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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
- IIT JAM
- Biotechnology (BT)
Parents: aabbccddee x aa bbccddee i. no. of gametes? ii. p(aabbccddee)?
Parents: AABBCCDdEe x Aa BbCcDdEe i. No. of gametes? ii. P(AABbccddEE)?
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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 variations in the Gametes would be ABC, ABc, AbC,Abc only. Thus 4 gametes & for both AA we take them in together with others such as Bb & Cc not separately that's why it's Just 4 not 5.
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i was my bad 😅sorry and thank you for clearing my doubt
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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 can take that only one gamete Will be formed - AB. Even if try & Form other gametes for this then it would be all same so we consider only one gamete for them & Heterozygotes have variations.
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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
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 to study.