Question

In fruit flies, red eyes are dominant (R) and white eyes are recessive (r).


– What is the probability that a cross between two heterozygous red-eyed flies will produce a homozygous white-eyed fly?



– What is the probability that a cross between a homozygous white-eyed fly and a heterozygous red-eyed fly will produce a homozygous white-eyed fly?

make simple and easy no big words

Answer 1

To solve the questions, we can use a simple method called a Punnett square, which helps us see the possible combinations of genes from the parents.

1. **Cross between two heterozygous red-eyed flies (Rr x Rr)**:

- Parent 1: Rr
- Parent 2: Rr

The possible combinations of alleles from these parents are:

- RR (homozygous red)
- Rr (heterozygous red)
- Rr (heterozygous red)
- rr (homozygous white)

The results can be summarized as follows:
- 1 RR (red)
- 2 Rr (red)
- 1 rr (white)

So, out of 4 total possible combinations, only **1** (the rr) is for the homozygous white-eyed flies.

The probability of getting a homozygous white-eyed fly (rr) is:
\[
\frac{1}{4} = 25\%
\]

2. **Cross between a homozygous white-eyed fly and a heterozygous red-eyed fly (rr x Rr)**:

- Parent 1: rr
- Parent 2: Rr

The possible combinations of alleles from these parents are:

- Rr (heterozygous red)
- rr (homozygous white)

The results can be summarized as follows:
- 1 Rr (red)
- 1 rr (white)

So, out of 2 total possible combinations, only **1** (the rr) is for the homozygous white-eyed flies.

The probability of getting a homozygous white-eyed fly (rr) is:
\[
\frac{1}{2} = 50\%
\]

**Summary**:
- The probability of two heterozygous red-eyed flies producing a homozygous white-eyed fly is **25%**.
- The probability of a homozygous white-eyed fly and a heterozygous red-eyed fly producing a homozygous white-eyed fly is **50%**.