Evolution II.2 Answers.
1. (4 pts) Contrast the predictions of blending inheritance
for F1 and F2 generations with those observed under
Mendelian inheritance.
Blending inheritance predicts both F1 and F2 hybrids to be intermediate between the two parents. For F1 hybrids, Mendelian inheritance predicts either (a) all individuals like one of the parents (dominance), or (b) individuals intermediate
(incomplete dominance). For the F2 hybrids,
Mendelian inheritance predicts either (a) both parental phenotypes (dominance) or both intermediate and parental phenotypes (incomplete dominance).
2. (2 pts) When would you expect independent assortment
to break down?
When the genes in question are “linked,” i.e., on
the same chromosome and close to each other.
Historically, deviation from independent assortment was the basis for the construction of genetic maps, i.e., the closer together a pair of genes,
the less likely a crossing over between them.
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3. (6 pts) Continue the preceding table into the 5th generation.
Selfing Leads to Heterozygote Elimination.
(Assume Four Seeds per Cross)
Cross / Genotype
AA
Aa
aa
4th Generation
28 (AA x AA)
112
0
0
8 (Aa x Aa)
8
16
8
28 (aa x aa)
0
0
112
4th Generation
120
16
120
Total
(.46875)
(.0625)
(.46875)
5th Generation
120 (AA x AA)
480
0
0
16 (Aa x Aa)
16
32
16
120 (aa x aa)
0
0
480
5th Generation
496
32
496
Total
(.484375)
(.03125)
(.484375)
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A simpler approach is as follows:
 In any generation, half the offspring of 𝑨𝒂 × 𝑨𝒂
crosses are heterozygotes, which are the only
source of heterozygotes under selfing. It follows
that the frequency of heterozygotes in the nth
generation, 𝒑𝑨𝒂 (𝒏) =
𝟏 𝒏
(𝟐) .
 The remaining individuals are homozygotes and
since their numbers are equal, the frequency of
each homozygote is
𝒑𝑨𝑨 (𝒏) = 𝒑𝒂𝒂 (𝒏) = [𝟏 −
𝟏 𝒏
(𝟐) ] /𝟐.
 Example: For the 5th generation,
𝟏 𝟓
𝟏
𝒑𝑨𝒂 (𝟓) = ( ) =
=. 𝟎𝟑𝟏𝟐𝟓
𝟐
𝟑𝟐
(𝟏−. 𝟎𝟑𝟏𝟐𝟓) . 𝟗𝟔𝟖𝟕𝟓
𝒑𝑨𝑨 (𝟓) = 𝒑𝒂𝒂 (𝟓) =
=
=. 𝟒𝟖𝟒𝟑𝟕𝟓
𝟐
𝟐
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4. (8 pts) Suppose there are three alleles, A1, A2, A3, with
gene frequencies p, q, r. How many genotypes are
there? What are their H-W frequencies?
There are six genotypes: 𝑨𝟏 𝑨𝟏 , 𝑨𝟏 𝑨𝟐 , 𝑨𝟏 𝑨𝟑 , etc. The
proportions are p2 : 2pq : 2pr : q2 : 2qr : r2.
5. (8 pts) Suppose the genotypic frequencies, 𝑝𝐴𝐴 , etc., are
0.10, 0.50 and 0.40. a. What are the gene frequencies?
b. Is the population in Hardy-Weinberg equilibrium?
a. AA homozygotes have two copies of gene A,
Aa heterozygotes have one copy and aa homozygotes have none. Then
𝟐 × 𝟎. 𝟏𝟎 + 𝟎. 𝟓𝟎 . 𝟕𝟎
𝒑𝑨 =
=
=. 𝟑𝟓
𝟐
𝟐
Likewise, aa homozygotes have two copies of
gene a, heterozygotes have one copy and AA
homozygotes, none. Then
𝟐 × 𝟎. 𝟒𝟎 + 𝟎. 𝟓𝟎 𝟏. 𝟑𝟓
𝒑𝒂 =
=
=. 𝟔𝟓
𝟐
𝟐
Alternatively, it may be observed that
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𝒑𝒂 = 𝟏 − 𝒑𝑨 = 𝟏−. 𝟑𝟓 =. 𝟔𝟓
b. H-W frequencies as follows:
𝒑𝑨𝑨 =. 𝟑𝟓𝟐 =. 𝟏𝟐𝟐𝟓; 𝒑𝑨𝒂 = 𝟐 ×. 𝟑𝟓 ×. 𝟔𝟓 =. 𝟒𝟓𝟓;
𝒑𝒂𝒂 =. 𝟔𝟓𝟐 =. 𝟒𝟐𝟐𝟓. Thus there is an excess of
heterozygotes (.50 vs. .445) and an insufficiency of homozygotes (.10 vs. .1225 and .40
vs. .445). Depending on population size, these
deviations might or might not be statistically
significant.
6. (2 pts) What might one conclude if a population is not in
H-W equilibrium?
Other forces – selection, migration, mutation, etc.
– are affecting gene frequencies.
7. (4 pts) In the case of heterozygote superiority, is p*
stable or unstable?
Stable. If the heterozygote is superior to both
homozygotes, p increases if 𝒑 < 𝒑∗ (but not
equal to 0 or 1) and decreases if 𝒑 > 𝒑∗ (but not
equal to 0 or 1). In other words, 𝒑 𝒑∗.
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8. (4 pts) In the case of heterozygote inferiority, is p* stable or unstable?
Unstable. The argument is as given above but
with the signs reversed. In other words, 𝒑 moves
away from 𝒑∗ , unless, of course, 𝒑 equals 0 or 1.
9.
(6 pts) Using Eq (6), compute the equilibrium frequency
of HbS.
q*  (1  p*)  1 
1  .14
.86
 1
 .12
(1  .88)  (1  .14)
.12  .86
10. (8 pts) Do you expect the frequency of HbS to be higher
or lower among American than African blacks? Give two
reasons why or why not?
Lower in U.S. due to (1) eradication of malaria –
there is no longer selection for heterozygotes –
and (2) intermarriage with individuals of nontropical descent.
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11. (8 pts) Tay-Sachs disease (TSD) occurs in frequencies of
.03-.04 in Jews of Eastern European extraction. TSD is a
neurodegenerative disorder caused by a single autosomal mutation. In its most common form, it is almost invariably fatal by age 4. Assume the disease only presents in homozygotes (two copies of the mutation) and
a mutation rate of 2 X 10-6.
a. Compute the equilibrium frequency of TSD according
to Eq (10). Hint: What is the value of the coefficient of
selection, s ?
−𝟔
𝝁
𝟐
×
𝟏𝟎
𝒒∗ ≈ √ = √
= 𝟏. 𝟒𝟏 × 𝟏𝟎−𝟑 ≪. 𝟎𝟑−. 𝟎𝟒
𝒔
𝟏
b. What do you conclude?
The observed frequency is an order of magnitude
greater than that predicted by mutation-selection
balance. Something else involved. Possibilities include heterozygote advantage and a small population “bottleneck” in recent past.
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12. (4 pts) Parent-offspring comparisons in humans would
yield high heritability for life-time earning and religion –
i.e., the children of the rich tend to be rich, etc. Does
this necessitate the existence of “poverty genes”?
Explain.
Of course not. Environmental factors are often
involved and, in many cases, predominate.
13. (6 pts) In assessing whether or not intelligence has a
heritable basis, twin studies are often used. Design such
a study. Hint. Not all twins are the same.
Compare correlations between monozygotic
(identical) and dizygotic (non-identical) twins
reared apart and together. Also compare
correlations between siblings, and parent-child
correlations. The table below gives data compiled
from multiple studies. The bottom-line conclusion
is that both heredity and environment are
important.
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IQ correlations from multiple studies compiled by Devlin, B.
et al. 1997. The heritability of IQ. Nature. 388: 468-471. Column 0 is the average of the various studies weighted by
sample size. Columns I-IV are the correlations predicted by
models that allow for various genetic, maternal and environmental effects. Monozygotic twins manifest the highest correlations, but even here, environment plays a role – correlations are higher if the children are reared together. If only
genetics mattered, monozygotic twins would manifest a correlation of 1.0; dizygotic twins, sibling and parent-child, a
correlation of 0.5 and adoptive parent-child, a correlation of
0.
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15. (6 pts) Asume the average IQ in a population is 100
and h2 = 0.4. What is the expected IQ of the daughter
of parents whose IQs are 120 and 110? What about
the IQ of a child born to parents with IQs of 90?
Use Eq (11) with 115 and 90 for Tp. Yields 106,
96.
16. (6 pts) Comparing the offspring’s expected IQ to that
of the parents and to the population mean, what do
you conclude?
Offspring will be intermediate between parents
and the population average. This is sometimes
referred to as “reversion to the mean”.
17. (2 pts) Which of the three modes (stabilizing,
directional, disruptive) of selection always increases
phenotypic variance? Which mode shifts trait
frequency distributions?
Disruptive; directional.
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18. (2 pts) Which of the three modes of selection is often a
consequence of trade-offs and could be responsible for
evolutionary stasis?
Stabilizing.
19. (8 pts) How might you account for the fact that the
observed mean birth weight of humans is slightly less
than the optimum?
One possibility is parent-offspring conflict. What’s
good for junior’s fitness is not necessarily good
for mom’s. Specifically, Mom has multiple shots at
reproduction; each of her children has only one
shot at life.
20. (8 pts) The following species nest in Alaska where
summer nights can be cold: golden eagle, great horned
owl, kingbird, rufous hummingbird, yellow-shafted
flicker. One of these goes torpid (reduced metabolic rate
and body temperature) at night. Which species do you
imagine does this? Explain.
Rufous hummingbird. It’s the smallest, has the
largest surface area to volume ratio and rate of
heat loss. It is therefore the most likely to :run
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out of gas” if it attempts to maintain its body
temperature at night.
21. (8 pts) Small cliff swallows are more agile fliers than
large individuals and therefore more likely to be able to
capture insects on the fly. How does this affect selection
for body size in this species?
Works against selection for large body size to
reduce death from hypothermia. One can imagine
selection in different years for small or large body
size depending on the weather.
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