Problem LS4 Set 1
Correct numerical results are not answers to the problems. Assign symbols for genes and
phenotypes for alleles (e.g. Yellow (Y) gene, Y allele = yellow and y allele = g reen) that you use
in your answer. You are encouraged to use a Punnett square, where appropriate, in your
answer. Write out fractions and do math with labels to explain your numerical answer (e.g. 1/2
boys x 1/2 affected = 1/4 affected boys).
Problem 2.15. How many genetically different gametes could be formed by a man/woman with
the following genotypes?
(a) Aa bb CC DD
2 (A or a) 1 (b) 1 (C) 1 (D) = 2 gamete genotype
(b) AA Bb Cc dd
1 (A) 2 (B or b) 2 (C or c) 1 (d) = 4 gamete genotype
(c) A a Bb cc Dd
2 (A or a) 2 (B or b) 1 (c) 2 (D or d) = 8 gamete genotype
(d) A a Bb Cc Dd
2 (A or a) 2 (B or b) 2 (C or c) 2 (D or d) = 16 gamete genotype
Problem 2.12. You purchased a black stallion of unknown genotype. You mate him to a red
mare, and she delivers twin foals, one red and one black.
(a) Assuming that alternative alleles of a single gene are involved, what can you and what cant
you say about the coat color alleles in these horses?
Coat color gene: B = black allele, R = red allele
It is ambiguous whether B is dominant to R, or R is dominant to B.
Either [(Stallion) B R x (Mare) RR] or [(Stallion) BB x (Mare) RB] would just as
likely give [black (BR) and red (RR) foals] or [black (BB) and red ( RB) foals],
respectively
(b) How would your belief in how coat color is inherited be influence by the knowledge that
mating the same black stallion and red mare in the following season produced twin foals that
were both black?
The new information would have no effect on our belief about coat color. If R were
dominant, or if B were dominant, the stallion and mare would produce black and red foals in
equal proportions.
mare
R
stallion
B
BR
OR
R
RR
mare
R
B
stallion
B
RB
BB
(c) Among these horses, what cross, or crosses, could provide a definitive answer to how coat
color is inherited?
Because the dominant phenotype is being expressed by the heterozygote among these
horses, crossing either two red horses or two black horses it is ambiguous as to which-would represent the monohybrid cross and could uncover the recessive phenotype in 1/4 of
offspring that have the opposite coat color.
Problem 2.11. A mong native Americans, two types of earwax (cerumen) are seen, dry and
sticky. Inheritance of this trait can be observed in the types of offspring produced by different
kinds of matings:
Offspring
Parents
Sticky x Sticky
Sticky x Dry
Dry x Dry
Number of matings
10
8
12
Sticky
32
21
0
Dry
6
9
42
(a) How is earwax type inherited?
E = gene for type of earwax; E = sticky allele; and e = dry allele
E (sticky) x E (sticky)
E (sticky) x ee (dry)
ee (dry) x ee (dry)
(b) Why is there no 3:1 or 1:1 ratios in the data shown in the chart?
The genotype of parents with sticky earwax is ambiguous, either E E or Ee. Crosses with
sticky parents are combinations of cross with EE and Ee parents. So, the 1:0 segregation in
crosses including the homozygous dominant EE parents dilutes the number of recessive ee
offspring that are obtained from the 3:1 and 1:1 segregation ratios in crosses with the E e
parents.
Problem 2.16. In the following three crosses with dominant & recessive alleles at four gene:
(a) What is the probability of producing offspring with the phenotype of one of the two parents?
(b) How many phenotypes? What proportion of offspring with each phenotype?
(c) How many genotypes?
a. aa Bb Cc dd x Aa bb Cc DD
(a)
Parent phenotypes:
Probability{a B C d} = (1/2 aa)(1/2 Bb)(3/4 C)(0) = 0 --none
Probability{A b C D} = (1/2 Aa)(1/2 bb)(3/4 C)(1 Dd) = 3/16
(b)
2 (A or a)2 (B or b)2 (C or c)1 (D) = 8 phenotypes
Probability{a B C D} = (1/2 aa)(1/2 Bb)(3/4 C)(1 Dd) = 3/16
Probability{a b C D} = (1/2 aa)(1/2 bb)(3/4 C)(1 Dd) = 3/16
Probability{a B c D} = (1/2 aa)(1/2 Bb)(1/4 cc)(1 D d) = 1/16
Probability{a b c D} = (1/2 aa)(1/2 bb)(1/4 cc)(1 Dd) = 1/16
Probability{A B C D} = (1/2 Aa)(1/2 Bb)(3/4 C)(1 Dd) = 3/16
Probability{A b C D} = (1/2 Aa)(1/2 bb)(3/4 C)(1 Dd) = 3/16
Probability{A B c D} = (1/2 Aa)(1/2 Bb)(1/4 cc)(1 D d) = 1/16
Probability{A b c D} = (1/2 Aa)(1/2 bb)(1/4 cc)(1 D d) = 1/16
Or
abcD
aBcD
abCD
aBCD
abcD
aa bb cc DD
abcD
aa Bb cc DD
aBcD
aa bb Cc DD
abCD
aa Bb Cc DD
aBCD
ab
aa bb
ab
aa Bb
aB
aa bb
ab
aa Bb
aB
CD
Cc DD
CD
Cc DD
CD
CC DD
CD
CC DD
CD
aBCD=3
abCD=3
aBcD=1
abcD=1
(c)
Ab
Aa bb
Ab
Aa Bb
AB
Aa bb
Ab
Aa Bb
AB
cD
cc DD
cD
cc DD
cD
Cc DD
CD
Cc DD
CD
Ab
Aa bb
Ab
Aa Bb
AB
Aa bb
Ab
Aa Bb
AB
CD
Cc DD
CD
Cc DD
CD
CC DD
CD
CC DD
CD
ABCD=3
AbCD=3
ABcD=1
AbcD=1
2 (Aa, aa)2 (Bb, bb)3 (CC, Cc, cc)1 (D d) = 12 genotypes
b. A a BB CC Dd x Aa bb cc Dd
(a)
Parent phenotypes:
Probability{ A B C D} = (3/4 A)(1 B b)(1 Cc)(3/4 D ) = 9/16
Probability{ A b c D} = (3/4 A)(0 bb)(0 cc)(3/4 D ) = 0 --none
(b)
2 (A or a)1 (B)1 (C) 2 (D or d) = 4 phenotypes
Probability{ a B C d} = (1/4 aa)(1 Bb)(1 Cc)(1/4 dd) = 1/16
Probability{ a B C D} = (1/4 aa)(1 Bb)(1 Cc)(3/4 Dd) = 3/16
Probability{ A B C d} = (3/4 Aa)(1 Bb)(1 Cc)(1/4 dd) = 3/16
Probability{ A B C D} = (3/4 Aa)(1 Bb)(1 Cc)(3/4 Dd) = 9/16
AbcD
Abcd
abcD
abcd
aB
Aa Bb
AB
Aa Bb
AB
aa Bb
aB
aa Bb
aB
Cd
Cc Dd
CD
Cc dd
Cd
Cc Dd
CD
Cc dd
Cd
aB
Aa Bb
AB
Aa Bb
AB
aa Bb
aB
aa Bb
aB
CD
Cc DD
CD
Cc Dd
CD
Cc DD
CD
Cc Dd
CD
AB
AA Bb
AB
AA Bb
AB
Aa Bb
AB
Aa Bb
AB
Cd
Cc DD
CD
Cc dd
Cd
Cc Dd
CD
Cc dd
Cd
ABCD=9
aBCD=3
ABCd=3
aBCd=1
(c)
3 (AA, Aa, aa)1 (Bb)1 (Cc)3 (DD, D d, dd) = 9 genotypes
c. AA Bb cc dd x A a Bb Cc Dd
AB
AA Bb
AB
AA Bb
AB
Aa Bb
AB
Aa Bb
AB
CD
Cc DD
CD
Cc Dd
CD
Cc DD
CD
Cc Dd
CD
(a)
A B c d: (1 A)(3/4 B )(1/2 cc)(1/2 dd) = 3/16
A B C D: (1 A)(3/4 B )(1/2 Cc)(1/2 Dd) = 3/16
(b)
1 (A)2 (B or b)2 (C or c)2 (D or d) = 8 phenotypes
A b c d: (1 A)(1/4 bb)(1/2 cc)(1/2 dd) = 1/16
A b C d: (1 A)(1/4 bb)(1/2 Cc)(1/2 dd) = 1/16
AB
A B C D aa bb
ab
A B C d AA BB
AB
A B c D AA BB
AB
A B c d AA BB
AB
A b C D AA Bb
AB
A b C d AA Bb
AB
A b c D AA Bb
AB
A b c d AA Bb
AB
a B C D Aa BB
AB
a B C d Aa BB
AB
a B c D Aa BB
AB
a B c d Aa BB
AB
a b C D Aa Bb
AB
a b C d Aa Bb
AB
a b c D Aa Bb
AB
a b c d Aa Bb
AB
cd
cc DD
cD
Cc dd
Cd
cc Dd
cD
Cc dd
Cd
Cc Dd
CD
Cc dd
Cd
cc Dd
cD
cc dd
cd
Cc Dd
CD
Cc dd
Cd
cc Dd
cD
cc dd
cd
Cc Dd
CD
Cc dd
Cd
cc Dd
cD
cc dd
cd
Ab
aa bb
ab
AA Bb
AB
AA Bb
AB
AA Bb
AB
AA bb
Ab
AA bb
Ab
AA bb
Ab
AA bb
Ab
Aa Bb
AB
Aa Bb
AB
Aa Bb
AB
Aa Bb
AB
Aa bb
Ab
Aa bb
Ab
Aa bb
Ab
Aa bb
Ab
cd
Cc DD
CD
Cc dd
Cd
cc Dd
cD
cc dd
cd
Cc Dd
CD
Cc dd
Cd
cc Dd
cD
cc dd
cd
Cc Dd
CD
Cc dd
Cd
cc Dd
cD
cc dd
cd
Cc Dd
CD
Cc dd
Cd
cc Dd
cD
cc dd
cd
ABCD=3
ABCd=3
ABcD=3
ABcd=3
AbCD=1
AbCd=1
AbcD=1
Abcd=1
or
(c)
A b C D: (1 A)(1/4 bb)(1/2 Cc)(1/2 D d) = 1/16
A b c D: (1 A)(1/4 bb)(1/2 cc)(1/2 D d) = 1/16
A B c d: (1 A)(3/4 B )(1/2 cc)(1/2 dd) = 3/16
A B C d: (1 A)(3/4 B )(1/2 Cc)(1/2 dd) = 3/16
A B C D: (1 A)(3/4 B )(1/2 Cc)(1/2 Dd) = 3/16
A B c D: (1 A)(3/4 B )(1/2 cc)(1/2 D d) = 3/16
2 (AA, Aa)3 (BB, B b, bb)2 (Cc, cc)2 (Dd, dd) = 24 genotypes
Problem 2.18. Galactosemia is a recessive human disease that is treatable by restricting lactose
and glucose in the diet. Susan Smithers and her husband are both heterozygous for the
galactosemia gene.
a. Susan is pregnant with twins. If she has fraternal (nonidentical) twins, what is the
probability both of the twins will be girls who have galactosemia?
First child
(1/2 girls)(1/4 affected)
Second child
(1/2 girls)(1/4 affected) = 1/64
b. If the twins are identical, what is the probability that both will be girls and have galactosemia?
(1/2 both girls)(1/4 both affected) = 1/8
For parts cg, assume that none of the children is a twin.
c. If Susan and her husband have four children, what is the probability that none of the four will
have galactosemia?
(3/4 unaffected)(3/4 unaffected)(3/4 unaffected)(3/4 unaffected) = 81/256
d. If the couple has four children, what is the probability that at least one child will have
galactosemia?
At least one child = NOT all four children unaffected =
1 (3/4 3/4 3/4 3/4) = 1 (3/4)4 = 1 (81/256) = 175/256 = about 2/3
e. If the couple has four children, what is the probability that the first two will have galactosemia
and second two will not?
(1/4 affected)(1/4 affected)(3/4 unaffected)(3/4 unaffected) = 9/256
f. If the couple has three children, what is the probability that two of the children will have
galactosemia and one will not, regardless of the order?
1st unaffected: (3/4 unaffected)(1/4 affected)(1/4 affected)
2nd unaffected: (1/4 unaffected)(3/4 affected)(1/4 affected)
3rd unaffected: (1/4 unaffected)(1/4 affected)(3/4 affected)
Total
=
=
=
=
3/64 +
3/64 +
3/64
9/64 14%
Or, = [(all children)!/(affected)!(unaffected)!](1/4)(affected)(3/4)(unaffected) =
[3!/2!1!](1/4)2(3/4)1
=
3(1/16)(3/4) = 9/64 = 14%
g. If the couple has four children with galactosemia, what is the probability that their next child
will have galactosemia?
The probability is the same for all children and is independent for each child, so the
probability is 1/4.
Problem 2.21. To raise money for her textbooks, a UCLA student bred guinea pigs, starting with
a male with smooth black fur and a female with rough white fur. She was sorry to see that the
first F1 litter of eight contained only rough black animals because smooth white animals sell for a
higher price. To her disappointment, the second F1 litter from those same parents contained
only seven rough black animals. Soon the first litter had begun to produce F2 offspring, and they
showed a variety of coat types. Before long, the student had 125 F2 guinea pigs. Eight of them
had smooth white coats, 25 had smooth black coats, 23 were rough and white, and 69 were
rough and black.
(a) Assign gene labels and alleles in your explanation for how coat color and texture
characteristics are inherited? Theoretically, how many F2 g uinea pigs with each phenotype were
expected?
All 15 F1 animals had black fur like the male parent and rough texture like the female
parent. So, black is dominant to white, and rough is dominant to smooth. The fact that all
animals were the same suggests that the parents were pure-bred homozygotes for fur color
and texture.
Fur color and texture are explained by the inheritance of two independent genes:
C = fur color gene: The dominant C allele gives black fur, and the recessive c allele gives
white fur.
T = fur texture gene: The dominant T allele gives rough fur, and the recessive t allele gives
smooth fur.
A 3:1 segregation for each gene in F2 tells us that the F1 parents were both heterozygous.
Observed: black fur = 94, white fur = 31
Observed: rough fur = 92, smooth fur = 33
Expected: dominant phenotype = 94, recessive phenotype = 31 (3:1 ratio, 125 animals)
Also, 9:3:3:1 ratio, independent assortment in dihybrid offspring:
Expected: 70 black rough fur, 23 black smooth fur, 23 white rough fur, 8 white smooth fur.
(b) Which F2 animal would be mated with a rough black F1 in a testcross to confirm that the F1 is
a dihydrid? What possible phenotypes and proportions of offspring should the student expect
from a testcross with a rough black F2?
The white smooth F2 are homozygous recessive (cc tt) for both traits and would be used in a
testcross.
If rough black F2 is CC TT, then all offspring will be rough black (Cc Tt )
If F2 is Cc TT, then offspring will be 1 rough black (Cc Tt):1 rough white (cc Tt)
If F2 is CC Tt, then offspring will be 1 rough black (Cc Tt):1 smooth black ( Cc tt)
If F2 is Cc Tt, then offspring will be 1 rough black (Cc Tt):1 smooth black ( Cc tt): 1 rough
white (cc Tt):1 smooth white (cc tt )
Problem 2.27. For each of the following matings between fruit flies (Drosophila melanogaster),
assign gene symbols, give genotypes of each of the parents and give expected phenotypic ratios
of offspring for each gene.
Cross
1
Male
Wings
Eyes
Tiny
Oval
x
Female
Wings
Eyes
Tiny
Oval
2
Normal
Narrow
x
Tiny
Oval
3
Normal
Narrow
x
Normal
Oval
4
Normal
Narrow
x
Normal
Oval
Offspring
78 Tiny wings, oval eyes
24 Tiny wings, narrow eyes
45 Normal wings, oval eyes
40 Normal wings, narrow eyes
38 Tiny wings, oval eyes
44 Tiny wings, narrow eyes
35 Normal wings, oval eyes
29 Normal wings, narrow eyes
10 Tiny wings, oval eyes
11 Tiny wings, narrow eyes
62 Normal wings, oval eyes
19 Tiny wings, oval eyes
T = wingsize gene: dominant T allele gives normal wings, and recessive t allele gives tiny
wings.
N = eye shape gene: dominant N allele give oval eyes, and recessive n allele gives narrow
eyes.
1. tt Nn (male) x tt Nn (female) -3:1 segregation for eye shape
2. Tt nn (male) x tt Nn (female) -1:1 segregation for both eye shape and wingsize
3. Tt nn (male) x Tt Nn (female) -- 1:1 segregation for eye shape
3:1 segregation for wingsize
4. Tt nn (male) x Tt NN (female) -- 3:1 segregation for wingshape
Problem 2.25. The following chart shows the results of different matings between jimsonweed
plants that had either purple or white flowers and spiny or smooth pods. (a) Assign gene
symbols for each trait and indicate the phenotypes associated with dominant and recessive
alleles. (b) Indicate the genotypes of the parents for each of the crosses and the ratio of
phenotypes expected in offspring by those genotypes.
Parents
a. purple spiny x purple spiny
b. purple spiny x purple smooth
c. purple spiny x white spiny
d. purple spiny x white spiny
e. purple smooth x purple smooth
f. white spiny x white spiny
(a)
(b)
Purple
Spiny
94
40
34
89
0
0
Offspring
White
Purple
Spiny
Smooth
32
28
0
38
30
0
92
31
0
36
45
0
White
Smooth
11
0
0
27
11
16
C = color gene: Dominant C allele is purple; and, recessive c allele is white.
P = pod shape gene: Dominant P allele is spiny; and, recessive p allele is smooth.
a. Cc Pp x Cc Pp
9 purple spiny : 3 white spiny : 3 purple smooth : 1 white smooth
b. CC Pp x CC pp 1 purple spiny : 1 purple smooth
c. Cc PP x cc PP
1 purple spiny : 1 white spiny
d. Cc Pp x cc Pp
3 purple spiny : 3 white spiny : 1 purple smooth : 1 white smooth
e. Cc pp x Cc pp
3 purple smooth : 1 white smooth
f. cc Pp x cc Pp
3 white spiny : 1 white smooth
Problem 2.10. In humans, a dimple in the chin is a dominant characteristic.
A man who does not have a chin dimple has children with a woman with a chin dimple whose
mother lacked a chin dimple.
(a) What proportion of their children would be expected to have a chin dimple?
A = dimple chin gene: dominant A allele gives a dimple, recessive a allele give no dimple.
Man without dimple must be aa, and a dimpled woman could be AA or Aa but this dimpled
woman is Aa because the womans mother without a dimple must have been aa.
A man with a chin dimple and a woman who lacks a chin dimple produce a child who lacks a
dimple.
(b) What are the possible genotypes for the man?
A man with a chin dimple could be AA or Aa, but this man must be Aa because the child
lacks a dimple and must be aa, which means that the child inherited an a allele from his dad.
A man with a chin dimple and a nondimpled woman produce eight children, all have the chin
dimple.
(c) Can you be certain of the mans genotype? Why or why not? What genotype is more likely,
and why?
A man with a chin dimple can be either AA or Aa. With a nondimpled woman, who must be
aa, the man will have either all dimpled children, if he is AA, or a 1:1 ratio of dimpled and
nondimpled children, if he is Aa.
The chance that he is Aa, and has eight dimpled children =
(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = (1/2)8 = 1/256 = 0.4 percent
chance.
Although we cannot be absolutely certain, we can be very confident that the man is AA
because the chance of having this many dimpled children while being Aa is very low.
Problem 2.13. A pea plant that is tall, has green pods, and has purple flowers that are terminal
is crossed to a plant from a pure-breeding strain that is dwarf, has yellow pods, and has white
flowers that are axial. In the F1, all plant have axial flowers, 1/2 are tall while 1/2 are dwarf, all
plants have green pods, and 1/2 have purple flowers while 1/2 have white flowers.
(a) Assign gene names and alleles for the four traits. What are the genotypes of the parents?
T = tall gene: dominant T allele gives tall plants, recessive t allele gives dwarf plants.
A = flower position gene: dominant A allele give axial flowers, recessive a allele give
terminal flowers.
C = flower color: dominant C allele give purple flowers, recessive c allele give white flowers.
P = pod color: dominant P allele give green, recessive p allele gives yellow pods.
Parental genotypes are Tt aa Cc PP and tt AA cc pp.
(b) What phenotypes do you expect to see in the F2 f rom self-fertilized F1 plants that are tall,
have purple flowers that develop into green pods? And, in what proportion?
F1 plant is the trihybrid Tt Cc Pp.
F2 plants are
3/43/43/4 = 27/64
3/43/41/4 = 9/64
3/41/43/4 = 9/64
1/43/43/4 = 9/64
3/41/41/4 = 3/64
1/43/41/4 = 3/64
1/41/43/4 = 3/64
1/41/41/4 = 1/64
{Or an eight-by-eight Punnett square}
T C P : tall, purple flowers, green pods
T C pp : tall, purple flowers, yellow pods
T cc P : tall, white flowers, green pods
tt C P : dwarf, purple flowers, green pods
T cc pp : tall, white flowers, yellow pods
tt C pp : dwarf, purple flowers, yellow pods
tt cc P : dwarf, purple flowers, green pods
tt cc pp : dwarf, purple flowers, yellow pods