Answers to Genetics Problems
BI164
I.
May, 2005
a. These results indicate 100% of the offspring are colored and the
proportion of mice with normal coordination to waltzing mice is 3:1. We
know that the white mice have the cc genotype. Since all of the F1 are
colored, the colored parents must be CC. To produce the 3:1 ratio of
normal to waltzing behavior, we expect the parents are all heterozygotes
(Vv). Therefore the cross is CC Vv x cc Vv.
b. The results here suggest a phenotypic ratio of 9:3:3:1. We expect these
proportions in a dihybrid cross of heterozygotes. Therefore, the cross is
Cc Vv x Cc Vv.
c. We know the genotype of the white, waltzing mice (cc vv). The
offspring occur in a phenotypic ratio of 1:1:1:1. This result suggests a test
cross with mates heterozygous for both traits. (Note that this result is a
confirmation of Mendel’s Law of Independent Assortment). Therefore the
cross is Cc Vv x cc vv.
II.
For two heterozygote parents (Aa), 1/4 of all offspring would be expected
to show the recessive trait of albinism.
The chance of a child of these parents having normal skin color is 3/4. To
determine the probability of one normal and two albino children, we
multiply the probabilities of these independent events, if the normal child
was born in each of the three birth orders:
3/4 x 1/4 x 1/4 = 3/64
1/4 x 3/4 x 1/4 = 3/64
1/4 x 1/4 x 3/4 = 3/64
The sum of these is the total probability: 9/64
(Note: This is more of a math problem than a genetics problem. We got these
problem sets out of a genetics textbook, which gave the answer as 3/64. A
brilliant Colby student pointed out that, if the probabilities were calculated that
way, then they wouldn’t sum to 1.)
III.
The man must have genotype Cc ff because he has cataracts (at least one C
allele) but his father did not (cc) and he does not have fragile bones (ff).
On the other hand, the woman must be cc Ff because she has no cataracts
(cc) and her father did not have fragile bones (ff).
The cross then is cc Ff x Cc ff. By inspection, you can see that half of the
children will have cataracts (Cc) and half will have normal eyes (cc). In
like fashion, half of the children will have fragile bones (Ff) and half will
have normal bones (ff).
We can answer all four questions at once when we realize that each
combination of traits is equally likely.
1/2 x 1/2 = 1/4 regular eyes, regular bones
1/2 x 1/2 = 1/4 cataracts, regular bones
1/2 x 1/2 = 1/4 regular eyes, fragile bones
1/2 x 1/2 = 1/4 cataracts, fragile bones
IV.
This example involves epistasis. We know that the F1 generation will be
heterozygotic for both traits (Cc Gg). When we cross Cc Gg x Cc Gg, we
normally expect a 9:3:3:1 ratio. For this example, we should see:
9/16 C_ G_
3/16 C_ gg
3/16 cc G_
1/16 cc gg
However, we know that the presence of at least one C allele causes the
fruit to be white. Thus, the C_ G_ genotypes and C_ gg phenotypes will
all be white (12/16 cases).
If a plant has the cc genotype, the G locus determines the color. 3/16 of
the fruit will be yellow and 1/16 will be green.
The expected results are therefore: 12 white: 3 yellow: 1 green
V.
Let’s begin with the observation that the great-grandmother (I-1) must be
homozygous recessive (we can use A and a to represent the two alleles).
One of her children shows the recessive trait. Therefore, the greatgrandfather (I-2) must be Aa. II-1, II-3 and II-7 have to be Aa; they had to
inherit an a allele from their mother and the A allele from their father. II-5
must be aa.
The four crosses below all involve parents that do not show the trait.
These parents can be either carriers (Aa) or homozygote dominants (AA).
To calculate the chance of any of these crosses producing affected
children, we have to know three probabilities: the probability the mother
is a heterozygote; the probability the father is a heterozygote; the
probability that two heterozygotes will produce an affected child. The
first two probabilities must be determined from the pedigree. The last
probability is 1/4, as you know from Mendelian monohybrid crosses.
III-1 x III-12 - III-1 has a 50% chance of being Aa, III-12 is definitely Aa
The three probabilities listed above are multiplied: 1/2 x 1 x 1/4 = 1/8
III-4 x III-14 - both parents have a 50% chance of being Aa
Chances of aa child – 1/2 x 1/2 x 1/4 = 1/16
III-6 x III-13
III-6 has a brother that shows the trait. Her father was definitely Aa. That
means her mother had to be a carrier too. Since III-6 does not show the
trait, only three possibilities remain (Aa, aA or AA). Thus, III-6 has a 2/3
chance of being Aa. III-13 is definitely a heterozygote.
Chances of aa child - 2/3 x 1 x 1/4 = 1/6
IV-1 x IV-2
IV-1’s mother has a 2/3 chance of being Aa. Therefore, the chance that IV1 is Aa is 2/3 x 1/2
IV-2’s mother was certainly Aa. Therefore, the chance that IV-2 is Aa is
1/2
Chances of aa child = 1/3 x 1/2 x 1/4 = 1/24
VI.
From the Hardy-Weinberg model, the frequency of a heterozygote is 2 x
frequency of allele A x frequency of allele B
For these examples, frequency of A1A2 = 2(0.6)(0.3) = 0.36
Frequency of A1A3 = 2(0.6)(0.1) = 0.12
Frequency of A2A3 = 2(0.3)(0.1) = 0.06
VII.
Frequency of homozygote recessives in the class = 14/25 = 0.56 = q2
Solve for q by taking the square root = 0.748
p = 1.000 – 0.748 = 0.252
Frequency of homozygote dominants = p2 = (0.252)2 = 0.064
We multiply the 25 students by 0.064 to yield the expect number of TT
students – 1.59
We round 1.59 to 2 and therefore predict that only 2 of the 11 tasters in the
class were homozygote dominants and 9 were heterozygotes.
Alternatively, we could determine the frequency of heterozygotes by 2pq
= 2(0.748)(0.252) = 0.377
Multiply the number of students (25) by 0.377 = 9.42, which we round to 9
students