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# Mendelian populations consist of a group of interbreeding individuals

Mendelian populations consist of a group of interbreeding individuals that form a gene pool. To describe the population, genotypic and allele (gene) frequencies are often computed under the assumptions of the Hardy-Weinberg Law. Utilizing this assumption, let A and a represent dominant and recessive alleles whose respective frequencies are p and q in a given interbreeding population at equilibrium.
a. If 16% of the individual in the population have recessive phenotypes, what percentage of the total number of recessive genes exist in the heterozygous condition?
b. If 1% of the individuals were homozygous recessive what percentage of the recessive genes would occur in heterozygotes?
c. What is the allele frequency of p and q in b?
Hello,
I am working on this and will get back to you momentarily.
DM
Customer: replied 4 years ago.

ok thank you

Mendelian populations consist of a group of interbreeding individuals that form a gene pool. To describe the population, genotypic and allele (gene) frequencies are often computed under the assumptions of the Hardy-Weinberg Law. Utilizing this assumption, let A and a represent dominant and recessive alleles whose respective frequencies are p and q in a given interbreeding population at equilibrium.
a. If 16% of the individual in the population have recessive phenotypes, what percentage of the total number of recessive genes exist in the heterozygous condition?
p2 + 2pq + q2 = 1 and p + q = 1
p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p2 = percentage of homozygous dominant individuals
q2 = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals
1. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 16%.
That means that q2 = 16% or 0.16
Therefore q = the square root of 0.16 = 0. 4 (40%) = the frequency of the recessive allele
Meaning that if p + q = 1, p=0.6 and q=0.4
Meaning that q2 = 0.16 and 2pq = 2 (0.6*0.4) = 0.48
Total number of recessive genes = 2pq +q2 = 0.48 + 0.16 = 0.64
Percent of total number of recessive genes in heterozygous condition = 0.48/0.64 = 75%
2. b. If 1% of the individuals were homozygous recessive what percentage of the recessive genes would occur in heterozygotes?

If q2 = 1% = 0.01
q = 1
p = 0.9
p2 = 0.81
2pq = 1.8
Recessive genes = q2 + 2pq = 0.01 + 1.8 = 1.81
Recessive genes in heterozygotes = 1.8/1.81 = 99%

c. What is the allele frequency of p and q in b?

q = 1
p = 0.9
Customer: replied 4 years ago.

Thank you very much. I was really stuck!! I appreciate the details.

My pleasure. Please let me know if you have more questions and please click Accept if you found this helpful.
DM