Question

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.

Parent phenotypes:   Progeny phenotypes

burnsi × burnsi 39 burnsi, 6 pipiens

burnsi × pipiens 23 burnsi, 33 pipiens

burnsi × pipiens 196 burnsi, 210 pipiens

a. On the basis of these results, what is the most likely mode of inheritance of the burnsi phenotype?

b. Give the most likely genotypes of the parent in each cross.

c. 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.

Answer 1

Pipiens phenotype produce normal spots in frog and burnsi phenotype lacks spots on back in frog. So the three crosses given in question are :

Parent phenotypes:   Progeny phenotypes

Cross1 : burnsi x burnsi 39 burnsi, 6 pipiens

Cross 2 : burnsi × pipiens 23 burnsi, 33 pipiens

Cross 3 : burnsi × pipiens 196 burnsi, 210 pipiens

Ans a) as we see in cross 1: when 2 burnsi are crossed it produces pipiens also this means pipiens allele is present in the burnsi parent but is recessive. As we know recessive characters are expressed in Homozygous condition only but none of the parent is pipiens so this means the parent is heterozygous. And the dominant character is expressed phenotypically this shows that burnsi is dominant over pipiens.

In cross 2, when a burnsi and pipiens is crossed it produces both burnsi and pipiens progenies. And this suggests that one is Homozygous (pipiens) and other is heterozygous (burnsi).

In cross 3, when a burnsi is crossed with pipiens it produces results similar to cross 2. So burnsi parent is heterozygous and pipiens is Homozygous and only this type of cross produces both burnsi and pipiens progenies.

So, the inheritance of burnsi(B) character is dominant, lacks spots on back.

Ans b) as genotypes are predicted above in ans a, so the genotypes of parents in each cross are :

B = burnsi and b= pipiens

Cross 1 : parent 1(burnsi) = Bb and parent 2(burnsi) = Bb

Cross 2: parent 1(burnsi) = Bb and parent 2( pipiens) = bb

Cross 3: parent 1(burnsi)= Bb and parent 2(pipiens) = bb

Ans c) chi-square test for each cross

Cross 1: total progenies = 39+6 =45

Expected ratio is 3:1(phenotypic) as heterozygote parents are crossed.

Burnsi progenies = 3/4 * 45 = 33.75

Pipiens progenies = 1/4 * 45 = 11.25

	Observed-O	expected-E	O-E	(O-E)^2	(O-E)^2/E
Burnsi	39	33.75	5.25	27.5625	27.56/33.75 =0.81
Pipiens	6	11.25	-5.25	27.5625	27.5625/11.25 =2.45
Total					3.26

Degree of freedom(df)= (r-1)*(c-1)

= 2-1 * 2-1 = 1*1 = 1.

At 1 df chi-square value is 2.706 and probability is 0.1 and 0.05 at this chi-square value. So the result we get in cross 1 is a result of by chance.

For cross 2 : Bb x bb

For this expected ratio is 2:2 or 1:1.

Total progenies= 23+33 = 56

Expected = burnsi = 1/2 * 56 = 28

Pipiens = 1/2 * 56 = 28.

	Observed-O	Expected-E	O-E	(O-E)^2	(O-E)^2/E
Burnsi	23	28	-5	25	25/28= 0.892
Pipiens	33	28	5	25	25/28= 0.892
Total					1.784

df = (2-1)*(2-1) = 1

P value is greater than 0.05.

In cross 3 : burnsi = Bb

Pipiens= bb

Total progenies = 196+210 =406

Expected ratio = 1:1

Expected progenies = burnsi = 1/2*406= 203

Pipiens = 1/2 *406 = 203

	Observed-O	Expected-E	O-E	(O-E)^2	(O-E)^2/E
Burnsi	196	203	-4	16	16/203 =0.24
Pipiens	210	203	10	100	100/203 = 0.24
Total					0.48

Df = (2-1)*(2-1) = 1

Here again P is greater than 0.05

0.48 > 0.05

So , all the three crosses and the predicted genotypes of parents shows that burnsi is dominant over pipiens.

Thank you....

Expect a ???...