The genetic makeup of a mouse population is provided in the table above. At the start of an experiment, researchers remove all the gray mice from the population, resulting in a population that is not in hardy- weinberg equilibrium. Calculate the frequency of the recessive allele in the remaining population of mice. Give your answer as a decimal to two places.
Answers
According to me Bihar and chattisgarh
The question is incomplete as it lacks the data which has been provided in the attached figure.
Answer:
Frequency of Recessive allele= 0.30
Explanation:
When a population is in a Hardy-Weinberg equilibrium then the
dominant allele is represented as-p
recessive allele is represented as-q
And mathematical expression for genotype is P²+q²+2pq.
In the given question, the gray mice are not selected that is
bb is not selected and therefore population has pp and 2pq genotypes.
Total population= 2059, alleles for 2059 oragnisms= 2 x 2059 = 4118
To calculate the recessive allele frequency in remaining population
= No. of copies of a particular allele / Total no. of all alleles for that gene in a population
= q²+2pq/ total alleles
= 0+ 1218/4118
= 1218/4118
= 0.30
To learn more:
1. Hardy-Weinberg equilibrium: https://brainly.in/question/1799931
