YOU MUST SHOW YOUR WORK TO GET CREDIT! (You might want to try to apply the proba
ID: 186255 • Letter: Y
Question
YOU MUST SHOW YOUR WORK TO GET CREDIT! (You might want to try to apply the probability to this question rather than using punnet analysis to answer this question....It should faster) 6. (4 pts) For the following cross: Aa Bb Cc Dd Ee Ff X AA Bb Cc Dd ee Ff, what is the probability of obtaining an individual who is: A. Aa Bb Cc Dd Ee Ff B. A-B-C-D-E-F- C. Assuming that all the genes assort independently, how many different phenotypes will be present among the progeny for the cross described above? SHOW YOUR WORK! EXPLAIN HOW YOU DETERMINED THIS ANSWER FOR FULL CREDIT. D. Assuming that all the genes assort independently, how many different genotypes will be present among the progeny for the cross described above? SHOW YOUR WORK! EXPLAIN HOW YOU DETERMINED THIS ANSWER FOR FULL CREDIT.Explanation / Answer
Answer:
AaBbccDdEeff x AABbCcDdeeFf ------Parents
Aa x AA = AA (1/2)& Aa(1/2)
Bb x Bb = BB (1/4), Bb (1/2) & bb(1/4)
cc x Cc = Cc (1/2) & cc(1/2)
Dd x Dd = DD (1/4), Dd (1/2) & dd (1/4)
Ee x ee = Ee (1/2) & ee (1/2)
ff x Ff = Ff (1/2) & ff (1/2)
A).
The probability of AaBbCcDdEeff = ½ * ½ * ½ * ½ * ½ * ½ = 1/64
B).
A_B_C_D_E_F_ = 1 * ¾ * ½ * ¾ * ½ * ½ = 9/128
C).
AaBbccDdEeff x AABbCcDdeeFf
Aa x AA = A_ = 1 phenotype
Bb x Bb = B_ & bb = 2 phenotypes
cc x Cc = Cc & cc = 2 phenotypes
Dd x Dd = D_ & dd =2 phenotypes
Ee x ee = Ee & ee = 2 phenotypes
ff x Ff = Ff & ff = 2 phenotypes
Total number of phenotypes = 1 * 2 * 2 * 2 * 2 * 2 = 32 different types of phenotypes
D).
AaBbccDdEeff x AABbCcDdeeFf
Aa x AA = AA & Aa = 2 genotypes
Bb x Bb = BB, Bb & bb = 3 genotypes
cc x Cc = Cc & cc = 2 genotypes
Dd x Dd = DD, Dd & dd = 3 genotypes
Ee x ee = Ee & ee = 2 genotypes
ff x Ff = Ff & ff = 2 genotypes
Total number of genotypes = 2 * 3 * 2 * 3 * 2 * 2 = 144 different types of phenotypes
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