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Hyde Chapter 2 – Solutions
3
2
MENDEL’S PRINCIPLES
CHAPTER SUMMARY QUESTIONS
2.
Perform a testcross by crossing the tall pea plant of unknown genotype with a
homozygous recessive dwarf plant (dd). If the offspring consist only of tall pea
plants, then the plant in question would be homozygous (cross: DD dd Dd). If
the offspring consists of both tall and dwarf pea plants, then the plant in question
would be heterozygous (cross: Dd dd 1/2 Dd:1/2 dd).
4.
Cross
(P generation)
Number of different
F1 gametes
Number of different
F2 genotypes
Number of different
F2 phenotypes
Degrees of freedom
in chi-square test
6.
AABB aabb
4
AABBCC aabbcc
8
AABBCCDD aabbccdd
16
9
27
81
4
8
16
3
7
15
a. The offspring are in an approximate 3:1 phenotypic ratio of red eyes to brown
eyes. The 3:1 ratio suggests that this is a monohybrid cross and that red eye is
dominant and brown eye is recessive. The cross would be Bb Bb.
b. There are two phenotypes in approximately a 1:1 ratio in the offspring. We don’t
know which allele is dominant from this cross. The two phenotypes indicate one
gene, and the 1:1 ratio indicates a mating of a heterozygote and homozygous
recessive individuals (the testcrossing of a heterozygote). Therefore, the cross
can be depicted as Aa aa.
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8.
Hyde Chapter 2—Solutions
In the table, D = dominant; R = recessive; M = mutant; + = wild type.
Allele
y+
Hw
Ax+
Co
rv+
dow
M (2)e+
J
tuf +
bur
Dominant
or Recessive
D
Mutant
or Wild Type
+
D
R
D
D
R
R
D
D
R
M
+
M
+
M
+
M
+
M
Alternative
Allele
y
Hw +
Ax
Co+
rv
dow+
M(2)e
J+
tuf
bur +
Dominant
or Recessive
R
R
D
R
R
D
D
R
R
D
10. 1. b, 2. a, 3. b, 4. c (one can pick the ace of diamonds).
EXERCISES AND PROBLEMS
12. a. AA Aa, aa Aa, AA aa, and Aa Aa.
b. AA AA and aa aa.
14. The 3:1 phenotypic ratio of the F2 generation is composed of a 1:2:1 genotypic ratio
that corresponds to 1/4 DD, 1/2 Dd, and 1/4 dd. The 1/4 dwarf F2 (dd), when selfed,
produce all dwarf progeny (dd). The tall F2 (3/4 of total F2), when selfed, fall into
two categories: 1/4 (DD, 1/3 of the tall F2) produces all tall, and 1/2 (Dd, 2/3 of the
tall F2) produces tall and dwarf progeny in a 3:1 ratio.
We can compute the relative proportions of genotypes and phenotypes in the F3
generation using the following table.
Fraction of
Genotype in F2
F2 Genotype Generation
DD
Dd (dwarf)
DD (tall)
Dd (tall)
F3 genotypic
ratio
F3 phenotypic
ratio
1/4
1/4
1/2
F3 Genotype
Dd
dd
1 1/4 = 1/4
1 1/4 = 1/4
1/4 1/2 = 1/2 1/2 =
1/8
1/4
1/4 1/2 =
1/8
3/8 DD
3/8 dd
5/8 tall
1/4 Dd
3/8 dwarf
Hyde Chapter 2 – Solutions
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Overall, the F3 are 3/8 DD (tall), 2/8 Dd (tall), and 3/8 dd dwarf.
16. When a decahybrid is selfed, it would produce 210 = 1024 different gametes; 1/(210)2
or approximately 0.000001 of the F2 would be homozygous recessive; 310 = 59,049
different genotypes yielding 210 = 1024 different phenotypes would appear.
If the decahybrid were testcrossed, it would produce 210 different gametes; 1/210 =
0.00098 of the F2 would be homozygous recessive; 210 different genotypes and
phenotypes would appear.
18. The round, yellow F2 plants are made up of four genotypes; the round, green of two
genotypes; the wrinkled, yellow of two genotypes; and the wrinkled, green of one
genotype. Testcrossing all these genotypes produces the following results:
W–G–: 1/16 WWGG wwgg all WwGg (round, yellow)
2/16 WwGG wwgg 1/2 WwGg (round, yellow), 1/2 wwGg (wrinkled,
yellow)
2/16 WWGg wwgg 1/2 WwGg (round, yellow), 1/2 Wwgg (round, green)
4/16 WwGg wwgg 1/4 WwGg (round, yellow), 1/4 Wwgg (round, green),
1/4 wwGg (wrinkled, yellow), 1/4 wwgg (wrinkled, green)
W–gg: 1/16 WWgg wwgg all Wwgg (round, green)
2/16 Wwgg wwgg 1/2 Wwgg (round, green), 1/2 wwgg (wrinkled, green)
wwG–: 1/16 wwGG wwgg all wwGg (wrinkled, yellow)
2/16 wwGg wwgg 1/2 wwGg (wrinkled, yellow), 1/2 wwgg (wrinkled,
green)
wwgg: 1/16 wwgg wwgg all wwgg (wrinkled, green)
20. Examine each trait separately. There are 104 long:34 short; and 69 brown:69 red.
Length appears in a 3:1 ratio; therefore, long is dominant and the 3:1 ratio in the
progeny tells us that each parent is heterozygous. Eye color appears in a 1:1 ratio.
We can’t conclude which allele is dominant; all we can conclude is that one parent is
a recessive homozygote and that one parent is a heterozygote. (We will assume that
+
+
red is the wild type.) If bw = red, bw = brown, s = long, and s = short, one possible
+
+
+
way to indicate the cross is bw bw s s bwbw s s.
22. You could set up a Punnett square and count the boxes; but unfortunately, this is an
8 8 matrix that yields 64 squares to count. A very tedious proposition! Set up the
cross: DdGgWw DdGgWw and look at one gene at a time.
a. The chance of getting a dominant trait from a monohybrid cross is 3/4.
Therefore, the chance of all three dominant traits together is 3/4 3/4 3/4 =
27/64.
b. The chance of getting a recessive trait from a monohybrid cross is 1/4.
Therefore, the chance for all three recessives is 1/4 1/4 1/4 = 1/64.
c. The chance of dwarf is 1/4, the chance of green is 1/4, and the chance of round
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Hyde Chapter 2—Solutions
is 3/4. Therefore the total chance producing the dwarf, green, round plant is 1/4
1/4 3/4 = 3/64.
+
+
Let e = ebony, e = wild type, b = black, and b = wild type. The first cross is
+ +
+ +
+
+
ee b b e e bb. This will yield only e e b b. Thus, all F1 flies will exhibit a
wild type phenotype.
+
+
b. Selfing of e e b b results in the following
+
+
wild type
9/16 e – b –
+
3/16 e – bb
black
+
ebony
3/16 ee b –
1/16 ee bb
ebony, black
Since it is difficult to distinguish black and ebony, 7/16 will be dark-bodied and
9/16 will be normal.
+
+
+ +
c. The first backcross is e e b b ee b b . Progeny = 1/2 ebony:1/2 normal.
+
+
+ +
The second backcross is e e b b e e bb. Progeny = 1/2 black:1/2 normal.
In each case we have a testcross situation for only one gene.
24. a.
26. The F1 progeny are AaBbCcDdEe. The chance of getting any individual with a
particular homozygous genotype is (1/4)5. Since we are looking for two different
possibilities, we have 2(1/4)5 = 2/1024 = 1/512.
28. This cross can be set up as a dihybrid cross: AaPp AaPp, where a is the recessive
albinism allele and p is the recessive PKU allele.
a. The chance of a recessive genotype (aa) from two heterozygotes is 1/4.
b. The probability of having albinism (aa) but not PKU (P–) is 1/4 3/4 =
3/16.
The probability of having PKU (pp) but not albinism (A–) is 1/4 3/4 =
3/16.
For an either/or situation, we add the probabilities, 3/16 + 3/16 = 6/16 = 3/8.
c. For two independent events to occur simultaneously, we multiply their
individual probabilities, 1/4 1/4 = 1/16.
30. We calculate the frequency of deaths from cancer as 300/900 = 1/3, and the deaths
due to heart disease as 200/900 = 2/9.
a. 1/3
b. For death by cancer or heart disease, we add the probabilities, 1/3 + 2/9 = 5/9.
32. The cross is Aa aa. Therefore, there is a one-half chance of either a taster or
nontaster offspring. We can imagine five different birth order scenarios that give one
taster child (symbolized as T) and four nontasters (symbolized as N) children. T, N,
N, N, N (where the taster child is the first-born); N, T, N, N, N (where the taster
child is the second-born); N, N, T, N, N; N, N, N, T, N; and N, N, N, N, T. The
probability of each scenario is 1/2 1/2 1/2 1/2 1/2 = 1/32. The probability
Hyde Chapter 2 – Solutions
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that one of five children will be a taster is given by the sum rule, 1/32 + 1/32 + 1/32
+ 1/32 + 1/32 = 5/32.
34. Hypothesis: WwGg WwGg produces W–G–:W–gg:wwG–:wwgg in a 9:3:3:1 ratio.
Critical chi-square at 0.05, 3 df, = 7.815.
Observed numbers (O)
Expected ratio
Expected numbers (E)
O–E
(O – E)2
(O – E)2/E
Offspring
W–G– W–gg wwG–
315
108
101
9/16
3/16
3/16
312.75 104.25 104.25
2.25
3.75
–3.25
5.06
14.06 10.56
0.016
0.135 0.101
wwgg
32
1/16
34.75
–2.75
7.56
0.218
Total
556
556
0.470 = 2
Since this chi-square, 0.470, is less than the critical chi-square, we fail to reject (that
is, accept) our hypothesis of two-locus genetic control with dominant alleles at each
locus.
36. a.
This is a dihybrid cross. From the F1 progeny, we can determine that longwinged and tan-bodied must be dominant to short-winged and dark-bodied,
respectively. Because this is a dihybrid cross, we expect the F2 progeny should
be in a 9:3:3:1 ratio.
b. Our null hypothesis: This is a dihybrid cross with dominant alleles at each
locus.
Observed numbers (O)
Expected ratio
Expected numbers (E)
O–E
(O – E)2
(O – E)2/E
Long,
Tan
84
9/16
90
–6
36
0.400
Offspring
Long, Short,
Dark
Tan
27
35
3/16
3/16
30
30
–3
5
9
25
0.300 0.833
Short,
Dark
14
1/16
10
4
16
1.600
Total
160
160
3.133 = 2
Critical chi-square at 0.05, 3 df, = 7.815. The chi-square for our null hypothesis,
3.133, is less than the critical chi-square. Therefore, we fail to reject (that is,
accept) our hypothesis.
38. a.
b.
c.
d.
1/2 1/2 1/2 1/2 1/2 = 1/32.
(1/2 1/2 1/2 1/2 1/2) + (1/2 1/2 1/2 1/2 1/2) = 2/32 or 1/16.
P(A– bb C– D– E–) = 3/4 1/2 1/2 3/4 1/1 = 9/64.
The probability of an individual with a parental phenotype is 9/64 + 9/64 = 9/32.
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Hyde Chapter 2—Solutions
e.
Therefore, the remaining 1 – 9/32 = 23/32 will be phenotypically unlike either
parent.
P = 0, because all offspring of this cross will have the dominant E trait.
40. You should change your choice because the box you chose originally has a
1/3 chance of containing the prize, whereas the remaining box has a 2/3 chance of
containing the prize. The 1/3 chance of your choice is set by the fact that there were
three equally likely choices at the beginning. When your friend eliminated an empty
box, she left two choices: your original box and the third box. Since the probability
of your original choice has not changed, the probability that the remaining box
contains the prize must be 2/3 to give a combined probability of 1.0.