Hi, can someone one elaborate what I have done wrong in my calculations(according to ans key it is supposed to be 0.823 for WT and 0.1717 for parasites)? Basically I have tried this:
*Also sorry for typos here, friend of mine just translated it for me
@eudaimon the problem says: virulence of Nectria Haematococa is determined by the presence of dominant allele of gene P, which is located on the so called “B” chromosome. Ultimately, B chromosome is known for being unstable - resulting in strains either with one more copy of it or none at all. Nectria strains without dominant allele are known to reside as parasites (WT do not live parasitic lifestyle). Diploid cell with 2 more additional B chromosome, which was resulted by the cross of 2 pure line (WT and parasitic), was shifted to meiosis. Also it is known that there is a 6% chance of chromosome nondisjunction in meiosis 1, and there is no nondisjunction event in meiosis 2. Try to calculate the phenotypic ratio of gametes.
To solve this problem, we need to consider the possible outcomes of meiosis in the diploid cell with additional B chromosomes. Let’s denote the dominant allele that confers virulence as “P” and its absence (or recessive allele) as “p”. A diploid cell with two additional B chromosomes would have a genotype of “PPpp”, where “PP” are the alleles on the additional B chromosomes and “pp” are the alleles on the regular set of chromosomes.
During meiosis, these chromosomes can segregate in different ways, leading to different gametes. If there is no nondisjunction, the gametes will receive one B chromosome and one regular chromosome, resulting in either “Pp” or “pp” gametes. However, with a 6% chance of nondisjunction, we can have gametes with two B chromosomes, one B chromosome, or no B chromosomes at all.
Let’s calculate the expected phenotypic ratio of gametes, considering the outcomes of the first meiotic division with and without nondisjunction:
No nondisjunction (94% chance):
Nondisjunction (6% chance):
Since “P” is dominant, any gamete with at least one “P” will be virulent. Now, let’s calculate the proportions:
For virulent gametes (WT):
Adding these together gives the proportion of virulent gametes (WT):
[ WT = 0.47 + 0.015 + 0.015 = 0.50 ]
For non-virulent gametes (parasites):
Adding these together gives the proportion of non-virulent gametes (parasites):
[ Parasites = 0.47 + 0.015 + 0.015 = 0.50 ]
It seems there might be a misunderstanding or a miscalculation if the answer key suggests different proportions, or perhaps the problem setup implies some additional details that haven’t been considered here. The calculations above assume that all types of gametes are equally viable and that the presence of the “P” allele is the only factor determining virulence. If there are additional complexities not mentioned in the problem statement, these would need to be factored into the calculations.
this is the reason why AI shouldn’t approach any genetic q 
Oh, yeah, and also my answer appears to be right (srry for being late, apparently it was just insular q where all approximiate values are welcome)
to be fair, the problem set up wasn’t clear even to me. I didn’t understand what was meant by “chromosome B being unstable” and the sentence as is “either with on more copy of it or none at all” doesn’t make sense.
yeah, the text is kinda rusty (it was partially google translated from cantonese ). Also, I am pretty sure this “unstable” part just simply implies that B chromosome is “mobile” (making aneuploid strains of Nectrian possible)
