Download Practice Questions [Lectures 5-12](128 KB pdf file)

MCB142/IB163
Mendelian and Population Genetics
9/19/02
Practice questions lectures 5-12 (Hardy Weinberg, chi-square test, Mendel’s second law, gene
interactions, linkage maps, mapping human diseases, non-random mating, selection, genetic
drift, evolutionary genetics)
Question 1
You are studying an X-linked trait. There are two alleles, one showing complete dominance over the
other. In females, 84 percent show the dominant phenotype. What percentage of the males will show
the dominant phenotype, assuming that individuals in this large population are in Hardy Weinberg
proportions, and that the allele frequencies are equal in males and females?
Question 2
Human albinism is an autosomal recessive trait. Suppose that you find a village in the Andes where 1/4
of the population is albino. If the population size is 1000 and the population is in Hardy-Weinberg
proportions with respect to this trait, how many individuals are expected to be heterozygotes?
Question 3
An RFLP probe shows four different morphs (alleles) in a population of three-spined sticklebacks (a
fish-Gasterosteus aculeatus). How many different heterozygous genotypes will there be in this
population with regard to this RFLP?
Question 4
Suppose that you are a genetic counselor, and a couple seeks your advice about BHT tasting (people
who can taste BHT in processed food are recessive homozygotes for the “taster” allele and non-tasters
are homozygous dominant or heterozygous for the “non-taster” allele). Both prospective parents are
non-tasters, but a careful analysis of the husband’s pedigree reveals that he is a carrier of the taster
allele. The proportion of BHT tasters in the population is 16 percent. What is the probability that the
couple’s first child will be a non-taster of BHT assuming the population is in Hardy Weinberg
proportions?
Question 5
In a population under study, the frequency of an autosomal recessive disease is 1 percent. Assuming
Hardy Weinberg proportions, what is the probability of two heterozygotes marrying, and what
proportion of affected children will be born from such matings?
Question 6
In a Pygmy group in central Africa, the frequencies of alleles determining the ABO blood groups were
estimated as 0.2 for IA, 0.1 for IB, and 0.7 for i (allele O). Assuming Hardy Weinberg proportions,
determine the expected frequencies of the blood types A, B, AB, and O. (Alleles A and B are
codominant, and O is recessive.)
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Question 7
In a test of Hardy Weinberg proportions for a gene with 3 alleles, and hence 6 genotype classes,
determine whether the chi-square value given below for a number of different cases is compatible with
the null hypothesis, or rejects the null hypothesis of Hardy Weinberg proportions, in which case give
the P value. You will need to determine the degrees of freedom (df) in each case.
(a) The chi-square value is 12.5 and all expected values for the 6 genotype classes are >5.
(b) The chi-square value is 4.3, one genotype class with expected value <5 was combined with
another, for a total of 5 genotype classes.
(c) The chi-square value is 20.5, two genotype classes with expected values <5 were combined
together (the expected value was then >5).
(d) Describe what the P value means in each of the cases where the null hypothesis was rejected.
Question 8
A plant of genotype C/C ; d/d is crossed to c/c ; D/D and an F1 testcrossed to c/c ; d/d.
(a) If the genes are on separate chromosomes, what percentage of c/c d/d offspring will be seen?
(b) If the genes are linked, and 20 map units apart, what percentage of c/c d/d offspring will be seen?
Question 9
In Drosophila, the two genes w and sn are X-linked and 25 map units apart. A female fly of genotype
w+sn+/w sn is crossed to a male from a wild-type line. What percent of male progeny will be w+sn?
Question 10
Three genes, each represented by a normal (dominant) and a mutant (recessive) allele, are on separate
chromosomes and assort independently of one another. The F2 of a cross of the homozygous (A/A ;
B/B ; C/C) normal strain with the triply mutant strain (a/a ; b/b ; c/c) will produce 3/4 normal to 1/4
mutant for each character. Using these figures, and without a checkerboard (Punnett square), determine
what fractions of the F2 will be of the phenotypes indicated below. Show the component probabilities.
(a) A/- ; B/- ; c/c
(b) a/a ; b/b ; c/c
(c) A/- ; B/- ; C/(d) a/a ; B/- ; c/c
Question 11
A female fruit fly of genotype A/a ; B/b is crossed to a male fruit fly of genotype a/a ; b/b. The
progeny of this cross were
Genotype
Number of individuals
A/a ; B/b
38
a/a ; b/b
37
A/a ; b/b
12
a/a ; B/b
13
(a) What gametes were produced by parent 1 and in what proportions?
(b) Do these proportions represent independent assortment of the two genes?
(c) What can be deduced from these proportions?
(d) Draw a diagram or explain the origin of the two rarest progeny genotypes.
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Question 12
In mice the allele for color expression is C (c = albino). Another gene determines color (B = black and
b = brown). Yet another gene modifies the amount of color so that D = normal amount of color and d =
dilute (milky) color. Two mice that are C/c ; B/b ; D/d are mated. What proportion of progeny will be
dilute brown (assume complete dominance at each locus and independent assortment of the genes)?
Question 13
In pigs, when a pure-breeding red is crossed to a pure-breeding white the F1 are all red. However, the
F2 shows 9/16 red, 1/16 white, and 6/16 are a new color, sandy.
(a) Provide a genetic hypothesis for this observation
(b) How would you test this hypothesis?
Question 14
The genes Dm/dm, Lu/lu, and Sc/sc are inherited autosomally in humans. In a large population, you
find women who are phenotypically dominant for all three traits whose parents consisted of one
phenotypically dominant individual (for all three traits) and one phenotypically recessive individual
(for all three traits). You observe the progeny of all such women whose husbands have a recessive
phenotype for these three traits, and obtain the following data
Phenotypes
Number of individuals
Dm Lu Sc
407
Dm Lu sc
4
Dm lu sc
22
Dm lu Sc
72
dm Lu Sc
18
dm lu Sc
3
dm Lu sc
70
dm lu sc
404
1,000
(a) What is the map order of the three loci?
(b) What are the map distances between the three markers?
Question 15
Most genetic variation for many human loci lies within local populations rather than between
populations or races. What does this observation tell you about human genetic evolution?
Question 16
Most genetic variation for many human loci lies within local populations rather than between
populations or races. There are some contradictions to this general trend, such as hair and skin color,
and sickle cell anemia. What do these exceptions tell you about the evolution of those traits?
Question 17
Compare and contrast the neutral, balance, and evolutionary lag schools.
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Question 18
For each of the following sets of genotype relative fitness values, predict whether a stable
polymorphism will be maintained (in which case give the equilibrium frequency of the two alleles) or
whether one or the other of the alternate alleles will go to fixation (in which case state which allele will
become fixed in the population). Assume that the population size is very large and that selection is the
only force acting to alter the frequencies of the alleles at this locus.
AA
AB
BB
(a)
1.00
0.98
0.10
(b)
0.80
1.00
0.6
(c)
1.0
0.6
0.6
Question 19
At a locus with two alleles A and B, the homozygous AA individual has a fitness of 0.5 relative to the
heterozygote AB, while the BB genotype is lethal. Under this selection scheme
(a) what is the equilibrium allele frequency of A?
(b) what are the three genotype frequencies expected at zygote formation, assuming random mating?
Question 20
What will determine the rate of a molecular clock for a gene that codes for a functional protein?
Question 21
To what extent is evolution a chance process and to what extent a directed process?
Question 22
In the thousands of years that dogs have been domesticated, breeders have developed numerous breeds
with many desirable traits. In many breeds of dogs, however, undesirable characteristics have appeared
as well. The undesirable features include single gene traits such as retinal degeneration or deafness.
Why do these undesirable traits persist?
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MCB142/IB163
Mendelian and Population Genetics
9/19/02
Answers to Practice questions lectures 5-12
Question 1
Let allele A represent the dominant allele, with frequency p in both males and females, and a the
recessive allele, with frequency q, with p + q =1. We are given the information that in females 84%
have the dominant phenotype. This means that 16% have the recessive phenotype aa, with expected
frequency q2, so using the square root formula (we can’t use allele counting since we cannot
distinguish the heterozygotes from the dominant homozygote class) under the assumption that the
population is in Hardy Weinberg proportions, the estimate of q = 0.4, hence p = 0.6. In males, the
frequency of the dominant phenotype is equal to the allele frequency, hence the answer is 0.6.
Question 2
We represent the autosomal recessive allele for albinism by a, with frequency q, and the dominant
allele is represented by A, with frequency p, with p + q =1. We are told to assume Hardy Weinberg
proportions, thus we can use the square root formula to estimate q. Since q2 = 0.25, the estimate of q =
0.5. The expected proportion of heterozygous individuals is 2pq = 0.5, thus in the population of size
1,000 we expect 500 individuals to be heterozygotes, and hence carriers of the disease predisposing
allele.
Question 3
Denoting the four alleles as A, B, C, and D, then there are 6 heterozygous genotypes, AB, AC, AD,
BC, BD, and CD. (There are 4 homozygous genotypes possible: AA, BB, CC, and DD.)
Question 4
Denote the recessive taster allele by t (frequency q), so that tasters are tt (frequency q2). We are given
that the husband is heterozygous, hence Tt. We are given that the mother is a non-taster, so she is
either TT or Tt. Assuming Hardy Weinberg proportions, we estimate q using the square root formula:
given that q2 = 0.16, then the estimate is q = 0.4, hence p = 0.6. The frequency of homozygous TT
individuals in the population is p2 = 0.36, and the frequency of heterozygotes Tt is 2pq = 2(0.6)(0.4) =
0.48. Given that the woman is a non-taster, i.e., she is either TT or Tt, the conditional probability that
she is TT is 0.36/0.84 = 0.4286, and the probability she is Tt, is 0.48/ 0.84 = 0.5714. We have to
consider both possible genotypes of the mother, since a non-taster child is possible both if she is TT or
Tt. Thus the required probability is 1 (the probability the man is Tt) x 0.4286 (the probability the
woman is TT given she is a non-taster) x 1 (the probability that their first child will be a non-taster (TT
or Tt), given these genotypes) (= 0.4286) + 1 (the probability the man is Tt) x 0.5714 (the probability
the woman is Tt given she is a non-taster) x 3/4 (the probability that their first child will be a nontaster (TT or Tt) given these genotypes) (= 0.4286), thus the required probability is 0.8572 (= 0.4286 +
0.4286, note that it is a coincidence of the given allele frequencies that the probabilities from both
possible genotypes of the woman are the same, usually these would be different).
A quicker way to get the answer, is to calculate the probability their first child will be a taster = 1 (the
probability the man is Tt) x 0.5714 (the probability the woman is Tt given she is a non-taster) x 1/4
(the probability that their first child will be a taster (tt), given these genotypes = 0.1428, and thus the
required probability is 1 – 0.1428 = 0.8572.
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Question 5
Assuming Hardy Weinberg proportions, we use the square root formula to calculate the allele
frequency q of the recessive allele, denoted by a: given q2 =0.01, q = 0.1, hence p = 0.9. The number of
heterozygous carriers (Aa) is thus 2pq = 0.18, and the frequency of heterozygous x heterozygous
matings (assuming mating is at random) is (2pq) x (2pq) = 0.0324. Since ¼ of the children from such a
mating will be affected (frequency = 0.0324/4 = 0.0081), this represents 81% of the total cases which
overall have a frequency of 1%.
Question 6
f(A) = (0.2)2 + 2(0.2)(0.7) = 0.32
f(B) = = (0.1)2 + 2(0.1)(0.7) = 0.15
f(AB) = 2(0.2)(0.1) = 0.04
f(OO) = (0.7)2 = 0.49
(check that the sum of the genotype frequencies adds to 1)
Question 7
(a) df = 3 (=6 – 1 –2, 6 classes minus 1 since we know the total count, minus 2 allele frequency
estimates, the third allele frequency is known once we know 2 of the allele frequencies), the chisquare value of 12.5 is between the P<0.01 and P<0.001 values for a chi-square distribution with 3
df (see table from handout), so we reject the null hypothesis with P<0.01.
(b) df = 2 (5 – 1 –2), the chi-square value of 4.3 with 2 df is compatible with the null hypothesis, so
these data are not significant.
(c) df = 2 (5 –1 –2), the chi-square of 20.5 with 2 df is very significantly different from the null
hypothesis (fit to Hardy Weinberg proportions) with P<0.001.
(d) The P value represents the type 1 error, which is the probabaility that you have rejected the null
hypothesis when it was true, i.e., the probability you have made a mistake in rejecting the null
hypothesis. So in (a) the type 1 error is 0.01, while in (c) it is smaller at 0.001.
Question 8
(a) All F1 individuals will be C/c D/d and will produce equal numbers (since the genes are on different
chromosomes) of the 4 gametic types CD, Cd, cD, and cd. Since the testcross individual only
produces gametes of type cd, the proportion of c/c d/d offspring will be 25%.
(b) All F1 individuals will again be C/c D/d. However, since the genes are now linked (r = 20 cM =
20% recombination) apart, and we know the parental gametic types are Cd and cD, the 4 gametic
types CD, Cd, cD, and cd are expected in proportions r/2, (1-r)/2, (1-r)/2, and r/2 respectively,
where r = 0.2. Since the testcross individual only produces gametes of type cd, the proportion of
c/c d/d offspring will be 10%.
Question 9
The female will produce four gametes: the 2 parental w+sn+ and w sn will occur with equal frequencies
of (1-r)/2 where r = 0.25, i.e., each with 37.5% frequency, while the 2 non-parental (recombinant)
gametes w+sn and w s+ will each occur with a frequency of 12.5% (check total number of gametes =
100%). Since males have only one X chromosome which they receive from their mothers, the percent
of male progeny that will be w+sn is 12.5%.
Question 10
(a) ¾ x ¾ x ¼ = 9/64
(b) ¼ x ¼ x ¼ = 1/64
(c) ¾ x ¾ x ¾ = 27/64
(d) ¼ x ¾ x ¼ = 3/64
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Question 11
(a) The proportions were 38% AB, 37% ab, 12% Ab, and 13% aB.
(b) No, because we expect 25% of each under independent assortment, and these numbers are very
different from that. (One could do the chi-square test to show lack of independent assortment, but it
is not asked for in this question.)
(c) The two genes must be linked; the %R (RF) = (12 + 13) / 100 = 25%, i.e., the two loci are 25 map
units (25 cM) apart on the same chromosome.
(d) The female parent received gametes AB and ab from her two parents (she is AB/ab), the male
parent received ab gametes from both his parents (he is ab/ab), and the two rarest progeny classes
represent recombinant classes from the female parent.
Question 12
We need to determine the probability of a C/- ; bb ; dd offspring from this cross = ¾ x ¼ x ¼ = 3/64.
Question 13
(a) The presence of a dominant allele at either of two loci (A/- ; b/b or a/a ; B/-) gives sandy, where
red is determined by the dominant alleles of both loci (A/- ; B/-), and white by the recessive
genotype at both loci (a/a ; b/b).
(b) Do testcrosses of a large number of the sandy offspring with the white offspring. Under this
hypothesis, the sandy offspring are A/A ; b/b (1/16) (gametes Ab) or A/a ; b/b (2/16) (gametes ½
Ab and ½ ab) or a/a ; B/B (1/16) (gametes aB) or a/a ; B/b (2/16) (gametes ½ aB and ½ ab), i.e.,
three types of gametes with equal frequencies: Ab, aB, and ab, so the offspring from the testcrosses
with (a/a ; b/b) should be equal proportions of
A/a ; b/b (sandy)
a/a ; B/b (sandy)
a/a ; b/b (white),
i.e., 2/3 sandy and 1/3 white.
(One should also perform other crosses to confirm the model, including testcrosses and crosses of
the putative genotypes of specific sandy offspring.)
Question 14
(a) The non-recombinant gametes from the mother are Dm Lu Sc and dm lu sc, while the double
recombinant gametes (the least frequent class) are Dm Lu sc and dm lu Sc, thus the order of the
three loci is Dm Sc Lu or Lu Sc Dm.
(b) The Dm Sc recombination distance is (4 + 22 + 18 + 3) / 1,000 = 4.7% (= 4.7 mu = 4.7 cM).
The Sc Lu recombination distance is (4 + 70 + 72 + 3) / 1,000 = 14.9%.
The Dm Sc recombination distance is (22 + 72 + 18 + 70 + 4 x 2 + 3 x 2) / 1,000 = 19.6%.
Correct answer must include double recombinants for Dm Sc, and check that the recombination
distances are consistent, i.e., 4.7 + 14.9 = 19.6.
Question 15
It suggests that humans have had persistent migration between populations that has prevented localized
differentiation for most genes
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Question 16
They most likely reflect adaptations to local conditions. Dark skin is more adaptive in warm sunny
climates, whereas light skin may be more adaptive when exposure to sun is minimal. The sickle cell
anemia allele provides protection against malaria (in heterozygotes and homozygotes), and so is at
high frequency in malarial environments, despite the deleterious effects of the allele in homozygous
individuals with sickle cell anemia. Other genes which do not contribute to local adaptation would not
show differentiation (the majority of genes).
Question 17
The neutral school maintains that much of the genetic variation in populations represents allelic
variants that are selectively equivalent to each other and polymorphisms seen in a population represent
a balance of new mutations entering the population, some of which will replace older alleles by drift
(chance) effects. In this way genetic changes accumulate between distinct species.
The balance school maintains that much of the genetic variation in populations has adaptive
significance.
The evolutionary lag school argues that much of the variation seen in populations may represent
transient variation, as selectively advantageous alleles replace older alleles in a ‘race’ to keep up with
changes in the environment.
It is difficult, if not impossible, to know what fraction of the genetic variation in populations falls into
each of these three categories, and all we can say is that each contributes to genetic variation.
Question 18
(a) A will be fixed.
(b) Balanced polymorphism, f(A) = p1 = s2/(s1 + s2) = 0.4/0.6 = 2/3 (s1 = 0.2, s2 = 0.4), f(B) = p2 =
s1/(s1 + s2) = 0.2/0.6 = 1/3.
(c) A will be fixed.
Question 19
(a) Using the formula to determine equilibrium allele frequencies with heterozygote advantage, s1 =
0.5, and s2 = 1.0, so the equilibrium frequency of A = p1 = s2/(s1 + s2) = 1/1.5 = 2/3 (and the
equilibrium frequency of B = 0.5/1.5 = 1/3).
(b) At zygote formation, the expected genotype frequencies are the Hardy Weinberg proportions 4/9
AA, 4/9 AB, and 1/9 BB.
Question 20
The degree to which changes in the protein will decrease the fitness of the organism. If the protein is
very sensitive to change (cannot function optimally with even small changes in its amino acid
composition) then the molecular clock will be extremely slow. If the protein can function adequately
even with many changes, perhaps to sites that are not immediately associated with the active site on
the protein, then the molecular clock will be relatively fast.
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Question 21
The ultimate source of genetic variation is mutation, so the building blocks of evolution come from
chance events. Also, mutations which are selectively equivalent to existing alleles in the population
can become established in the population by genetic drift (chance) effects. Thus, even without direct
selection, evolution will occur in populations, since they are always of finite size and subject to drift
effects. Evolution is not a directed process in terms of having a purpose or striving for improvement.
Since there are many conceivable ways to succeed at surviving and reproducing, natural selection
lacks direction in a specific sense. But there are more ways of failing to survive and reproduce than
there are to succeed at any point in time, and therefore natural selection provides some direction in
moving populations away from traits that fail and towards the traits that succeed
Question 22
Development of a new breed of any plant or animal frequently includes inbreeding, and this may
expose rare recessive alleles in the gene pool. This is a special problem when breeders start with only a
few animals and create entire stocks from their descendants. Correlated responses to selection also
contribute to the development of undesirable characteristics when selecting for desirable
characteristics. Selection for a nicely shaped head or a particular leg length in a dog may have
undesirable effects on other skeletal features. Genetic linkage and association (linkage disequilibrium)
of desirable alleles at some loci with undesirable alleles at other loci contribute greatly to the
correlated selection of two or more different traits.
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