Download Practice Questions, Lectures 6-13 (259 KB pdf file)

MCB142/IB163 Thomson
Mendelian and Population Genetics
Sep.16, 2003
Practice questions lectures 6-13
Mendelian and Population Genetics Equations, Tables, Formulas, etc.
confidence limits of an allele frequency estimate: denoting by p the true frequency of allele A,
the variance of the allele frequency estimated by the method of gene (allele) counting is
p(1-p)/(2n) where n is the sample size. An estimate of this variance is obtained using this
formula and the estimated allele frequency. The standard deviation (sd) is the square root of
the variance. The 95% confidence interval for the range of the true allele frequency p is twice
the sd on either side of the estimate.
Theoretical Chi-square values with different probabilities:
Degrees
Probability
Of Freedom
0.05*
0.01**
0.001***
1
3.84
6.64
10.83
2
5.99
9.21
13.82
3
7.82
11.35
16.27
4
9.49
13.28
18.47
5
11.07
15.09
20.52
degrees of freedom (df): when the expected values are independent of any features of the observed
data, then the df is given by the number of genotype classes considered (with expected value
>=5) minus 1.
When the observed data is used in calculations to determine the expected values, e.g., for testing
HWP the allele frequencies must be estimated from the observed data to determine the expected
HWP for that data set, then the df is as above minus the number of independent parameters
estimated (which is k-1 for a k allele system).
linkage disequilibrium (LD): Consider two genes with two alleles each (A, a and B, b), and
hence four gametic types (referred to as haplotypes): AB, Ab, aB, and ab. The frequencies of
the four gametes (denoted f(AB) etc.) can be described in terms of the allele frequencies
pA (pa= 1 - pA) and pB (pb = 1 - pB) at the two loci, and the LD parameter D, as follows:
f (AB) = pA pB + D, f (Ab) = pA pb - D, f (aB) = pa pB - D, f (ab) = pa pb + D,
where D = f(AB) - pA pB.
heterozygote advantage: the heterozygote has a higher relative fitness than both homozygotes,
this leads to a balanced polymorphism. If we denote the relative fitness values of the 3
genotypes in a 2-allele system as follows:
A1A1
A1A2
A2A2
relative fitness
1 - s1
1
1 - s2
then the equilibrium frequency for allele A1 is p1* = s2/(s1 + s2), and
that for allele A2 is p2* = s1/(s1 + s2), with p1* + p2* = 1.
Page 1
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 mating.
Among all the cases of the disease in the population, what percentage of these cases resulted
from heterozygous 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.)
Page 2
Question 7
In a Northern European population, the frequencies of alleles at the HLA –A gene are:
Alleles
A1
A2
A3
A11 A29
Frequencies 0.2
0.3
0.1
0.3
0.1.
Assume Hardy Weinberg proportions (HWP) to answer all the following questions.
In a population of size 1,000 individuals:
(a) how many are expected to be heterozygotes A2A29?
(b) how many are expected to be heterozygotes A1AX, where AX denotes all alleles except A1?
(c) How many total homozygotes are expected, i.e., counting all homozygous classes?
Question 8
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 9
Perform a chi-square goodness of fit test of Hardy Weinberg proportions (HWP) for the
genotype data given below for a single gene with 2 codominant alleles.
Determine the following quantities (a.-c.) and describe (d.) (show all work):
(a) The total chi-square value
(b) The degrees of freedom (df)
(c) Whether the chi-square value indicates that the observed genotype values are compatible
with the null hypothesis of HWP, or if they are significantly different from HWP, in which
case give the P value.
(d) Briefly describe what a P value represents when a test is significant, i.e., the null hypothesis
is rejected.
Observed data
Genotypes
Observed (O)
AA
29
AB
62
BB
9
Total = 100
Page 3
Question 10
In samples from three different populations, the following genotypes were observed for the
codominant alleles A1 and A2:
Population
1
2
3
A1 A1
20
30
72
A1 A2
40
8
122
A2A2
40
12
91
(a). Calculate the frequency of alleles A1 and A2 in each population.
(b). What are the Hardy-Weinberg frequencies and expected genotypic numbers for the three
populations.
(c). Calculate the chi-square value for test to fit to Hardy-Weinberg proportions for these
populations, and indicate if the populations are statistically different from Hardy-Weinberg
proportions.
Question 11
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 12
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 13
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
Number of individuals
Genotype
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.
Page 4
Question 14
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 15
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 16
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
Number of individuals
Phenotypes
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?
Page 5
Question 17
In corn, the alleles
+
+
c and c result in colored (c ) versus colorless (c) seeds,
+
+
wx and wx in nonwaxy (wx ) versus waxy (wx) endosperm, and
+
+
sh and sh in plump (sh ) versus shrunken (sh) endosperm.
Plants grown from seeds heterozygous for each of these pairs of alleles (from crosses of a pure
breeding colorless waxy parent and pure breeding shrunken parent) were testcrossed with plants
homozygous for colorless, waxy, shrunken seeds.
The progeny seeds were as follows:
colorless, waxy, plump
334
colored, nonwaxy, shrunken
339
colorless, nonwaxy, plump
105
colored, waxy, shrunken
98
colorless, nonwaxy, shrunken
55
colored, waxy, plump
61
colorless, waxy, shrunken
5
colored, nonwaxy, plump
3
Total 1,000
(a) Determine the map order of the three genes, and the recombination fraction (RF) between
each of the three pairs of genes.
(b) If the three genes were on three separate chromosomes, what numbers (or ratios) would be
expected for each of the 8 classes of progeny seeds?
Question 18
For the following 3 examples (I) – (III) of mapping disease genes in humans, indicate whether
they are an example of:
(A) positional cloning
(B) functional cloning
(C) candidate gene approach
(The same answer may apply to more than one example.)
(I) Sickle-cell anemia (an autosomal recessive disease) was demonstrated to be due to a
mutation in the β chain of the hemoglobin gene (Hb-β) and the Hb-β gene was later mapped
(II) In a screen of genetic markers across the genome, Huntington disease (an autosomal
dominant disease with variable age of onset) was shown to be linked to an RFLP marker on
chromosome 4.
(III) Genes of the immune response HLA complex are studied for linkage and association with
autoimmune and infectious diseases and cancers.
Page 6
Question 19
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 20
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 21
Compare and contrast the neutral, balance, and evolutionary lag schools.
Question 22
What are the consequences on the frequency of the heterozygote class of complete assortative
mating among three genotypes, A1 A1 , A1 A2, and A2A2?
Question 23
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 t 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 24
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 with this equilibrium
allele frequency, assuming random mating?
Question 25
What will determine the rate of a molecular clock for a gene that codes for a functional protein?
Page 7
Question 26
A molecular biologist determines the DNA sequence for a particular gene in 4 different species (labeled a,
b, c, and d) and finds the following nucleotide differences between each pair of species
a
b
c
d
a
b
c
d
-
5
-
10
10
-
30
30
30
-
From the fossil record she knows that species a and c diverged 20 Myr. ago. If the rate of nucleotide
substitutions is constant over years and among lineages, how long ago did species c and d diverge?
Question 27
To what extent is evolution a chance process and to what extent a directed process, i.e., how
does random change contribute to evolution? How, if at all, is evolution a directed process?
Question 28
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?
Page 8
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 nontaster) x 3/4 (the probability that their first child will be a non-taster (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 nontaster) 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.
Page 9
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 of all affected children 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) The expected HWP of the heterozygote A1A29 is 2(0.3)(0.1) = 0.06,
In a population of size 1,000 there would be 60 A2A29 heterozygotes expected under HWP.
(b) The frequency of AX is 0.8 (= 1 – 0.2), so the expected HWP is 2(0.2)(0.8) = 0.32
In a population of 1,000 there would be 320 A1AX heterozygotes expected under HWP.
2
(c) The frequency of all homozygotes expected under HWP is Σpi where pi is the frequency of
the ith allele and summation is over all alleles, i = 1-5 in this case,
2
2
2
2
2
= (0.2) + (0.3) + (0.1) + (0.3) + (0.1) = 0.24,
In a population of 1,000 there would be 240 total homozygotes expected under HWP.
Question 8
(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
chi-square 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.
Page 10
Question 9
Allele frequency estimates by allele (gene) counting
p(A) = [29 + (62)/2]/100, OR = 0.29 + 0.62/2, OR = [2(29) + 62]/200 = 0.6,
q(B) = 0.4
Expected (E)
36
48
16
Total = 100
(O – E)
-7
+14
-7
Sum = 0s
2
1.36
4.08
3.06
(O – E) /E
(a) Chi-square = 8.51
(b) df = 1 (= 3 – 1 – 1), since 1 allele frequency is estimated
(c) significant, P<0.01
(d) The P value is the type 1 error, the probability you have rejected the
null hypothesis (fit to HWP in this case) when in fact it is true,
i.e., it is the probability of getting a value this extreme under the
null hypothesis (the probability you are wrong in rejecting the null hypothesis).
Question 10
(a) and (b)
Population
1
2
3
(c)
1
2
3
p
0.4
0.68
0.467
q
0.6
0.32
0.533
p2
0.16
0.462
0.218
2pq
0.48
0.434
0.498
p2 N
16
23.1
62.1
2pqN
48
21.8
141.9
q2 N
36
5.1
80.9
X2
2.78
20.1
5.63
q2
0.36
0.102
0.284
n.s
p<0.001
p<0.05
Question 11
(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%.
Page 11
Question 12
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 13
(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 14
We need to determine the probability of a C/- ; bb ; dd offspring from this cross = ¾ x ¼ x ¼ =
3/64.
Question 15
(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.)
Page 12
Question 16
(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 17
+
c colored and c colorless
+
wx nonwaxy and wx waxy
+
sh plump and sh shrunken
colorless, waxy, plump
334
colored, nonwaxy, shrunken
339
colorless, nonwaxy, plump
105
colored, waxy, shrunken
98
colorless, nonwaxy, shrunken
55
colored, waxy, plump
61
colorless, waxy, shrunken
5
colored, nonwaxy, plump
3
Total 1,000
+
c wx sh
+
+
c wx sh
+
+
c wx sh
+
c wx sh
+
c wx sh
+
+
c wx sh
c wx sh
+
+
+
c wx sh
+
c sh wx
+
+
c sh wx
+
+
c sh wx
+
c sh wx
+
c sh wx
+
+
c sh wx
c sh wx
+
+
+
c sh wx
(a) (i) Map order
The parental gametes (most common)
and also from given parental information are
+
colorless, waxy, plump (c wx sh ) and
+
+
colored, nonwaxy, shrunken (c wx sh
The double recombinant class (two rarest classes) are
colorless, waxy, shrunken (c wx sh) and
+
+
+
colored, nonwaxy, shrunken (c wx sh )
Thus the middle gene is shrunken, and gene order is
colorless (c), shrunken (sh), waxy (wx), or vice versa.
Correct answer re gene order can also be obtained by working out the 3 pairwise recombination
fractions without correcting for double recombinants, which gives the largest distance for the
pair c wx, if this approach is used, the corrected recombination distance including twice the
double recombinants should also then be worked out for next part of question.
Page 13
(ii) Recombination distances
c sh pairwise RF fraction:
+
colorless (c), plump (sh )
+
colored (c ), shrunken (sh)
colorless (c), shrunken (sh)
+
+
colored (c ), plump (sh )
334 + 105 = 439
339 + 98 = 437
55 + 5 =
60
61 + 3 =
64
Total
1,000
RF = (60 + 64)/1,000 = 124/1,000 = 0.124 = 12.4% (= 12.4 mu = 12.4 cM)
wx sh pairwise RF fraction:
+
55 + 339 = 394
nonwaxy (wx ), shrunken (sh)
+
+
105 + 3 =
108
nonwaxy (wx ), plump (sh )
waxy (wx), shrunken (sh)
5 + 98 = 103
waxy, plump
334 + 61 = 395
Total
1,000
RF = (108 + 103)/1,000 = 211/1,000 = 0.211 = 21.1% (= 21.1 mu = 21.1 cM)
c wx pairwise RF fraction:
+
55 + 105 = 160
colorless (c), nonwaxy (wx )
colorless (c), waxy (wx)
5 + 334 =
339
+
98 + 61 =
159
colored (c ), waxy (wx)
+
+
339 + 3 =
342
colored (c ), nonwaxy (wx )
Total
1,000
To determine RF need to include double recombinants with the single recombinant class,
RF = (160 + 2(5) + 159 + 2(3))/1,000 = 335/1,000 = 0.335 (= 33.5% = 33.5 mu = 33.5 cM),
and check that 12.4 + 21.1 = 33.5 cM.
(b) 3 genes unlinked
If all 3 genes were unlinked, we expect equal numbers in each of the 8 classes,
i.e., 125 in each class (1:1:1:1:1:1:1:1 ratio).
Question 18
(I) B
(II) A
(III) C
Functional cloning (disease – function – gene- map)
Positional cloning (disease – map – gene – function)
Candidate gene – genes with a known or proposed function with the potential to influence the
disease phenotype are investigated for a direct role in disease.
Question 19
It suggests that humans have had persistent migration between populations that has prevented
localized differentiation for most genes
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Question 20
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 21
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 22
This mating is equivalent to self-fertilization. The heterozygous class would decrease by half
each generation.
Question 23
(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 24
(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.
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Question 25
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.
Question 26
60 Myr ( = (30/10) x 20 Myr
Question 27
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 28
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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