BIOL 151-Homework 7, Population Genetics need to follow the procedure Please sub
ID: 254252 • Letter: B
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BIOL 151-Homework 7, Population Genetics need to follow the procedure Please submit this homework assignment to Blackboard. To submit your assignment you below: After completing your homework assignment (using pencil and paper) take pictures of each page of the assignment and copy the pictures into one Word file, OR scan your homework assignment and create a Word or PDF file. You may discuss the problems with other You must show all of your work to get credit for the homework problems. students in this class (in fact working in groups on these problems is one of the best ways to learn) the answers of other students. Anybody that plagiarizes or copies another student's assignment will receive a zero will be reported to the University as described in the Student code of conduct (see syllabus for additional information.) Write your answer in your own words. You will be given questions very similar to these on the lecture test, so to do them! , but do not plagiarize and know how Problem 1: ah khorthorm cath her s gen RR results in red coat color, Rr is roan and rr is white. a. Given that 127 red, 106 roan and 67 white animals were found in the Central Valley of California, calculate the b. If we assume that the cattle are mating randomly, what is the expected frequency of each genotype in the next c. The rancher has observed that white shorthorn cattle are sterile (unable to reproduce). What are the frequencies of d. If none of the white cattle are mating, what are the expected frequencies of genotypes for the offspring in the next e. Is the breeding population of cattle -ie that does not include the white individuals- in Hardy Weinberg Equilibrium? frequencies of the R allele and the r allele in the gene pool of the population. generation? the R and r alleles in the part of the population that is capable of reproducing? generation Why or why not?Explanation / Answer
a) Number of red (RR) = 127
Number of roan (Rr) = 106
Number of white (rr) = 67
Total = 127+106+67 = 300
Frequency of R allele (p)= [(2*Number of RR)+(1*Number of Rr)]/(2*Total) = [(2*127)+(1*106)]/(2*300)=0.6
Frequency of r allele (q) = [(2*Number of rr)+(1*Number of Rr)]/(2*Total) = [(2*67)+(1*106)]/(2*300)=0.4
b. As per Hardy Weinberg equilibrium,
p2+2pq+q2=1
where p=allele frequency 1; q=allele frequency 2
Expected frequency of RR (p2) =0.6*0.6=0.36
Expected frequency of rr (q2) =0.4*0.4=0.16
Expected frequency of Rr (2pq)=2*0.6*0.4=0.48
c. When white cattles are sterile, it will not participiate in the next generation. So the total population reduces to 233 (300-67).
Frequency of R allele (pnew)= [(2*127)+106]/(2*233)=0.77
Frequency of r allele (qnew)= [(2*0)+106]/(2*233)=0.23
d. The new genotype frequency of RR (pnew2)=0.77*0.77=0.59
The new genotype frequency of Rr(2pnewqnew)=2*0.77*0.23=0.36
The new genotype frequency of rr (qnew2)=0.23*0.23=0.05
e. pnew2 + 2pnewqnew + qnew2 =0.59+0.36+0.05 =1
So it is in Hardy Weinberg equilibrium
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