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Medelian Genetics question
Justine_5420
#1 Posted : Tuesday, June 30, 2020 12:07:25 AM
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For question 30 in the Biological sciences section bank, could someone please explain to me how the probability that, of the red F1 beetles, both were heterozygous is 2/3?

INSTR_Rheann_72
#2 Posted : Wednesday, July 01, 2020 12:05:57 AM
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Hi there,

Would you be able to link the question?

Thank you
Justine_5420
#3 Posted : Wednesday, July 01, 2020 5:46:24 PM
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https://www.mcatofficial...on-bank#quiz/take/28180

In a species of beetle, red body color is dominant to brown. Two red beetles are crossed and produce 31 red and 9 brown offspring (F1 generation). If two red F1 beetles are crossed, what is the probability that both red and brown beetles will appear in the F2 generation? (Note: Assume Mendelian inheritance patterns.)
INSTR_Rheann_72
#4 Posted : Friday, July 03, 2020 2:12:38 AM
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The F1 generation will have 1 RR 2 Rr and 1 rr. So you could be crossing RRxRR (no brown in F2) RRxRr (no brown in F2) RR x other Rr (no brown in F2) or Rr x Rr (3 red to 1 brown in F2)
Justine_5420
#5 Posted : Tuesday, July 07, 2020 8:09:19 PM
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Hello,

Thank you for getting back to me. I understand those genotypic ratios but for some reason am still lacking the logic to generate a 2/3 ratio for the probability both were heterozygotes in the F1 generation? Isn't this probability, based on mendillian genetics, 1/2 if you do a Rr x Rr cross?
INSTR_Katerina_102
#6 Posted : Wednesday, July 08, 2020 2:18:22 AM
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Hi,

I can't actually see the link that you have posted, but usually when I see a 2/3 ratio, the impression that I get is that one of the homozygous variants is lethal. For example, here rr or RR could be lethal, leaving you with a ratio of 2/3, because you would never see the offspring of the lethal genotype grow up.
Justine_5420
#7 Posted : Friday, July 10, 2020 11:36:38 PM
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Hi,

Sorry I'm not sure how to link to the question, but it is question 30 in the Biology section Bank. I can copy the question here - there is no associated images or diagrams:

"In a species of beetle, red body color is dominant to brown. Two red beetles are crossed and produce 31 red and 9 brown offspring (F1 generation). If two red F1 beetles are crossed, what is the probability that both red and brown beetles will appear in the F2 generation? (Note: Assume Mendelian inheritance patterns.)"

I don't think any of the genotypes are lethal as this is not stated in the question stem. Any help in answering this would be appreciated, if you use a punnette square would you mind showing how you set up your punnette square?
INSTR_Katerina_102
#8 Posted : Monday, July 13, 2020 6:27:02 PM
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Not going to lie, this question haunted me for a couple days as it has a lot to juggle in terms of restrictions and extra information. I apologize for the delay between now and my poorer reply before.

So the best thing I've come up with reference to the question is:

1."In a species of beetle, red body color is dominant to brown. Two red beetles are crossed and produce 31 red and 9 brown offspring (F1 generation).


Your parent genotypes: Rr and Rr

F1 genotypes : RR, Rr, Rr and rr.

2.If two red F1 beetles are crossed, what is the probability that both red and brown beetles will appear in the F2 generation? (Note: Assume Mendelian inheritance patterns.)"

If you cross two red F1 beetles, you are only considering the 31 red offspring which should be roughly 2/3 Rr and 1/3 RR. This is how you get the 3 in the denominator by eliminating the possibility of using rr artificially.


3.From here, you can either have an RR x Rr cross or an Rr x Rr cross.

What you want is the chance that you will get brown and red offspring, which is any cross that is not involving RR as one of the parents.

As a result, I can calculate the probability of red and brown offspring (PRB) by the following equation:

PRB = 1 - PR

Where PR is the probability of red only offspring.

PR can be described as 2 different crosses:

RR x Rr which has a probability of (1/3)*(2/3) = 2/9
and RR x RR which has a probability of (1/3)*(1/3) = 1/9

PR is the sum of these two independant possibilities, so we get 3/9 = 1/3 probability of red only offspring.

Therefore PRB = 1 - (1/3) = 2/3.

Needless to say, this took me a couple of days to chew on and come up with a somewhat reasonable answer. I definitely wouldn't have gotten this on the MCAT. I would say it's a question I would have flagged and guessed, and likely gotten wrong.

Please let me know if this is still unclear, I hope this helps!

Katt
INSTR_Katerina_102
#9 Posted : Wednesday, July 15, 2020 1:54:03 PM
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Hi,

Sorry for the double posting - I think this question's answer is actually wrong, because I missed the chance of selecting a red beetle second, which actually changes the odds to 4/9 that you would get red and brown offspring.

I have attached the logic here.

https://ibb.co/my3K1M5
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