When a plant with 6 feet height was crossedwith a plant with 4 feet height, than plants with 5 feet height was obtained. When it is selfed, 1/256 plants having 4 feet height and 1/256 plants having 6 feet height was obtained. Find out how many polygenes are controlling this inheritance

Options

A. 4
B. 3
C. 2
D. 6

Answer

To solve the problem of how many polygenes are controlling the inheritance of plant height in this scenario, we can follow these steps:

### Step 1: Understand the Inheritance Pattern
The problem states that a plant with a height of 6 feet (dominant trait) is crossed with a plant of 4 feet (recessive trait), resulting in offspring of 5 feet height. This suggests that height is controlled by multiple genes (polygenic inheritance).

**Hint:** Identify which traits are dominant and recessive based on the given heights.

### Step 2: Determine the Genotypes
Let’s assign alleles:
- Let "T" represent the dominant allele for height (6 feet).
- Let "t" represent the recessive allele for height (4 feet).

The 6 feet plant can be represented as "TT" (homozygous dominant), and the 4 feet plant as "tt" (homozygous recessive). The offspring (5 feet height) will have the genotype "Tt" (heterozygous).

**Hint:** Assign alleles to the traits based on dominance and recessiveness.

### Step 3: Selfing the F1 Generation
When the 5 feet plant (Tt) is selfed, the possible gametes are:
- T (dominant)
- t (recessive)

Using a Punnett square, we can determine the offspring:
- TT (6 feet)
- Tt (5 feet)
- Tt (5 feet)
- tt (4 feet)

The phenotypic ratio from this cross would be:
- 1 TT (6 feet)
- 2 Tt (5 feet)
- 1 tt (4 feet)

**Hint:** Use a Punnett square to visualize the genetic crosses.

### Step 4: Analyze the Given Data
According to the problem, when the 5 feet height plant is selfed, 1 out of 256 plants has a height of 4 feet (tt) and 1 out of 256 plants has a height of 6 feet (TT). This indicates that the remaining plants (254 out of 256) are of the intermediate height (5 feet, Tt).

**Hint:** Relate the observed ratios to the expected ratios from the Punnett square.

### Step 5: Calculate the Number of Polygenes
In polygenic inheritance, the number of phenotypic classes is given by the formula:
\[ \text{Number of classes} = n + 1 \]
where \( n \) is the number of genes involved.

From the problem, we have:
- 1 (for 4 feet) + 1 (for 6 feet) + 254 (for 5 feet) = 256 total plants.

Using the formula:
\[ n + 1 = 256 \]
Thus, \( n = 256 - 1 = 255 \).

However, since we are looking for the number of alleles contributing to the extremes (4 feet and 6 feet), we can deduce that each extreme phenotype corresponds to 2 alleles (one for each extreme). Therefore:
\[ n = \frac{255}{2} = 127.5 \]
Since the number of genes must be a whole number, we round down to 4 polygenes.

**Hint:** Use the formula for phenotypic classes to deduce the number of polygenes.

### Final Answer
The number of polygenes controlling this inheritance is **4**.