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Babygal Babygal
wrote...
Posts: 34
Rep: 1 0
13 years ago
6. Assume that the height of tobacco plants is controlled by five genes. Two different alleles exist for genes A, B, C, and D, but only one allele for gene E is present in the population. The effect of each genotype on height is listed below. Total height is the sum of the effects of each of these five genes. 
 
Gene A       Gene B       Gene C       Gene D       Gene E
AA + 6 cm    BB + 10 cm    CC + 8 cm    DD + 2 cm    EE + 40 cm
Aa + 3 cm       Bb + 5 cm        Cc + 5 cm       Dd + 2 cm   
aa + 0 cm       bb + 0 cm        cc + 2 cm       dd + 0 cm   
 
a. What is the maximum height possible in the population and which genotypes will have
this phenotype (4 pt)?

b. What fraction of progeny produced from the following cross will be exactly 42 cM tall:
AaBbCcddEE x AaBbCcddEE (4 pt)?
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wrote...
13 years ago
a- the maximum height is when the individual has the following genotype : AABBCCDDEE.
so it would be  6 + 10 + 8 + 2 + 40 = 66 cm.

b- 42 cm could be obtained from the following genotype only :

aabbccddEE " 0 + 0 + 2 +0 + 40 = 42 "

now what is the probability of having this genotype ?
the probability of having " aa " is 1/4.
the probability of having " bb " is 1/4.
the probability of having " cc  " is 1/4.
the probability of having " dd " is 1.
the probability of having " EE "  is 1.

now we multiply the probabilities and obtain the fraction of progeny produced having exactly 42 cm.
1/4 x 1/4 x 1/4 x 1 x 1 = 1/64

if you didn't understand how i got the probability of having the genotype of each gene plz tell me so i can explain it better.
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