BIOL276
Dr. Young
Name _______________________
Due _________________________
Genetics Problems Set #1 Answer Key
For problems in genetics, if no particular order is specified, you can assume that a
specific order is not required.
1.
2.
What is the probability of drawing one card at random from a deck and obtaining the
following? (A deck has 52 cards. Each card has a suit—spade, heart, diamond, or club—
and a number [1-10] or a face—jack, queen, king. Thus, there are 13 cards of each suit
and four cards of each number or face.) Express your answers as both a fraction and a
percentage.
a.
The queen of diamonds P(queen) = 4/52, and P(diamond) = 13/52. Using
the Multiplication Rule, (4/52)(13/52) = 1/52 = 0.019 = 1.9%
b. Any heart P(heart) = 13/52 = .25 = 25%
c.
A 10 (any suit) P(10 of any suit) = P(10) = 4/52 = .077 = 7.7%
d. Any even-numbered card There are 5 even-numbered cards per suit (2,
4, 6, 8, 10), and there are 4 suits. Therefore, (4)(5)/52 = 20/52 = .38 = 38%
e.
An even-numbered spade There are only five even-numbered cards that
are also spades. Therefore, 5/52 = 9.6%
In a family of seven children, what is the probability of obtaining the following numbers
of boys and girls?
a.
All boys (!)7 = 1/128
b.
All children of the same sex The children could be all boys or all girls:
(!)7 chance of being all boys and (!)7 chance of being all girls.
1/128 + 1/128 = 2/128 or 1/64 chance of being either all boys or all
girls.
Parts c–e require the use of the binomial expansion. Let a equal the
probability of being a girl and b equal the probability of being a boy. The
probabilities of a and b are !.
(a + b)7 = a7 + 7a6b + 21a5b2 + 35a4b3 + 35a3b4 + 21a2b5 + 7ab6 + b7
c.
d.
e.
Six girls and one boy The probability for part (c) is provided for by the
term 7a6b. Because the probabilities of a and b are !, then the
overall probability is 7(!)6(!) = 7/128.
Four boys and three girls This probability is provided for by the term
35a3b4. The overall probability is 35(!)3(!)4 = 35/128.
Four girls and three boys Using the term 35a4b3, we see that the overall
probability is 35(!)4(!)3 = 35/128.
3.
Two gene loci, A and B, are unlinked (and thus assort independently), and alleles A and B
are dominant over alleles a and b. Indicate the probabilities of producing the following.
a.
An AB gamete from an AaBb individual? 1/4
b. An AB gamete from an AABb individual? 1/2
c.
An AABB zygote from a cross AaBb ! AaBb? 1/16
d. An AaBb zygote from a cross AaBb ! AABB? 1/4
e.
An Aabb zygote from a cross AaBb ! AAbb? 1/4
f.
An AB phenotype from a cross AaBb ! AaBb? 9/16
g.
An AB phenotype from a cross aabb ! AABB? 1 (100%)
h. An aB phenotype from a cross AaBb ! AaBB? 1/4
4.
A man has either an AaBB or AABb genotype with equal probability.
a.
Assume these genes are unlinked (i.e., assorting independently). What is the
overall probability that the man will produce an Ab gamete? Only the AABb
genotype can produce Ab gametes, and the expected probability of
Ab gametes produced will be 1/2. Because the parent has an equal
probability (50%) of having either genotype (AaBB or AABb), the final
probability of the parent producing an Ab gamete is (1/2) " (1/2) = 1/4
= 0.25 = 25%.
b. What is the probability of the man producing an AB gamete? For genotype
AaBB, the probability of generating an AB gamete is 1/2. For
genotype AABb, the probability of generating an AB gamete is 1/2.
Again, because both genotypes have equal probabilities, the final
probability of the parent generating an AB gamete is (1/2) " (1/2) +
(1/2) " (1/2) = 2/4 = 0.5 = 50%.
5.
In snapdragons, the allele for red flowers is incompletely dominant over the allele for
white flowers, and thus heterozygotes have pink flowers. What ratios of snapdragon
flower colors would you expect to see among progeny generated from the following
crosses?
a.
red ! white 100% pink
b. red ! pink 50% pink; 50% red
c.
white ! pink 50% pink; 50% white
d. white ! white 100% white
e.
pink ! pink 25% red; 50% pink; 25% white
f.
red ! red 100% red
6.
Two pea plants with purple flowers are crossed. Among the offspring, 63 have purple
flowers, and 17 have white flowers. With a chi-square test, compare the observed
numbers with a 3:1 ratio and determine the probability that the difference between
observed and expected could be a result of chance. Plugging the numbers into the
chi-square equation: (63 – 60)2/60 + (17 – 20)2 /20 = 0.15 + 0.45 = 0.6 (the chisquare statistic). Degrees of freedom = the # of classes (phenotypes) – 1 =
2 – 1 = 1. Looking at the chi-square table for one degree of freedom and
using the standard critical value of 5%, we do not reject the hypothesis and
thus conclude that the observed differences between observed and
expected progeny numbers are a result of chance. We can further
conclude that both purple parents are heterozygous at the single gene
locus controlling flower color (at least for purple and white).
7.
In a plant species, you notice that purple and yellow leaf colors, as well as hairy and
smooth stems, segregate. You cross a plant with purple leaves and hairy stems to a plant
with yellow leaves and hairy stems, and generate the progeny indicated below:
Class
1
2
3
4
Total:
a.
b.
a.
Phenotypes
68 yellow leaves, hairy stems
66 purple leaves, hairy stems
22 purple leaves, smooth stems
25 yellow leaves, smooth stems
181
Indicate the most likely genotypes for each parent. Using P for the leaf
color locus and S for the stem type locus: PpSs " ppSs.
Propose a hypothesis to explain the progeny results. Based on this hypothesis,
what ratios are expected for each of the four classes of progeny? A possible
hypothesis is that each trait is controlled by independently
assorting single gene pairs, and that one allele in each pair exhibits
complete dominance over the other. Class 1 = 3/8, class 2 = 3/8,
class 3 = 1/8, class 4 = 1/8.
Using the chi-square method, test your hypothesis and indicate whether you
accept or reject it. (68 – 68)2/68 + (66 – 68)2/68 + (22 – 23)2/23 + (25 –
23)2/23 = 0.276. This number (the chi-square statistic) is very small;
therefore you confidently accept your hypothesis.
8.
In animals, the inability to make the pigment melanin results in albinism, a recessive
somatic condition. Two normal parents, who have decided to have three children, have a
first child that is albino (genotype aa). Determine the following probabilities:
a.
That the second and third children are also albino. Because the first child is
albino, both parents must be carriers for the recessive albino allele.
The probability of each child having the albino genotype aa equals
1/4 = 0.25 = 25%. Using the multiplicative rule, the probability of the
next two children being albino is (1/4) (1/4) = 1/16 = 0.0625 (6.25%).
b.
That the second and third children are normal. The probability of the
couple having any child with normal pigmentation (genotypes AA
or Aa) is 3/4 (75%). Therefore, using the multiplicative rule, the
probability of the next two children having normal pigmentation is
(3/4) (3/4) = 9/16 = 0.563 = 56.3%.
9.
In sailfin mollies (fish), gold color is due to an allele (g) that is recessive to the allele for
normal color (G). A gold fish is crossed with a normal fish. Among the offspring, 88 are
normal and 82 are gold.
a.
What are the most likely genotypes of the parents in this cross? Because we
know that gold (g) is recessive to normal (G), we can figure out that
the gold parent fish has the aa genotype. We also know that the
normal parent fish must have at least one G allele, but because
there are gold offspring (gg), we can predict that the normal parent
must be heterozygous Gg
In summary: gold fish (gg) " normal fish (Gg) !88 normal fish (Gg)
82 gold fish (gg)
b.
Assess the plausibility of your hypothesis by performing a chi-square test. If
the parents are gold (gg) " normal (Gg), we would expect the
offspring to have a phenotypic ratio of ! gold to ! normal. By
using the goodness-of-fit chi-square test, we can determine if the
deviations are likely to have been produced by chance.
Phenotype
Gold
Normal
Total
Observed (O)
82
88
170
Expected (E)
85
85
170
(O–E)2/E or (! 2)
0.11
0.11
0.21
Degrees of freedom (df) = (number of phenotypic classes) – 1.
Because there are two phenotypic classes, the degrees of freedom
(df) is 1.
From the chi-square table, we can see that the calculated chisquare value falls between 0.211 ( P of .9) and 0.51 ( P of .975). The
probability is relatively high that differences between what we
expected and what we observed was generated by chance and that
our parents are as predicted, (gg " Gg).
10.
In guinea pigs, the allele for black coat color (B) is dominant over the allele for white coat
color (b). At an independently assorting locus, an allele for rough coat (R) is dominant
over an allele for smooth coat (r). A guinea pig that is homozygous for black color and
rough coat is crossed with a guinea pig that has a white and smooth coat. In a series of
matings, the F1 are crossed with guinea pigs having white, smooth coats. From these
matings, the following phenotypes appear in the offspring: 24 black, rough guinea pigs;
26 black, smooth guinea pigs; 23 white, rough guinea pigs; and 5 white, smooth guinea
pigs.
a.
Using a chi-square test, compare the observed numbers of progeny with those
expected from the cross.
Phenotype
Black, smooth
Black, rough
White, rough
White, smooth
Total
Observed (O)
24
26
23
5
78
Expected (E)
19.5
19.5
19.5
19.5
19.5
(O–E)2/E or (! 2)
1.04
2.17
0.63
10.9
14.74
Degrees of freedom = 4 – 1 = 3. The chi-square value falls below
12.838 for a probability value less than .005 or 0.5 % that random
chance produced the observed ratio of guinea pigs.
b.
11.
What conclusions can you draw from the results of the chi-square test? From
the chi-square value, we can see that it is unlikely that random
variations produced the observed ratio. Some other phenomena
must be acting.
(How’s your logic?) Marta collects her laundry from the dryer and places it in a basket.
She has nine pairs of socks in the load. If she randomly pulls socks from the basket, what
is the probability that she will pull nine different socks from the basket before she pulls a
sock that makes a pair?
(18/18) x (16/17) x (14/16) x (12/15) x (10/14) x (8/13) x (6/12) x (4/11) x (2/10)
= 10,321,920/980,179,200 = 0.0105 = 1.05%