Designer Genes C

Mr.Epithelium
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Re: Designer Genes C

Post by Mr.Epithelium » December 29th, 2020, 11:48 am

WangwithaTang wrote:
December 29th, 2020, 11:02 am
Julia has a sibling with Tay Sachs (a recessive trait). Neither Julia, nor her parents has the
disease, and none of them has been carrier tested. However, Julia’s husband, Steve, fathered a
child with Tay Sachs in a previous marriage. Based on this incomplete information, calculate the
probability that if Julia and Steve have a child, the child will have Tay Sachs.
A. 1/4
B. 1/6
C. 1/9
D. 1/18
E. 1/24

I was taking a past heredity test for practice (University of Florida 2019 B), and I can't figure out why the answer is 1/6. Aren't both Julia and Steve guaranteed heterozygotes, making it 1/4?
This problem is very tricky!
Based on the prompt, Julia's genotype is unknown (we only know her phenotype- that she is not affected). Since we know that her parents are unaffected but had an affected child, both of her parents are heterozygous (carrier of Tay Sachs).
Because Julia is not affected, she can be homozygous dominant or heterozygous. Out of the unaffected genotypes, there is a 2/3 chance for her to be a carrier (draw the Punnett square and cross out the homozygous recessive box for better visualization).
Now for Steve: Steve can be heterozygous or homozygous recessive- he has at least one recessive allele to pass down to his previous affected child. However, we know that Steve is unaffected due to Tay Sachs being a lethal recessive disease (it leads to death at a young age). This means that he must be a carrier.
Thus, there is an overall 2/3 (chance that Julia is a carrier) * 1/4 (probability of having an affected child with two heterozygous parents) = 1/6.
These users thanked the author Mr.Epithelium for the post (total 4):
WangwithaTang (December 29th, 2020, 11:59 am) • gz839918 (December 30th, 2020, 9:19 am) • donutsandcupcakes (January 2nd, 2021, 7:19 am) • gaun22 (January 28th, 2021, 7:17 am)
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ledwards003
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Re: Designer Genes C

Post by ledwards003 » January 19th, 2021, 5:19 am

WangwithaTang wrote:
December 29th, 2020, 11:02 am
Julia has a sibling with Tay Sachs (a recessive trait). Neither Julia, nor her parents has the
disease, and none of them has been carrier tested. However, Julia’s husband, Steve, fathered a
child with Tay Sachs in a previous marriage. Based on this incomplete information, calculate the
probability that if Julia and Steve have a child, the child will have Tay Sachs.
A. 1/4
B. 1/6
C. 1/9
D. 1/18
E. 1/24

I was taking a past heredity test for practice (University of Florida 2019 B), and I can't figure out why the answer is 1/6. Aren't both Julia and Steve guaranteed heterozygotes, making it 1/4?
I understand where the confusion would come from, but it is not actually guaranteed that Julia is a carrier. Her parents are both carriers, so Julia is either a carrier or is fully dominant. She has a 2/3 chance of being a carrier (because we know she can’t be recessive, as she doesn’t have the disease, so there is a 2/3 chance she is a carrier and a 1/3 chance she has both dominant alleles). Steve IS a carrier, as we know because he has a child with Tay Sachs. This means there is a 1/6 chance of the child having Tay Sachs, at least the way I understand it.

Does that make sense? If not I can try to elaborate a little differently.

(I’ve realized I said “I can’t elaborate” when I was first typing this, that’s my bad)
Last edited by ledwards003 on January 20th, 2021, 6:14 am, edited 2 times in total.
These users thanked the author ledwards003 for the post (total 2):
Mr.Epithelium (January 19th, 2021, 8:03 am) • WangwithaTang (January 22nd, 2021, 8:48 pm)
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