added to an aqueous
solution of ethanoic acid, causes both the pH and the percentage ionization of CH
3COOH to decrease ?
A NaCH 3COO
B NaCl
*
C CH 3COOH
D NaNO 3
E AgCl
40 A compound of carbon, hydrogen and oxygen is found
to be 52.13% carbon by mass, 13.13% hydrogen by mass, and 34.74% oxygen by mass. What is the simplest formula of the compound?
A C5H8O
B C3H4O3
* C C2H6O
D CH 2O2
E CHO
Δh1
Gas Patm
mercury (Hg) Δh2 H, 1.008 g mol−1
C, 12.01 g mol−1
O, 16.00 g mol−1 |
https://uwaterloo.ca/chemistry/sites/default/files/uploads/documents/chem-13-news-exam-solution-2007.pdf | 9 |
© 2007 UNIVERSITY OF WATERLOO CHEM 13 NEWS EXAM / 9
1
1A
18
8A
1
H
1.008
2
2A
13
3A
14
4A
15
5A
16
6A
17
7A 2
He
4.003
3
Li
6.941 4
Be
9.012 5
B
10.816
C
12.01 7
N
14.01 8
O
16.009
F
19.0010
Ne
20.18
11
Na
22.99 12
Mg
24.31
3
3B
4
4B
5
5B
6
6B
7
7B
8
←
9
8B
10
→
11
1B
12
2B 13
Al
26.9814
Si
28.09 15
P
30.97 16
S
32.0717
Cl
35.4518
Ar
39.95
19
K
39.10 20
Ca
40.08 21
Sc
44.96 22
Ti
47.88 23
V
50.94 24
Cr
52.00 25
Mn
54.9426
Fe
55.8527
Co
58.9328
Ni
58.6929
Cu
63.5530
Zn
65.3831
Ga
69.7232
Ge
72.59 33
As
74.92 34
Se
78.9635
Br
79.9036
Kr
83.80
37
Rb
85.47 38
Sr
87.62 39
Y
88.91 40
Zr
91.22 41
Nb
92.91 42
Mo
95.94 43
Tc
(98) 44
Ru
101.145
Rh
102.946
Pd
106.447
Ag
107.948
Cd
112.449
In
114.850
Sn
118.7 51
Sb
121.8 52
Te
127.653
I
126.954
Xe
131.3
55
Cs
132.9 56
Ba
137.3 57
La
138.9 72
Hf
178.5 73
Ta
180.9 74
W
183.9 75
Re
186.276
Os
190.277
Ir
192.278
Pt
195.179
Au
197.080
Hg
200.681
Tl
204.482
Pb
207.2 83
Bi
209.0 84
Po
(209) 85
At
(210) 86
Rn
(222)
87
Fr
(223) 88
Ra
226 89
Ac
227.0 104
Rf 105
Db 106
Sg 107
Bh 108
Hs 109
Mt 110
Uun 111
Uuu 112
Uub 113
Uut
58
Ce
140.1 59
Pr
140.9 60
Nd
144.2 61
Pm
(145) 62
Sm
150.463
Eu
152.0064
Gd
157.365
Tb
158.966
Dy
162.567
Ho
164.968
Er
167.3 69
Tm
168.9 70
Yb
173.071
Lu
175.0
90
Th
232.0 91
Pa
231.0 92
U
238.0 93
Np
237.094
Pu
(244) 95
Am
(243) 96
Cm
(247) 97
Bk
(247) 98
Cf
(251) 99
Es
(252) 100
Fm
(257) 101
Md
(258) 102
No
(259) 103
Lr
(260) DATA SHEET
CHEM 13 NEWS EXAM 2006
DETACH CAREFULLY
C o n s t a n t s : C o n v e r s i o n f a c t o r s :
NA = 6.022 × 1023 mol−1 1 atm = 101.325 kPa = 760 torr = 760 mm Hg
R = 0.082058 atm L K−1 mol−1 0oC = 273.15 K
= 8.3145 kPa L K−1 mol−1
= 8.3145 J K−1 mol−1
Kw = 1.0×10−14 (at 298 K)
F = 96 485 C mol−1
Equations: PV = nRT k t1/2 = 0.693 pH = pK a + log ( [base] / [acid] ) −± −=24
2bb a cx
a |
https://www.press.muni.cz/media/3019066/answers_to_all_questions.pdf | 1 | WC -1
Answers to All
Questions and Problems
Chapter 1
1.1 In a few sentences, what were Mendel’s key ideas about
inheritance?
ANS: Mendel postulated transmissible factors—genes—to
explain the inheritance of traits. He discovered that
genes exist in different forms, which we now call alleles.
Each organism carries two copies of each gene. During
reproduction, one of the gene copies is randomly incor-
porated into each gamete. When the male and female
gametes unite at fertilization, the gene copy number is
restored to two. Different alleles may coexist in an organ-
ism. During the production of gametes, they separate
from each other without having been altered by
coexistence.
1.2 Both DNA and RNA are composed of nucleotides. What
molecules combine to form a nucleotide?
ANS: Each nucleotide consists of a sugar, a nitrogen-containing
base, and a phosphate.
1.3 Which bases are present in DNA? Which bases are pres-
ent in RNA? Which sugars are present in each of these
nucleic acids?
ANS: The bases present in DNA are adenine, thymine, gua-
nine, and cytosine; the bases present in RNA are adenine,
uracil, guanine, and cytosine. The sugar in DNA is
deoxyribose; the sugar in RNA is ribose.
1.4 What is a genome?
ANS: A genome is the set of all the DNA molecules that are
characteristic of an organism. Each DNA molecule
forms one chromosome in a cell of the organism.
1.5 The sequence of a strand of DNA is ATTGCCGTC. If
this strand serves as the template for DNA synthesis,
what will be the sequence of the newly synthesized
strand?
ANS: TAACGGCAG
1.6 A gene contains 141 codons. How many nucleotides are
present in the gene’s coding sequence? How many amino
acids are expected to be present in the polypeptide
encoded by this gene?ANS: There are 3 × 141 = 423 nucleotides in the gene’s cod-
ing sequence. Its polypeptide product will contain 141
amino acids.
1.7 The template strand of a gene being transcribed is CTT-
GCCAGT. What will be the sequence of the RNA made
from this template?
ANS: GAACGGUCT
1.8 What is the difference between transcription and
translation?
ANS: T ranscription is the production of an RNA chain using a
DNA chain as a template. T ranslation is the production
of a chain of amino acids—that is, a polypeptide—using
an RNA chain as a template.
1.9 RNA is synthesized using DNA as a template. Is DNA
ever synthesized using RNA as a template? Explain.
ANS: Sometimes, DNA is synthesized from RNA in a process
called reverse transcription. This process plays an impor-
tant role in the life cycles of some viruses.
1.10 The gene for a-globin is present in all vertebrate species.
Over millions of years, the DNA sequence of this gene
has changed in the lineage of each species. Consequently,
the amino acid sequence of a-globin has also changed in
these lineages. Among the 141 amino acid positions in
this polypeptide, human a-globin differs from shark
a-globin in 79 positions; it differs from carp a-globin in
68 and from cow a-globin in 17. Do these data suggest
an evolutionary phylogeny for these vertebrate species?
ANS: The human and cow a-globins are least different; there-
fore, on the assumption that differences in a-globin
refl ect the degree of phylogenetic relationship, the
human and the cow are the most closely related organ-
isms among those mentioned. The next closest “relative”
of humans is the carp, and the most distant relative is the
shark.
1.11 Sickle-cell anemia is caused by a mutation in one of the
codons in the gene for b-globin; because of this mutation,
the sixth amino acid in the b-globin polypeptide is a
valine instead of a glutamic acid. A less severe type of ane-
mia is caused by a mutation that changes this same codon
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to one specifying lysine as the sixth amino acid in the
b-globin polypeptide. What word is used to describe the
two mutant forms of this gene? Do you think that an indi -
vidual carrying these two mutant forms of the b-globin
gene would suffer from anemia? Explain.
ANS: The two mutant forms of the b-globin gene are properly
described as alleles. Because neither of the mutant alleles
can specify a “normal” polypeptide, an individual who
carries each of them would probably suffer from
anemia.
1.12 Hemophilia is an inherited disorder in which the blood-
clotting mechanism is defective. Because of this defect,
people with hemophilia may die from cuts or bruises,
especially if internal organs such as the liver, lungs, or
kidneys have been damaged. One method of treatment
involves injecting a blood-clotting factor that has been
purified from blood donations. This factor is a protein
encoded by a human gene. Suggest a way in which mod -
ern genetic technology could be used to produce this
factor on an industrial scale. Is there a way in which the
inborn error of hemophilia could be corrected by human
gene therapy?
ANS: The gene for the human clotting factor could be isolated
from the human genome and transferred into bacteria,
which could then be grown in vats to produce large
amounts of the gene’s protein product. This product could
be isolated from the bacteria, purified, and then injected
into patients to treat hemophilia. Another approach would
be to transfer a normal copy of the clotting factor gene
into the cells of people who have hemophilia. If expressed
properly, the transferred normal gene might be able to
compensate for the mutant allele these people naturally
carry. For this approach to succeed, the normal clotting
factor gene would have to be transferred into the cells that
produce clotting factor, or into their precursors.
Chapter 2
2.1 Carbohydrates and proteins are linear polymers. What
types of molecules combine to form these polymers?
ANS: Sugars combine to form carbohydrates; amino acids
combine to form proteins.
2.2 All cells are surrounded by a membrane; some cells are
surrounded by a wall. What are the differences between
cell membranes and cell walls?
ANS: Cell membranes are made of lipids and proteins; they
have a fluid structure. Cell walls are made of more rigid
materials such as cellulose.
2.3 What are the principal differences between prokaryotic
and eukaryotic cells?
ANS: In a eukaryotic cell, the many chromosomes are con -
tained within a membrane-bounded structure called the
nucleus; the chromosomes of prokaryotic cells are not
contained within a special subcellular compartment. Eukaryotic cells usually possess a well-developed inter -
nal system of membranes and they also have membrane-
bounded subcellular organelles such as mitochondria
and chloroplasts; prokaryotic cells do not typically have
a system of internal membranes (although some do), nor
do they possess membrane-bounded organelles.
2.4 Distinguish between the haploid and diploid states.
What types of cells are haploid? What types of cells are
diploid?
ANS: In the haploid state, each chromosome is represented
once; in the diploid state, each chromosome is repre -
sented twice. Among multicellular eukaryotes, gam -
etes are haploid and somatic cells are diploid.
2.5 Compare the sizes and structures of prokaryotic and
eukaryotic chromosomes.
ANS: Prokaryotic chromosomes are typically (but not always)
smaller than eukaryotic chromosomes; in addition, pro -
karyotic chromosomes are circular, whereas eukaryotic
chromosomes are linear. For example, the circular chro -
mosome of E. coli , a prokaryote, is about 1.4 mm in cir -
cumference. By contrast, a linear human chromosome
may be 10–30 cm long. Prokaryotic chromosomes also
have a comparatively simple composition: DNA, some
RNA, and some protein. Eukaryotic chromosomes are
more complex: DNA, some RNA, and a lot of protein.
2.6 With a focus on the chromosomes, what are the key
events during interphase and M phase in the eukaryotic
cell cycle?
ANS: During interphase, the chromosomes duplicate. During
M phase (mitosis), the duplicated chromosomes, each
consisting of two identical sister chromatids, condense
(a feature of prophase), migrate to the equatorial plane of
the cell (a feature of metaphase), and then split so that
their constituent sister chromatids are separated into dif -
ferent daughter cells (a feature of anaphase); this last
process is called sister chromatid disjunction.
2.7 Which typically lasts longer, interphase or M phase? Can
you explain why one of these phases lasts longer than the
other?
ANS: Interphase typically lasts longer than M phase. During
interphase, DNA must be synthesized to replicate all the
chromosomes. Other materials must also be synthesized
to prepare for the upcoming cell division.
2.8 In what way do the microtubule organizing centers of
plant and animal cells differ?
ANS: The microtubule organizing centers of animal cells have
distinct centrosomes, whereas the microtubule organiz -
ing centers of plant cells do not.
2.9 Match the stages of mitosis with the events they encom -
pass: Stages: (1) anaphase, (2) metaphase, (3) prophase,
and (4) telophase. Events: (a) reformation of the nucleo -
lus, (b) disappearance of the nuclear membrane,
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(c) condensation of the chromosomes, (d) formation of
the mitotic spindle, (e) movement of chromosomes to
the equatorial plane, (f) movement of chromosomes to
the poles, (g) decondensation of the chromosomes, (h)
splitting of the centromere, and (i) attachment of micro -
tubules to the kinetochore.
ANS: (1) Anaphase: (f), (h); (2) metaphase: (e), (i); (3) prophase:
(b), (c), (d); (4) telophase: (a), (g).
2.10 Arrange the following events in the correct temporal
sequence during eukaryotic cell division, starting with
the earliest: (a) condensation of the chromosomes,
(b) movement of chromosomes to the poles, (c) duplica -
tion of the chromosomes, (d) formation of the nuclear
membrane, (e) attachment of microtubules to the kineto -
chores, and (f) migration of centrosomes to positions on
opposite sides of the nucleus.
ANS: (c), (f), (a), (e), (b), (d).
2.11 In human beings, the gene for b-globin is located on chro -
mosome 11, and the gene for a-globin, which is another
component of the hemoglobin protein, is located on chro -
mosome 16. Would these two chromosomes be expected to
pair with each other during meiosis? Explain your answer.
ANS: Chromosomes 11 and 16 would not be expected to pair
with each other during meiosis; these chromosomes are
heterologues, not homologues.
2.12 A sperm cell from the fruit fly Drosophila melanogaster
contains four chromosomes. How many chromosomes
would be present in a spermatogonial cell about to enter
meiosis? How many chromatids would be present in a
spermatogonial cell at metaphase I of meiosis? How
many would be present at metaphase II?
ANS: There are eight chromosomes in a Drosophila spermato -
gonial cell about to enter meiosis. There are 16 chroma -
tids in a Drosophilia spermatogonial cell at metaphase I of
meiosis. There are eight chromatids in a Drosophilia cell
at metaphase II of meiosis.
2.13 Does crossing over occur before or after chromosome
duplication in cells going through meiosis?
ANS: Crossing over occurs after chromosomes have duplicated
in cells going through meiosis.
2.14 What visible characteristics of chromosomes indicate
that they have undergone crossing over during meiosis?
ANS: The chiasmata, which are visible late in prophase I of
meiosis, indicate that chromosomes have crossed over.
2.15 During meiosis, when does chromosome disjunction
occur? When does chromatid disjunction occur?
ANS: Chromosome disjunction occurs during anaphase I.
Chromatid disjunction occurs during anaphase II.
2.16 In Arabidopsis , is leaf tissue haploid or diploid? How
many nuclei are present in the female gametophyte? How many are present in the male gametophyte? Are
these nuclei haploid or diploid?
ANS: Leaf tissue is diploid. The female gametophyte contains
eight identical haploid nuclei. The male gametophyte
contains three identical haploid nuclei.
2.17 From the information given in T able 2.1 in this chapter,
is there a relationship between genome size (measured in
base pairs of DNA) and gene number? Explain.
ANS: Among eukaryotes, there does not seem to be a clear
relationship between genome size and gene number. For
example, humans, with 3.2 billion base pairs of genomic
DNA, have about 20,500 genes, and Arabidopsis plants,
with about 150 million base pairs of genomic DNA, have
roughly the same number of genes as humans. However,
among prokaryotes, gene number is rather tightly cor -
related with genome size, probably because there is so
little nongenic DNA.
2.18 Are the synergid cells in an Arabidopsis female gameto -
phyte genetically identical to the egg cell nestled between
them?
ANS: Yes.
2.19 A cell of the bacterium Escherichia coli , a prokaryote, con -
tains one chromosome with about 4.6 million base pairs
of DNA comprising 4288 protein-encoding genes. A cell
of the yeast Saccharomyces cerevisiae , a eukaryote, contains
about 12 million base pairs of DNA comprising 6268
genes, and this DNA is distributed over 16 distinct chro -
mosomes. Are you surprised that the chromosome of a
prokaryote is larger than some of the chromosomes of a
eukaryote? Explain your answer.
ANS: It is a bit surprising that yeast chromosomes are, on aver -
age, smaller than E. coli chromosomes because, as a rule,
eukaryotic chromosomes are larger than prokaryotic
chromosomes. Yeast is an exception because its genome—
not quite three times the size of the E. coli genome—is
distributed over 16 separate chromosomes.
2.20 Given the way that chromosomes behave during meiosis,
is there any advantage for an organism to have an even
number of chromosome pairs (such as Drosophila does),
as opposed to an odd number of chromosome pairs (such
as human beings do)?
ANS: No, there is no advantage associated with an even num -
ber of chromosomes. As long as the chromosomes come
in pairs, they will be able to synapse during prophase I
and then disjoin during anaphase I to distribute the
genetic material properly to the two daughter cells.
2.21 In flowering plants, two nuclei from the pollen grain par -
ticipate in the events of fertilization. With which nuclei
from the female gametophyte do these nuclei combine?
What tissues are formed from the fertilization events?
ANS: One of the pollen nuclei fuses with the egg nucleus in the
female gametophyte to form the zygote, which then
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develops into an embryo and ultimately into a sporo -
phyte. The other genetically functional pollen nucleus
fuses with two nuclei in the female gametophyte to form
a triploid nucleus, which then develops into a triploid
tissue, the endosperm; this tissue nourishes the develop -
ing plant embryo.
2.22 The mouse haploid genome contains about 2.9 × 109
nucleotide pairs of DNA. How many nucleotide pairs of
DNA are present in each of the following mouse cells:
(a) somatic cell, (b) sperm cell, (c) fertilized egg, (d) pri -
mary oocyte, (e) first polar body, and (f) secondary
spermatocyte?
ANS: (a) 5.8 × 109 nucleotide pairs (np); (b) 2.9 × 109 np;
(c) 5.8 × 109 np; (d) 11.6 × 109 np; (e) 5.8 × 109 np;
and (f) 5.8 × 109 np
2.23 Arabidopsis plants have 10 chromosomes (five pairs) in
their somatic cells. How many chromosomes are present
in each of the following: (a) egg cell nucleus in the female
gametophyte, (b) generative cell nucleus in a pollen
grain, (c) fertilized endosperm nucleus, and (d) fertilized
egg nucleus?
ANS: (a) 5, (b) 5, (c) 15, (d) 10.
Chapter 3
3.1 On the basis of Mendel’s observations, predict the results
from the following crosses with peas: (a) a tall (dominant
and homozygous) variety crossed with a dwarf variety;
(b) the progeny of (a) self-fertilized; (c) the progeny from
(a) crossed with the original tall parent; (d) the progeny
of (a) crossed with the original dwarf parent.
ANS: (a) All tall; (b) 3/4 tall, 1/4 dwarf; (c) all tall; (d) 1/2 tall,
1/2 dwarf.
3.2 Mendel crossed pea plants that produced round seeds
with those that produced wrinkled seeds and self-fertil -
ized the progeny. In the F2, he observed 5474 round
seeds and 1850 wrinkled seeds. Using the letters W and
w for the seed texture alleles, diagram Mendel’s crosses,
showing the genotypes of the plants in each generation.
Are the results consistent with the Principle of
Segregation?
ANS: Round ( WW ) × wrinkled ( ww) → F1 round ( Ww); F1
self-fertilized → F2 3/4 round (2 WW ; 1 Ww), 1/4 wrin -
kled ( ww). The expected results in the F2 are 5493 round,
1831 wrinkled. T o compare the observed and expected
results, compute c2 with one degree of freedom;
(5474 − 5493)2/5493 = (1850 − 1831)2/1831 = 0.263,
which is not significant at the 5% level. Thus, the results
are consistent with the Principle of Segregation.
3.3 A geneticist crossed wild, gray-colored mice with white
(albino) mice. All the progeny were gray. These progeny
were intercrossed to produce an F2, which consisted of
198 gray and 72 white mice. Propose a hypothesis to explain these results, diagram the crosses, and compare
the results with the predictions of the hypothesis.
ANS: The data suggest that coat color is controlled by a single
gene with two alleles, C (gray) and c (albino), and that C
is dominant over c. On this hypothesis, the crosses are
gray ( CC) × albino ( cc) → F1 gray ( Cc); F1 × F1 → 3/4
gray (2 CC: 1 Cc), 1/4 albino ( cc). The expected results in
the F2 are 203 gray and 67 albino. T o compare the
observed and expected results, compute c2 with one
degree of freedom: (198 − 203)2/203 + (67 − 72)2/72 =
0.470, which is not significant at the 5% level. Thus, the
results are consistent with the hypothesis.
3.4 A woman has a rare abnormality of the eyelids called
ptosis, which prevents her from opening her eyes com -
pletely. This condition is caused by a dominant allele,
P. The woman’s father had ptosis, but her mother had
normal eyelids. Her father’s mother had normal
eyelids.
(a) What are the genotypes of the woman, her father, and
her mother?
(b) What proportion of the woman’s children will have
ptosis if she marries a man with normal eyelids?
ANS: (a) Woman’s genotype Pp, father’s genotype Pp, mother’s
genotype pp; (b) ½
3.5 In pigeons, a dominant allele C causes a checkered pat -
tern in the feathers; its recessive allele c produces a plain
pattern. Feather coloration is controlled by an indepen -
dently assorting gene; the dominant allele B produces
red feathers, and the recessive allele b produces brown
feathers. Birds from a true-breeding checkered, red vari -
ety are crossed to birds from a true-breeding plain,
brown variety.
(a) Predict the phenotype of their progeny.
(b) If these progeny are intercrossed, what phenotypes
will appear in the F2 and in what proportions?
ANS: (a) Checkered, red ( CC BB ) × plain, brown ( cc bb) → F1
all checkered, red ( Cc Bb ); (b) F2 progeny: 9/16 check -
ered, red ( C- B-), 3/16 plain, red ( cc B-), 3/16 checkered,
brown ( C- bb), 1/16 plain, brown ( cc bb).
3.6 In mice, the allele C for colored fur is dominant over the
allele c for white fur, and the allele V for normal behavior
is dominant over the allele v for waltzing behavior, a
form of dis-coordination. Given the genotypes of the
parents in each of the following crosses:
(a) Colored, normal mice mated with white, normal mice
produced 29 colored, normal, and 10 colored, waltzing
progeny
(b) Colored, normal mice mated with colored, normal
mice produced 38 colored, normal, 15 colored, waltzing,
11 white, normal, and 4 white, waltzing progeny
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(c) Colored, normal mice mated with white, waltzing
mice produced 8 colored, normal, 7 colored, waltzing,
9 white, normal, and 6 white, waltzing progeny.
ANS: (a) colored, normal ( CC Vv ) × white, normal ( cc Vv )
(b) colored, normal ( Cc Vv ) × colored, normal ( Cc Vv );
(c) colored, normal ( Cc Vv ) × white, waltzing ( cc vv).
3.7 In rabbits, the dominant allele B causes black fur and the
recessive allele b causes brown fur; for an independently
assorting gene, the dominant allele R causes long fur and
the recessive allele r (for rex) causes short fur. A homozy -
gous rabbit with long, black fur is crossed with a rabbit
with short, brown fur, and the offspring are intercrossed.
In the F2, what proportion of the rabbits with long, black
fur will be homozygous for both genes?
ANS: Among the F2 progeny with long, black fur, the geno -
typic ratio is 1 BB RR : 2 BB Rr : 2 Bb RR : 4 Bb Rr ; thus, 1/9
of the rabbits with long, black fur are homozygous for
both genes.
3.8 In shorthorn cattle, the genotype RR causes a red coat,
the genotype rr causes a white coat, and the genotype
Rr causes a roan coat. A breeder has red, white, and
roan cows and bulls. What phenotypes might be
expected from the following matings and in what
proportions?
(a) Red × red
(b) Red × roan
(c) Red × white
(d) Roan × roan.
ANS: (a) All red; (b) 1/2 red, 1/2 roan; (c) all roan; (d) 1/4 red,
1/2 roan, 1/4 white
3.9 How many different kinds of F1 gametes, F2 genotypes,
and F2 phenotypes would be expected from the following
crosses:
(a) AA × aa;
(b) AA BB × aa bb ;
(c) AA BB CC × aa bb cc ?
(d) What general formulas are suggested by these
answers?
ANS:
F1 Gametes F2 Genotypes F2 Phenotypes
(a) 2 3 2
(b) 2 × 2 = 4 3 × 3 = 9 2 × 2 = 4
(c) 2 × 2 × 2 = 8 3 × 3 × 3 = 27 2 × 2 × 2 = 8
(d) 2n 3n 2n, where n is the
number of genes 3.10 A researcher studied six independently assorting genes in
a plant. Each gene has a dominant and a recessive allele:
R black stem, r red stem; D tall plant, d dwarf plant; C full
pods, c constricted pods; O round fruit, o oval fruit; H
hairless leaves, h hairy leaves; W purple flower, w white
flower. From the cross (P1) Rr Dd cc Oo Hh Ww × (P2)
Rr dd Cc oo Hh ww,
(a) How many kinds of gametes can be formed by P1?
(b) How many genotypes are possible among the prog -
eny of this cross?
(c) How many phenotypes are possible among the
progeny?
(d) What is the probability of obtaining the Rr Dd cc Oo
hh ww genotype in the progeny?
(e) What is the probability of obtaining a black, dwarf,
constricted, oval, hairy, purple phenotype in the
progeny?
ANS: (a) 2 × 2 × 1 × 2 × 2 × 2 = 32; (b) 3 × 2 × 2 × 2 × 3 ×
2 = 144; (c) 2 × 2 × 2 × 2 × 2 × 2 = 64; (d) (1/2) ×
(1/2) × (1/2) × (1/2) × (1/4) × (1/2) = 1/128; (e) (3/4) ×
(1/2) × (1/2) × (1/2) × (1/4) × (1/2) = 3/256.
3.11 For each of the following situations, determine the
degrees of freedom associated with the c2 statistic and
decide whether or not the observed c2 value warrants
acceptance or rejection of the hypothesized genetic ratio.
Hypothesized Ratio Observed− c2
(a) 3:1 7.0
(b) 1:2:1 7.0
(c) 1:1:1:1 7.0
(d) 9:3:3:1 5.0
ANS: (a) 1, reject; (b) 2, reject; (c) 3, accept; (d) 3, accept.
3.12 Mendel testcrossed pea plants grown from yellow, round
F1 seeds to plants grown from green, wrinkled seeds and
obtained the following results: 31 yellow, round; 26
green, round; 27 yellow, wrinkled; and 26 green, wrin -
kled. Are these results consistent with the hypothesis that
seed color and seed texture are controlled by indepen -
dently assorting genes, each segregating two alleles?
ANS: On the hypothesis, the expected number in each class
is 27.5; c2 with three degrees of freedom is calculated as
(31 − 27.5)2/27.5 + (26 − 27.5)2/27.5 + (27 − 27.5)2/27.5 +
(26 − 27.5)2/27.5 = 0.618, which is not significant at the
5% level. Thus, the results are consistent with the hypoth -
esis of two independently assorting genes, each segregating
two alleles.
3.13 Perform a chi-square test to determine if an observed
ratio of 30 tall to 20 dwarf pea plants is consistent with
an expected ratio of 1:1 from the cross Dd × dd.
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ANS: c2 = (30 − 25)2/25 + (20 − 25)2/25 = 2, which is less
than 3.84, the 5 percent critical value for a chi-square
statistic with one degree of freedom; consequently, the
observed segregation ratio is consistent with the expected
ratio of 1:1.
3.14 Seed capsules of the Shepherd’s purse are either triangu -
lar or ovoid. A cross between a plant with triangular seed
capsules and a plant with ovoid seed capsules yielded F1
hybrids that all had triangular seed capsules. When these
F1 hybrids were intercrossed, they produced 80 F2 plants,
72 of which had triangular seed capsules and 8 of which
had ovoid seed capsules. Are these results consistent with
the hypothesis that capsule shape is determined by a sin -
gle gene with two alleles?
ANS: If capsule shape is determined by a single gene with two
alleles, the F2 plants should segregate in a 3:1 ratio. T o
test for agreement between the observed segregation
data and the expected ratio, compute the expected num -
ber of plants with either triangular or ovoid seed cap -
sules: (3/4) × 80 = 60 triangular and (1/4) × 80 = 20
ovoid; then compute a c2 statistic with one degree of
freedom: c2 = (72 − 60)2/60 + (8 − 20)2/20 = 9.6, which
exceeds the critical value of 3.84. Consequently, the data
are inconsistent with the hypothesis that capsule shape is
determined by a single gene with two alleles.
3.15 Albinism in humans is caused by a recessive allele a.
From marriages between people known to be carriers
(Aa) and people with albinism ( aa), what proportion of
the children would be expected to have albinism? Among
three children, what is the chance of one without albi -
nism and two with albinism?
ANS: Half the children from Aa × aa matings would have
albinism. In a family of three children, the chance that
one will be unaffected and two affected is 3 × (1/2)1 ×
(1/2)2 = 3/8.
3.16 If both husband and wife are known to be carriers of the
allele for albinism, what is the chance of the following
combinations in a family of four children: (a) all four
unaffected; (b) three unaffected and one affected; (c) two
unaffected and two affected; (d) one unaffected and three
affected?
ANS: (a) (3/4)4 = 81/256; (b) 4 × (3/4)3 × (1/4)1 = 108/256; (c)
6 × (3/4)2 × (1/4)2 = 54/256; (d) 4 × (3/4)1 × (1/4)3 =
12/256.
3.17 In humans, cataracts in the eyes and fragility of the bones
are caused by dominant alleles that assort independently.
A man with cataracts and normal bones marries a woman
without cataracts but with fragile bones. The man’s
father had normal eyes, and the woman’s father had nor -
mal bones. What is the probability that the first child of
this couple will (a) be free from both abnormalities; (b)
have cataracts but not have fragile bones; (c) have fragile
bones but not have cataracts; (d) have both cataracts and
fragile bones?ANS: Man ( Cc ff ) × woman ( cc Ff ). (a) cc ff, (1/2) × (1/2) = 1/4;
(b) Cc ff, (1/2) × (1/2) = 1/4; (c) cc Ff, (1/2) × (1/2) = 1/4;
(d) Cc Ff , (1/2) × (1/2) = 1/4.
3.18 In generation V in the pedigree in Figure 3.15, what is
the probability of observing seven children without the
cancer-causing mutation and two children with this
mutation among a total of nine children?
ANS: 9!/(7! 2!) × (1/2)7 × (1/2)2 = 0.07
3.19 If a man and a woman are heterozygous for a gene, and if
they have three children, what is the chance that all three
will also be heterozygous?
ANS: (1/2)3 = 1/8
3.20 If four babies are born on a given day: (a) What is the
chance that two will be boys and two will be girls?
(b) What is the chance that all four will be girls? (c) What
combination of boys and girls among four babies is most
likely? (d) What is the chance that at least one baby will
be a girl?
ANS: (a) 4 × (1/2)2 × (1/2)2 = 4/16; (b) (1/2)4 = 1/16; (c) 2
boys, girls; (d) 1 − probability that all four are boys = 1
− (1/2)4 = 15/16.
3.21 In a family of six children, what is the chance that at least
three are girls?
ANS: (20/64) + (15/64) + (6/64) + (1/64) = 42/64
3.22 The following pedigree shows the inheritance of a domi -
nant trait. What is the chance that the offspring of the
following matings will show the trait: (a) III-1 × III-3;
(b) III-2 × III-4?
I
II
III2 1
34 2 1
34 2 1
ANS: (a) zero; (b) 1/2
3.23 The following pedigree shows the inheritance of a reces -
sive trait. Unless there is evidence to the contrary, assume
that the individuals who have married into the family do
not carry the recessive allele. What is the chance that the
offspring of the following matings will show the trait: (a)
III-1 × III-12; (b) II-4 × III-14; (c) III-6 × III-13; (d)
IV-1 × IV-2?
I
12
1
12342II
III3
56784 7
14 15 16 178 5
109
2 111 12 136
ANS: (a) (1/2) × (1/4) = 1/8; (b) (1/2) × (1/2) × (1/4) = 1/16;
(c) (2/3) × (1/4) = 1/6; (d) (2/3) × (1/2) × (1/2) × (1/4)
= 1/24
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3.24 In the following pedigrees, determine whether the trait is
more likely to be due to a dominant or a recessive allele.
Assume the trait is rare in the population.
I
12
1
1223II
III
IV
V
(a)5
346
2345 1
1
I
II
III
IV12
1 234
(b)5
45 12 36 78
12 3
ANS: (a) Recessive; (b) dominant.
3.25 In pedigree ( b) of Problem 3.24, what is the chance that
the couple III-1 and III-2 will have an affected child?
What is the chance that the couple IV-2 and IV-3 will
have an affected child?
ANS: For III-1 × III-2, the chance of an affected child is 1/2.
For IV-2 × IV-3, the chance is zero.
3.26 Peas heterozygous for three independently assorting
genes were intercrossed.
(a) What proportion of the offspring will be homozy -
gous for all three recessive alleles?
(b) What proportion of the offspring will be homozy -
gous for all three genes?
(c) What proportion of the offspring will be homozy -
gous for one gene and heterozygous for the other two?
(d) What proportion of the offspring will be homozy -
gous for the recessive allele of at least one gene?
ANS: (a) (1/4)3 = 1/64; (b) (1/2)3 = 1/8; (c) 3 × (1/2)1 ×
(1/2)2 = 3/8; (d) 1 − probability that the offspring is not
homozygous for the recessive allele of any gene = 1 − (3/4)3
= 37/64.
3.27 The following pedigree shows the inheritance of a reces -
sive trait. What is the chance that the couple III-3 and
III-4 will have an affected child?
I
12
12II
III
24 13ANS: 1/2
3.28 A geneticist crosses tall pea plants with short pea plants.
All the F1 plants are tall. The F1 plants are then allowed
to self-fertilize, and the F2 plants are classified by height:
62 tall and 26 short. From these results, the geneticist
concludes that shortness in peas is due to a recessive
allele ( s) and that tallness is due to a dominant allele ( S).
On this hypothesis, 2/3 of the tall F2 plants should be
heterozygous Ss. T o test this prediction, the geneticist
uses pollen from each of the 62 tall plants to fertilize the
ovules of emasculated flowers on short pea plants. The
next year, three seeds from each of the 62 crosses are
sown in the garden and the resulting plants are grown to
maturity. If none of the three plants from a cross is short,
the male parent is classified as having been homozygous
SS; if at least one of the three plants from a cross is short,
the male parent is classified as having been heterozygous
Ss. Using this system of progeny testing, the geneticist
concludes that 29 of the 62 tall F2 plants were homozy -
gous SS and that 33 of these plants were heterozygous Ss.
(a) Using the chi-square procedure, evaluate these results
for goodness of fit to the prediction that 2/3 of the tall F2
plants should be heterozygous.
(b) Informed by what you read in A Milestone in Genet -
ics: Mendel’s 1866 Paper, which you can find in the Stu -
dent Companion Site, explain why the geneticist’s
procedure for classifying tall F2 plants by genotype is not
definitive.
(c) Adjust for the uncertainty in the geneticist’s classifi -
cation procedure and calculate the expected frequencies
of homozygotes and heterozygotes among the tall
F2 plants.
(d) Evaluate the predictions obtained in (c) using the chi-
square procedure.
ANS: (a) The observed numbers, expected numbers, and chi-
square calculation are laid out in the following table:
Observed Expected (Obs − Exp)2
/Exp
Dominant
homozygotes
(SS)29 62 × 1/3 = 20.7 3.33
Heterozygotes
(Ss)33 62 × 2/3 = 41.3 1.67
T otal 62 62 5.00
The total chi-square value is greater than the critical
value for a chi-square statistic with one degree of free -
dom (3.84). Therefore, we reject the hypothesis that the
expected proportions are 1/3 and 2/3.
(b) The problem with the geneticist’s classification
procedure is that it allows for a heterozygote to be
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misclassified as a homozygote if none of its three prog -
eny shows the recessive (short) phenotype. The proba -
bility of this event is 1/2 for any one offspring—therefore
(1/2)3 = 1/8 for all three offspring.
(c) The predicted frequencies must take into account the
probability of misclassifying a heterozygote as a homo -
zygote. The frequency of heterozygotes expected a priori
(2/3) must be decreased by the probability of misclassifi -
cation (1/8); thus, the predicted frequency of heterozy -
gotes is 62 × (2/3) × (1 − 1/8) = 62 × (7/12) = 36.2.
The predicted frequency of homozygotes is obtained by
subtraction: 62 – 36.2 = 25.8.
(d) The chi-square calculation is (29 − 25.8)2/25.8 + (33
− 36.2)2/36.2 = 0.68, which is much less than the critical
value for a chi-square statistic with one degree of free -
dom. Therefore, we tentatively accept the idea that
adjusting for the probability of misclassification explains
the observed data.
3.29 A researcher who has been studying albinism has identi -
fied a large group of families with four children in which
at least one child shows albinism. None of the parents in
this group of families shows albinism. Among the chil -
dren, the ratio of those without albinism to those with
albinism is 1.7:1. The researcher is surprised by this
result because he thought that a 3:1 ratio would be
expected on the basis of Mendel’s Principle of Segrega -
tion. Can you explain the apparently non-Mendelian
segregation ratio in the researcher’s data?
ANS: The researcher has obtained what appears to be a non-
Mendelian ratio because he has been studying only fami -
lies in which at least one child shows albinism. In these
families, both parents are heterozygous for the mutant
allele that causes albinism. However, other couples in the
population might also be heterozygous for this allele but,
simply due to chance, have failed to produce a child with
albinism. If a man and a woman are both heterozygous
carriers of the mutant allele, the chance that a child they
produce will not have albinism is 3/4. The chance that
four children they produce will not have albinism is
therefore (3/4)4 = 0.316. In the entire population of fami -
lies in which two heterozygous parents have produced a
total of four children, the average number of affected
children is 1. Among families in which two heterozygous
parents have produced at least one affected child among a
total of four children, the average must be greater than 1.
T o calculate this conditional average , let us denote the
number of children with albinism by x, and the probabil -
ity that exactly x of the four children have albinism by
P(x). The average number of affected children among
families in which at least one of the four children is
affected—that is, the conditional average—is therefore
SxP(x)/(1 − P(0)), where the sum starts at x = 1 and ends
at x = 4. We start the sum at x = 1 because we must
exclude those cases in which none of the four children is
affected. The divisor (1 − P(0)) is the probability that the
couple has had at least one affected child among their four children. Now P(0) = 0.316 and SxP(x) = 1. There -
fore, the average we seek is simply 1/(1 − 0.316) = 1.46.
If, in the subset of families with at least one affected child,
the average number of affected children is 1.46, then the
average number of unaffected children is 4 – 1.46 = 2.54.
Thus, the expected ratio of unaffected to affected chil -
dren in these families is 2.54:1.46, or 1.74:1, which is
what the researcher has observed.
Chapter 4
4.1 What blood types could be observed in children born to
a woman who has blood type M and a man who has
blood type MN?
ANS: M and MN.
4.2 In rabbits, coloration of the fur depends on alleles of the
gene c. From information given in the chapter, what phe -
notypes and proportions would be expected from the fol -
lowing crosses: (a) c+c+ × cc; (b) c+c × c+c; (c) c+c h × c+c ch;
(d) cc ch × cc; (e) c+c h × c+c; (f) c hc × cc?
ANS: (a) All wild-type; (b) 3/4 wild-type, 1/4 albino; (c) 3/4
wild-type, 1/4 chinchilla; (d) 1/2 chinchilla, 1/2 albino;
(e) 3/4 wild-type, 1/4 Himalayan; (f) 1/2 Himalayan, 1/2
albino.
4.3 In mice, a series of five alleles determines fur color. In
order of dominance, these alleles are as follows: AY, yel-
low fur but homozygous lethal; AL, agouti with light
belly; A+, agouti (wild-type); at, black and tan; and a,
black. For each of the following crosses, give the coat
color of the parents and the phenotypic ratios expected
among the progeny: (a) AYAL × AYAL; (b) AYa × ALat;
(c) ata × AYa; (d) ALat × ALAL; (e) ALAL × AYA+; (f) A+at ×
ata; (g) ata × aa; (h) AYAL × A+at; and (i) AYaL × AYA+.
ANS:
Parents Offspring
(a) Yellow × yellow 2 yellow: 1 light belly
(b) Yellow × light belly 2 yellow: 1 light belly:
1 black and tan
(c) Black and tan × yellow 2 yellow: 1 black and tan:
1 black
(d) Light belly × light belly All light belly
(e) Light belly × yellow 1 yellow: 1 light belly
(f) Agouti × black and tan 1 agouti: 1 black and tan
(g) Black and tan × black 1 black and tan: 1 black
(h) Yellow × agouti 1 yellow: 1 light belly
(i) Yellow × yellow 2 yellow: 1 light belly
4.4 In several plants, such as tobacco, primrose, and red clo -
ver, combinations of alleles in eggs and pollen have been
found to influence the reproductive compatibility of the
plants. Homozygous combinations, such as S1 S1, do not
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develop because S1 pollen is not effective on S1 stigmas.
However, S1 pollen is effective on S2 S3 stigmas. What
progeny might be expected from the following crosses
(seed parent written first): (a) S1 S2 × S2 S3; (b) S1 S2 × S3 S4;
(c) S4 S5 × S4 S5; and (d) S3 S4 × S5 S6? .
ANS: (a) S1 S2, S1 S3, S2 S3; (b) S1 S3, S1 S4, S2 S3, S2 S4; (c) S4 S5;
(d) S3 S5, S3 S6, S4 S5, S4 S6.
4.5 From information in the chapter about the ABO blood
types, what phenotypes and ratios are expected from the
following matings: (a) IA IA × IB IB; (b) IA IB × ii; (c) IA i ×
IB i; and (d) IA i × ii;
ANS: (a) All AB; (b) 1 A: 1 B; (c) 1 A: 1 B: 1 AB: 1 O; (d) 1 A: 1 O.
4.6 A woman with type O blood gave birth to a baby, also
with type O blood. The woman stated that a man with
type AB blood was the father of the baby. Is there any
merit to her statement?
ANS: No. The woman must be ii; if her mate is IA IB; they could
not have an ii child.
4.7 A woman with type AB blood gave birth to a baby with
type B blood. T wo different men claim to be the father.
One has type A blood, the other has type B blood. Can
the genetic evidence decide in favor of either?
ANS: No. The woman is IAIB. One man could be either IAIA or
IAi; the other could be either IBIB or IBi. Given the uncer -
tainty in the genotype of each man, either could be the
father of the child.
4.8 The flower colors of plants in a particular population
may be blue, purple, turquoise, light blue, or white. A
series of crosses between different members of the popu -
lation produced the following results:
Cross Parents Progeny
1 Purple × blue All purple
2 Purple × purple 76 purple, 25 turquoise
3 Blue × blue 86 blue, 29 turquoise
4 Purple × turquoise 49 purple, 52 turquoise
5 Purple × purple 69 purple, 22 blue
6 Purple × blue 50 purple, 51 blue
7 Purple × blue 54 purple, 26 blue,
25 turquoise
8 T urquoise × turquoise All turquoise
9 Purple × blue 49 purple, 25 blue,
23 light blue
10 Light blue × light blue 60 light blue,
29 turquoise, 31 white
11 T urquoise × white All light blue
12 White × white All white
13 Purple × white All purple How many genes and alleles are involved in the inheri -
tance of flower color? Indicate all possible genotypes for
the following phenotypes: (a) purple; (b) blue; (c) tur -
quoise; (d) light blue; (e) white.
ANS: One gene with four alleles. (a) purple: cp cp, cp cb; cp ct, cp cw;
(b) blue: cb cb, cb ct, cb cw; (c) turquoise: ct ct, ct cw; (d) light
blue: ct cw; (e) white: cw cw.
4.9 A woman who has blood type O and blood type M mar -
ries a man who has blood type AB and blood type MN. If
we assume that the genes for the A-B-O and M-N blood-
typing systems assort independently, what blood types
might the children of this couple have, and in what
proportions?
ANS: The woman is ii LMLM; the man is IAIB LMLN; the blood
types of the children will be A and M, A and MN, B and
M, and B and MN, all equally likely.
4.10 A Japanese strain of mice has a peculiar, uncoordinated
gait called waltzing, which is due to a recessive allele, v.
The dominant allele V causes mice to move in a coordi -
nated manner. A mouse geneticist has recently isolated
another recessive mutation that causes uncoordinated
movement. This mutation, called tango , could be an allele
of the waltzing gene, or it could be a mutation in an
entirely different gene. Propose a test to determine
whether the waltzing and tango mutations are alleles, and
if they are, propose symbols to denote them.
ANS: Cross homozygous waltzing with homozygous tango . If
the mutations are alleles, all the offspring will have an
uncoordinated gait; if they are not alleles, all the off -
spring will be wild-type. If the two mutations are alleles,
they could be denoted with the symbols v (waltzing ) and
vt (tango ).
4.11 Congenital deafness in human beings is inherited as a
recessive condition. In the following pedigree, two deaf
individuals, each presumably homozygous for a recessive
mutation, have married and produced four children with
normal hearing. Propose an explanation.
I
II
III
IV
ANS: The individuals III-4 and III-5 must be homozygous for
recessive mutations in different genes; that is, one is aa
BB and the other is AA bb ; none of their children is deaf
because all of them are heterozygous for both genes
(Aa Bb ).
4.12 In the fruit fly, recessive mutations in either of two inde -
pendently assorting genes, brown and purple , prevent the
synthesis of red pigment in the eyes. Thus, homozygotes
for either of these mutations have brownish-purple eyes.
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However, heterozygotes for both of these mutations
have dark red, that is, wild-type eyes. If such double het -
erozygotes are intercrossed, what kinds of progeny will
be produced, and in what proportions?
ANS: 9/16 dark red, 7/16 brownish purple.
4.13 The dominant mutation Plum in the fruit fly also causes
brownish-purple eyes. Is it possible to determine by
genetic experiments whether Plum is an allele of the
brown or purple genes?
ANS: No. The test for allelism cannot be performed with dom -
inant mutations.
4.14 From information given in the chapter, explain why mice
with yellow coat color are not true-breeding.
ANS: The allele for yellow fur is homozygous lethal.
4.15 A couple has four children. Neither the father nor the
mother is bald; one of the two sons is bald, but neither of
the daughters is bald.
(a) If one of the daughters marries a nonbald man and
they have a son, what is the chance that the son will
become bald as an adult?
(b) If the couple has a daughter, what is the chance that
she will become bald as an adult?
ANS: The mother is Bb and the father is bb. The chance that a
daughter is Bb is 1/2. (a) The chance that the daughter
will have a bald son is (1/2) × (1/2) = 1/4. (b) The chance
that the daughter will have a bald daughter is zero.
4.16 The following pedigree shows the inheritance of ataxia,
a rare neurological disorder characterized by uncoordi -
nated movements. Is ataxia caused by a dominant or a
recessive allele? Explain.
I
II
III
IV
ANS: Dominant. The condition appears in every generation
and nearly every affected individual has an affected par -
ent. The exception, IV-2, had a father who carried the
ataxia allele but did not manifest the trait—an example of
incomplete penetrance.
4.17 Chickens that carry both the alleles for rose comb ( R)
and pea comb ( P) have walnut combs, whereas chickens
that lack both of these alleles (i.e., they are genotypically
rr pp) have single combs. From the information about
interactions between these two genes given in the chap -
ter, determine the phenotypes and proportions expected
from the following crosses: (a) RR Pp × rr Pp ;
(b) rr PP × Rr Pp ;
(c) Rr Pp × Rr pp ;
(d) Rr pp × rr pp .
ANS: (a) 3/4 walnut, 1/4 rose; (b) 1/2 walnut, 1/2 pea; (c) 3/8
walnut, 3/8 rose, 1/8 pea, 1/8 single; (d) 1/2 rose, 1/2
single.
4.18 Rose-comb chickens mated with walnut-comb chickens
produced 15 walnut-, 14 rose-, 5 pea-, and 6 single-comb
chicks. Determine the genotypes of the parents.
ANS: Rr pp × Rr Pp .
4.19 Summer squash plants with the dominant allele C bear
white fruit, whereas plants homozygous for the recessive
allele c bear colored fruit. When the fruit is colored, the
dominant allele G causes it to be yellow; in the absence
of this allele (i.e., with genotype gg), the fruit color is
green. What are the F2 phenotypes and proportions
expected from intercrossing the progeny of CC GG and
cc gg plants? Assume that the C and G genes assort
independently.
ANS: 12/16 white, 3/16 yellow, 1/16 green.
4.20 The white Leghorn breed of chickens is homozygous for
the dominant allele C, which produces colored feathers.
However, this breed is also homozygous for the domi -
nant allele I of an independently assorting gene that
inhibits coloration of the feathers. Consequently, Leg -
horn chickens have white feathers. The white Wyandotte
breed of chickens has neither the allele for color nor the
inhibitor of color; it is therefore genotypically cc ii. What
are the F2 phenotypes and proportions expected from
intercrossing the progeny of a white Leghorn hen and a
white Wyandotte rooster?
ANS: 13/16 white, 3/16 colored.
4.21 Fruit flies homozygous for the recessive mutation scarlet
have bright red eyes because they cannot synthesize
brown pigment. Fruit flies homozygous for the recessive
mutation brown have brownish-purple eyes because they
cannot synthesize red pigment. Fruit flies homozygous
for both of these mutations have white eyes because they
cannot synthesize either type of pigment. The brown and
scarlet mutations assort independently. If fruit flies that
are heterozygous for both of these mutations are inter -
crossed, what kinds of progeny will they produce, and in
what proportions?
ANS: 9/16 dark red (wild-type), 3/16 brownish purple, 3/16
bright red, 1/16 white.
4.22 Consider the following hypothetical scheme of determi -
nation of coat color in a mammal. Gene A controls the
conversion of a white pigment P0 into a gray pigment
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P1; the dominant allele A produces the enzyme necessary
for this conversion, and the recessive allele a produces an
enzyme without biochemical activity. Gene B controls
the conversion of the gray pigment P1 into a black pig -
ment P2; the dominant allele B produces the active
enzyme for this conversion, and the recessive allele b
produces an enzyme without activity. The dominant
allele C of a third gene produces a polypeptide that com -
pletely inhibits the activity of the enzyme produced by
gene A; that is, it prevents the reaction P0→P1. Allele c of
this gene produces a defective polypeptide that does not
inhibit the reaction P0→P1. Genes A, B, and C assort
independently, and no other genes are involved. In the F2
of the cross AA bb CC × aa BB cc , what is the expected
phenotypic segregation ratio?
ANS: 9 black: 3 gray: 52 white.
4.23 What F2 phenotypic segregation ratio would be expected
for the cross described in the preceding problem if the
dominant allele, C, of the third gene produced a product
that completely inhibited the activity of the enzyme pro -
duced by gene B—that is, prevented the reaction P1→P2,
rather than inhibiting the activity of the enzyme pro -
duced by gene A?
ANS: 9 black: 39 gray: 16 white.
4.24 The Micronesian Kingfisher, Halcyon cinnamomina , has a
cinnamon-colored face. In some birds, the color contin -
ues onto the chest, producing one of three patterns: a
circle, a shield, or a triangle; in other birds, there is no
color on the chest. A male with a colored triangle was
crossed with a female that had no color on her chest, and
all their offspring had a colored shield on the chest.
When these offspring were intercrossed, they produced
an F2with a phenotypic ratio of 3 circle: 6 shield: 3 tri -
angle: 4 no color. (a) Determine the mode of inheritance
for this trait and indicate the genotypes of the birds in all
three generations. (b) If a male without color on his chest
is mated to a female with a colored shield on her chest
and the F1 segregate in the ratio of 1 circle: 2 shield:
1 triangle, what are the genotypes of the parents and
their progeny?
(a) The simplest explanation for the inheritance of the
trait is recessive epistasis combined with incomplete
dominance, summarized in the following table:
Genotype Phenotype Frequency in F2
AA B- Circle 3/16
Aa B- Shield 6/16
aa B- T riangular 3/16
A- bb No color 3/16
aa bb No color 1/16 (b) Father’s genotype: Aa bb ; mother’s genotype: Aa BB
Circle Shield Triangle
Progeny
genotypes:AA Bb Aa Bb aa Bb
4.25 In a species of tree, seed color is determined by four
independently assorting genes: A, B, C, and D. The
recessive alleles of each of these genes ( a, b, c, and d) pro -
duce abnormal enzymes that cannot catalyze a reaction
in the biosynthetic pathway for seed pigment. This path -
way is diagrammed as follows:
White precursor Yellow OrangeAB
Red
BlueC
D
When both red and blue pigments are present, the seeds
are purple. T rees with the genotypes Aa Bb Cc Dd and Aa
Bb Cc dd were crossed.
(a) What color are the seeds in these two parental
genotypes?
(b) What proportion of the offspring from the cross will
have white seeds?
(c) Determine the relative proportions of red, white, and
blue offspring from the cross.
ANS: (a) Purple × red; (b) proportion white ( aa) = 1/4;
(c) proportion red ( A- B- C- dd) = (3/4)(3/4)(3/4)(1/2) =
27/128, proportion white ( aa) = 1/4 = 32/128, propor -
tion blue ( A- B- cc Dd ) = (3/4)(3/4)(1/4)(1/2) = 9/128.
4.26 Multiple crosses were made between true-breeding lines
of black and yellow Labrador retrievers. All the F1 prog -
eny were black. When these progeny were intercrossed,
they produced an F2 consisting of 91 black, 39 yellow,
and 30 chocolate. (a) Propose an explanation for the
inheritance of coat color in Labrador retrievers. (b) Pro -
pose a biochemical pathway for coat color determination
and indicate how the relevant genes control coat
coloration.
ANS: (a) Because the F2 segregation is approximately 9 black:
3 chocolate: 4 yellow, coat color is determined by epista -
sis between two independently assorting genes: black =
B- E- ; chocolate = bb E- ; yellow = B- ee or bb ee .
(b) Yellow pigment— E brown pigment— B black
pigment.
4.27 T wo plants with white flowers, each from true-breeding
strains, were crossed. All the F1 plants had red flowers.
When these F1 plants were intercrossed, they produced
an F2 consisting of 177 plants with red flowers and
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142 with white flowers. (a) Propose an explanation for
the inheritance of flower color in this plant species.
(b) Propose a biochemical pathway for flower pigmenta -
tion and indicate which genes control which steps in this
pathway.
ANS: (a) Because the F2 segregation is approximately 9 red: 7
white, flower color is due to epistasis between two inde -
pendently assorting genes: red = A- B- and white = aa
B-, A- bb, or aa bb . (b) Colorless precursor— A colorless
product— B red pigment.
4.28 Consider the following genetically controlled biosyn -
thetic pathway for pigments in the flowers of a hypo -
thetical plant:
P0Gene A
Enzyme A
P1Gene B
Enzyme B
P2 P3Gene C
Enzyme C
Assume that gene A controls the conversion of a white
pigment, P0, into another white pigment, P1; the
dominant allele A specifies an enzyme necessary for
this conversion, and the recessive allele a specifies a
defective enzyme without biochemical function. Gene
B controls the conversion of the white pigment, P1,
into a pink pigment, P2; the dominant allele, B, pro -
duces the enzyme necessary for this conversion, and
the recessive allele, b, produces a defective enzyme.
The dominant allele, C, of the third gene specifies an
enzyme that converts the pink pigment, P2, into a red
pigment, P3; its recessive allele, c, produces an altered
enzyme that cannot carry out this conversion. The
dominant allele, D, of a fourth gene produces a poly -
peptide that completely inhibits the function of
enzyme C; that is, it blocks the reaction P2→P3. Its
recessive allele, d, produces a defective polypeptide
that does not block this reaction. Assume that flower
color is determined solely by these four genes and that
they assort independently. In the F2 of a cross between
plants of the genotype AA bb CC DD and plants of the
genotype aa BB cc dd , what proportion of the plants
will have (a) red flowers? (b) pink flowers? (c) white
flowers?
ANS: (a) Proportion red = (3/4)3 × (1/4) = 27/256; (b) pro -
portion pink = (3/4)4 + [(3/4)2 × (1/4)] = 117/256; (c)
proportion white = 1 − 144/256 = 112/256.
4.29 In the following pedigrees, what are the inbreeding coef -
ficients of A, B, and C?AB
Offspring of first
cousins once
removed
C
Offspring of second
cousinsOffspring of half–first
cousins
ANS: FA = (1/2)5 = 1/32; FB = 2 × (1/2)6 = 1/32; FC = 2 ×
(1/2)7 = 1/64.
4.30 A, B, and C are inbred strains of mice, assumed to be
completely homozygous. A is mated to B and B to C.
Then the A × B hybrids are mated to C, and the off -
spring of this mating are mated to the B × C hybrids.
What is the inbreeding coefficient of the offspring of this
last mating?
ANS: From the following pedigree, the inbreeding coefficient
is (1/2)3 (1 + FC) + (1/2)4(1 + FB) = 3/8 because FB =
FC = 1.
A
A × BB × C
(A × B) × CBC
4.31 Mabel and Frank are half siblings, as are Tina and Tim.
However, these two pairs of half siblings do not have any
common ancestors. If Mabel marries Tim and Frank
marries Tina and each couple has a child, what fraction
of their genes will these children share by virtue of
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common ancestry? Will the children be more or less
closely related than first cousins?
ANS: The pedigree is shown below:
Frank Mabel Tina Tim
The coefficient of relationship between the offspring of
the two couples is obtained by calculating the inbreeding
coefficient of the imaginary child from a mating between
these offspring and multiplying by 2: [(1/2)5 × 2] × 2 =
1/8. This is the same degree of relatedness as first
cousins.
4.32 Suppose that the inbreeding coefficient of I in the fol -
lowing pedigree is 0.25. What is the inbreeding coeffi -
cient of I’s common ancestor, C?
IC
ANS: FI = (1/2)3(1 + FC) = 0.25; thus, FC = 1.
4.33 A randomly pollinated strain of maize produces ears that
are 24 cm long, on average. After one generation of self-
fertilization, the ear length is reduced to 20 cm. Predict
the ear length if self-fertilization is continued for one
more generation.
ANS: The mean ear length for randomly mated maize is
24 cm and that for maize from one generation of self-
fertilization is 20 cm. The inbreeding coefficient of the
offspring of one generation of self-fertilization is 1/2,
and the inbreeding coefficient of the offspring of two
generations of self-fertilization is (1/2)(1 +1/2) = 3/4.
Mean ear length ( Y ) is expected to decline linearly with
inbreeding according to the equation Y = 24 − b F1
where b is the slope of the line. The value of b can be
determined from the two values of Y that are given. The
difference between these two values (4 cm) corresponds
to an increase in F from 0 to 1/2. Thus, b = 4/(1/2) =
8 cm, and for F = 3/4, the predicted mean ear length is
Y = 24 − 8 × (3/4) = 18 cm.Chapter 5
5.1 What are the genetic differences between male- and
female-determining sperm in animals with heteroga -
metic males?
ANS: The male-determining sperm carries a Y chro mo-
some; the female-determining sperm carries an X
chromosome.
5.2 A male with singed bristles appeared in a culture of Dro-
sophila . How would you determine if this unusual pheno -
type was due to an X-linked mutation?
ANS: Cross the singed male to wild-type females and then
intercross the offspring. If the singed bristle phenotype is
due to an X-linked mutation, approximately half the F2
males, but none of the F2 females, will show it.
5.3 In grasshoppers, rosy body color is caused by a recessive
mutation; the wild-type body color is green. If the gene
for body color is on the X chromosome, what kind of
progeny would be obtained from a mating between a
homozygous rosy female and a hemizygous wild-type
male? (In grasshoppers, females are XX and males are
XO.)
ANS: All the daughters will be green and all the sons will be
rosy.
5.4 In the mosquito Anopheles culicifacies , golden body ( go) is a
recessive X-linked mutation, and brown eyes ( bw) is a
recessive autosomal mutation. A homozygous XX female
with golden body is mated to a homozygous XY male
with brown eyes. Predict the phenotypes of their F1
offspring. If the F1 progeny are intercrossed, what kinds
of progeny will appear in the F2 and in what
proportions?
ANS: The cross is go/go +/+ female × +/Y bw/bw male → F1:
go/+ bw/+ females (wild-type eyes and body) and go/Y
bw/+ males (golden body, wild-type eyes). An intercross
of the F1 offspring yields the following F2 phenotypes in
both sexes.
Body Eyes Genotype Proportion
Golden Brown go/go or Y bw/bw (1/2) × (1/4)
= 1/8
Golden Wild-
typego/go or Y +/bw
or +(1/2) × (3/4)
= 3/8
Wild-
typeBrown +/go or Y bw/bw (1/2) × (1/4)
= 1/8
Wild-
typeWild-
type+/go or Y +/bw
or +(1/2) × (3/4)
= 3/8
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5.5 What are the sexual phenotypes of the following geno -
types in Drosophila : XX, XY, XXY, XXX, XO?
ANS: XX is female, XY is male, XXY is female, XXX is female
(but barely viable), XO is male (but sterile).
5.6 In human beings, a recessive X-linked mutation, g, causes
green-defective color vision; the wild-type allele, G,
causes normal color vision. A man (a) and a woman (b),
both with normal vision, have three children, all married
to people with normal vision: a color-defective son (c),
who has a daughter with normal vision (f); a daughter
with normal vision (d), who has one color-defective son
(g) and two normal sons (h); and a daughter with normal
vision (e), who has six normal sons (i). Give the most
likely genotypes for the individuals (a–i) in this family.
(a)
(c)
(f) (i)6(d) (e)(b)
(h)
(g)G/Y
g/Y
G/Y G/Y g/YG/gg/G G/GG/g
ANS: (a) XGY; (b) XGXg; (c) XgY; (d) XGXg; (e) XGXG; (f) XGXg;
(g) XgY; (h) XGY; (i) XGY
5.7 If both father and son have defective color vision, is it
likely that the son inherited the trait from his father?
ANS: No. Defective color vision is caused by an X-linked
mutation. The son’s X chromosome came from his
mother, not his father.
5.8 A normal woman, whose father had hemophilia, marries
a normal man. What is the chance that their first child
will have hemophilia?
ANS: The risk for the child is P(woman transmits mutant
allele) × P(child is male) = (1/2) × (1/2) = 1/4.
5.9 A man with X-linked color blindness marries a woman
with no history of color blindness in her family. The
daughter of this couple marries a normal man, and their
daughter also marries a normal man. What is the chance
that this last couple will have a child with color blindness?
If this couple has already had a child with color blindness,
what is the chance that their next child will be color blind?
ANS: The risk for the child is P(mother is C/c) × P(mother
transmits c) × P(child is male) = (1/2) × (1/2) × (1/2) =
1/8; if the couple has already had a child with color
blindness, P(mother is C/c) = 1, and the risk for each
subsequent child is 1/4.
5.10 A man who has color blindness and type O blood has
children with a woman who has normal color vision and
type AB blood. The woman’s father had color blindness. Color blindness is determined by an X-linked gene, and
blood type is determined by an autosomal gene.
(a) What are the genotypes of the man and the woman?
(b) What proportion of their children will have color
blindness and type B blood?
(c) What proportion of their children will have color
blindness and type A blood?
(d) What proportion of their children will be color blind
and have type AB blood?
ANS: (a) The man is X c Y ii; the woman is X+ Xc IA IB. (b) Prob -
ability color blind = 1/2; probability type B blood = 1/2;
combined probability = (1/2) × (1/2) = 1/4. (c) Proba -
bility color blind = 1/2; probability type A blood = 1/2;
combined probability (1/2) × (1/2) = 1/4. (d) 0.
5.11 A Drosophila female homozygous for a recessive X-linked
mutation that causes vermilion eyes is mated to a wild-
type male with red eyes. Among their progeny, all the
sons have vermilion eyes, and nearly all the daughters
have red eyes; however, a few daughters have vermilion
eyes. Explain the origin of these vermilion-eyed
daughters.
ANS: Each of the rare vermilion daughters must have resulted
from the union of an X( v) X(v) egg with a Y-bearing
sperm. The diplo-X eggs must have originated through
nondisjunction of the X chromosomes during oogenesis
in the mother. However, we cannot determine if the
nondisjunction occurred in the first or the second mei -
otic division.
5.12 In Drosophila , vermilion eye color is due to a recessive
allele ( v) located on the X chromosome. Curved wings
are due to a recessive allele ( cu) located on one autosome,
and ebony body is due to a recessive allele ( e) located on
another autosome. A vermilion male is mated to a curved,
ebony female, and the F1 males are phenotypically wild-
type. If these males were backcrossed to curved, ebony
females, what proportion of the F2 offspring will be wild-
type males?
ANS: P(male) = 1/2; P(male transmits first wild-type autosome)
= 1/2; P(male transmits other wild-type autosome) = 1/2;
therefore, combined proportion, P(wild-type male) = 1/8
5.13 A Drosophila female heterozygous for the recessive
X-linked mutation w (for white eyes) and its wild-type
allele w 1 is mated to a wild-type male with red eyes.
Among the sons, half have white eyes and half have red
eyes. Among the daughters, nearly all have red eyes;
however, a few have white eyes. Explain the origin of
these white-eyed daughters.
ANS: Each of the rare white-eyed daughters must have resulted
from the union of an X( w) X(w) egg with a Y-bearing
sperm. The rare diplo-X eggs must have originated
through nondisjunction of the X chromosomes during
the second meiotic division in the mother.
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5.14 In Drosophila , a recessive mutation called chocolate (c)
causes the eyes to be darkly pigmented. The mutant phe -
notype is indistinguishable from that of an autosomal
recessive mutation called brown (bw). A cross of choco -
late-eyed females to homozygous brown males yielded
wild-type F1 females and darkly pigmented F1 males. If
the F1 flies are intercrossed, what types of progeny are
expected, and in what proportions? (Assume that the
double mutant combination has the same phenotype as
either of the single mutants alone.)
ANS: 3/8 wild-type (red), 5/8 brown for both male and female
F2 progeny.
5.15 Suppose that a mutation occurred in the SRY gene on the
human Y chromosome, knocking out its ability to pro -
duce the testis-determining factor. Predict the pheno -
type of an individual who carried this mutation and a
normal X chromosome.
ANS: Female.
5.16 A woman carries the androgen-insensitivity mutation
(ar) on one of her X chromosomes; the other X carries
the wild-type allele ( AR). If the woman marries a normal
man, what fraction of her children will be phenotypically
female? Of these, what fraction will be fertile?
ANS: Three-fourths will be phenotypically female (genotypi -
cally ar/AR , AR/AR , or ar/Y). Among the females, 2/3
(ar/AR and AR/AR ) will be fertile; the ar/Y females will
be sterile.
5.17 Would a human with two X chromosomes and a Y chro -
mosome be male or female?
ANS: Male.
5.18 In Drosophila , the gene for bobbed bristles (recessive allele
bb, bobbed bristles; wild-type allele+, normal bristles) is
located on the X chromosome and on a homologous seg -
ment of the Y chromosome. Give the genotypes and
phenotypes of the offspring from the following crosses:
(a) Xbb Xbb × Xbb Y+;
(b) Xbb Xbb × Xbb Y+;
(c) X+ Xbb × X+ Ybb;
(d) X+ Xbb × Xbb Y+.
ANS: (a) 1/2 Xbb Xbb bobbed females, 1/2 Xbb Y+ wild-type
males; (b) 1/2 X+ Xbb wild-type females, 1/2 Xbb Ybb
bobbed males; (c) 1/4 X+ X+ wild-type females, 1/4 X+
Xbb wild-type females, 1/4 X+ Ybb wild-type males, 1/4 Xbb
Ybb bobbed males; (d) 1/4 X+ Xbb wild-type females, 1/4
Xbb Xbb bobbed females, 1/4 X+ Y+ wild-type males, 1/4
Xbb Y+ wild-type males.
5.19 Predict the sex of Drosophila with the following chromo -
some compositions (A = haploid set of autosomes):
(a) 4X 4A;
(b) 3X 4A; (c) 2X 3A;
(d) 1X 3A;
(e) 2X 2A;
(f) 1X 2A.
ANS: (a) Female; (b) intersex; (c) intersex; (d) male: (e) female;
(f) male.
5.20 In chickens, the absence of barred feathers is due to a
recessive allele. A barred rooster was mated with a non -
barred hen, and all the offspring were barred. These F1
chickens were intercrossed to produce F2 progeny,
among which all the males were barred; half the females
were barred and half were nonbarred. Are these results
consistent with the hypothesis that the gene for barred
feathers is located on one of the sex chromosomes?
ANS: Yes. The gene for feather patterning is on the Z chromo -
some. If we denote the allele for barred feathers as B and
the allele for nonbarred feathers as b, the crosses are as
follows: B/B (barred) male × b/W (nonbarred) female →
F1: B/b (barred) males and B/W (barred) females. Inter -
crossing the F1 produces B/B (barred) males, B/b (barred)
males, B/W (barred) females, and b/W (nonbarred)
females, all in equal proportions.
5.21 A Drosophila male carrying a recessive X-linked mutation
for yellow body is mated to a homozygous wild-type
female with gray body. The daughters of this mating all
have uniformly gray bodies. Why are not their bodies a
mosaic of yellow and gray patches?
ANS: Drosophila does not achieve dosage compensation by
inactivating one of the X chromosomes in females.
5.22 What is the maximum number of Barr bodies in the
nuclei of human cells with the following chromosome
compositions:
(a) XY;
(b) XX;
(c) XXY;
(d) XXX;
(e) XXXX;
(f) XYY?
ANS: a) Zero; (b) one; (c) one; (d) two; (e) three; (f) zero.
5.23 Males in a certain species of deer have two nonhomolo -
gous X chromosomes, denoted X1 and X2, and a Y chro -
mosome. Each X chromosome is about half as large as
the Y chromosome, and its centromere is located near
one of the ends; the centromere of the Y chromosome is
located in the middle. Females in this species have two
copies of each of the X chromosomes and lack a Y chro -
mosome. How would you predict the X and Y chromo -
somes to pair and disjoin during spermatogenesis to
produce equal numbers of male- and female-determin -
ing sperm?
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ANS:
Metaphase Anaphase
Y
X1 X2
Since the centromere is at the end of each small X chro -
mosome but in the middle of the larger Y, both X1 and X2
pair at the centromere of the Y chromosome during meta -
phase so that the two X chromosomes disjoin together
and segregate from the Y chromosome during anaphase.
5.24 A breeder of sun conures (a type of bird) has obtained
two true-breeding strains, A and B, which have red eyes
instead of the normal brown found in natural popula -
tions. In Cross 1, a male from strain A was mated to a
female from strain B, and the male and female offspring
all had brown eyes. In Cross 2, a female from strain A was
mated to a male from strain B, and the male offspring
had brown eyes and the female offspring had red eyes.
When the F1 birds from each cross were mated brother
to sister, the breeder obtained the following results:
PhenotypeProportion in
F2 of Cross 1Proportion in
F2 of Cross 2
Brown male 6/16 3/16
Red male 3/16 5/16
Brown female 3/16 3/16
Red female 5/16 5/16
Provide a genetic explanation for these results.
ANS: Color is determined by an autosomal gene (alleles A
and a) and a sex-linked gene (alleles B and b) on the Z
chromosome (females are ZW and males are ZZ) and
the recessive alleles are mutually epistatic—that is, aa, bb,
or bW birds have red eyes, and A- B- or A- BW birds
have brown eyes.
Cross 1
P Strain A red male × Strain B red female
aa BB AA bW
F1Aa Bb
Brown males× Aa BW
Brown females
F2A- Bb
Brown males (6/16)A- BW
Brown females (3/16)
aa Bb
Red males (2/16)A- bW
Red females (4/16)
aa BW
Red females (1/16)Cross 2
P Strain A red female × Strain B red male
aa BW AA bb
F1Aa Bb
Brown males× Aa bW
Red females
F2A- Bb
Brown males (3/16)A- bW
Brown females (3/16)
A- bb
Red males (3/16)A- bW
Red females (3/16)
aa b-
Red males 2/16aa -W
Red females (2/16)
5.25 In 1908, F . M. Durham and D. C. E. Marryat reported
the results of breeding experiments with canaries. Cin -
namon canaries have pink eyes when they first hatch,
whereas green canaries have black eyes. Durham and
Marryat crossed cinnamon females with green males
and observed that all the F1 progeny had black eyes,
just like those of the green strain. When the F1 males
were crossed to green females, all the male progeny
had black eyes, whereas all the female progeny had
either black or pink eyes, in about equal proportions.
When the F1 males were crossed to cinnamon females,
four classes of progeny were obtained: females with
black eyes, females with pink eyes, males with black
eyes, and males with pink eyes—all in approximately
equal proportions. Propose an explanation for these
findings.
ANS: Eye color in canaries is due to a gene on the Z chromo -
some, which is present in two copies in males and one
copy in females. The allele for pink color at hatching
(p) is recessive to the allele for black color at hatching
(P). There is no eye color gene on the other sex chro -
mosome ( W), which is present in one copy in females
and absent in males. The parental birds were genotypi -
cally p/W (cinnamon females) and P/P (green males).
Their F1 sons were genotypically p/P (with black eyes
at hatching). When these sons were crossed to green
females (genotype P/W), they produced F2 progeny
that sorted into three categories: males with black eyes
at hatching ( P/-, half the total progeny), females with
black eyes at hatching ( P/W, a fourth of the total prog -
eny), and females with pink eyes at hatching ( p/W, a
fourth of the total progeny). When these sons were
crossed to cinnamon females (genotype p/W), they
produced F2 progeny that sorted into four equally fre -
quent categories: males with black eyes at hatching
(genotype P/p), males with pink eyes at hatching (gen -
otype p/p), females with black eyes at hatching
(genotype P/W), and females with pink eyes at hatch -
ing (genotype p/W).
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Chapter 6
6.1 In the human karyotype, the X chromosome is approxi -
mately the same size as seven of the autosomes (the
so-called C group of chromosomes). What procedure
could be used to distinguish the X chromosome from the
other members of this group?
ANS: Use one of the banding techniques.
6.2 In humans, a cytologically abnormal chromosome 22,
called the “Philadelphia” chromosome because of the
city in which it was discovered, is associated with
chronic leukemia. This chromosome is missing part of
its long arm. How would you denote the karyotype of
an individual who had 46 chromosomes in his somatic
cells, including one normal 22 and one Philadelphia
chromosome?
ANS: 46, XX, del(22)(q) or 46, XY, del(22)(q), depending on
the sex chromosome constitution.
6.3 During meiosis, why do some tetraploids behave more
regularly than triploids?
ANS: In allotetraploids, each member of the different sets
of chromosomes can pair with a homologous partner
during prophase I and then disjoin during anaphase I.
In triploids, disjunction is irregular because homol o-
gous chromosomes associate during prophase I either
by forming bivalents and univalents or by forming
trivalents.
6.4 The following table presents chromosome data on four
species of plants and their F1 hybrids:
Meiosis I Metaphase
Species or
F1 HybridRoot Tip
Chromosome
NumberNumber of
BivalentsNumber of
Univalents
A 20 10 0
B 20 10 0
C 10 5 0
D 10 5 0
A × B 20 0 20
A × C 15 5 5
A × D 15 5 5
C × D 10 0 10
(a) Deduce the chromosomal origin of species A.
(b) How many bivalents and univalents would you expect
to observe at meiotic metaphase I in a hybrid between
species C and species B?
(c) How many bivalents and univalents would you expect
to observe at meiotic metaphase I in a hybrid between
species D and species B?ANS: (a) Species A is an allotetraploid with a genome from
each of species C and species D; (b) 0 bivalents and
15 univalents; (c) 0 bivalents and 15 univalents.
6.5 A plant species A, which has seven chromosomes in its
gametes, was crossed with a related species B, which has
nine. The hybrids were sterile, and microscopic observa -
tion of their pollen mother cells showed no chromosome
pairing. A section from one of the hybrids that grew vig -
orously was propagated vegetatively, producing a plant
with 32 chromosomes in its somatic cells. This plant was
fertile. Explain.
ANS: The fertile plant is an allotetraploid with 7 pairs of chro -
mosomes from species A and 9 pairs of chromosomes
from species B; the total number of chromosomes is
(2 × 7) + (2 × 9) = 32.
6.6 A plant species X with n = 5 was crossed with a related
species Y with n = 7. The F1 hybrid produced only a few
pollen grains, which were used to fertilize the ovules of
species Y. A few plants were produced from this cross,
and all had 19 chromosomes. Following self-fertilization,
the F1 hybrids produced a few F2 plants, each with
24 chromosomes. These plants were phenotypically dif -
ferent from either of the original species and were highly
fertile. Explain the sequence of events that produced
these fertile F2 hybrids.
ANS: The F1 hybrid had 5 chromosomes from species X and 7
chromosomes from species Y, for a total of 12. When this
hybrid was backcrossed to species Y, the few progeny
that were produced had 5 + 7 = 12 chromosomes from
the hybrid and 7 from species Y, for a total of 19. This
hybrid was therefore a triploid. Upon self-fertilization,
a few F2 plants were formed, each with 24 chromosomes.
Presumably the chromosomes in these plants consisted
of 2 × 5 = 10 from species X and 2 × 7 = 14 from spe -
cies Y. These vigorous and fertile F2 plants were there -
fore allotetraploids.
6.7 Identify the sexual phenotypes of the following
genotypes in human beings: XX, XY, XO, XXX, XXY,
XYY.
ANS: XX is female, XY is male, XO is female (but sterile), XXX
is female, XXY is male (but sterile), and XYY is male.
6.8 If nondisjunction of chromosome 21 occurs in the divi -
sion of a secondary oocyte in a human female, what is the
chance that a mature egg derived from this division will
receive two number 21 chromosomes?
ANS: 1/2
6.9 A Drosophila female homozygous for a recessive X-linked
mutation causing yellow body was crossed to a wild-type
male. Among the progeny, one fly had sectors of yellow
pigment in an otherwise gray body. These yellow sectors
were distinctly male, whereas the gray areas were female.
Explain the peculiar phenotype of this fly.
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ANS: The fly is a gynandromorph, that is, a sexual mosaic. The
yellow tissue is X( y)/O and the gray tissue is X( y)/X(+).
This mosaicism must have arisen through loss of the
X chromosome that carried the wild-type allele, pre -
sumably during one of the early embryonic cleavage
divisions.
6.10 The Drosophila fourth chromosome is so small that flies
monosomic or trisomic for it survive and are fertile. Sev -
eral genes, including eyeless (ey), have been located on this
chromosome. If a cytologically normal fly homozygous
for a recessive eyeless mutation is crossed to a fly mono -
somic for a wild-type fourth chromosome, what kinds of
progeny will be produced, and in what proportions?
ANS: Approximately half the progeny should be disomic ey/+
and half should be monosomic ey/O. The disomic prog -
eny will be wild-type, and the monosomic progeny will
be eyeless.
6.11 A woman with X-linked color blindness and T urner syn -
drome had a color-blind father and a normal mother. In
which of her parents did nondisjunction of the sex chro-
mosomes occur?
ANS: Nondisjunction must have occurred in the mother. The
color blind woman with T urner syndrome was produced
by the union of an X-bearing sperm, which carried the
mutant allele for color blindness, and a nullo-X egg.
6.12 In humans, Hunter syndrome is known to be an X-linked
trait with complete penetrance. In family A, two pheno -
typically normal parents have produced a normal son,
a daughter with Hunter and T urner syndromes, and a son
with Hunter syndrome. In family B, two phenotypically
normal parents have produced two phenotypically nor -
mal daughters and a son with Hunter and Klinefelter syn -
dromes. In family C, two phenotypically normal parents
have produced a phenotypically normal daughter,
a daughter with Hunter syndrome, and a son with Hunter
syndrome. For each family, explain the origin of the child
indicated in italics.
ANS: The daughter with T urner and Hunter syndromes in
family A must have received her single X chromosome
from her mother, who is heterozygous for the mutation
causing Hunter syndrome. The daughter did not receive
a sex chromosome from her father because sex chromo -
some nondisjunction must have occurred during meiosis
in his germline. The son with Klinefelter syndrome in
family B is karyotypically XXY, and both of his X chro -
mosomes carry the mutant allele for Hunter syndrome.
This individual must have received two mutant
X chromosomes from his heterozygous mother due to X
chromosome nondisjunction during the second meiotic
division in her germline. The daughter with Hunter syn -
drome in family C is karyotypically XX, and both of her
X chromosomes carry the mutant allele for Hunter syn -
drome. This individual received the two mutant X chromosomes from her heterozygous mother through
nondisjunction during the second meiotic division in the
mother’s germline. Furthermore, because the daughter
did not receive a sex chromosome from her father, sex
chromosome nondisjunction must have occurred during
meiosis in his germline too.
6.13 Although XYY men are phenotypically normal, would
they be expected to produce more children with sex
chromosome abnormalities than XY men? Explain.
ANS: XYY men would produce more children with sex chro -
mosome abnormalities because their three sex chromo -
somes will disjoin irregularly during meiosis. This
irregular disjunction will produce a variety of aneuploid
gametes, including the XY, YY, XYY, and nullo sex chro -
mosome constitutions.
6.14 In a Drosophila salivary chromosome, the bands have a
sequence of 1 2 3 4 5 6 7 8. The homologue with which
this chromosome is synapsed has a sequence of 1 2 3 6 5
4 7 8. What kind of chromosome change has occurred?
Draw the synapsed chromosomes.
ANS: The animal is heterozygous for an inversion:
213
21366 44
78
7855
6.15 Other chromosomes have sequences as follows: (a) 1 2 5
6 7 8; (b) 1 2 3 4 4 5 6 7 8; (c) 1 2 3 4 5 8 7 6. What kind
of chromosome change is present in each? Illustrate how
these chromosomes would pair with a chromosome
whose sequence is 1 2 3 4 5 6 7 8.
ANS: (a) Deletion:
12 567 81234
34567 8
(b) Duplication:
1 234 56 8 71 2344
456 8 7
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(c) A terminal inversion:
12345687
123467
8 5
6.16 In plants translocation heterozygotes display about 50 per -
cent pollen abortion. Why?
ANS: In translocation heterozygotes, only alternate segrega -
tion leads to euploid gametes, and the frequency of alter -
nate segregation is typically around 5 percent.
6.17 One chromosome in a plant has the sequence A B C D E
F , and another has the sequence M N O P Q R. A recip -
rocal translocation between these chromosomes pro -
duced the following arrangement: A B C P Q R on one
chromosome and M N O D E F on the other. Illustrate
how these translocated chromosomes would pair with
their normal counterparts in a heterozygous individual
during meiosis.
ANS:
AB CO NM
AB CO NMDEF
DEF
RQP
RQP
6.18 In Drosophila , the genes bw and st are located on chromo -
somes 2 and 3, respectively. Flies homozygous for bw
mutations have brown eyes, flies homozygous for st
mutations have scarlet eyes, and flies homozygous for
bw and st mutations have white eyes. Doubly heterozy -
gous males were mated individually to homozygous bw;
st females. All but one of the matings produced four
classes of progeny: wild-type, and brown-, scarlet- and
white-eyed. The single exception produced only wild-
type and white-eyed progeny. Explain the nature of this
exception.
ANS: The exceptional male, whose genotype is bw/+ st/+, is
heterozygous for a translocation between chromosomes
2 and 3. It is not possible to determine whether the
translocation is between the two mutant chromosomes
or between the two wild-type chromosomes, that is,
whether it is T( bw; st) or T( +; +); however, it clearly is not between a mutant chromosome and a wild-type
chromosome, that is, T( bw; +) or T( +; st). If it were, the
progeny would be either brown or scarlet, not either
wild-type or white.
6.19 A phenotypically normal boy has 45 chromosomes, but
his sister, who has Down syndrome, has 46. Suggest an
explanation for this paradox.
ANS: The boy carries a translocation between chromosome 21
and another chromosome, say chromosome 14. He also
carries a normal chromosome 21 and a normal chromo -
some 14. The boy’s sister carries the translocation, one
normal chromosome 14, and two normal copies of chro -
mosome 21.
6.20 Distinguish between a compound chromosome and a
Robertsonian translocation.
ANS: A compound chromosome is composed of segments
from the same pair of chromosomes, as when two X
chromosomes become attached to each other. A Rob -
ertsonian translocation involves a fusion of segments
from two different pairs of chromosomes. These seg -
ments fuse at or near the centromeres, usually with the
loss of the short arms of each of the par tic i pat ing
chromosomes.
6.21 A yellow-bodied Drosophila female with attached-X
chromosomes was crossed to a white-eyed male. Both
of the parental phenotypes are caused by X-linked
recessive mutations. Predict the phenotypes of the
progeny.
ANS: All the daughters will be yellow-bodied and all the sons
will be white-eyed.
6.22 A man has attached chromosomes 21. If his wife is cyto -
logically normal, what is the chance their first child will
have Down syndrome?
ANS: Zygotes produced by this couple will be either trisomic
or monosomic for chromosome 21. Thus, 100 percent of
their viable children will develop Down syndrome.
6.23 Analysis of the polytene chromosomes of three popula -
tions of Drosophila has revealed three different banding
sequences in a region of the second chromosome:
Population Banding Sequence
P1 1 2 3 4 5 6 7 8 9 10
P2 1 2 3 9 8 7 6 5 4 10
P3 1 2 3 9 8 5 6 7 4 10
Explain the evolutionary relationships among these
populations.
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ANS: The three populations are related by a series of
inversions:
P11 2 3 4 5 6 7 8 9 10
P21 2 3 9 8 7 6 5 4 10
P31 2 3 9 8 5 6 7 4 10
6.24 Each of six populations of Drosophila in different geo -
graphic regions had a specific arrangement of bands in
one of the large autosomes:
(a) 12345678
(b) 12263478
(c) 15432678
(d) 14322678
(e) 16223478
(f) 154322678
Assume that arrangement (a) is the original one. In what
order did the other arrangements most likely arise, and
what type of chromosomal aberration is responsible for
each change?
ANS: Arrangement (a) produced (c) by inversion of segment
2345; (c) produced (f) by a duplication of band 2; (f) pro -
duced (d) by a deletion of band 5; (d) produced (e) by
inversion of segment 43226; (e) produced (b) by inver -
sion of segment 622.
6.25 The following diagram shows two pairs of chromosomes
in the karyotypes of a man, a woman, and their child.
The man and the woman are phenotypically normal, but
the child (a boy) suffers from a syndrome of abnormali -
ties, including poor motor control and severe mental
impairment. What is the genetic basis of the child’s
abnormal phenotype? Is the child hyperploid or hypo -
ploid for a segment in one of his chromosomes?
Mother Father Child
ANS: The mother is heterozygous for a reciprocal transloca -
tion between the long arms of the large and small chro -
mosomes; a piece from the long arm of the large
chromosome has been broken off and attached to the
long arm of the short chromosome. The child has inher -
ited the rearranged large chromosome and the normal
small chromosome from the mother. Thus, because the
rearranged large chromosome is deficient for some of its
genes, the child is hypoploid. 6.26 A male mouse that is heterozygous for a reciprocal
translocation between the X chromosome and an auto -
some is crossed to a female mouse with a normal karyo -
type. The autosome involved in the translocation
carries a gene responsible for coloration of the fur. The
allele on the male’s translocated autosome is wild-type,
and the allele on its nontranslocated autosome is
mutant; however, because the wild-type allele is domi -
nant to the mutant allele, the male’s fur is wild-type
(dark in color). The female mouse has light color in her
fur because she is homozygous for the mutant allele of
the color-determining gene. When the offspring of the
cross are examined, all the males have light fur and all
the females have patches of light and dark fur. Explain
these peculiar results.
ANS: The phenotype in the female offspring is mosaic because
one of the X chromosomes is inactivated in each of their
cells. If the translocated X is inactivated, the autosome
attached to it could also be partially inactivated by a
spreading of the inactivation process across the translo -
cation breakpoint. This spreading could therefore inacti -
vate the color-determining gene on the translocated
autosome and cause patches of tissue to be phenotypi -
cally mutant.
6.27 In Drosophila , the autosomal genes cinnabar (cn) and
brown (bw) control the production of brown and red eye
pigments, respectively. Flies homozygous for cinnabar
mutations have bright red eyes, flies homozygous for
brown mutations have brown eyes, and flies homozy -
gous for mutations in both of these genes have white
eyes. A male homozygous for mutations in the cn and bw
genes has bright red eyes because a small duplication
that carries the wild-type allele of bw (bw+) is attached
to the Y chromosome. If this male is mated to a karyo -
typically normal female that is homozygous for the cn
and bw mutations, what types of progeny will be
produced?
ANS: The sons will have bright red eyes because they will
inherit the Y chromosome with the bw+ allele from their
father. The daughters will have white eyes because they
will inherit an X chromosome from their father.
6.28 In Drosophila , vestigial wing ( vg), hairy body ( h), and eye -
less ( ey) are recessive mutations on chromosomes 2, 3,
and 4, respectively. Wild-type males that had been irradi -
ated with X rays were crossed to triply homozygous
recessive females. The F1 males (all phenotypically wild-
type) were then testcrossed to triply homozygous reces -
sive females. Most of the F1 males produced eight classes
of progeny in approximately equal proportions, as would
be expected if the vg, h, and ey genes assort indepen -
dently. However, one F1 male produced only four classes
of offspring, each approximately one-fourth of the total:
(1) wild-type, (2) eyeless, (3) vestigial, hairy, and (4) ves -
tigial, hairy, eyeless. What kind of chromosome aberra -
tion did the exceptional F1 male carry, and which
chromosomes were involved?
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ANS: A reciprocal translocation between chromosomes 2
and 3. One translocated chromosome carries the wild-
type alleles of vg and h on chromosomes 2 and 3, respec -
tively, and the other carries the recessive mutant alleles
of these genes. The chromosome that carries the ey gene
(chromosome 4) is not involved in the rearrangement.
6.29 Cytological examination of the sex chromosomes in a man
has revealed that he carries an insertional translocation. A
small segment has been deleted from the Y chromosome
and inserted into the short arm of the X chromosome; this
segment contains the gene responsible for male differen -
tiation ( SRY). If this man marries a karyotypically normal
woman, what types of progeny will the couple produce?
ANS: XX zygotes will develop into males because one of their X
chromosomes carries the SRY gene that was translocated
from the Y chromosome. XY zygotes will develop into
females because their Y chromosome has lost the SRY gene.
Chapter 7
7.1 Mendel did not know of the existence of chromosomes.
Had he known, what change might he have made in his
Principle of Independent Assortment?
ANS: If Mendel had known of the existence of chromosomes, he
would have realized that the number of factors determining
traits exceeds the number of chromosomes, and he would
have concluded that some factors must be linked on the
same chromosome. Thus, Mendel would have revised the
Principle of Independent Assortment to say that factors on
different chromosomes (or far apart on the same chromo -
some) are inherited independently.
7.2 From a cross between individuals with the genotypes Cc
Dd Ee × cc dd ee , 1000 offspring were produced. The class
that was C- D- ee included 351 individuals. Are the genes
c, d, and e on the same or different chromosomes? Explain.
ANS: The class represented by 351 offspring indicates that at
least two of the three genes are linked.
7.3 If a is linked to b, and b to c, and c to d, does it follow
that a recombination experiment would detect linkage
between a and d? Explain.
ANS: No. The genes a and d could be very far apart on the
same chromosome—so far apart that they recombine
freely, that is, 50 percent of the time.
7.4 Mice have 19 autosomes in their genome, each about
the same size. If two autosomal genes are chosen ran -
domly, what is the chance that they will be on the same
chromosome?
ANS: 1/19
7.5 Genes on different chromosomes recombine with a fre -
quency of 50 percent. Is it possible for two genes on the
same chromosome to recombine with this frequency?
ANS: Yes, if they are very far apart. 7.6 If two loci are 10 cM apart, what proportion of the cells
in prophase of the first meiotic division will contain a
single crossover in the region between them?
ANS: 20%
7.7 Genes a and b are 20 cM apart. An a+ b+/a+ b+ individual
was mated with an a b/a b individual.
(a) Diagram the cross and show the gametes produced by
each parent and the genotype of the F1.
(b) What gametes can the F1 produce, and in what
proportions?
(c) If the F1 was crossed to a b/a b individuals, what off -
spring would be expected, and in what proportions?
(d) Is this an example of the coupling or repulsion link -
age phase?
(e) If the F1 were intercrossed, what offspring would be
expected, and in what proportions?
ANS: (a) Cross: a+ b+/a+ b+ × a b/a b. Gametes: a+ b+ from one
parent, a b from the other. F1: a+ b+/a b
(b) 40% a+ b+, 40% a b, 10% a+ b, 10% a b+
(c) F2 from testcross: 40% a+ b+/a b, 40% a b/a b, 10% a+
b/a b, 10% a b+/a b
(d) Coupling linkage phase
(e) F2 from intercross:
Sperm
40% a+ b+
a+ b+/a+ b+a+ b+/a+ ba+ b+/a b+a+ b+/a b
a b/a+ b+a b/a+ ba b/a b+a b/a b
a+ b/a+ b+a+ b/a+ ba+ b/a b+a+ b/a b
a b+/a+ b+a b+/a+ ba b+/a b+a b+/a b10% a b+10% a+ b10% a b+40% a+ b+
10% a+ b16% 16% 4% 4%
40% a b40% a b
16% 16% 4% 4%
Eggs
4% 4% 1% 1%
4% 4% 1% 1%
Summary of phenotypes:
a+ and b+ 66%
a+ and b 9%
a and b+ 9%
a and b 16%
7.8 Answer questions (a)–(e) in the preceding problem under
the assumption that the original cross was a+ b/a+ b ×
a b+/a b+.
ANS: (a) Cross: a+ b/a+ b × a b+/a b+ Gametes: a+ b from one
parent, a b+ from the other F1: a+ b/a b+
(b) 40% a+ b, 40% a b+, 10% a+ b+, 10% a b
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(c) F2 from testcross: 40% a+ b/a b , 40% a b+/a b, 10%
a+ b+/a b, 10% a b/a b
(d) Repulsion linkage phase
(e) F2 from intercross:
Sperm
40% a+ b
a+ b/a+ ba+ b/a+ b+a+ b/a b a+ b/a b+
a b+/a+ ba b+/a+ b+a b+/a b a b+/a b+
a+ b+/a+ ba+ b+/a+ b+a+ b+/a b a+ b+/a b+
a b/a+ ba b/a+ b+a b/a b a b/a b+10% a b10% a+ b+10% a b 40% a+ b
10% a+ b+16% 16% 4% 4%
40% a b+40% a b+
16% 16% 4% 4%
Eggs
4% 4% 1% 1%
4% 4% 1% 1%
Summary of phenotypes:
a+ and b+ 51%
a+ and b 24%
a and b+ 24%
a and b 1%
7.9 If the recombination frequency in the previous two
problems were 40 percent instead of 20 percent, what
change would occur in the proportions of gametes and
testcross progeny?
ANS: Coupling heterozygotes a+ b+/a b would produce the
following gametes: 30% a+ b+, 30% a b, 20% a+ b, 20%
a b+; repulsion heterozygotes a+ b/a b+ would produce the
following gametes: 30% a+ b, 30% a b+, 20% a+ b+, 20%
a b. In each case, the frequencies of the testcross progeny
would correspond to the frequencies of the gametes.
7.10 A homozygous variety of maize with red leaves and nor -
mal seeds was crossed with another homozygous variety
with green leaves and tassel seeds. The hybrids were then
backcrossed to the green, tassel-seeded variety, and the
following offspring were obtained: red, normal 124; red,
tassel 126; green, normal 125; green, tassel 123. Are the
genes for plant color and seed type linked? Explain.
ANS: No. The leaf color and tassel seed traits assort
independently.
7.11 A phenotypically wild-type female fruit fly that was
heterozygous for genes controlling body color and
wing length was crossed to a homozygous mutant male
with black body (allele b) and vestigial wings (allele
vg). The cross produced the following progeny: gray
body, normal wings 126; gray body, vestigial wings 24;
black body, normal wings 26; black body, vestigial
wings 124. Do these data indicate linkage between the
genes for body color and wing length? What is the
frequency of recombination? Diagram the cross,
showing the arrangement of the genetic markers on
the chromosomes.ANS: Yes. Recombination frequency = (24 + 26)/(126 + 24 +
26 + 124) = 0.167. Cross:
b vgfemale×male
b+ vg+
126 124 24 26b vg
b+ vg+b vg
b+ vgb vg
b vg+b vg
b vg
b vg
b vg
7.12 Another phenotypically wild-type female fruit fly het -
erozygous for the two genes mentioned in the previous
problem was crossed to a homozygous black, vestigial
male. The cross produced the following progeny: gray
body, normal wings 23; gray body, vestigial wings 127;
black body, normal wings 124; black body, vestigial wings
26. Do these data indicate linkage? What is the frequency
of recombination? Diagram the cross, showing the
arrangement of the genetic markers on the
chromosomes.
ANS: Yes. Recombination frequency = (23 + 26)/(23 + 127 +
124 + 26) = 0.163. Cross:
b+ vgfemale male
b vg+
23 26 127 124b vg
b+ vg+b vg
b+ vgb vg
b vg+b vg
b vg
b vg
b vg×
7.13 In rabbits, the dominant allele C is required for colored
fur; the recessive allele c makes the fur colorless (albino).
In the presence of at least one C allele, another gene
determines whether the fur is black ( B, dominant) or
brown ( b, recessive). A homozygous strain of brown rab -
bits was crossed with a homozygous strain of albinos.
The F1 were then crossed to homozygous double reces -
sive rabbits, yielding the following results: black 34;
brown 66; albino 100. Are the genes b and c linked? What
is the frequency of recombination? Diagram the crosses,
showing the arrangement of the genetic markers on the
chromosomes.
ANS: Yes. Recombination frequency is estimated by the fre -
quency of black offspring among the colored offspring:
34/(66 + 34) = 0.34. Cross:
C b c B
C bc B
C b c b
c Bc b
C b
c bc bc bc bc B c b C B
brown albino albino black
66 100 34×
×
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7.14 In tomatoes, tall vine ( D) is dominant over dwarf ( d),
and spherical fruit shape ( P) is dominant over pear
shape ( p). The genes for vine height and fruit shape are
linked with 20 percent recombination between them.
One tall plant (I) with spherical fruit was crossed with
a dwarf, pear-fruited plant. The cross produced the
following results: tall, spherical 81; dwarf, pear 79;
tall, pear 22; dwarf spherical 17. Another tall plant
with spherical fruit (II) was crossed with the dwarf,
pear-fruited plant, and the following results were
obtained: tall, pear 21; dwarf, spherical 18; tall, spheri -
cal 5; dwarf, pear 4. Diagram these two crosses, show -
ing the genetic markers on the chromosomes. If the
two tall plants with spherical fruit were crossed with
each other, that is, I × II, what phenotypic classes
would you expect from the cross, and in what
proportions?
ANS: Plant I has the genotype D P/d p , and when crossed to a
d p/d p plant produces four classes of progeny:
D P d p D p d P
d p d p d p d p
81 79 22 17
Plant II has the genotype D p/d P , and when crossed to a
d p/d p plant produces four classes of progeny:
D p d P D P d p
d p d p d p d p
21 18 5 4
If the two plants are crossed ( D P/d p × D p/d P ), the
phenotypes of the offspring can be predicted from the
following table.
Gametes from plant I
Gametes
from
plant IID p
D p/D P
0.16
d p/d p
0.04D P/d p
0.04d P/d p
0.16D p/d p
0.16
d p/d P
0.01D P/d P
0.01d P/d P
0.04D p/d P
0.04
d p/D p
0.01D P/D p
0.01d P/D p
0.04D p/D p
0.04
D P/D P
0.04
d p/D P
0.04d P/D P
0.16D p
D P0.40
0.40
0.400.40 0.10
0.10
0.100.10D p d pd P
d P
d p
Summary of phenotypes:
T all, spherical 0.54
T all, pear 0.21
Dwarf, spherical 0.21
Dwarf, pear 0.04
7.15 In Drosophila , the genes sr (stripe thorax) and e (ebony
body) are located at 62 and 70 cM, respectively, from
the left end of chromosome 3. A striped female homozygous for e+ was mated with an ebony male
homozygous for sr+. All the offspring were phenotypi -
cally wild-type (gray body and unstriped).
(a) What kind of gametes will be produced by the F1
females, and in what proportions?
(b) What kind of gametes will be produced by the F1
males, and in what proportions?
(c) If the F1 females are mated with striped, ebony
males, what offspring are expected, and in what
proportions?
(d) If the F1 males and females are intercrossed, what off -
spring would you expect from this intercross, and in
what proportions?
ANS: (a) The F1 females, which are sr e+/sr+ e, produce four
types of gametes: 46% sr e+, 46% sr+ e, 4% sr e, 4% sr+ e+.
(b) The F1 males, which have the same genotype as the F1
females, produce two types of gametes: 50% sr e+, 50%
sr+ e; remember, there is no crossing over in Drosophila
males. (c) 46% striped, gray; 46% unstriped, ebony; 4%
striped, ebony; 4% unstriped, gray. (d) The offspring
from the intercross can be obtained from the follow ing
table.
Sperm
sr e+
0.46 0.23 0.23
0.23 0.23
0.002 0.002
0.002 0.002sr+ e+
0.04sr e
0.04sr e+/sr e+sr e+/sr+ e0.50 0.50sr e+sr+ e
sr+ e
0.46sr+ e/sr e+sr+ e/sr+ e Eggs
sr e/sr e+sr e/sr+ e
sr+ e+/sr e+sr+ e+/sr+ e
Summary of phenotypes:
Striped, gray 0.25
Unstriped, gray 0.50
Striped, ebony 0
Unstriped, ebony 0.25
7.16 In Drosophila , genes a and b are located at positions 22.0
and 42.0 on chromosome 2, and genes c and d are
located at positions 10.0 and 25.0 on chromosome 3.
A fly homozygous for the wild-type alleles of these four
genes was crossed with a fly homozygous for the reces -
sive alleles, and the F1 daughters were backcrossed to
their quadruply recessive fathers. What offspring would
you expect from this backcross, and in what
proportions?
ANS: Because the two chromosomes assort independently, the
genetic makeup of the gametes (and, therefore, of the
backcross progeny) can be obtained from the following
table.
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Chromosome 3 in gametes
Chromo-
some 2
in
gametesc d
a ba b c da b c+ d+
a+ b+ c d
a+ b c d
a b+ c da b+ c+ d+a b+ c d+a b+ c+ da+ b+ c+ d+
a+ bcd+a+ b+ c+ d+a+ b+ c d+a+ b+ c+ d
a+ b c+ da b c+ d a b c d+0.425
0.40 0.17
0.17
0.0425
0.04250.0425
0.04250.0075 0.0075
0.0075 0.00750.170.17 0.03 0.03
0.03 0.03 0.40
0.10
0.100.425 0.075 0.075c+ d+
a+ b+
a+ b
a b+c+ d c d+
7.17 The Drosophila genes vg (vestigial wings) and cn (cinnabar
eyes) are located at 67.0 and 57.0, respectively, on chro -
mosome 2. A female from a homozygous strain of vesti -
gial flies was crossed with a male from a homozygous
strain of cinnabar flies. The F1 hybrids were phenotypi -
cally wild-type (long wings and dark red eyes).
(a) How many different kinds of gametes could the F1
females produce, and in what proportions?
(b) If these females are mated with cinnabar, vestigial
males, what kinds of progeny would you expect, and in
what proportions?
ANS: (a) The F1 females, which are cn vg+/cn+ vg, produce four
types of gametes: 45% cn vg+, 45% cn+ vg, 5% cn+ vg+,
5% cn vg . (b) 45% cinnabar eyes, normal wings; 45%
reddish-brown eyes, vestigial wings; 5% reddish-brown
eyes, normal wings; 5% cinnabar eyes, vestigial wings.
7.18 In Drosophila , the genes st (scarlet eyes), ss (spineless bris-
tles), and e (ebony body) are located on chromosome 3,
with map positions as indicated:
st ss e
44 58 70
Each of these mutations is recessive to its wild-type allele
(st+, dark red eyes; ss+, smooth bristles; e+, gray body).
Phenotypically wild-type females with the genotype st ss
e+/st+ st+ ss+ e were crossed with triply recessive males.
Predict the phenotypes of the progeny and the frequen -
cies with which they will occur assuming (a) no interfer -
ence and (b) complete interference.
ANS: In the following enumeration, classes 1 and 2 are paren -
tal types, classes 3 and 4 result from a single crossover
between st and ss, classes 5 and 6 result from a single
crossover between ss and e, and classes 7 and 8 result
from a double crossover, with one of the exchanges
between st and ss and the other between ss and e.
Class Phenotypes(a) Frequency
with No
Interference(b) Frequency
with Complete
Interference
1 Scarlet,
spineless0.3784 0.37
2 Ebony 0.3784 0.37
3 Scarlet, ebony 0.0616 0.07
4 Spineless 0.0616 0.075 Scarlet,
spineless, ebony0.0516 0.06
6 Wild-type 0.0516 0.06
7 Scarlet 0.0084 0
8 Spineless, ebony 0.0084 0
7.19 In maize, the genes Pl for purple leaves (dominant over Pl
for green leaves), sm for salmon silk (recessive to Sm for
yellow silk), and py for pigmy plant (recessive to Py for
normal-size plant) are on chromosome 6, with map posi -
tions as shown:
pl sm py
45 55 65
Hybrids from the cross Pl sm py / Pl sm py × pl Sm Py / pl
Sm Py were testcrossed with pl sm py / pl sm py plants. Pre -
dict the phenotypes of the offspring and their frequen -
cies assuming (a) no interference and (b) complete
interference.
ANS: In the enumeration below, classes 1 and 2 are parental
types, classes 3 and 4 result from a single crossover
between Pl and Sm, classes 5 and 6 result from a single
crossover between Sm and Py, and classes 7 and 8 result
from a double crossover, with one of the exchanges
between Pl and Sm and the other between Sm and Py.
Class Phenotypes(a) Frequency
with No
Interference(b) Frequency
with Complete
Interference
1 Purple, salmon,
pigmy0.405 0.40
2 Green, yellow,
normal0.405 0.40
3 Purple, yellow,
normal0.045 0.05
4 Green, salmon,
pigmy0.045 0.05
5 Purple, salmon,
normal0.045 0.05
6 Green, yellow,
pigmy0.045 0.05
7 Purple, yellow,
pigmy0.005 0
8 Green, salmon,
normal0.005 0
7.20 In maize, the genes Tu, j2, and gl3 are located on chro -
mosome 4 at map positions 101, 106, and 112, respec -
tively. If plants homozygous for the recessive alleles of
these genes are crossed with plants homozygous for the
dominant alleles, and the F1 plants are testcrossed to tri -
ply recessive plants, what genotypes would you expect,
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and in what proportions? Assume that interference is
complete over this map interval.
ANS: In the following enumeration, classes 1 and 2 are paren -
tal types, classes 3 and 4 result from crossing over
between Tu and j2, and classes 5 and 6 result from cross -
ing over between j2 and Gl3; only the chromosome from
the triply heterozygous F1 plant is shown. Because inter -
ference is complete, there are no double crossover
progeny.
Class Genotype Frequency
1 tu j2 gl3 0.445
2 Tu J2 Gl3 0.445
3 tu J2 Gl3 0.025
4 Tu j2 gl3 0.025
5 tu j2 Gl3 0.030
6 Tu J2 gl3 0.030
7.21 A Drosophila geneticist made a cross between females
homozygous for three X-linked recessive mutations ( y,
yellow body; ec, echinus eye shape; w, white eye color) and
wild-type males. He then mated the F1 females to triply
mutant males and obtained the following results:
Females Males Number
+ + + / y ec w + + + 475
y ec w / y ec w y ec w 469
y + + / y ec w y + + 8
+ ec w / y ec w + ec w 7
y + w / y ec w y + w 18
+ ec + / y ec w + ec + 23
+ + w / y ec w + + w 0
y ec + / y ec w y ec + 0
Determine the order of the three loci y, ec, and w, and
estimate the distances between them on the linkage map
of the X chromosome.
ANS: The double crossover classes, which are the two that
were not observed, establish that the gene order is
y—w—ec. Thus, the F1 females had the genotype y w ec /+
+ +. The distance between y and w is estimated by the
frequency of recombination between these two genes:
(8 + 7)/1000 = 0.015; similarly, the distance between w
and ec is (18 + 23)/1000 = 0.041. Thus, the genetic map
for this segment of the X chromosome is y—1.5 cM— w—
4.1 cM— ec.
7.22 A Drosophila geneticist crossed females homozygous for
three X-linked mutations ( y, yellow body; B, bar eye
shape; v, vermilion eye color) to wild-type males. The F1
females, which had gray bodies and bar eyes with dark
red pigment, were then crossed to y B+ v males, yielding
the following results:Phenotype
546
244
160
50Yellow, bar
vermilion}Yellow, vermilion
bar }Yellow
bar, vermilion}Yellow, bar, vermilion
wild-type }Number
Determine the order of these three loci on the X chro -
mosome and estimate the distances between them.
ANS: The last two classes, consisting of yellow, bar flies and ver -
milion flies, with a total of 50 progeny, result from double
crossovers. Thus, the order of the genes is y—v—B , and
the F1 females had the genotype y v B/+ + +. The dis -
tance between y and v is the average number of crossovers
between them: (244 + 50)/1000 = 29.4 cM; likewise, the
distance between v and B is (160 + 50)/1000 = 21.0 cM.
Thus, the genetic map is y—29.4 cM— v—21.0 cM— B.
7.23 Female Drosophila heterozygous for three recessive
mutations e (ebony body), st (scarlet eyes), and ss (spineless
bristles) were testcrossed, and the following progeny
were obtained:
Phenotype Number
Wild-type 67
Ebony 8
Ebony, scarlet 68
Ebony, spineless 347
Ebony, scarlet, spineless 78
Scarlet 368
Scarlet, spineless 10
Spineless 54
(a) What indicates that the genes are linked?
(b) What was the genotype of the original heterozygous
females?
(c) What is the order of the genes?
(d) What is the map distance between e and st?
(e) What is the map distance between e and ss?
(f) What is the coefficient of coincidence?
(g) Diagram the crosses in this experiment.
ANS: (a) T wo of the classes (the parental types) vastly outnum -
ber the other six classes (recombinant types); (b) st +
+/+ ss e; (c) st—ss—e; (d) [(145 + 122) × 1 + (18) ×
2]/1000 = 30.3 cM; (e) (122 + 18)/1000 = 14.0 cM;
(f) (0.018)/(0.163 × 0.140) = 0.789. (g) st + +/+ ss e
females × st ss e /st ss e males → two parental classes and
six recombinant classes.
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7.24 Consider a female Drosophila with the following X chro -
mosome genotype:
w dor+
w+ dor
The recessive alleles w and dor cause mutant eye colors
(white and deep orange, respectively). However, w is epi -
static over dor; that is, the genotypes w dor / Y and w
dor/ w dor have white eyes. If there is 40 percent recom -
bination between w and dor, what proportion of the sons
from this heterozygous female will show a mutant phe -
notype? What proportion will have either red or deep
orange eyes?
ANS: The female will produce four kinds of gametes: 30% w +,
30% + dor, 20% w dor , and 20% + +; thus, 80% of the
progeny will be mutant (either white or deep orange)
and 50% will be pigmented (either red or deep orange).
7.25 In Drosophila , the X-linked recessive mutations prune (pn)
and garnet (g) recombine with a frequency of 0.4. Both of
these mutations cause the eyes to be brown instead of
dark red. Females homozygous for the pn mutation were
crossed to males hemizygous for the g mutation, and the
F1 daughters, all with dark red eyes, were crossed with
their brown-eyed brothers. Predict the frequency of sons
from this last cross that will have dark red eyes.
ANS: The F1 females are genotypically pn +/+ g. Among their
sons, 40 percent will be recombinant for the two X-linked
genes, and half of the recombinants will have the wild-
type alleles of these genes. Thus, the frequency of sons
with dark red eyes will be 1/2 × 40% = 20%.
7.26 Assume that in Drosophila there are three genes x, y, and
z, with each mutant allele recessive to the wild-type
allele. A cross between females heterozygous for these
three loci and wild-type males yielded the following
progeny:
Females + + + 1010
Males + + + 39
+ + z 430
+ y z 32
x + + 27
x y + 441
x y z 31
T otal: 2010
Using these data, construct a linkage map of the three
genes and calculate the coefficient of coincidence.
ANS: Ignore the female progeny and base the map on the male
progeny. The parental types are + + z and x y +. The
two missing classes ( + y + and x + z) must represent
double crossovers; thus, the gene order is y—x—z . The
distance between y and x is (32 + 27)/1000 = 5.9 cM and that between x and z is (31 + 39)/1000 = 7.0 cM. Thus,
the map is y—5.9 cM— x—7.0 cM— z. The coefficient of
coincidence is zero.
7.27 In the nematode Caenorhabditis elegans , the linked genes
dpy (dumpy body) and unc (uncoordinated behavior) recom -
bine with a frequency P. If a repulsion heterozygote car -
rying recessive mutations in these genes is self-fertilized,
what fraction of the offspring will be both dumpy and
uncoordinated?
ANS: (P/2)2
7.28 In the following testcross, genes a and b are 20 cM apart,
and genes b and c are 10 cM apart: a + c / + b + × a b
c / a b c. If the coefficient of coincidence is 0.5 over this
interval on the linkage map, how many triply homozy -
gous recessive individuals are expected among 1000
progeny?
ANS: 5.
7.29 Drosophila females heterozygous for three recessive
mutations, a, b, and c, were crossed to males homozygous
for all three mutations. The cross yielded the following
results:
Phenotype Number
+ + + 75
+ + c 348
+ b c 96
a + + 110
a b + 306
a b c 65
Construct a linkage map showing the correct order of
these genes and estimate the distances between them.
ANS: From the parental classes, + + c and a b +, the heterozy -
gous females must have had the genotype + + c/a b +.
The missing classes, + b + and a + c, which would rep -
resent double crossovers, establish that the gene order
is b—a—c. The distance between b and a is (96 + 110)/
1000 = 20.6 cM and that between a and c is (65 +
75)/1000 = 14.0 cM. Thus, the genetic map is b—20.6
cM— a—14.0 cM— c.
7.30 A Drosophila second chromosome that carried a recessive
lethal mutation, l(2)g14 , was maintained in a stock with a
balancer chromosome marked with a dominant muta -
tion for curly wings. This latter mutation, denoted Cy, is
also associated with a recessive lethal effect—but this
effect is different from that of l(2)g14 . Thus, l(2)g14 /Cy
flies survive, and they have curly wings. Flies without the
Cy mutation have straight wings. A researcher crossed
l(2)g14 / Cy females to males that carried second chromo -
somes with different deletions (all homozygous lethal)
balanced over the Cy chromosome (genotype Df /Cy).
Each cross was scored for the presence or absence of
progeny with straight wings.
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Cross
1
1n o23456789101112Location of deletion Non-Curly
progeny
2n o
3 yes
4 yes
5n o
In which band is the lethal mutation l(2)g14 located?
ANS: The lethal mutation resides in band 7.
7.31 The following pedigree, described in 1937 by C. L.
Birch, shows the inheritance of X-linked color blindness
and hemophilia in a family. What is the genotype of II-2?
Do any of her children provide evidence for recombina -
tion between the genes for color blindness and
hemophilia?
Key to phenotypes:
Normal
HemophilicColor blind
12 3I
12
II
III1 23
ANS: II-1 has the genotype C h/c H , that is, she is a repulsion
heterozygote for the alleles for color blindness ( c) and
hemophilia ( h). None of her children are recombinant
for these alleles.
7.32 The following pedigree, described in 1938 by B. Rath,
shows the inheritance of X-linked color blindness and
hemophilia in a family. What are the possible genotypes
of II-1? For each possible genotype, evaluate the chil -
dren of II-1 for evidence of recombination between the
color blindness and hemophilia genes.
I
12
12II
III
12 3 4Key to phenotypes:
Normal
Hemophilic
Color vision uncer tainColor blind and hemophilicColor blind
ANS: II-1 is either (a) C h/c H or (b) c h/C H . Her four sons
have the genotypes c h (1), C h (2), c H (3), and C H (4). If
II-1 has the genotype C h/c H , sons 1 and 4 are recombi -
nant and sons 2 and 3 are nonrecombinant. If II-1 has
the genotype c h/C H , sons 2 and 3 are recombinant and sons 1 and 4 are nonrecombinant. Either way, the fre -
quency of recombination is 0.5.
7.33 A normal woman with a color-blind father married a
normal man, and their first child, a boy, had hemophilia.
Both color blindness and hemophilia are due to X-linked
recessive mutations, and the relevant genes are separated
by 10 cM. This couple plans to have a second child. What
is the probability that it will have hemophilia? Color
blindness? Both hemophilia and color blindness? Nei -
ther hemophilia nor color blindness?
ANS: The woman is a repulsion heterozygote for the alleles
for color blindness and hemophilia—that is, she is C h/c
H. If the woman has a boy, the chance that he will have
hemophilia is 0.5 and the chance that he will have color
blindness is 0.5. If we specify that the boy have only one
of these two conditions, then the chance that he will
have color blindness is 0.45. The reason is that the boy
will inherit a nonrecombinant X chromosome with a
probability of 0.9, and half the nonrecombinant X chro -
mosomes will carry the mutant allele for color blindness
and the other half will carry the mutant allele for hemo -
philia. The chance that the boy will have both conditions
is 0.05, and the chance that he will have neither condi -
tion is 0.05. The reason is that the boy will inherit a
recombinant X chromosome with a probability of 0.1,
and half the recombinant X chromosomes will carry
both mutant alleles and the other half will carry neither
mutant allele.
7.34 T wo strains of maize, M1 and M2, are homozygous for
four recessive mutations, a, b, c, and d, on one of the
large chromosomes in the genome. Strain W1 is
homozygous for the dominant alleles of these muta -
tions. Hybrids produced by crossing M1 and W1 yield
many different classes of recombinants, whereas
hybrids produced by crossing M2 and W1 do not yield
any recombinants at all. What is the difference between
M1 and M2?
ANS: M2 carries an inversion that suppresses recombination in
the chromosome.
7.35 A Drosophila geneticist has identified a strain of flies with
a large inversion in the left arm of chromosome 3. This
inversion includes two mutations, e (ebony body) and cd
(cardinal eyes), and is flanked by two other mutations, sr
(stripe thorax) on the right and ro (rough eyes) on the left.
The geneticist wishes to replace the e and cd mutations
inside the inversion with their wild-type alleles; he plans
to accomplish this by recombining the multiply mutant,
inverted chromosome with a wild-type, inversion-free
chromosome. What event is the geneticist counting on
to achieve his objective? Explain.
ANS: A two-strand double crossover within the inversion;
the exchange points of the double crossover must lie
between the genetic markers and the inversion
breakpoints.
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Chapter 8
8.1 By what criteria are viruses living? nonliving?
ANS: Viruses reproduce and transmit their genes to progeny
viruses. They utilize energy provided by host cells and
respond to environmental and cellular signals like other
living organisms. However, viruses are obligate parasites;
they can reproduce only in appropriate host cells.
8.2 How do bacteriophages differ from other viruses?
ANS: Bacteriophages reproduce in bacteria; other viruses
reproduce in protists, plants, and animals.
8.3 In what ways do the life cycles of bacteriophages T4 and
l differ? In what aspects are they the same?
ANS: Bacteriophage T4 is a virulent phage. When it infects a
host cell, it reproduces and kills the host cell in the pro -
cess. Bacteriophage lambda can reproduce and kill the
host bacterium—the lytic response—just like phage T4,
or it can insert its chromosome into the chromosome of
the host and remain there in a dormant state—the lyso -
genic response.
8.4 How does the structure of the l prophage differ from the
structure of the l chromosome packaged in the l head?
ANS: The mature (packaged) lambda chromosome and the
lambda prophage are circular permutations of one
another (see Figure 8.5).
8.5 In what way does the integration of the l chromosome
into the host chromosome during a lysogenic infection
differ from crossing over between homologous
chromosomes?
ANS: The insertion of the phage l chromosome into the host
chromosome is a site-specific recombination process
catalyzed by an enzyme that recognizes specific sequences
in the l and E. coli chromosomes. Crossing over between
homologous chromosomes is not sequence specific. It
can occur at many sites along the two chromosomes.
8.6 Geneticists have used mutations that cause altered pheno -
types such as white eyes in Drosophila , white flowers and
wrinkled seeds in peas, and altered coat color in rabbits to
determine the locations of genes on the chromosomes of
these eukaryotes. What kinds of mutant phenotypes have
been used map genes in bacteria?
ANS: Three main types of bacterial mutants have been used to
map genes in bacteria; these include mutants unable to
utilize specific sugars as energy sources (such as lactose),
mutants unable to synthesize essential metabolites (these
are called auxotrophs), and mutants resistant to drugs
and antibiotics. Wild-type bacteria can use almost any
sugar as an energy source, can grow on minimal media,
and are killed by antibiotics, whereas mutants in genes
controlling these processes result in different growth
characteristics. These growth phenotypes can be used to
map genes in bacteria. 8.7 You have identified three mutations— a, b, and c—in
Streptococcus pneumoniae . All three are recessive to their
wild-type alleles a+, b+, and c+. You prepare DNA from a
wild-type donor strain and use it to transform a strain
with genotype a b c. You observe a+b+ transformants and
a+c+ transformants, but no b+c+ transformants. Are these
mutations closely linked? If so, what is their order on the
Streptococcus chromosome?
ANS: The a, b, and c mutations are closely linked and in the
order b—a—c on the chromosome.
8.8 A nutritionally defective E. coli strain grows only on a
medium containing thymine, whereas another nutrition -
ally defective strain grows only on a medium containing
leucine. When these two strains were grown together, a
few progeny were able to grow on a minimal medium
with neither thymine nor leucine. How can this result be
explained?
ANS: There are two possible explanations. One possibility is
that a spontaneous mutation caused reversion of either
auxotrophic strain to the prototrophic condition. Because
this requires only one mutation in one cell, this is a pos -
sibility, although rare. Another, more likely, possibility is
that conjugation occurred between the E. coli parental
auxotrophic strains. During conjugation, genes from the
parental strains recombined. Because each parent had a
wild-type gene copy for either thymine or leucine, recom -
binant progeny containing the wild-type copy of each
gene would be able to synthesize both nutrients and grow
on minimal medium.
8.9 Assume that you have just demonstrated genetic recom -
bination (e.g., when a strain of genotype a b+ is present
with a strain of genotype a+ b, some recombinant geno -
types, a+ b+ and a b, are formed) in a previously unstudied
species of bacteria. How would you determine whether
the observed recombination resulted from transforma -
tion, conjugation, or transduction?
ANS: Perform two experiments: (1) determine whether the
process is sensitive to DNase and (2) determine whether
cell contact is required for the process to take place. The
cell contact requirement can be tested by a U-tube
experiment (see Figure 8.9). If the process is sensitive to
DNase, it is similar to transformation. If cell contact is
required, it is similar to conjugation. If it is neither sensi -
tive to DNase nor requires cell contact, it is similar to
transduction.
8.10 (a) What are the genotypic differences between F- cells,
F+ cells, and Hfr cells? (b) What are the phenotypic dif -
ferences? (c) By what mechanism are F− cells converted
into F+ cells? F+ cells to Hfr cells? Hfr cells to F+ cells?
ANS: (a) F− cells, no F factor present; F+ cells, autonomous F
factor; Hfr cells, integrated F factor (see Figure 8.14).
(b) F+ and Hfr cells have F pili; F− cells do not. (c) F−
cells are converted into F+ cells by the conjugative trans -
fer of F factors from F+ cells. Hfr cells are formed when
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Subsets and Splits