Help with a biology NSAA question

Hello Vera8. I can help you to respond to your questions incase you need help.

Hello! I'm just a bit confused with a NSAA question and would really appreciate any help Link to the paper: https://www.undergraduate.study.cam.ac.uk/sites/www.undergraduate.study.cam.ac.uk/files/publications/nsaa_past_paper_2020_0.pdf

Q57 (pg 144) I put C but the answer is E, and I don't understand why the first statement is correct. I thought that for S the theoretical number of genotypes would be 6 x 6 so 36, and likewise 5 x 5 so 25 for R. So I get a difference of 11, not 6

Thank you in advance!

Hey,
Your method to calculate the theoretical number of genotypes is correct, but you forgot to subtract the repeats (for example AB and BA are the same thing). If you do that you should get the right answer! I literally just scribbled down all the possibilities in a table thing, but a more mathsy explanation is that this is a combination with repetition (more on this here). Hope that helped and good luck for your NSAA

"The theoretical number of different genotypes in population S is 6 more than in population R."

Population S has 6 alleles for the gene, while population R has 5 alleles. To calculate the theoretical number of different genotypes, you can use the formula n(n+1)/2, where n is the number of alleles. In this case:

For population R: 5(5+1)/2 = 5(6)/2 = 30/2 = 15 genotypes.
For population S: 6(6+1)/2 = 6(7)/2 = 42/2 = 21 genotypes.
The theoretical number of different genotypes in population S is 21, which is indeed 6 more than in population R (15). So, statement 1 is correct.

"The theoretical number of different homozygous genotypes in population P, Q, R, S is 3, 4, 5, 6."

To calculate the theoretical number of different homozygous genotypes, you can use the formula n, where n is the number of alleles. Based on the given information:

For population P: 3 homozygous genotypes.
For population Q: 4 homozygous genotypes.
For population R: 5 homozygous genotypes.
For population S: 6 homozygous genotypes.
So, statement 2 is correct.

"In populations P, Q, and R, there are more different homozygous combinations than there are different heterozygous combinations."

We've already determined the number of different homozygous combinations for populations P, Q, and R (3, 4, and 5, respectively). Now, let's calculate the number of different heterozygous combinations using the formula n(n-1)/2, where n is the number of alleles:

For population P: 3(3-1)/2 = 3(2)/2 = 6/2 = 3 heterozygous combinations.
For population Q: 4(4-1)/2 = 4(3)/2 = 12/2 = 6 heterozygous combinations.
For population R: 5(5-1)/2 = 5(4)/2 = 20/2 = 10 heterozygous combinations.
In all three populations (P, Q, R), there are more different heterozygous combinations than homozygous combinations. So, statement 3 is incorrect.

Based on the analysis, the correct answers are:

Statement 1 is correct.
Statement 2 is correct.
Statement 3 is incorrect.
Therefore, the correct answer is B) 1 only

Hello, thank you for your help- I get why I have to divide my answers by 2 now (the link you put is very useful thanks). I’m just thinking though- if I divide 36 by 2 I get 18 and if I divide 25 by 2 it’s 12.5 which gives a difference of 5.5 so I think I must have gone wrong somewhere but not sure where 🤔

Hi - no problem!
when subtracting the repeats, I don’t think you can divide by 2…

for example for population R with 5 alleles (A,B,C,D,E)
‘you could have:
AA, AB, AC, AD, AE
BA, BB, BC, BD, BE
CA, CB, CC, CD, CE
DA, DB, DC, DD, DE
EA, EB, EC, ED, EE

the ones in bold are the ones that are repeats, I’ve already got them before so we have to subtract them because they’re the same thing
it forms a triangle kinda shape
so the calculation would be 25 - 10 = 15
And doing the same thing for the one with 6 alleles it would be 36 - 15 = 21
and then 21 - 15 = 6
But yeah this method is really long and complicated so the above poster’s formula might be a better approach

I see, thank you so much for your help I really appreciate it! Have a great day

FWIW I think the other reply was from someone posting a chatgpt response. The numbers n(n+1)/2 are the triangular numbers which corresponds to the numbers given by Icosagon. Even if you remember a formula, its better to undersand what it represents.

Oh thanks, I had no idea! I also calculated the number of combinations of alleles for S using the nCr button on my calculator which gave me 15, and 10 for R. I then added another 6 to 15 (21) and 5 to R (15) which gives a difference of 6. Not a clue why it works though, I think I better do some reading up about probabilities, combinations and permutations...

Thanks everyone for the help though

Not a clue about the biology and Id not really think of it nCr ((n+1)C2 more precisely). Its simply as Icosagon demonstrated

AA, AB, AC, AD, AE
AB, BB, BC, BD, BE
CA, CB, CC, CD, CE
DA, DB, DC, DD, DE
EA, EB, EC, ED, EE,

The bolds are repeats which you dont want to count but the diagnonls only occur once. So think of it as a rectangle n*(n+1) where the diagonals are repeated, then simply divide by 2, so overall n(n+1)/2

AA, AA, AB, AC, AD, AE
AB, BB, BB, BC, BD, BE
CA, CB, CC, CC, CD, CE
DA, DB, DC, DD, DD, DE
EA, EB, EC, ED, EE, EE,