Question

Show the work for your calculations in order to receive credit for the question

1.) In eggplants, purple color fruit is incompletely dominant over white color fruit. A cross between two heterozygotes produces the following phenotypic ratio: purple fruit: 169, violet fruit: 407, white fruit: 188. (total 7 pts)

a. Write the parental genotypes and phenotypes. ____________________
b. What phenotype distribution would you expect in the progeny? (You must do a Punnett square analysis to get this.) (Show your work for the Punnett square here) (1pt)

c. Do a chi-square analysis to determine whether the progeny fits the expected distribution. Remember: analysis includes interpretation, not just calculation of a chi-square value.

Answer 1

The alleles are co-dominant as heterozygotes give violet color instead of white or purple. Consider the alleles to be P and p.

a. The parents are heterozygous. Hence, they are heterozygotes.

Genotype= Pp

Phenotype= violet

b.

		Pp
		P	p
Pp	P	PP Purple	Pp Violet
	p	Pp Violet	Pp white

The genotypes are PP, Pp and pp

The phenotypes obtained for fruits are 1 purple: 2 Violet: 1 white/

c. Null hypothesis: there is no difference between the expected and observed values. The phenotypes obtained are according to Hardy-Weinberg equation.

The phenotypic ratio is 1:2:1

Total number= 169+407+188= 764

Expected purple fruits= 764/4= 191

Expected white fruits= 191

Expected violet fruits= 191X 2= 382

Phenotypes	Observed	Expected	(O-E)	(O-E)2	(O-E)2/E
Purple fruits	169	191	-22	484	2.534031
Violet fruits	407	382	25	625	1.636126
White fruits	188	191	-3	9	0.04712
Total	764	764	0	1118	Chi Square value = 4.217277

Degree of Freedom (df) =Number of phenotypes-1= 3-1= 2

At df of 2, the critical chi-square value from Chi-square table at p<0.05 is 5.991. As the Chi-square obtained for the monohybrid cross (4.22) is less than the critical chi-square value (5.99), the null hypothesis is rejected. Null hypothesis is accepted only when the chi square value is higher than the critical chi-square value at p<0.05. Thus, there is a significant difference between the expected and observed values for the phenotypes. Hence, the monohybrid cross is not in Hardy Weinberg equilibrium. The progeny does not fit the expected distribution.