# Introduction To Genetic Analysis, Eleventh Edition Test Bank

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Chapter 2
Single-Gene Inheritance
MULTIPLE-CHOICE QUESTIONS
Sections 2.1. and 2.2. (Single gene inheritance, The chromosomal basis of single-gene
inheritance patterns)
1. If a plant of genotype A/a is selfed, and numerous offspring are scored, what proportion of the
progeny is expected to have homozygous genotypes?
A) 0
B) 25%
C) 50%
D) 75%
E) 100%
Answer: C
2. What is the maximum number of heterozygous genotypes that could be produced by
monohybrid self?
A) 1
B) 2
C) 3
D) 4
E) 6
Answer: A
3. A plant is heterozygous at three loci. How many different gamete genotypes can it theoretically
produce with respect to these three loci?
A) 2
B) 3
C) 4
D) 8
E) 16
Answer: D
4. In mountain rabbits, the EL-1 gene is located on chromosome 3. Four alleles of this gene have
been identified in the population. With respect to EL-1, what is the maximum number of
genotypes in the progeny of a single cross between two mountain rabbits?
A) 1
B) 2
C) 3
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D) 4
E) 6
Answer: D
5. A wild-type strain of haploid yeast is crossed to a mutant strain with phenotype d. What
phenotypic ratios will be observed in the progeny?
A) all wild type
B) 75% wild type and 25% mutant (d)
C) 50% wild type and 50% mutant (d)
D) 25% wild type and 75% mutant (d)
E) all mutant (d)
Answer: C
Section 2.3. (The molecular basis of Mendelian inheritance patterns)
6. Mice (Mus musculus) have 40 chromosomes per diploid cell (2n = 40). How many doublestranded DNA molecules and how many chromosomes are there in a mouse cell that is in the G2
stage of the cell cycle?
A) 40 DNA molecules and 20 chromosomes
B) 40 DNA molecules and 40 chromosomes
C) 40 DNA molecules and 80 chromosomes
D) 80 DNA molecules and 40 chromosomes
E) 80 DNA molecules and 80 chromosomes
Answer: D
7. A mutation occurs in a germ cell of a pure-breeding, wild-type male mouse prior to DNA
replication. The mutation is not corrected, and the cell undergoes DNA replication and a normal
meiosis producing four gametes. How many of these gametes will carry the mutation?
A) 1
B) 2
C) 3
D) 4
E) It is impossible to predict.
Answer: B
8. What is the mechanism that ensures Mendelโs first law of segregation?
A) formation of chiasmata
B) formation of the kinetochore
C) pairing of homologous chromosomes
D) segregation of homologous chromosomes during meiosis I
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E) segregation of sister chromatids during meiosis II
Answer: D
9. A laboratory mouse homozygous for an RFLP marker is mated to a wild mouse that is
heterozygous for that marker. One of the heterozygous individuals resulting from this cross is
mated back to the wild parent. What proportion of the offspring will have the same RFLP pattern
as the original laboratory mouse?
A) none of the offspring
B) 1/4
C) 1/2
D) 3/4
E) all of the offspring
Answer: C
10. The diagram below shows a part of the biochemical pathway responsible for fruit color in
peppers (Caspicum annuum). Enzyme 1 is responsible for catalyzing the reaction that turns the
colorless precursor into yellow pigment, whereas Enzyme 2 catalyzes the step that turns the
yellow pigment into red pigment. A breeder crosses a pure-breeding plant that makes yellow
peppers to a pure-breeding plant that makes red peppers. What proportion of the offspring will
make red peppers?
Enzyme 1
Enzyme 2
Colorless precursor
Yellow pigment
Red pigment
A) all of the offspring
B) 3/4
C) 1/2
D) 1/4
E) none of the offspring
Answer: A
11. The wild-type eye color in the fruit fly Drosophila melanogaster is dark red, as a result of a
mixture of bright red and brown pigments. โEnzyme Aโ is encoded by the โaโ gene, and is
required to synthesize the bright red pigment. A lack of red pigment results in a somewhat brown
eye color. You cross two fruit flies who are heterozygous for a recessive mutation that completely
inactivates the โaโ gene. What proportion of their offspring will have a recessive eye color
phenotype?
A) all of the offspring
B) 3/4
C) 1/2
D) 1/4
E) none of the offspring
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Answer: D
Section 2.4. (Some genes discovered by observing segregation ratios)
12. In pet rabbits, brown coat color is recessive to black coat color. A black female rabbit gives
birth to four black-coated and three brown-coated baby rabbits. What can be deduced about the
genotype of the baby rabbitsโ father?
A) He could be heterozygous black/brown or homozygous brown.
B) He could be heterozygous black/brown or homozygous black.
C) He must be heterozygous black/brown.
D) He must be homozygous black.
E) He must be homozygous brown.
Answer: A
13. โDumpyโ is a commonly used mutant phenotype in the nematode worm C. elegans. Two
โDumpyโ individuals are crossed to each other, and this cross produces 210 โDumpyโ and 68
wild-type individuals. If one of the โDumpyโ individuals used in this cross was mated with a wild
type, what โDumpyโ: wild-type ratio would we observe in the offspring?
A) 0:1
B) 1:0
C) 1:1
D) 1:3
E) 3:1
Answer: C
14. A female rabbit of phenotype cโฒ is crossed to a male rabbit with cch. The F1 is comprised of
five rabbits with a cโฒ phenotype, two with cch phenotype, and three with c phenotype. Of the
phenotypically c rabbits, two are females and are backcrossed to their father. This cross produces
only rabbits with cch phenotype. These results suggest that:
A) c could be dominant or recessive to cโฒ.
B) c is dominant to cโฒ, but recessive to cch.
C) c is dominant to cch, but recessive to cโฒ.
D) c is dominant to both cโฒ and cch.
E) c is recessive to both cโฒ and cch.
Answer: E
15. A plant with small red flowers is crossed to a plant with large white flowers. The resulting F 1
is comprised of 75 plants with small red flowers and 72 plants with small white flowers.
If flower color and flower size are controlled by a single gene each, what can be concluded from
these results?
A) Flower color is controlled by a sex-linked gene.
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B) Red color and small size are dominant to white color and large size, respectively.
C) Small size is dominant to large size, but we canโt determine which color is dominant.
D) We canโt determine which color and which size are dominant.
E) White color and small size are dominant to red color and large size.
Answer: C
16. A dominant gene b+ is responsible for the wild-type body color of Drosophila; its recessive
allele b produces black body color. A testcross of a heterozygous b+/b female by a black b/b
male gave 52 black and 58 wild-type progeny. If a black female from these progeny were crossed
with a wild-type brother, what phenotypic ratios would be expected in their offspring?
A) All males will be wild type, and all females will be black.
B) All progeny will be black.
C) All progeny will be wild type.
D) 75% will be wild type, 25% will be black.
E) 50% will be wild type, 50% will be black.
Answer: E
Section 2.5. (Sex-linked single-gene inheritance patterns)
17. A very common type of redโgreen colorblindness in humans is caused by a mutation in a
gene located on the X chromosome. Knowing that the mutant allele is recessive to the wild type,
what is the probability that the son of a woman whose father is colorblind is going to also be
colorblind?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: C
18. A phenotypically normal woman is heterozygous for the recessive Mendelian allele causing
phenylketonuria, a disease caused by the inability to process phenylalanine in food. She is also
heterozygous for a recessive X-linked allele causing redโgreen colorblindness. What percentage
of her eggs will carry the dominant allele that allows normal processing of phenylalanine and the
X-linked recessive allele that causes colorblindness?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: B
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19. A rare, curly winged mutant of Drosophila was found in nature. A mating of this fly with a
true-breeding, normal laboratory stock produced progeny in the ratio 1 curly winged to 1 normal
(both sexes had the same ratio). All curly winged progeny of this cross, mated with normal
progeny of the same cross, again yielded 1 curly winged to 1 normal fly. When mated with one
another, the curly winged progeny of the first cross yielded a progeny of 623 curly : 323 normal.
This ratio strongly suggests which of the following?
A) Curly and normal are in the 3:1 ratio expected from intercrossing monohybrid genotypes for a
recessive mutant allele (curly).
B) Curly and normal are in the 3:1 ratio expected from intercrossing monohybrid genotypes for a
dominant mutant allele (curly).
C) The curly winged parent of the curly ๏ด curly cross is homozygous.
D) Flies homozygous for the curly allele are lethal and never survive.
E) The gene for curly is sex-linked.
Answer: D
20. A female Drosophila with the mutant phenotype a is crossed to a male who has the mutant
phenotype b. In the resulting F1 generation all females are wild-type and all males have the a
mutant phenotype. Based on these results, we can conclude that the mode of inheritance of the
phenotypes of interest is:
A) autosomal for a and X-linked for b.
B) dominant for a and recessive for b.
C) recessive for a and dominant for b.
D) recessive for both a and b.
E) X-linked for a and autosomal for b.
Answer: D
21. A recessive X-linked gene mutation is known to generate premature baldness in males but is
without effect in women. If a heterozygous female marries an affected male, what proportion of
all their children is expected to be prematurely bald?
A) 1/4
B) 1/8
C) 1/16
D) 1/32
E) 1/216
Answer: A
Section 2.6. (Human pedigree analysis)
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22. You have three jars of gumballs. The first jar has 100 white gumballs and 25 green, the
second jar has 50 white and 150 blue, and the third jar contains 500 white and 10 red.
If you randomly draw one gumball from each jar, what is the probability for all white gumballs?
A) 0.196 or 19.6%
B) 0.109 or 10.9%
C) 0.056 or 5.6%
D) 0.567 or 56.7%
E) This is impossible (0% chance).
Answer: A
23. You have three jars of gumballs. The first jar has 100 white gumballs and 25 green, the
second jar has 50 white and 150 blue, and the third jar contains 500 white and 10 red.
If you randomly draw one gumball from each jar, what is the probability for all white OR all
colored gumballs?
A) 0.199 or 19.9%
B) 0.112 or 11.2%
C) 0.058 or 5.8%
D) 0.589 or 58.9%
E) This is impossible (0% chance).
Answer: A
24. You have three jars of gumballs. The first jar has 100 white gumballs and 25 green, the
second jar has 50 white and 150 blue, and the third jar contains 500 white and 10 red.
If you randomly draw one gumball from each jar, what is the probability for at least one white
gumball?
A) 0.997 or 99.7%
B) 0.85 or 85%
C) 0.69 or 69 %
D) 0.034 or 3.4%
E) This is impossible (0% chance).
Answer: A
25. The following pedigree concerns the autosomal recessive disease phenylketonuria (PKU).
The couple marked A and B are contemplating having a baby but are concerned about the baby
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having PKU. What is the probability of the first child having PKU? Unless you have evidence to
the contrary, assume that a person marrying into the pedigree (i.e., not a descendant of the two
parents at the top of the pedigree) is not a carrier. The filled-in individuals have PKU.
A) 0
B) 1/12
C) 1/4
D) 3/4
E) 9/64
Answer: B
26. The following pedigree depicts the inheritance of a rare hereditary disease affecting muscles.
What is the most likely mode of inheritance of this disease?
A) autosomal dominant
B) autosomal recessive
C) X-linked dominant
D) X-linked recessive
E) Y-linked
Answer: D
27. The following pedigree shows the inheritance of attached earlobes (black symbols) and
unattached earlobes (white symbol). Both alternative phenotypes are quite common in human
populations.
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If the phenotypes are determined by alleles of one gene, then attached earlobes are inherited as:
A) an autosomal dominant trait.
B) an autosomal recessive trait.
C) a dominant trait that could be either autosomal or X-linked.
D) a recessive trait that could be either autosomal or X-linked.
E) an X-linked dominant trait.
Answer: A
28. In the human pedigree shown below, black symbols indicate individuals suffering from a rare
genetic disease, whereas white symbols represent people who do not have the disease.
Based on the pedigree, what is the most likely mode of inheritance of this rare genetic disease?
A) autosomal dominant
B) autosomal recessive
C) X-linked dominant
D) X-linked recessive
E) Y-linked
Answer: C
29. The following pedigree shows the inheritance of a mild, but very rare condition in Siberian
Husky dogs. If individuals 1 and 2 are crossed, what is the probability that they will produce an
affected pup?
A)
B)
C)
D)
E)
1/36
1/16
4/36
4/16
16/36
Answer: C
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30. What is the probability that individual A is a heterozygous with respect to the condition
depicted in the pedigree?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Answer: E
31. What is the most likely mode of inheritance of the exceptionally rare condition represented in
the pedigree below, and why?
A) impossible to determine, because the condition is so rare
B) recessive, because it is only present in one generation, but we do not have enough
information to tell whether itโs X-linked or autosomal
C) recessive, because unaffected parents have an unaffected child, and autosomal, because there
are more autosomes than there are X chromosomes
D) X-linked recessive, because this would require the smallest number of rare alleles in the
pedigree
E) X-linked recessive, because it only affects a male, and his parents are unaffected
Answer: D
32. A couple is both heterozygous for the autosomal recessive disease cystic fibrosis (CF). What
is the probability that their first child will either be a boy or have CF?
A) 6/8
B) 5/8
C) 3/8
D) 2/8
E) 1/8
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Answer: B
33. Cystic fibrosis is an autosomal recessive condition. If the parents of a boy with cystic fibrosis
have two more children, what is the probability that both of these children will be unaffected?
A) 1/16
B) 3/16
C) 4/16
D) 9/16
E) 16/16
Answer: D
MATCHING QUESTIONS
34. Below are a list of crosses and a list of progeny phenotypic ratios. Match each cross with the
expected progeny phenotypic ratio. Write the letter corresponding to the progeny phenotypic
ratio in the space provided on the right of each cross. Each progeny ratio may be used
multiple times. If a cross has no corresponding progeny ratio, write an โXโ in the space
provided.
List of crosses
1)
monohybrid self (diploids)
2)
monohybrid testcross (diploids)
3)
mutant ๏ด wild-type in a haploid
4)
homozygous dominant ๏ด homozygous recessive (diploid)
5)
mutant 1 ๏ด mutant 2 in a haploid
_d__
โaโโ
โaโโ
โeโโ
โaโโ
List of phenotypic ratios observed in the progeny
a)
b)
c)
d)
e)
1:1
1:2:1
2:1:1
3:1
1:0
35. Four patterns of inheritance and four pedigrees are shown below. Assume that individuals
marrying into the family are homozygous for the wild-type allele. Match each of the
inheritance patterns with a pedigree. If there is no pedigree to match an inheritance pattern,
write โXโ beside that inheritance pattern.
A) autosomal dominant ______III____
B) autosomal recessive ______II____
C) X-linked dominant _______X___
D) X-linked recessive ______IV____
E) Y-linked
_______I___
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OPEN-ENDED QUESTIONS
Sections 2.1 and 2.2 (Single gene inheritance, The chromosomal basis of single-gene
inheritance patterns)
36. Mendel studied the inheritance of phenotypic characters determined by alleles of seven
different genes. It is an interesting coincidence that the pea plant has seven pairs of
chromosomes (n = 7). What is the probability that, by chance, Mendelโs seven genes would
each be located on a different chromosome? You may assume that the peaโs chromosomes
are all the same size.
Answer: 6!/76 = 6.12 * 10๏ญ3
Take each gene in turn. The probability is 1 that the first gene falls on a chromosome. The
probability that the second gene falls on any of the remaining six chromosomes is 6/7, the
next is 5/7, etc. The overall probability is the product of all these.
37. In Labrador retrievers, black color coat (B/โ) is dominant to brown color coat (b/b). A
breeder crosses two black individuals who have previously produced some brown puppies. If
the cross produces six puppies:
a) what is the probability that the first born will be brown?
b) what is the probability that four of them will be brown and two will be black?
c) what is the probability that at least one of them will be brown?
Answer: a) Both parents must be heterozygotes B/b because they have previously produced
brown puppies. The probability that they produce a brown puppy is therefore ยผ.
b) Each pup has ยพ chance of being black and ยผ chance of being brown. The order in which
the brown and black puppies are born does not matter, so there are 15 different permutations
of 4 brown + 2 black (5!). Hence, the probability is:
15[(1/4)(1/4)(1/4)(1/4)(3/4)(3/4)] = 135/4096 = 3.3%
c) In this case, the only instance that does not satisfy the condition is the case in which all
puppies are black. The probability of this event is (3/4)6 = 729/4096 = 17.8%. Therefore, the
probability of obtaining at least one brown puppy is 1 ๏ญ (729/4096) = 82.2%.
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38. In a particular species of plants, flower color is dimorphic: some individuals have red
flowers, whereas others have yellow flowers. If flower color is controlled by a single gene
with two alleles (cred and cyellow):
a) what would be the simplest way to determine which allele is dominant?
b) what will be the genotypic ratio in the offspring of a cross between a monohybrid and a
pure-breeding individual?
Answer: a) Crossing a pure-breeding red to a pure-breeding yellow individual, and assessing the
phenotype of the monohybrid produced. If it makes red flowers, then cred is dominant; if it
makes yellow flowers, then cyellow is dominant.
b) 1:1; half of the offspring will be heterozygous, and half will be homozygous like the purebreeding parent.
39. Suppose that red flower color (RR or Rr) is dominant to white flower color (rr) in a petunia.
A friend has a petunia plant with red flowers and wants to determine whether the plant is RR
or Rr.
a) What cross could you perform to help your friend determine the genotype of his petunia
plant?
b) How will this cross help you determine the genotype of your friendโs red-flowered
petunia? That is, how will the results from this cross differ if the red-flowered petunia is RR
versus Rr?
Answer: a) Perform a testcross (test the red petunia to a genotypically rr petunia).
b) You will observe different segregation in the testcross progeny, depending on the
genotype of the red petunia. If the red petunia is RR, then all testcross progeny will be red; if
the red petunia is Rr, then ยฝ of the testcross progeny will be red (Rr) and ยฝ will be white
(rr).
40. Suppose that a single gene controls fruit color in mango. Yellow fruit (Y) is dominant to red
fruit (y). Suppose a true-breeding yellow mango plant was crossed with a red-fruited plant,
and the resulting F1 was selfed. The F2 segregated as expected. If one of the yellow-fruited
plants was randomly selected and selfed, what is the probability that its progeny would
segregate for fruit color? Explain your logic.
Answer: The F2 consists of ยผ YY: ยฝ Yy : ยผ yy. Thus, the yellow-fruited plant that was randomly
picked could be either YY or Yy. There is a 1/3 chance that it was YY and 2/3 chance that it
was Yy. If a YY plant was selected and selfed, the progeny would not segregate for fruit
color. If a Yy plant was selected, the progeny would segregate for fruit color.
Section 2.3. (The molecular basis of Mendelian inheritance patterns)
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41. The wild-type flower color of a particular species of plant is blue. The diagram below shows
a simplified version of the biochemical pathways responsible for the synthesis of the blue
pigment. Suppose that gene โAโ codes for Enzyme A and gene โBโ for Enzyme B.
A friend provides you with a pure-breeding plant that makes colorless (white) flowers. What
genetic experiment(s) could you perform to determine whether your plant lacks Enzyme A or
Enzyme B? (Suppose that you have access to any pure-breeding lines that you need.)
Enzyme A
Enzyme B
Colorless precursor 1
Colorless precursor 2
Blue pigment
Answer: The โunknownโ white mutant can be crossed to a pure-breeding mutant that lacks
Enzyme 2 (genotype b/b); if the โunknownโ mutant lacks Enzyme 2, then the entire F1
should make only white (colorless) flowers, but if the โunknownโ mutant lacks Enzyme 1,
then the F1 should inherit a functioning A allele from the b/b parent and a functioning B allele
from the โunknownโ (a/a) parent, and therefore make blue flowers.
ALTERNATIVELY: A heterozygous A/a can be produced by crossing a wild type to a pure
line that lacks Enzyme A. This heterozygous can be crossed to the โunknownโ mutant; if a
1:1 of blue:white is observed in the offspring, then our โunknownโ mutant most likely lacks
Enzyme A and is therefore a/a.
42. Yellow leaves on a plant can be caused by genetic mutations, viruses, or unfavorable
environmental conditions. Suppose you find a plant that has yellow leaves, and you want to
determine if the cause of the phenotype is a genetic mutation or an environmental stress.
Design an experiment to differentiate between the different possibilities.
Answer: Cross the yellow plant with a normal plant. Self the resulting F 1 and look for a
consistent, predictable segregation pattern. For example, the presence of a 3 green :1 yellow
segregation ratio would suggest that the yellow phenotype was caused by a recessive
mutation.
Section 2.4. (Some genes discovered by observing segregation ratios)
43. Suppose that the length of a duckโs tail is determined by a single autosomal gene with two
alleles: L (long tail) and l (short tail). When a female duck with a long tail was backcrossed
to her father, she produced three ducklings with a long tail and three with a short tail.
a) What are the possible genotypes of the female duck and of her father?
b) What is the most likely genotype of the female duckโs father? (Justify your answer using
probabilities).
Answer: a) The presence of ducklings with the recessive phenotype among the offspring indicates
that both the mother (the โfemale duckโ) and the male used in the cross carry the recessive l
allele. The mother must be L/l as she has a long tail phenotype. The father could be L/l or l/l.
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b) l/l is more likely. The probability of the cross L/l ๏ด l/l producing a 1:1 ratio within an
offspring of six ducklings is [(1/2)6]*10 = ~15%, whereas the probability that the cross L/l ๏ด
L/l produce a 1:1 ratio in an offspring of six ducklings is [(3/4)3(1/4)3]*10 = ~6.6%.
44. Loppins are fictitious (but useful) diploid invertebrates that produce large offspring and
normally have long antennae. Short antennae mutants also exist. Unfortunately for the
geneticists working on these organisms, the malesโ antennae donโt fully develop until the
loppin equivalence of โmiddle age.โ A female with short antennae is crossed to a young male,
and all the females in their offspring have the short antennae mutant phenotype. A subset of
these F1 females are crossed to a middle-aged male with short antennae, and all the females
produced by these crosses have short antennae. However, all the crosses between the F 1
females and their brothers produce both short antennae and long antennae loppins in a ratio of
about 3:1. How can these results be explained? Provide the genotypes of as many individuals
as possible.
Answer: The 3:1 ratio obtained in the cross between brothers and sisters suggests that short
antennae (S) is dominant to long antennae (s), and that the F1 females and their brothers are
heterozygous (S/s). The young male used in the original cross is probably homozygous for the
long antennae allele (s/s); the middle aged male with short antennae is probably homozygous
(S/S, because all the progenies have short antennae). The cross between F1 females and the
middle aged male produces about 50% S/s and 50% S/S individuals; the cross between the F1
females and their brothers produces about 25% S/S, 25% s/s and 50% S/s, hence the observed
phenotypic ratios.
Section 2.5. (Sex-linked single-gene inheritance patterns)
45. Wild-type Drosophila melanogaster have a brown/grey body color. Mutants exist that have
a yellow body color. Several crosses were performed between phenotypically wild type and
yellow individuals, and the results of each cross are reported in the table below.
Deduce the mode of inheritance of the yellow body phenotype and genotypes of the parents
and offspring in the following crosses.
Parents
Female
a) wild type
b) yellow
c) yellow
d) wild type
Answer:
*
*
*
*
Male
yellow
yellow
wild type
yellow
Progeny
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
Females
Males
wild type yellow
wild type yellow
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
198
0
203
0
0
156
0
145
210
0
0
190
102
98
99
97
All the offspring in cross (a) are wild type; yellow is recessive to wild type;
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letโs define A as the dominant wild-type allele and a as the yellow mutant recessive. In all
the progenies we have roughly equal numbers of males and females, which is what is
expected. However, there is some sex bias and reciprocal crosses give different results: all
the sons of yellow females (homozygous a/a) are yellow; all the daughters of wild-type
males are wild type; this suggests sex-linkage.
In fact:
Progeny
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
Females
Males
Parents
wild type
yellow
wild type yellow
Female
Male
โโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโโ
a) XA/XA
* Xa/Y
XA/Xa
Xa/Xa
XA/Y
not possible
expected ratio
all WT, as observed
all WT, as observed
b) Xa/Xa
* Xa/Y
expected ratio
not possible Xa/Xa
all yellow, as observed
c) Xa/Xa
* XA/Y
expected ratio
XA/Xa
not possible not possible Xa/Y
all WT, as observed
all yellow, as observed
d) XA/Xa
* Xa/Y
expected ratio
XA/Xa
Xa/Xa
1:1, as observed
not possible Xa/Y
all yellow, as observed
XA/Y
Xa/Y
1:1, as observed
46. The black and yellow pigments in the coats of cats are controlled by an X-linked pair of
alleles. Females heterozygous for these alleles have areas of black and areas of yellow in
their coat (called tortoise-shell, or calico if there are also patches of white hair).
a) A calico cat has a litter of eight kittens: one yellow male, two black males, two yellow
females, and three calico females. Assuming there is a single father for the litter, what is his
probable color?
b) A yellow cat has a litter of four kittens: one yellow and three calico. Assuming there is a
single father for the litter, what is the probable sex of the yellow kitten?
c) How would you prove that XO cats are phenotypically female? What female kitten colors
(with respect to yellow, calico, and black) would you look for in which types of parental
color crosses?
Answer: a) Yellow (genotype: yw;Y chromosome) where yw = yellow; yw+ = black
b) Male. Since the father must be black (genotype yw+;Y chromosome), the only true yellow
progeny cannot have received a color gene from the father. It must be male, and must have
received its one X chromosome from its mother.
c) Look for female kittens that fail to express an allele they should have inherited from their
mothers: black female kittens from yellow mothers or yellow female kittens from black
mothers. These kittens should have an X chromosome from their fathers as usual; the fact
that they show no alleles from their mothers may suggest they developed from eggs without
an X chromosome and therefore that XO is female. Similarly, look for female kittens that
fail to express a color that should have been inherited from their father. Female progeny of
yellow tom cats should be either yellow or calico and of black tom cats, either black or
16
calico, depending on the allele inherited from the mother. A black daughter of a yellow tom
cat might come from a sperm lacking any sex chromosome. Chromosomal checks would be
required on these unexpected progeny.
47. A young woman is worried about having a child because her motherโs only sister had a son
with Duchenne muscular dystrophy (DMD). The young woman has no brothers or sisters.
(DMD is a rare X-linked recessive disorder.)
a) Draw the relevant parts of the pedigree of the family described above. (Be sure to include
the grandmother, the three women mentioned, and all their mates.)
b) State the most likely genotype of everyone in the pedigree.
c) Calculate the probability that the young womanโs first child will have DMD.
Answer: a) pedigree and b) genotypes
c) The grandmother must have been D/d. There is a 1/2 chance that the mother is D/d and, if
so, a further 1/2 chance that the woman herself is D/d. If she is, 1/2 of her sons will have
DMD. Since the probability of a son is also 1/2, the overall probability is 1/2 * 1/2 * 1/2 *
1/2 = 1/16.
Section 2.6. (Human pedigree analysis)
48. a) In families with four children, what proportion of the families will have at least one boy?
b) In families with two girls and one boy, what fraction of the families will have the boy as
the second child?
c) In families with four children, what fraction of the families will have the gender order
male-female-female-male?
Answer: a) 0.9375, since 1 ๏ญ Prob. of 4 girls, or 1 ๏ญ (.5)4 = 1 ๏ญ 0.0625. The frequency can be
calculated more laboriously by expanding the binomial (p + q)4 = p4 + 4p3q + 6p2q2 + 4pq3 +
q4 and calculating that 15/16 (0.9375) of the distribution has one boy.
b) 1/3, because the frequencies of MFF, FMF, and FFM families are equal.
c) Of four-child families, 6/16 have two boys and two girls; only 1/6 of such families will
have the birth order MFFM. Therefore, 1/16 will have that particular birth order. The same
answer can be derived as (0.5)4.
17
49. A man whose mother had cystic fibrosis (autosomal recessive) marries a phenotypically
normal woman from outside the family, and the couple considers having a child.
a) If the frequency of cystic fibrosis heterozygotes (carriers) in the general population is 1
in 25, what is the chance that the first child will have cystic fibrosis?
b) If the first child does have cystic fibrosis, what is the probability that the second child
will be normal?
Answer: a) The man must be a heterozygote, C/c. The probability that his wife is C/c is 1/25, and
if they are both C/c the probability of having an affected child is 1/4. Overall, the
probability is (1/25)(1/4) = 1/100.
b) The first child shows that both parents must have been C/c, so the probability that the
next child will be normal is 3/4.
50. Consider the following pedigree of a rare autosomal recessive disease. Assume all people
marrying into the pedigree do not carry the abnormal allele.
a) If individuals A and B have a child, what is the probability that the child will have the
disease?
b) If individuals C and D have a child, what is the probability that the child will have the
disease?
c) If the first child of C * D is normal, what is the probability that their second child will
have the disease?
d) If the first child of C * D has the disease, what is the probability that their second child
will have the disease?
Answer: a) Choosing M for unaffected and m for the disorder, male B must be M/m, and female
A has a 2/3 chance of being M/m. The overall chance of an affected child is 1 * 2/3 * 1/4 =
1/6.
b) If Cโs mother A is heterozygous, C stands a 1/2 chance of being heterozygous. Dโs
mother must be heterozygous, and D stands a 1/2 chance of inheriting that heterozygosity.
The overall chance of an affected child is 2/3 * 1/2 * 1 * 1/2 * 1/4 = 1/24.
c) The probability is still 1/24.
d) Now that we know individuals C and D must both be M/m, the chance of the second child
being m/m is 1/4.
18
51. Below is the pedigree of a family where some individuals are affected with a mild condition
of the skin.
a) Based on the pedigree, what is the most likely mode of inheritance of this condition, and
why?
b) Indicate the respective genotypes of each individual represented. For individuals who
could have two or more genotypes, calculate the relative probability of each possible
genotype.
c) What is the probability that individuals 1 and 2 will have an affected daughter?
Answer: a) Autosomal recessive; it is the only mode of inheritance whereby two unaffected
parents can have an affected daughter (as is the case for I-1 and I-2 and their first child).
b) A = WT; a = mild condition
The affected individuals are a/a; all four individuals in generation I are A/a; the unaffected
people in generation II have a probability of 2/3 of being A/a and 1/3 of being A/A; and
for individual โ2โ a few more calculations are required:
– if both her parents are A/A (probability of 1/9), then sheโs A/A;
– if one of her parents is A/A and the other A/a (probability of 4/9), then she has a 50%
chance of being A/A and 50% chance of being A/a;
– if both of her parents are carriers (probability of 4/9), then she has a 2/3 chance of
being A/a and 1/3 chance of being A/A.
– Overall, her probability of being A/A is (1/9) + (1/2)(4/9) + (4/9)(1/3) = 13/27
and her probability of being A/a is (1/2)(4/9) + (4/9)(2/3) = 14/27
c) (14/27)(1/2) = 7/27
52. In the late 1800s, Mendel defined two fundamental laws of transmission genetics; these were
subsequently used to establish chromosome theory as scientists examined visible chromosomes
in meiotic cells. Define these two laws, and diagram where in the process of meiosis these two
processes actually occur.
Answer: Students can diagram meiosis, which is a healthy review of their understanding of the
process, and identify within this process the observations of Mendel (working without knowledge
of meiosis). Mendelโs first law focused upon the segregation of genetic determinants during
meiosis. This is essentially the anaphase I-mediated process of reducing ploidy during meiosis I.
Mendelโs second law, independent assortment, occurs during meiosis I as homologous
chromosomes are lined up and assorted to meiocytes. This process is distinctly random in each
meiotic process and is key to genetic diversity within gamete production.
19

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