can u please provide step by step instructions on how to get the proper answer..
ID: 151058 • Letter: C
Question
can u please provide step by step instructions on how to get the proper answer..and a simple explanation for all parts of question #1 and #2 ....thank u
Genotype A1A1 A1A2 A2A2 Total Numbers observed 700 110 190 1,000 (a) What are the frequencies of the genotypes in this population? (b) What are the allele frequencies in the population? (c) Is the population in Hardy-Weinberg equilibrium? 2. cystic fibrosis (CF) is a single-locus recessive genetic disease of humans. The gen on chromosome 7, and affected individuals in Scotland occur at a frequency of 1 in 1.9 newboms: it is one of the commonest serious genetic diseases among Europeans. Affected individuals rarely survive to reproduce (a) Assuming random mating with respect to this gene. what is the frequency of mutant CF allele in the Scottish population? (b) Assuming random mating, what is the frequency of carriers (i.e., heterozygotes for the mutant allele? (c) If there is a balance between mutation at rate and selection pressure s for a deleterious recessive gene, he equilibrium gene frequency q" is expected to be Assuming the allele frequency in Scotland is at equilibrium under mutation- selection balance, what is the mutation rateExplanation / Answer
Ans1. Genotype frequency is the proportion of a particular genotype in the total population.
We can calculate the frequency of each genotype by dividing the number of individuals with that genotype by the total number of individuals the population.
(a) The total population is 1000.
Genotype frequency of A1A1 - 700/1000 =0.7
Genotype frequency A1A2 - 110/1000 = 0.11
Genotype frequency A2A2 - 190/1000 = 0.19
(b) Allele frequencies in the population.
Allele frequency = Number of copies of a particular allele in a population / total number of all alleles for that gene in a population.
There are two alleles A1, A2
There are two copies of the allele A1 in A1A1 genotype therefore no of A1 allele = 2 * 700 = 1400
allele A1 in A1A2 genotype = 110 and the total number of allele = 2 * 1000 = 2000
A1 = 2 * 700 + 110/ 2000 =1400+ 110 / 2000 = 0.75
A2 = 2 *190+ 110/ 2000 = 0.245
(c) Hardy Weinberg equations are p2 + 2pq + q2 = 1 and p+ q =1 ( A1A1 + A1A2 + A2A2 = 1)
We need to put the values of A1 and A2 in the equation
0.75 * 0.75 + 2 * 0.75 * 0.245 + 0.245 * 0.245 = 1
0.57 + 0.37 + 0.06 = 1
so the frequency of A1A1 genotype is 0.57
frequency of A1A2 genotype is 0.37
frequency of A2A2 genotype is 0.06
For the population to be in Hardy Weinberg equilibrium the genotype frequencies should be equal to the calculated above.
They are different from the data given above, so the population is not in Hardy Weinberg equilibrium.
In the second question, the cystic fibrosis frequency is not visible.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.