A researcher is performing crosses on mice. When he crosses mice that are hetero

Question

A researcher is performing crosses on mice. When he crosses mice that are heterozygous for a wild-type agouti allele and a dominant mutant yellow allele he observes a total of 102 yellow and 48 agouti in the subsequent generation. He suspects the yellow allele may be lethal in the homozygous dominant state. To test this hypothesis, he performs chi-square tests of both a standard monohybrid 3:1 ratio and a 2:1 ratio that would result from the lethal allele.

1. Consider the 3:1 ratio first. What would be the expected number of the dominant phenotype in a standard monohybrid cross with no lethal alleles involved?

2.Consider the 3:1 ratio again. What would be the expected number of the recessive phenotype in a standard monohybrid cross with no lethal alleles involved?

3. Calculate the chi-square value when your null hypothesis is the 3:1 ratio.

Using the same numbers from question 1.

4.Now consider the 2:1 ratio. What would be the expected number of the dominant phenotype in a cross where the homozygous dominant is lethal?

5.Consider the 2:1 ratio again. What would be the expected number of the recessive phenotype in a cross where the homozygous dominant is lethal?

6.Now calculate the chi-square value when your null hypothesis is the 2:1 ratio.

7. With a p-value of 0.05, what is the critical value we use to test our calculated chi-square value against in these cases?

8. Based on your answers to questions 3 & 6, which ratios can we reject?

9. Based on the results of the chi-square tests, should the researcher now be more confident that a lethal allele is involved?

10.

Another researcher is performing dihybrid crosses on mice. For a particular trait, he obtains three phenotypes in the F2 generation and suspects either dominant or recessive epistasis is involved.

Using a p-value of 0.05, what critical value would he use to compare his calculated chi-value against in these cases?

Explanation / Answer

1. Dominant phenotype = 112
2. Recessive phenotype = 38
3.

4. 2/3 X 150 = 100

5.   Recessive 1/3X100 = 50

6.

7. P value at 0.05 = 3.84
Degree of freedom is 1

8. Reject 3:1

9. Yes

Observed Expected O-E (O-E)2 (O-E)2/E Yellow 102 3/4 X 150 = 112 -10 100 0.89 Agouti 48 1/4 X 150 = 38 10 100 2.63 150 3.53

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