what portion of the F2 would have tails 6cm in length, how many offspring in the
ID: 48714 • Letter: W
Question
what portion of the F2 would have tails 6cm in length, how many offspring in the F2 would have 5cm long tails? What allele frequency and how does it relate to evolution? Does understanding is an allele frequencies change your view of evolution? Does evolution occur with a single directionality? Explain in relation to this experiment. The allele frequencies of the parents in our simulation were p = 0.75 and q-0.25 heterozygote has a selective advantage and therefore a fitness of W12 1. Assume the homozygotes have a fitness of W11 0.4 and W22-0.1, respectively. Calculate the allele frequencies after two rounds of selection. How do these calculations compare to the data you collected and the fitness values of the experiment? . The em Solving (2 pts) A selected, inbred rat strain had a tail measuring an average of 5.4 cm. An inbred, wild caught strain had 6.2 cm long tails. The selected and wild caught strains were crossed producing the F1 and F2 below Generation Fl F2 Progeny 20 250 Phenotype 5.8 cm 5 to 6.6 cm Assume that 4 genes control tail length. What is the average contribution per allele in the F2 generation?Explanation / Answer
Allele frquency it is measure of how many times means frequency of a allele occuring in a population. It is expressed as percentage or in proportion. The change in the percentae of the allele frequency changes the idea of evolution. It greatly explains the genetic drift.
Evolution usually occurs in the course of time and is very slow. But in some cases it may take very rapid and in single directionality. Eg: Snakes loosing the legs.
2. The average contribution per allele in the second generation is (6.6-5.0)/(2x4)= 0.2cm.
As the per allele proportion is 0.2 the rats may have tails of size 5.0, 5.2, 5.4, 5.8, 6.0, 6.2, 6.2, 6.6
the number of alleles are 8. so ((8!)/(5!)(3!))/(1/2)^5(1/2)^3= 56/256 =0.2187 portion of the F2 would have tails measuring 6cm.
offspring in the F2 would have 5cm long tails: (8!)/(0!8!)/(1/2)^0 (1/2)^8=0.97. Approximaely 1 will have 5cm lon tail
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.