MHT CET · BIOLOGY · STD 12 - 5. Origin and Evolution of Life

In a population of 10,000 cattle, 6400 have homozygous dominant genotype (AA) and 400 have homozygous recessive genotype (aa). Find out all number of cattle having heterozygous genotype (Aa) as per Hardy-Weinberg's principle.

A 1600
B 3200
C 160
D 320

Verified Solution

Answer & Solution

Correct Answer

(B) 3200

Step-by-step Solution

Detailed explanation

Hardy-Weinberg's principle for the genotypic frequencies-
\(\mathrm{p}^2\) + 2pq + \(\mathrm{q}^2\) =1
AA =\(\mathrm{p}^2\) aa =\(\mathrm{q}^2\) and for 2Aa = 2pq.
AA = 6400; aa = 400 Putting the values in equation, we will get 3200.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Identify the CORRECT match from the Columns I, II and III.

I	II	III
1. Interstitial cells	a. Cortex of ovary	i. Follicular fluid
2. Sertoli cells	b. Ovarian follicle	ii. Progesterone
3. Granulosa cells	c. Testis	iii. Attachment of sperm bundle
4. Cells of corpus luteum	d. Seminiferous tubules	iv. Testosterone

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Column I	Column II
Q.1. Amylase	a. coagulated milk protein
Q.2. Trypsin	b. dipeptides
Q.3. Chymotrypsin	c. disaccharides
Q.4. Rennin	d. polypeptides