Question

4) an AaBb parent and an aabb parent produce children in these ratios:

20% AaBb

30% Aabb

30% aaBb

20% aabb

a) If they had a total of 10 offspring, how many of each offspring type would the above percentages represent, and what chi-square value would you get if you assumed simple mendelian inheritance? The 5% chi-square significance cutoff for 3 degrees of freedom is 7.815. Using this, is the data a good fit to Mendelian expectations.

​​​​​b) Now suppose the parents were oak trees. Oak trees can produce about 10,000 offspring in a good year. If there were a total of 10,000 offspring, how many of each offspring type would the above percentages represent, and what chi-square value would you get if you assumed mendelian inheritance? Is the data a good fit to Mendelian expectations?

Answer 1

Answer-

According to the given question-

When we make a cross between AaBb i.e. heterozygous parent with the aabb i.e. homozygous recessive parent we get following progeny.

AaBb aabb

	ab	ab	ab	ab
AB	AaBb	AaBb	AaBb	AaBb
Ab	Aabb	Aabb	Aabb	Aabb
aB	aaBb	aaBb	aaBb	aaBb
ab	aabb	aabb	aabb	aabb

AaBb : Aabb : aaBb : aabb in the ratio of 1 :1 : 1 : 1 - According to mendal inheritence .

But according to question we get in the ratio of - 20% AaBb : 30% Aabb : 30% aaBb :20% aabb. so out of total 10 , which means that

AaBb = 2,:

Aabb = 3 ,

aaBb = 3 ,

aabb = 2

But expected frequency of offspring will be in the ratio of 1 : 1 i.e.

AaBb = 2.5

Aabb = 2.5

aaBb = 2.5

aabb = 2.5

Chi square characteristics-

Phenotype	Observed frequency (O)	Expected frequency (E)	O - E	(O-E)2	(O-E)2 / E
AaBb	2	2.5	2-2.5 = -0.5	0.25	0.25 / 2.5 = 0.1
Aabb	3	2.5	3- 2.5 = 0.5	0.25	0.25 / 2.5 = 0.1
aaBb	3	2.5	3- 2.5 = 0.5	0.25	0.25 / 2.5 = 0.1
aabb	2	2.5	2-2.5 = -0.5	0.25	0.25 / 2.5 = 0.1 X2 = 0.1 + 0.1 + 0. 1 + 0.1 = 0.4

X2 = 0.4

here degree of freedom = 4-1 = 3

For 3 degree of freedom the Critical value of X2  at 5 % or 0.05 significance is 7.815 but we have our chi square characteristic is 0.4 which is less than critical valve i.e. 0.4 < 7.815 and we accept the null hypothesis.

(b) Considering above condition for oak trees which can produce about 10,000 offspring then

ratio of - 20% AaBb , : 30% Aabb : 30% aaBb :and 20% aabb. so out of total 10000 , which means that

AaBb = 2000,:

Aabb = 3000 ,

aaBb = 3000 ,

aabb = 2000

But expected will be in the ratio of 1 : 1 which means that each have 2500 offspring. i.e.

AaBb = 2500,:

Aabb = 2500 ,

aaBb = 2500 ,

aabb = 2500

Phenotype	Observed frequency (O)	Expected frequency (E)	O - E	(O-E)2	(O-E)2 / E
AaBb	2000	2500	2000-2500 = -500	250000	250000 / 2500 = 100
Aabb	3000	2500	3000-2500 = 500	250000	250000 / 2500 = 100
aaBb	3000	2500	3000-2500 = 500	250000	250000 / 2500 = 100
aabb	2000	2500	2000-2500 = -500	250000	250000 / 2500 = 100 X2 = 100 + 100 + 100 + 100 = 400

X2 = 400

here degree of freedom = 4-1 = 3

For 3 degree of freedom the Critical value of X2  at 5 % or 0.05 significance is 7.815 but we have our chi square characteristic is 400 which is more than the  critical valve i.e. 400 > 7.815 and we reject the null hypothesis.