Huntington’s disease is a dominant genetic disorder, which means that having just one copy of the mutated gene is enough to develop the disease. In this case, we have a man who is heterozygous for the Huntington gene (), represented as ‘Hh’, and he marries a woman who is homozygous recessive (‘hh’).
a) Will the man or woman develop the disease?
The man, being ‘Hh’, has one normal allele (‘h’) and one mutated allele (‘H’). Since Huntington’s is a dominant disorder, he will develop the disease as he has the mutated allele. The woman, being ‘hh’, has two normal alleles and will not develop the disease.
b) What is the probability that their first child will develop the disease?
To find the probability of their first child developing Huntington’s disease, we can use a Punnett square. Let’s outline the alleles:
- Father’s alleles: H (dominant) and h (recessive)
- Mother’s alleles: h (recessive) and h (recessive)
Now, we can set up the Punnett square:
| h | h | |
|---|---|---|
| H | Hh (disease) | Hh (disease) |
| h | hh (no disease) | hh (no disease) |
From the Punnett square results, there are four possible combinations for their children:
- 2 out of 4 (50%) will be heterozygous (‘Hh’) and develop Huntington’s disease.
- 2 out of 4 (50%) will be homozygous recessive (‘hh’) and will not develop the disease.
Thus, the probability that their first child will develop Huntington’s disease is 50%.