J A Moore investigated the inheritance of spotting patterns

J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J.A. Moore. 1943. Journal of Heredity 34:3-7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. a. On the basis of these results, which phenotype is dominant? b. Give the most likely genotypes of the parents in each cross. c. For each of the three crosses above, use a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected on the basis of your proposed genotypes. In other words, use the Chi Square test to determine if the genotypes you assigned are likely. For each cross, state the Chi Square value, the approximate P value (use the Chi Square table in the text), and your conclusion in plain English.

Solution

The third cross reveals that burnsi (B) is dominatnt over pipiens (b) because there is no pipiens in offspring.

In third cross the genotype should be Bb x Bb; that will produce Bb progenies in high possibility.

In second cross Bb x bb will be the genotypes of parents because of high number of pipiens. To get the results the burnsi parent should have to heterozygote.

In first cross Burnsi x burnsi may have BB x Bb because of presence of both progenies and high number of burnsi.

For performing chi square test X² = (Observed- Expected)²/ Expected

According to Hardy-Weinberg Law; if alleles are B and b

(B+b)²= B² +b²+2Bb =1

First cross, expected frequency for Bb x Bb = BB, Bb, bb

So B = b= 0.25 and Bb = 0.50 are always as expected frequency. For, dominant genotype it should be 0.75 (B+Bb)

For 45 progenies:

Genotypes

Observed

Expected

(O-E)²

X²

BB +Bb

39

45 x 0.75 =34

25

0.74

bb

6

11

25

2.27

Total X²

3.01

For 3.01 consider the table of degree of freedom and X² value. The value 11.34 occurs. Now check it in the table of probability, in which the values are significant and rejecting null hypothesis. According that way we can find the values for other crosses.

Genotypes	Observed	Expected	(O-E)2	X2
BB +Bb	39	45 x 0.75 =34	25	0.74
bb	6	11	25	2.27
Total X2				3.01