Snouters are small critters that live on both sides of the Ohio River. A populat

Question

Snouters are small critters that live on both sides of the Ohio River. A population on the north side of the Ohio is fixed for an allele that codes for blue feet. The south side of the river is fixed for an allele for pink feet. Sixty blue-footed snouters from the north and 40 pink-footed snouters from the south colonize a newly formed island in the middle of the river. Please calculate the following. What are the genotype frequencies of the offspring of the next generation (First generation of offspring, F1 ), assuming random mating?

Explanation / Answer

An allele is a gene donated by a parent. The two parents' alleles in combination in an offspring make up its genotype. For example, one parent's blood-type alleles may be AA, giving the parent a phenotype of type A blood. The other parent has a genotype AB. The child can then have a genotype of AA or AB, but not BB. So the child's "phenotype" can be only type A or type AB blood. You can calculate genotype probabilities in children, knowing the parents' genotypes, and even calculate the allele and genotype frequencies of a population, knowing the phenotype frequencies. Suppose the parents have genotypes Rs and Rs, where s is a recessive trait that doesn't appear (manifest as a phenotype) unless the offspring's genotype is ss. Suppose for the sake of discussion that the phenotype for ss is not having earlobes. 2 Ascribe equal probability to the outcomes RR, Rs, sR, and ss, where the notation used temporarily here indicates which allele came from which parent: the first position indicating the allele donated from the first parent, in other words. Then because of the random nature in which alleles are passed on, the probability of RR in a child is ¼, the probability of Rs is ¼, the probability of sR is ¼, and the probability of ss is ¼. Find the probability of the child's not having earlobes by looking solely at the probability of the genotype ss, namely ¼. The probability of Rs and sR is irrelevant to the original question, because s is a recessive allele. R dominates it and s doesn't manifest when allele R is in the same organism. So the answer to the problem is ¼. If s were dominant, then you'd sum up the frequencies of all the genotypes with any s in it, which would equal ¾. Solving Allele and Genotype Frequencies From Population Phenotype Frequencies 4 Suppose rats turn white only if their alleles are both s. So the whiteness genotype is ss. Suppose further that the probability of a rat being white is 9%. Suppose that R is the dominant gene, and that you need to find the frequency of the R allele and the Rs genotype. 5 Let Pr(s) represent the probability of the s allele. Then the probability of the ss genotype has probability Pr(s)Pr(s). So Pr(s) must equal the square root of 9%, i.e. Pr(s)=0.3. 6 Solve for the probability of the R gene by taking the complement of the probability of the s gene. I.e. Pr(R)+Pr(s)=1, so Pr(R)=1-Pr(s)=0.7. So the probability of any randomly selected allele being R is 70%. 7 Solve for the probability of the genotype Rs by multiplying the probability of a given allele being R and the probability of a given allele being s. Then take into consideration that Rs and sR are equivalent genotypes. So you get 2*Pr(R)Pr(s) = 2*0.7*0.3 = 0.42.

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