Joe is 27, female, Caucasian, with a family history of...

Joe is 27, female, Caucasian, with a family history of Huntington's disease. She has been tested and found out that she carries the Huntingtin gene. She also found out that while she has only has affected allele, she has around 44 trinucleotide repeats on the affected allele. Joe gets pregnant by a man who does not carry the Huntington gene at all. a. Describe the pattern of inheritance Joe can expect for her child. Include a hand drawing of the appropriate family tree for all possible gene combinations for the child & give percentages of chance inheriting the gene given that specific combination. Then explain why or why not each combination would result in a child carrying the gene. b. What is the chance Joe's child will carry the gene (Give a percentage)? c. What is Joe's chance of developing Huntington's Disease (Give a percentage)? i. Around what age would you expect her to start showing symptoms?

Transcript

00:01 Hello students, the question asked is a woman has a father, so there is a woman, she's having a father who died of huntington disease.

00:18 So, this the father of this woman died of huntington disease.

00:24 Knowing that the huntington disease, knowing that the huntington undergoes autosomal dominant inheritance, what is the probability that she will develop the symptoms of the disease? so, we have to find what is the probability of the daughter of a man who died of huntington disease.

01:07 As we know that, also it is mentioned in the question that the huntington disease is autosomal dominant, so the father of this woman must have one affected allele, one affected and one normal allele.

01:28 So, the mother of this woman must be normal because it is not mentioned here, so we consider that she must be a normal.

01:43 So, in this scenario, if a father contains a one allele, affected allele and one normal and the mother is totally normal, we could find the percentage or fraction of the offspring which could be affected by using the punnett's square.

02:04 So, this father will give rise to two alleles, one will be affected, one is normal and the mother will be both normal.

02:15 So, the offspring is between these two parents will be i, a affected in normal allele, affected in normal allele, normal allele and normal allele, which means that 50 % will be affected, which means that 2 by 4, which is one half is affected, while 50 % is normal or unaffected, which is again 2 by 4, which is one half.

02:48 So, between these two parents, the fraction of offsprings which will be affected is 50%.

02:55 So, the question asked was what is the probability that the daughter of this man will develop the symptoms of this disease.

03:06 So, there is a 50 % chance.

03:09 Under the given options, we were having, there are certain, some alternative options and among the given ones, the option e is the correct option, which says that 50 % because this woman has a 50 % chance of getting the huntington's disease...

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