Bonus questions Question a (2 pts) A red, round, true-breeding tomato is bred wi
ID: 219645 • Letter: B
Question
Bonus questions Question a (2 pts) A red, round, true-breeding tomato is bred with a yellow, oval, true-breeding tomato, and the F1s are testcrossed to a homozygous-recessive tester. This results in the offspring below yellow oval 11 red round yellow round 7 red oval 14 8 a) Calculate the chi-squared value to test the likelihood of linkage. b) What is your conclusion? Question b (2 pts)An F1 guinea pig with the genotype CACBDWDo is mated to a CBC8DoDo guinea pig. The resulting progeny are evaluated and the following classes produced in these numbers CACBDoDo CBCBDWDO CACBDWDO CBCBDoDO 326 420 73 84 a. What were the genotypes of the parental true breeding guinea pigs that produced the F1? b. What is the distance between the genes in cM?Explanation / Answer
Answer:
Q1) a).
Chisquare value= 3.00
b).
Degrees of freedom = 4-1 = 3
A chi-square of 3.0 is greater than the critical value (7.81), so we accept the null hypothesis. Genes are not linked.
Q2).. a).
Hint: Parental combination of phenotypes are morein number than recombinant combinations when the genes are linked.
So parental combinations are CACBDODO& CBCBDWDO and recombinants are CACBDWDO& CBCBDODO.
Recombination frequency (RF) = (no. of recombinant progeny/Total progeny)100
RF = (157/903)100 = 17.39%
RF(%) = Distance between the genes (cM)
b). So the distance between the two genes = 17.39 cM
Phenotype Observed(O) Expected (E) O-E (O-E)2 (O-E)2/E yelow, oval 11 10.00 1.00 1.00 0.10 red, round 14 10.00 4.00 16.00 1.60 yellow, round 7 10.00 -3.00 9.00 0.90 red, oval 8 10.00 -2.00 4.00 0.40 Total 40 40.00 3.00