Exercise
8.3.5 Sickle cell anemia is an inherited blood disease in which the body makes
sickle-shaped red blood cells. It is caused by a single mutation from glutamic
acid to valine at position 6 in the protein Hemoglobin B. The gene for
hemoglobin \(\mathrm{B}\) is on human chromosome \(11 ;\) a single nucleotide
change in the codon for the glutamic acid, GAG, to GTG causes the change from
glutamic acid to valine. The location of a genetic variation is called a locus
and the different genetic values (GAG and GTG) at the location are called
alleles. People who have GAG on one copy of chromosome 11 and GTG on the other
copy are said to be heterozygous and do not have sickle cell anemia and have
elevated resistance to malaria over those who have GAG on both copies of
chromosome 11 . Those who have GTG on both copies of chromosome 11 are said to
be homozygous and have sickle cell anemia \(-\) the hydrophobic valine allows
aggregation of hemoglobin molecules within the blood cell causing a sickle-
like deformation that does not move easily through blood vessels.
Let \(A\) denote presence of \(\mathrm{GAG}\) and \(a\) denote presence of GTG on
chromosome \(11,\) and let \(A A,\) \(A a\) and \(a a\) denote the various presences
of those codons on the two chromosomes of a person (note:
\(A a=a A) ; A A, A a\) and aa label are the genotypes of the person with
respect to this locus. It is necessary to assume non-overlapping generations,
meaning that all members of the population are simultaneously born, grow to
sexual maturity, mate, leave offspring and die. Let \(P, Q\) and \(R\) denote the
frequencies of \(A A, A a\) and aa genotypes in a breeding population and let
\(p\) and \(q\) denote the frequencies of the alleles \(A\) and \(a\) among the
chromosomes in the same population. The frequencies \(P, Q,\) and \(R\) are
referred to as genotype frequencies and \(p\) and \(q\) are referred to as allele
frequencies. In a population of size, \(N\), there will be \(2 N\) chromosomes and
\(P \times 2 N+Q \times N\) of the chromosomes will be \(A\).
In a mating of \(A A\) with \(A a\) adults, the chromosome in the fertilized egg
(zygote) obtained from \(A A\) must be \(A\) and the chromosome obtained from \(A
a\) will be \(A\) with probability \(1 / 2\) and will be \(a\) with probability \(1 /
2\). Therefore, the zygote will be \(A A\) with probability \(1 / 2\) and will be
\(A a\) with probability \(1 / 2\)
a. Show that the allele frequencies \(p\) and \(q\) in a breeding population with
genotype frequencies \(P, Q\) and \(R\) are given by
$$
p=P+\frac{1}{2} Q
$$
and
$$
q=\frac{1}{2} Q+R
$$
b. Assume a closed population (no migration) with random mating and no
selection. Complete the table showing probabilities of zygote type in the
offspring for the various mating possibilities, the frequencies of the mating
possibilities, and the zygote genotype frequencies. Include zeros with the
zygote type probabilities but omit the zeros in the zygote genotype
frequencies. Random mating assumes that the selection of mating partners is
independent of the genotypes of the partners.
c. When the table is complete, you should see that
$$
\begin{aligned}
\Sigma_{A a} &=\frac{1}{2} P Q+P R+\frac{1}{2} Q P+\frac{1}{2}
Q^{2}+\frac{1}{2} Q R+R P+\frac{1}{2} Q R \\
&=P Q+2 P R+\frac{1}{2} Q^{2}+Q R=2 P\left(\frac{1}{2}
Q+R\right)+Q\left(\frac{1}{2} Q+R\right) \\
&=(2 P+Q)\left(\frac{1}{2} Q+R\right) \quad=\quad 2\left(P+\frac{1}{2}
Q\right)\left(\frac{1}{2} Q+R\right) \\
&=2 p q
\end{aligned}
$$
Show that
$$
\Sigma_{A A}=p^{2} \quad \text { and } \quad \Sigma_{a a}=q^{2}
$$
This means that under the random mating hypothesis, the zygote genotype
frequencies of the offspring population are determined by the allele
frequencies of the adults. This is referred to as the Hardy-Weinberg theorem.
If the probability of an egg growing to adult and contributing to the next
generation of eggs is the same for all eggs, independent of genotype, then the
allele frequencies, \(p\) and \(q,\) are constant after the first generation.
Random mating does not imply the promiscuity that might be imagined. It means
that the selection of mating partner is independent of the genotype of the
partner. In the United States, blood type would be a random mating locus;
seldom does a United States young person inquire about the blood type of an
attractive partner. In Japan, however, this seems to be a big deal, to the
point that dating services arranging matches also match blood type. The major
histocompatibility complex (MHC) of a young person would seem to be fairly
neutral; few people even know their MHC type. It has been demonstrated,
however, that young women are repulsed by the smell of men of the same MHC
type as their own \(^{4}\).
d. Show that in a closed random mating population with no selection, if the
frequency of \(A\) in the adults in one generation is \(\hat{p},\) then the
frequency of \(A\) in adults in the next generation will also be \(\hat{p}\).
e. Suppose that because of malaria, an \(A A\) type egg, either male or female,
has probability 0.8 of reaching maturity and mating and because of sickle cell
anemia an aa type has only 0.2 probability of mating, but that an \(A a\) type
has 1.0 probability of mating. This condition is called selection. Then the
distribution of genotypes in the egg and the mating populations will be
\(\begin{array}{lccc}\text { Genotype } & A A & A a & a a \\ \text { Egg } &
p^{2} & 2 p q & q^{2} \\ \text { Adult } & 0.8 p^{2} / F & 2 p q / F & 0.2
q^{2} / F\end{array}\)
$$
\text { where } \quad F=0.8 p^{2}+2 p q+0.2 q^{2}
$$
Find the frequency of \(A\) in the adult population. Note: This will also be the
frequency of \(A\) in the next egg population.
f. We call \(F(p)\) the balance of the population, and because \(p+q=1\)
$$
F=F(p)=0.8 p^{2}+2 p(1-p)+0.2(1-p)^{2}
$$
You will be asked in Exercise 8.3 .8 to show that when the probability of
reproduction depends on the genotype (selection is present), during succeeding
generations, allele frequency, \(p,\) moves toward the value of local maximum of
\(F\).
1\. Show that \(F(p)=1-0.2 p^{2}-0.8(1-p)^{2}\).
2\. Find the value \(\hat{p}\) of \(p\) that maximizes \(F(p)\).
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