Answer :
To solve this genetic probability problem, we need to look at two different genetic crosses and determine the chances of certain types of offspring.
1. Cross between a heterozygous male (Ww) and a homozygous recessive female (ww):
- The male parent has the genotype Ww, meaning he has one dominant allele (W) and one recessive allele (w).
- The female parent is homozygous recessive, with the genotype ww, meaning both alleles are recessive.
We can use a Punnett square to determine the possible genotypes of their offspring:
```
| W | w | <- Male's alleles
-----------------
w | Ww | ww |
-----------------
w | Ww | ww |
```
- When you fill in the Punnett square, the combinations are:
- W from male and w from female -> Ww (heterozygous)
- w from male and w from female -> ww (homozygous recessive)
- Out of the four possible combinations (two Ww and two ww), half of them (2 out of 4) are Ww. Thus, the probability of the offspring being heterozygous (Ww) is 50%.
2. Cross between a heterozygous individual (Ww) and a homozygous dominant (WW):
- The first parent is heterozygous with genotype Ww.
- The second parent is homozygous dominant with genotype WW.
We can create another Punnett square:
```
| W | W | <- Second parent's alleles
-----------------
W | WW | WW |
-----------------
w | Ww | Ww |
```
- The combinations here are:
- W from both parents -> WW (homozygous dominant)
- W from the second parent and w from the first parent -> Ww (heterozygous)
- Since all offspring are either WW or Ww, there are no homozygous recessive (ww) offspring. Thus, the probability of having a homozygous recessive offspring (ww) is 0%.
By analyzing these crosses step by step, we find that:
- There is a 50% chance of getting heterozygous offspring in the first cross.
- There is a 0% chance of getting homozygous recessive offspring in the second cross.
1. Cross between a heterozygous male (Ww) and a homozygous recessive female (ww):
- The male parent has the genotype Ww, meaning he has one dominant allele (W) and one recessive allele (w).
- The female parent is homozygous recessive, with the genotype ww, meaning both alleles are recessive.
We can use a Punnett square to determine the possible genotypes of their offspring:
```
| W | w | <- Male's alleles
-----------------
w | Ww | ww |
-----------------
w | Ww | ww |
```
- When you fill in the Punnett square, the combinations are:
- W from male and w from female -> Ww (heterozygous)
- w from male and w from female -> ww (homozygous recessive)
- Out of the four possible combinations (two Ww and two ww), half of them (2 out of 4) are Ww. Thus, the probability of the offspring being heterozygous (Ww) is 50%.
2. Cross between a heterozygous individual (Ww) and a homozygous dominant (WW):
- The first parent is heterozygous with genotype Ww.
- The second parent is homozygous dominant with genotype WW.
We can create another Punnett square:
```
| W | W | <- Second parent's alleles
-----------------
W | WW | WW |
-----------------
w | Ww | Ww |
```
- The combinations here are:
- W from both parents -> WW (homozygous dominant)
- W from the second parent and w from the first parent -> Ww (heterozygous)
- Since all offspring are either WW or Ww, there are no homozygous recessive (ww) offspring. Thus, the probability of having a homozygous recessive offspring (ww) is 0%.
By analyzing these crosses step by step, we find that:
- There is a 50% chance of getting heterozygous offspring in the first cross.
- There is a 0% chance of getting homozygous recessive offspring in the second cross.
