p? + 2pq + q? = 1 (to determine the genotype frequencies) and p+q =1 (to determine the allele frequencies) p = frequency of the dominant allele in the population q = frequency of the recessive allele...


p? + 2pq + q? = 1 (to determine the genotype frequencies)<br>and<br>p+q =1 (to determine the allele frequencies)<br>p = frequency of the dominant allele in the population<br>q = frequency of the recessive allele in the population<br>p = percentage of homozygous dominant individuals<br>q?:<br>= percentage of homozygous recessive individuals<br>2pq = percentage of heterozygous individuals<br>Homework Problems<br>For these problems you will need to use a calculator and can round<br>for the percentages.<br>the nearest who number<br>Example problem: You have sampled a population in which you know that the percentage of<br>the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:<br>A. The frequency of the

Extracted text: p? + 2pq + q? = 1 (to determine the genotype frequencies) and p+q =1 (to determine the allele frequencies) p = frequency of the dominant allele in the population q = frequency of the recessive allele in the population p = percentage of homozygous dominant individuals q?: = percentage of homozygous recessive individuals 2pq = percentage of heterozygous individuals Homework Problems For these problems you will need to use a calculator and can round for the percentages. the nearest who number Example problem: You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: A. The frequency of the "aa" genotype. Answer: 36% (given) B. The frequency of the "a" allele. Answer: The frequency of aa is 36%, which means that q? = 0.36. If q? = 0.36, then q = 0.6, because 0.62 is equal to 0.36. Since q equals the frequency of the a allele, then the frequency is 60%. C. The frequency of the "A" allele. Answer: Since q = 0.6, and p +q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%. D. The frequencies of the genotypes "AA" and "Aa." Answer: The frequency of AA is equal to p?, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p? is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48). E. The frequencies of the two possible phenotypes if "A" is completely dominant over "a." Answers: Because "A" is dominant over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is only occurs if the genotype is homozygous "aa." Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive phenotype is simply the frequency of aa. The dominant phenotype frequency, then, is 64% (16% + 48%) and, in the first part of this question above, you have already shown that the recessive phenotype frequency is 36%. 64% + 36% = 100% or .64 + .36 = 1.
2. Among Death Valley mice, the dark allele (D) is dominant over the light allele (d). During<br>a recent student field trip, the following mice were captured:<br>Using the Hardy-Weinberg formula, calculate the following.<br>A. The frequency of mice that express the recessive trait and are therefore<br>homozygous recessive (dd). This is the q? value.<br>B. The frequency of recessive alleles (d) in the population. This is the q value.<br>C. The frequency of dominant alleles (D) in the population. This is the p value.<br>D. The percentage of mice that are homozygous dominant (DD). This is the p? value.<br>E. The percentage of mice in the population that are heterozygous (Dd). This is 2pq.<br>F. Suggest a reason for the number of d alleles in the population.<br>

Extracted text: 2. Among Death Valley mice, the dark allele (D) is dominant over the light allele (d). During a recent student field trip, the following mice were captured: Using the Hardy-Weinberg formula, calculate the following. A. The frequency of mice that express the recessive trait and are therefore homozygous recessive (dd). This is the q? value. B. The frequency of recessive alleles (d) in the population. This is the q value. C. The frequency of dominant alleles (D) in the population. This is the p value. D. The percentage of mice that are homozygous dominant (DD). This is the p? value. E. The percentage of mice in the population that are heterozygous (Dd). This is 2pq. F. Suggest a reason for the number of d alleles in the population.
Jun 11, 2022
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