1. (8 pts) To study the inheritance of the dominant allele for fur () color in m
ID: 150247 • Letter: 1
Question
1. (8 pts) To study the inheritance of the dominant allele for fur () color in mice, two mice heterozygous for yellow fur were mated. In a series of crosses, a total of 56 mice with yellow fur and 31 mice with agouti fur encoded by the recessive allele () were obtained a. Use a Chi-square test to determine if the outcome of this cross is consistent with the 3:1 phenotypic ratio usually obtained in the F2 of a mating between heterozygotes Write out the genotypes for the Pl and the F1 generations P1 Fi For the F2 generation, use a Punnett square Phenotypic ratio: % ry yellow, State a null hypothesis: yy agouti Chi square calculation Ospring Observed Espected ellow fur TOTAL Show all work for Chi-square calculation la (continued) Degrees of freedom (dn Chi square () value: P value: P- Answer: Was the null hypothesis rejected or did it fail to be rejected? Why? b. If the null hypothesis was rejected, proceed to test again using the null hypothesis that the dominant allele is lethal in the homozygous condition Chi square calculation showing all work Degrees of freedom (dn Chi square (x) value P value: P Answer is this null hypothesis rejected or does it fail to be rejected? Why?Explanation / Answer
Answer:
1a).
YY (yellow) x (agouti) yy-----------------------------------P1
Yy (yellow) --------------------------------------F1
Heterozygous yellow (Yy) x Heterozygous yellow (Yy)---P2
Y
y
Y
YY (yellow)
Yy(yellow)
y
Yy (yellow)
yy (agouti)
F2 phenotypic ratio = 3(yellow) : 1(agouti)
Chi square calculations:
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Yellow fur
56
65.25
-9.25
85.56
1.311
Agouti fur
31
21.75
9.25
85.56
3.934
87
87
5.25
Degrees of freedom = number of phenotypes -1
Df = 2-1 = 1
Chi square value = 5.25
p value = 3.84
Answer: Chi square value of 5.25 is greater than the critical value, so the null hypothesis is rejected.
1b).
Yy (yellow) x (agouti) yy-----------------------------------P1
Yy (yellow) & yy(agouti)-----------------------------F1
Heterozygous yellow (Yy) x agouti (yy)-----------------P2
Y
y
Y
YY (yellow-died)
Yy(yellow)
y
Yy (yellow)
yy (agouti)
F2 phenotypic ratio = 2(yellow) : 1(agouti)
Chi square calculations:
Phenotype
Observed(O)
Expected (E)
O-E
(O-E)2
(O-E)2/E
Yellow fur
56
58
-2
4.00
0.069
Agouti fur
31
29
2
4.00
0.138
87
87
0.21
Degrees of freedom = number of phenotypes -1
Df = 2-1 = 1
Chi square value = 0.21
p value = 3.84
Answer: Chi square value of 0.21 is less than the critical value, so the null hypothesis is accepted.
Y
y
Y
YY (yellow)
Yy(yellow)
y
Yy (yellow)
yy (agouti)
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