If you crossed a 100cm tree tree (AABBCC) with a 40cm tree (aabbcc) and then cro
ID: 25194 • Letter: I
Question
If you crossed a 100cm tree tree (AABBCC) with a 40cm tree (aabbcc) and then crossed the F1 to get the F2 generation, what proportion of the F2 offspring will grow to be 70cm tall?a) 1/16
b) 20/64
c) 20/128
d) 10/64
e) 1/8
Everyone is telling me that the answer is B) 20/64, but I don't know how they came to that answer, can someone please tell me HOW to answer the question?
Explanation / Answer
Well, technically this question is quite oversimplifying things, but it appears to suggest that each dominant allele is supposed to contribute 10 cm in height. Therefore, any genotype that has one dominant allele of the six (two per gene) would be 50 cm tall, two alleles would be 60 etc. So.... 1) You are crossing two heterozygotes (AaBbCc), which results in 27 different genotypes in the following proportions based on a branch diagram: 1 AABBCC $ 2 AABBCc $ 1 AABBcc $ 2 AABbCC $ 4 AABbCc $ 2 AABbcc *$ 1 AAbbCC $ 2 AAbbCc *$ 1 AAbbcc 2 AaBBCC $ 4 AaBBCc $ 2 AaBBcc *$ 4 AaBbCC $ 8 AaBbCc *$ 4 AaBbcc 2 AabbCC *$ 4 AabbCc 2 Aabbcc 1 aaBBCC $ 2 aaBBCc *$ 1 aaBBcc 2 aaBbCC *$ 4 aaBbCc 2 aaBbcc 1 aabbCC 2 aabbCc 1 aabbcc The starred genotyped would have three dominant and three recessive alleles equating to the middle (70 cm) height. If you add them up its 20 out of 64. 2) To be 70 cm or taller, you need at least 3 dominant alleles. So take the 20 and add those genotypes with 4 or more dominants and you get ($): 42 of 64 3) You can just do individual Punnett Squares to determine the possible genotypes. For instance, for the first one, you will have all Aa's, all BB's, and half Cc's and half cc's as progeny. Therefore, you will have Aa=+10, BB=+20, Cc=+10 to cc=+0 ===> 70 to 80 cm. aaBbcc x AabbCc: 1/2 Aa and 1/2 aa = +0 to +10 1/2 Bb and 1/2 bb = +0 to +10 1/2 Cc and 1/2 cc = +0 to +10 Total = 40 to 70 cm