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micro01 micro01
wrote...
Posts: 68
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12 years ago
Having a spot of trouble with biology, the example used here is RrBb X RRBB
where R stands for the dominant variation of running mouse (able to run in a straight line) and r stands for waltzing mouse (only walks in circles). B = black hair b= brown hair.
I don't want an answer for this specific problem, just giving an example of what i'm having difficulty with.
Thanks so much!
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nyonisikhanyonisikha
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12 years ago
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12 years ago
Using your example, you would take the pairs of the pairs. So using the genotype of ONLY the first mouse you would take RrBb, and take the first allele for the running trait, and the first allele for the hair color trait and you would get "RB". Using that same process [still using the genotype of the first mouse only] you would get the four segregation's of RB, Rb, rB, and rb. Those would be the top or left side of the punnet square or a dihybrid cross. Now using the same process for the genotype of the second mouse, you will get RB, RB, RB, and RB. Notice they are all the same because the mouse is homozygous for both traits. That will be the other side of your punnet square. Now you set it up, and distribute the alleles as you would normally. You will notice that each column or row depending on which side you use for the mouse that is heterozygous will be that same, because all four segregation s for the homozygous mouse are the same. You will end up with the genotypes RRBB, RRBb, RrBB, and RrBb. Hope this helps!
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