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Answer :
Let's solve the two questions step by step:
1. First Scenario: A heterozygous male with the genotype Ww is mated with a homozygous recessive female with the genotype ww.
To find out the probability of the offspring being heterozygous (Ww), we need to construct a Punnett square. This will help us visualize the possible genotypes of the offspring:
- The male (Ww) can produce two types of gametes: W and w.
- The female (ww) can produce only w gametes.
Here's how they combine:
```
W w
-------------
w | Ww | ww |
w | Ww | ww |
-------------
```
From the Punnett square, we see that out of four offspring, two are Ww (heterozygous).
Therefore, the probability of the offspring being heterozygous is 0.5 (2 out of 4).
2. Second Scenario: A heterozygous individual with genotype Ww is crossed with a homozygous dominant individual with genotype WW.
Again, we make a Punnett square to find the probability of offspring being homozygous recessive (ww).
- The first parent (Ww) can produce gametes: W and w.
- The second parent (WW) produces only W gametes.
Here's how they combine:
```
W W
-------------
W | WW | WW |
w | Ww | Ww |
-------------
```
From this Punnett square, we can see that none of the offspring are homozygous recessive (ww).
Therefore, the probability of having a homozygous recessive offspring is 0.0 (0 out of 4).
Overall, the chance that the offspring will be heterozygous in the first scenario is 0.5, and the probability of having a homozygous recessive offspring in the second scenario is 0.0.
1. First Scenario: A heterozygous male with the genotype Ww is mated with a homozygous recessive female with the genotype ww.
To find out the probability of the offspring being heterozygous (Ww), we need to construct a Punnett square. This will help us visualize the possible genotypes of the offspring:
- The male (Ww) can produce two types of gametes: W and w.
- The female (ww) can produce only w gametes.
Here's how they combine:
```
W w
-------------
w | Ww | ww |
w | Ww | ww |
-------------
```
From the Punnett square, we see that out of four offspring, two are Ww (heterozygous).
Therefore, the probability of the offspring being heterozygous is 0.5 (2 out of 4).
2. Second Scenario: A heterozygous individual with genotype Ww is crossed with a homozygous dominant individual with genotype WW.
Again, we make a Punnett square to find the probability of offspring being homozygous recessive (ww).
- The first parent (Ww) can produce gametes: W and w.
- The second parent (WW) produces only W gametes.
Here's how they combine:
```
W W
-------------
W | WW | WW |
w | Ww | Ww |
-------------
```
From this Punnett square, we can see that none of the offspring are homozygous recessive (ww).
Therefore, the probability of having a homozygous recessive offspring is 0.0 (0 out of 4).
Overall, the chance that the offspring will be heterozygous in the first scenario is 0.5, and the probability of having a homozygous recessive offspring in the second scenario is 0.0.
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