Consider yourself a genetic counselor. Megan and Eric come to you for your services. They are planning on having their first child. Megan’s father died from Huntington’s disease. Eric’s family has no history of Huntington’s disease. They would like to have the following questions answered before they undertake the awesome responsibility of parenthood.
1. What is the probability Megan will develop the symptoms of Huntington’s disease? (Assume her mother was hh.) Show the sample space.
2. What is the probability that their first child will develop Huntington’s disease?
*ive tried to work it out and talk to my professor will not help me, please help!!!!
Apparently you tried to copy and past the questions -- but it didn't work. Please type the questions.
Then post the answers that you think should be answered. We'll be glad to comment on your answers.
1.What is the probability Megan will develop the symptoms of Huntington’s disease? (Assume her mother was hh.) Show the sample space.--50%
2.What is the probability that their first child will develop Huntington’s disease?--25% chance having disease, 75% chance doesnt have disease
3.If Megan and Eric have three children, what is the probability that at least one of them will have Huntington’s disease? Round to the nearest tenth of a percent.--each kid has same chance
4.If Megan and Eric have two children, what is the probability that at least one of the will have Huntington’s disease? Round to the nearest tenth of a percent.--each kid has same chance
Your first two answers are correct. However, the 3rd and 4th questions ask what is the probability that one of their children will have the disease. You're right that each child has the same chance -- 1 of 4. But the more children this couple has, the more chances are that at least one child will have the disease.
As a genetic counselor, I'm here to assist you with your questions. Let's go through each question step by step:
1. Probability of Megan developing Huntington's disease symptoms:
To determine the probability of Megan developing symptoms, we need to understand the inheritance pattern of Huntington's disease. Huntington's disease is an autosomal dominant disorder, which means that if one parent carries the faulty gene, there is a 50% chance for each child to inherit the disease.
In this case, Megan's father had Huntington's disease, so he must have carried the faulty gene. Assuming her mother was unaffected (hh), Megan's father would be heterozygous (Hh) for Huntington's disease. This means Megan would have a 50% chance of inheriting the faulty gene from her father.
Therefore, the probability of Megan developing symptoms would be 50% or 0.5.
Sample Space Visualization:
To visualize the sample space for this question, we can use a Punnett square. Since Megan is at risk, we will denote her genotype as Hh. Here's how the Punnett square would look like:
| H | h |
---------------------
H | HH | Hh |
---------------------
h | Hh | hh |
Both Megan's alleles (Hh) would be present in the sample space, with a 50% chance of inheriting Huntington's disease.
2. Probability of their first child developing Huntington's disease:
To determine the probability of their first child developing Huntington's disease, we need to consider the genotypes of both Megan and Eric.
If Megan has inherited the faulty gene (H), there is a 50% chance for each child of hers to inherit it. However, if Megan did not inherit the faulty gene (h), their child will not develop Huntington's disease.
Since Eric's family has no history of Huntington's disease, we can assume that he does not carry the faulty gene. Therefore, the probability of their first child developing Huntington's disease depends solely on Megan's genotype.
Let's consider two scenarios:
- Scenario 1: Megan inherited the faulty gene (H) from her father (50% probability)
- Scenario 2: Megan did not inherit the faulty gene (h) from her father (50% probability)
In Scenario 1, there is a 50% chance for Megan to pass on the faulty gene to their child, resulting in the child having a 50% chance of developing Huntington's disease.
In Scenario 2, Megan cannot pass on the faulty gene, meaning their child will not develop Huntington's disease (0% chance).
To calculate the overall probability, we need to account for both scenarios:
(Scenario 1 probability) x (chance of their child inheriting the faulty gene in Scenario 1) + (Scenario 2 probability) x (chance of their child inheriting the faulty gene in Scenario 2)
(0.5) x (0.5) + (0.5) x (0) = 0.25
Therefore, the probability of their first child developing Huntington's disease would be 0.25 or 25%.
I hope this explanation helps you understand the probabilities involved. If you have any further questions, feel free to ask!