Nisarg Asked a Question
May 29, 2021 4:29 pmpts 30 pts
Parents: AABBCCDdEe x Aa BbCcDdEe i. No. of gametes? ii. P(AABbccddEE)?
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  • Abhijeet Gaurav
    No problem it's absolutely fine. These types of confusions show that you are immersed in the subject so that's a very good sign. It tends to happen with time as you give more time ...
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  • Abhijeet Gaurav thankyou
    Gametes are in even in Number except for Monohybrid Homozygous. So I think it Will give more clarity.
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    Nisarg
    thank you sir it's clear now
  • Abhijeet Gaurav
    You seem to have got bit diverted. See for both AA it is same the set so it will be A. Now here A is Homozygous so 2^0 = 1 . But in the case of AABbCc it is trihybrid so here varia...
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    Nisarg
    i was my bad 😅sorry and thank you for clearing my doubt
  • Abhijeet Gaurav
    We consider only for Heterozygotes because they are the one with variations in the DNA. Whereas the Homozygous ones are the similar set of gene in them. As from the example AABB we...
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    Nisarg
    so in the case of AABbCc don't we have to count the number of gametes as 2^2=4, and 1 more from AA? so total number will be 5 or 4, because as per equation we end up with 4 gametes...
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  • Abhijeet Gaurav Best Answer
    Yes you are right. By mistake I wrote 256 instead of 32. Thanks for pointing out. I did correction for it now.
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    Nisarg
    sir in the equation of finding number of gametes why do we consider only the heterozygotes(n) not the homozygotes?
  • Abhijeet Gaurav thankyou
    Answer of the both are solved below. The Number of gametes for the Parent 1 would be -8 Parent 2 would be -32
    • cropped4479392773231841328.jpg
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    Nisarg
    sorry sir but didn't understand how we end up with 256 in parent 2 it shouldn't be 32?
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