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genetic problemsModerator: BioTeam
5 posts • Page 1 of 1
genetic problemsI have a question and I was hoping someone will answer me
I was hoping someone could answer me because its really giving me a headache...
gellybelly08, don't be worried and have a rest if you have headache
1. If we do the first cross, we obtain this result: P: Brown x Silver ---> F1: Brown. 2. If we do the second cross, we obtain this result: Brown (F1) x Brown (F1) ---> F2: 47 Brown and 15 Silver. 3. We can expect that it is a 3:1 segregation (but only it is a hypothesis that we have to prove), because if we do 47/15 and 15/15, we obtain 3'13 (that is aproximate to 3) and 1 respectively. 4. This 3:1 segregation means that it is a monohybrid cross and this: - In the first cross: P: Brown (AA) x Silver (aa) ---> F1: Brown (Aa). - In the second cross: Brown [F1] (Aa) x Brown [F1] (Aa) ---> AA, Aa and aa. A_ has a probability of 3/4 (1/4 AA + 2/4 Aa) and aa has a probability of 1/4. This accepts our hypothesis of 3:1 segregation that we expected before. NOTE: To accept completely the hypothesis that we expected about 3:1 segregation, you should do the Chi squared test. 5. So, as we have called (A,a) to this gen which controls colour, we say that: - Brown minks have this genotype: A_ (AA or Aa) [Dominant homozygous or heterozygous]. - Silver minks have this genotype: aa [Recesive homozygous]. Did I answered your doubt? Take a look to your notes (about monohybrid crosses) and tell me if you don't understand something. See you! ![]() ![]() ![]()
Re: genetic problems
from the fact that the F1 is only brown directly indicates that brown is dominant and silver is recessive and also that the only possibility is that the P-generation is BB x bb where B = brown and b = silver. --> otherwise you'll never get 100% brown in F1 and these are all Bb (heterozygous) --> then we get the next crossbreeding of F1: Bb x Bb. This leads to 25% BB, 50% Bb and 25% bb --> thus now you know that of the brown minks (25/75) * 100% = 33,33% = BB and 66,67% = Bb --> we have 47* 0,3333 = 15 homozygous brown minks and thus 32 heterozygous brown minks. --> all the silver minks are homozygous, since we determined bb to be silver: homozygous recessive.
5 posts • Page 1 of 1
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