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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 \(F(p)\).

Step by step solution

01

Show allele frequency formulas

We know from the problem statement that there are three genotypes: 1. Homozygous dominant: \(AA\)2. Heterozygous: \(Aa\)3. Homozygous recessive: \(aa\)The frequency of allele \(A\) in the population is the total number of \(A\) alleles divided by the total number of alleles. Given the genotype frequencies \(P\) for \(AA\), \(Q\) for \(Aa\), and \(R\) for \(aa\):- The number of \(A\) alleles is given by \(2P imes N + Q imes N\), so the frequency \(p\) of allele \(A\) is: \[p = P + \frac{1}{2} Q\].Similarly, the frequency of allele \(a\) is the total number of \(a\) alleles divided by the total number of alleles. The number of \(a\) alleles is \(2R imes N + Q imes N\). Thus, the frequency \(q\) of allele \(a\) is:\[q = R + \frac{1}{2} Q\].This establishes the allele frequencies.

02

Create mating possibilities table

In a closed population with random mating, we can list possible mating combinations:- \(AA\) males with \(AA\) females will produce only \(AA\) offspring.- \(AA\) with \(Aa\) will produce \(AA\) or \(Aa\) with equal probability.- \(Aa\) with \(Aa\) can produce \(AA\), \(Aa\), or \(aa\) with probabilities 1/4, 1/2, and 1/4 respectively.- Similarly, continue this for \(Aa\) with \(aa\), and \(aa\) with \(aa\).Combining these allows the creation of a probability table quantifying probabilities of each genotype resulting from each parental pair.

03

Establish expression for \\(\Sigma_{AA}\\)

Each \(AA\) zygote comes from combinations resulting in double dominant alleles:1. Both parents \(AA\): always \(AA\).2. \(AA\) with \(Aa\): \(1/2\) the time.3. \(Aa\) with \(Aa\): \(1/4\) the time.General expression combining all mating scenarios results in:\[\Sigma_{AA} = P^2 + \frac{1}{2}PQ + \frac{1}{4} Q^2 = P^2\], showing the relationship to allele frequencies.

04

Prove Hardy-Weinberg Equilibrium Equation

The Hardy-Weinberg Equilibrium implicates ψ, under random and non-selective mating, expresses genotype frequencies as functions of allele frequencies alone:\(\Sigma_{AA} = p^2, \Sigma_{Aa} = 2pq, \Sigma_{aa} = q^2\)Use previously concluded expressions, strategy of equating, and populations parameters leads evolutionary balance when no mutation, migration or selection exist.

05

Verify allele frequency over generations

From \(\Sigma_{AA} = p^2\), reinforce: No selection or evolutionary disruptors (mutation, migration), quantifying allele constancy across generations.Given allele frequency \(\hat{p}\) in initial adults transfers generation-wise ensuring perpetuation:\(\hat{p}_{next-generation} = \hat{p}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Allele Frequency

In genetics, the term "allele frequency" refers to how often a particular allele appears within a population's gene pool. Alleles are different versions of a gene, and knowing their frequency in a population helps geneticists understand the genetic diversity and potential for evolution within that population.
In the context of sickle cell anemia, two alleles exist: the normal allele denoted as "A" and the allele associated with sickle cell anemia denoted as "a". Allele frequency is calculated by assessing the total number of each allele compared to the total number of alleles in the population.

• The allele frequency of "A" is represented by the formula: \( p = P + \frac{1}{2} Q \)
• The allele frequency of "a" is represented by the formula: \( q = \frac{1}{2} Q + R \)

Here, \(P\), \(Q\), and \(R\) stand for the genotype frequencies of \(AA\), \(Aa\), and \(aa\) respectively, showing how allele frequencies can be derived directly from genotype frequencies.

Genotype Frequency

Genotype frequency describes how often a specific genotype occurs in a population. A genotype is a combination of two alleles, one inherited from each parent. In the case of sickle cell anemia, the genotypes of interest are \(AA\), \(Aa\), and \(aa\).
These frequencies are expressed as:

• \(P\) for \(AA\) - individuals do not have sickle cell anemia and do not have resistance to malaria.
• \(Q\) for \(Aa\) - these individuals are carriers and exhibit malaria resistance.
• \(R\) for \(aa\) - individuals with sickle cell anemia.

The Hardy-Weinberg principle predicts that if no evolution is occurring, then the genotype frequencies will remain in constant proportion to the allele frequencies, given by the equations for \(\Sigma_{AA}\), \(\Sigma_{Aa}\), and \(\Sigma_{aa}\) as \(p^2\), \(2pq\), and \(q^2\) respectively.

Sickle Cell Anemia

Sickle cell anemia is a genetic blood disorder caused by a mutation affecting hemoglobin, the protein responsible for transporting oxygen in the blood. In individuals with sickle cell anemia, red blood cells assume a crescent or "sickle" shape, which can lead to blockages in blood vessels and result in various health complications.
The disorder arises due to a specific genetic mutation in the hemoglobin gene located on chromosome 11. This mutation changes a single nucleotide, which substitutes the amino acid glutamic acid with valine in the hemoglobin molecule.
Key points include:

• Individuals with the genotype \(aa\) are homozygous recessive and have the disease.
• Heterozygous carriers \(Aa\) have some health benefits, such as resistance to malaria, leading to higher survival rates in areas affected by the disease.
• The presence of this genetic condition influences allele and genotype frequencies in populations affected by malaria due to this survival advantage.

Random Mating

Random mating is a fundamental assumption of the Hardy-Weinberg Equilibrium, ensuring that alleles are assorted independently of an individual's genotype. This means that every individual in a population has an equal opportunity of mating with another, irrespective of their genetic makeup.
When random mating occurs, the population's allele frequencies remain stable across generations, as shown in the equation proving that the allele frequency \(\hat{p}\) in one generation will be the same in the subsequent generation.
The absence of non-random mating ensures:

• No preference for specific genotypes.
• The maintenance of genetic variation.
• Predictability in genotype frequencies based on allele frequencies.

This principle is crucial in studies of population genetics as it provides a baseline for understanding how evolutionary forces might shift allele frequencies over time.

Selection in Genetics

Selection in genetics refers to the process where specific genotypes have differential success in survival and reproduction, which can cause changes in allele frequencies within a population over time.
In the context of sickle cell anemia, selection operates due to the balance between the deleterious effects of the disease and the protective advantage against malaria. This phenomenon is an example of heterozygote advantage, where the genotype \(Aa\) is favored in malaria-endemic regions because it offers protection without causing full-blown sickle cell anemia.
Key insights about selection include:

• Genotype \(AA\) has a lower probability of reaching maturity due to vulnerability to malaria.
• Genotype \(Aa\) is favored, achieving a probability of 1 in mating and survival due to malaria resistance.
• Genotype \(aa\) has the lowest survival probability due to the health complications associated with sickle cell anemia.

Natural selection thus directly impacts the evolutionary dynamics of allele frequencies, influencing population genetic variance and adaptation.