Trial Q (1 Viewer)

[Yusra366]

anyone know how to do this q?

 

[Eagle Mum]

The answer to part a) is 10.

The details for part b), is that if you have four different alleles, designated A1, A2, A3 and A4, the ten possible genotype combinations are A1/A1, A1/A2, A1/A3, A1/A4, A2/A2, A2/A3, A2/A4, A3/A3, A3/A4 and A4/A4.

Note: Not by coincidence, the number of genotypes for 'n' number of alleles is the 'n'th triangular number, because if you draw up a n x n table with all the possible alleles on the horizontal & vertical axes, the combinations in the top right corner and bottom left corners would be the same genotype even if the maternal vs paternal inheritance are not the same (ie. A1/A4 is the same genotype as A4/A1), so the unique genotypes in the table would form a triangle from one of these corners to, and including, the diagonal options.

 

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I’d just write four alleles and count the possible combinations. I’m sure your familiar with the 3 allele blood groups right? It works just like that.
View attachment 31285


This formula will work for some allele combinations
View attachment 31286

Superiority entensifies*

 

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Here is part c of a graphed represention around vaccine influences on antibody concentration: