When a yellow mouse were crossed to another yellow mouse, the `F_(1)` generation produces yellow and brown-black mice in the ration 2:1. The yellow mice are never homozygous. The reason is

Options

A. Homozygous yellow cannot survive due to lethal effect of genes
B. Yellow mice are not very suitable to live
C. There is no formation of zygotes with homozygous yellow constitution
D. None of the above

Answer

To solve the question regarding the inheritance pattern observed in yellow mice, let's break it down step by step.

### Step-by-Step Solution:

1. **Understanding the Cross**:
- We start with two yellow mice, which we denote as Yy (heterozygous yellow) since yellow mice are never homozygous.

2. **Setting Up the Punnett Square**:
- When we cross two heterozygous yellow mice (Yy x Yy), we can set up a Punnett square to determine the possible genotypes of the offspring.
- The possible combinations from the cross are:
- YY (homozygous yellow)
- Yy (heterozygous yellow)
- Yy (heterozygous yellow)
- yy (brown-black)

3. **Analyzing the Offspring**:
- The expected genotypic ratio from the Punnett square is:
- 1 YY : 2 Yy : 1 yy
- This translates to a phenotypic ratio of:
- 2 yellow (Yy) : 1 brown-black (yy)

4. **Identifying the Lethal Gene Effect**:
- The key observation is that the homozygous yellow (YY) does not survive. This is due to the lethality associated with the homozygous dominant genotype.
- Therefore, the YY genotype is lethal, leading to the death of that offspring, which is why we only see yellow and brown-black mice in a 2:1 ratio.

5. **Conclusion**:
- The reason yellow mice are never homozygous is that the homozygous yellow genotype (YY) cannot survive due to the lethal effect of the genes associated with the yellow color.

### Final Answer:
The yellow mice are never homozygous because homozygous yellow cannot survive due to the lethal effect of genes.

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