Question

In a plant, height varies from 6 to 36 cm. When 6 cm and 36cm plants were crossing all Pre 2006 In the F2 generation, a continuous raride of heights was observed Most were around 21 n were 6 cm. Find out how many gene pairs are involved in this mode of inheritance

• A. 3
• B. 4
• C. 2
• D. 5

Answer 1

Answer: (A)

3

Answer 2

The inheritance pattern described (continuous variation, extreme parental phenotypes in F2, intermediate F1) indicates polygenic inheritance. Let 'n' be the number of gene pairs. The frequency of the shortest phenotype (all recessive alleles) in the F2 generation is $(1/4)^n$. The problem states that "a few were 6 cm" (the shortest height). We look for the value of 'n' among the options for which $(1/4)^n$ represents "a few".

For n=3, frequency = $(1/4)^3 = 1/64$.

For n=4, frequency = $(1/4)^4 = 1/256$.

Both 1/64 and 1/256 could be considered "a few".

Alternatively, using the height range: Shortest height = 6 cm, Tallest height = 36 cm, Intermediate (F1) height = 21 cm.

Let base height = 6 cm and contribution per dominant allele = a.

Tallest height = $6 + 2na = 36 \implies 2na = 30$.

F1 height = $6 + na = 21 \implies na = 15$.

From $na=15$, if n=3, $3a=15 \implies a=5$. Then $2na = 2(3)(5) = 30$, which is consistent.

If n=4, $4a=15 \implies a=3.75$. Then $2na = 2(4)(3.75) = 30$, which is consistent.

Both n=3 and n=4 fit the height data. However, the frequency of the extreme phenotype (1/64 for n=3) is a more standard value used in problems to identify the number of genes in polygenic inheritance. Thus, n=3 is the most likely answer.