Question

in guinea pigs, the allele for black coat is (B) is dominant over the allele for a white coat (b). at an independently assorting locus, an allele for rough coat (R) is dominant over the allele for smooth coat (r). a guinea pig that is homozygous for black color and rough coat is crossed with a guinea pig with a white, smooth coat. in a fears of matings, the F1 are crossed with guinea pigs having a white, smooth coat. from these matings, the following phenotypes appear in the offspring: 35 black, rough coats; 22 black, smooth coats; 27 white, rough coats and 7 white, smooth coats.

a) use a chi squared test to tes and see if this ration is due to independent assortment.

b) what conclusions can you draw from the results of this chi- squared test

c) degrees of freedom

Answer 1

In guinea pigs, the allele for black coat is (B) is dominant over the allele for a white coat (b). At an independently assorting locus, an allele for rough coat (R) is dominant over the allele for smooth coat (r).

So, the genotype of guinea pig that is homozygous for black color and rough coat = BBRR,

So, the genotype of a guinea pig with a white, smooth coat. = bbrr ( need to be homozygous as these are recessive characters).

After F1 cross all the offsprings are heterozygous black color and rough coat (BbRr)

Now, the F1 progeny (BbRr) are crossed with guinea pig with a white, smooth coat (bbrr).

So, the Expected genotypes showing in the punnet square are:

	BR	Br	bR	br
br	BbRr	Bbrr	bbRr	bbrr
br	BbRr	Bbrr	bbRr	bbrr
br	BbRr	Bbrr	bbRr	bbrr
br	BbRr	Bbrr	bbRr	bbrr

From the punnet square w can see that the ratio is 1:1:1:1

a)

Genotype	Observed No.	Observed ratio	Expected ratio	Expected No.	[Obs No. – Exp No.]2 / Exp No
black, rough coats	35	5	1	91* ¼ = 22.75	6.6
black, smooth coats	22	3.14	1	91* ¼ = 22.75	0.025
white, rough coats	27	3.86	1	91* ¼ = 22.75	0.794
white, smooth coats	7	1	1	91* ¼ = 22.75	10.904
Total No.	91			Chi Square value:	18.323

From the chi square value we can see that it is a very poor fit with our hypothesis of independent. (For df = 3, chi square value 7.82 implies 5% significency)

b) From the result we can say that significency of this result is less than 0.01. As it is less than 0.05, the ratio is a poor fit.

c) Degrees Of Freedom = Number of phenotypes - 1.

here Number of phenotype = 4.

So, df = 3.