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Lecture 7

LIFESCI 4 Lecture Notes - Lecture 7: Centimorgan, Chromosome


Department
Life Sciences
Course Code
LIFESCI 4
Professor
Pham H.D.
Lecture
7

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LIFESCI 4 - Lecture 7 - Beadle and Tatum
Beadle and Tatum: One gene - One Enzyme
Clicker 1:
Factors A, B, C, D control the production of a black pigment. The genes encoding A, B, C, D, are
independently assorting. The alternate alleles that give abnormal function of these genes are a,
b, c, and d. A black AA BB CC DD is crossed with a colorless aa bb cc dd to give a black F1. The F1
is then backcrossed with a colorless aa bb cc dd. Assume that A, B, C, D act in a pathway as
follows:
A B C D
Colorless colorless grey brown black
1. What percentage of the F2 are brown?
A) 1/2
B) 1/4
C) 1/8
D) 1/16
E) None of the above
Explanation:
P
AA BB CC DD x aa bb cc dd
F1
From P(Aa Bb Cc Dd) x aa bb cc dd
- P(F2 = brown) = P(Aa Bb Cc dd) = ½ X ½ X ½ X ½ = 1/16
2. What percentage of the F2 are colorless?
A) 1/4
B) 3/8
C) 1/2
D) ¾
E) None of the above
Explanation:
- Colorless if either enzyme A or B is mutated
- Two ways to achieve colorless: either (aa __ __ __ ) or (A_ bb __ __)
- Do not need to consider (aa bb __ __)
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- Once allele (a) is mutated by being recessive homozygous (aa), enzyme A
does not function and the chain is “broken” the remaining alleles do
not need to be considered
- In another words, (aa bb __ __) is the same as (aa __ __ __)
- P(colorless) = P(aa __ __ __) + P(A_ bb __ __) = ½ + (½)(½) = ½ + ¼ = ¾
Clicker 2:
Factors A, B, C, D control the production of a black pigment. The genes encoding A, B, C, D, are
independently assorting. The alternate alleles that give abnormal function of these genes are a,
b, c, and d. A black AA BB CC DD is crossed with a colorless aa bb cc dd. Genes A, B, C
independently assort, but genes C and D are linked with a 20 map unit separation. Assume
that A, B, C, D act in a pathway as follows:
A B C D
Colorless colorless grey brown black
1. What percentage of the F2 are black?
A) 1/4
B) 1/8
C) 1/10
D) 1/20
E) None of the above
Explanation:
- For F2 to be black, all enzymes must be functioning = phenotype (A_B_C_D)
- A, B, C are independently assorting, so do the probability normally
- P(A_B_C_) = (½)(½)(½)
- C and D are linked:
C_________D
c__20m.u__d
- The probability of crossing over, resulting in (C--d) is 0.2
- The probability of NOT crossing over (C--D) is 1 - 0.2 = 0.8
- P(C--D) = 0.8
- P(A_B_C) x P(C--D) = (½)(½)(½)(0.8) = 1/10
Review Clicker:
You believe that genes A and B are on different chromosome and therefore should
independently assort. To test this you cross a purebred line that is homozygous for the
dominant alleles of A and B with a purebred line that is homozygous for the recessive alleles a
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