NS FL# 5, B/B, Passage 10, recessive inheritance confusion

[icamel]

So I can see why TEX11 inheritance isn't sex-linked because the gene is on chromosome 12, but then why does the son exhibit the phenotype when he only gets one mutant allele from his mother and one wild-type allele from his father. Shouldn't he be heterozygous for the gene too? In the passage it says that TEX11 has a recessive mode of inheritance. To me the only way the son could exhibit the trait if it is an X-Linked gene and he gets the mutant X gene from his mother and Y from his dad. Help!

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[Nugester]

Non-disjunction. Received both chromosomes from mom, as seen in uniparental disomy, for example.

-Both from mom and one from dad, lost dads (trisomy rescue)
-Both from mom and dad gave nothing

 

[icamel]

Don't you think the passage should've stated that though? It seems like an important information. It seems like a tricky passage if they expect for us to figure that out.

 

[icamel]

But the mother is heterozygous for the gene so even if the son got both from the mother, then he would have the same banding patterns as his mother since he'll have both the wild allele and the mutant allele. His banding pattern shows that he only has the wild allele.

 

[Nugester]

I think you are confused. He got BOTH mutant alleles from his mother. The patient's father is wild-type, so his gel has one band as both normal chromosomes are there. His mom is heterozygous, so she she have two bands; one normal band (top) for one chromosome and the "mutant" split-band (bottom) for the abnormal chromosome. If the patient has both copies of his mom's chromosomes, he will only have the bottom "mutant" split bands, as seen in D. The patient doesn't have the normal chromosome, otherwise he'd have a band like his father's.

This is a tricky problem. However, it tests the concept of non-disjunction in meiosis. And since they mention the TEX11 interacts with SYCP2 and SYNAPTOMEAL complex, it makes since there is an issue with chromosome/tid separation.

 

[icamel]

What I'm trying to get at is his mom has a wild-type allele too so if it is a non-disjunction and he received both of his mother's allele for that gene then he too would have a wild-type allele and thus would exhibit the same bands as his mother.

But clearly, from the answers the son doesn't have the wild-type allele but only the mutant allele so where does he get his second mutant allele from? or does he only have one allele of the mutant type. I don't think there is any viable autosomal monosomy.

Dad: TT (homozygous)
MOM: Tt (heterozygous)
SON: tt (homozygous) <--- where does the son get the second recessive allele?

 

[Nugester]

You mentioned his mom is heterozygous and has the mutant allele. The only possible way for the patient to have both mutant alleles is A) dad isn't dad or B) from his mom!

Re-read what happens during meiosis I and how chromosomes double and non-disjunction and such. For an easier example, with trisomy 21 works applied to this case: (2) 21s from mom, and (1) 21 from dad. Then a thing such as a trisomy rescue can occur where you lose dad's copy and end up with only mom's copy.

Or something similar as monosomy rescue, which might be a little bit better to understand. Results from non-disjunction resulting in monsomy (mom's allele), but it rescues itself by copying the entire mutant allele, generating the second chromosome.

Both cases above can result in uniparental disomy and where the patient ends up with both mutant alleles from a SINGLE parent.

And viable monosomies may occur depending on the level of mosaicism, but that I imagine would be beyond the scope of this question.

 

[aldol16]

Just FYI, your name is visible in your screenshots. You should get rid of that if you want to protect your privacy.