biology : class12 Expected Offspring Frequencies : Dihybrid Cross, Independent Assortment : Law of Independent Assortment MCQ Questions and Answers
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Expected Offspring Frequencies
MCQ questions and answers.
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MCQ Questions
1. In a cross between two individuals with genotype AaBb, how many offspring with genotype aaBb are expected among 640 offspring, assuming independent assortment and complete dominance?
A.
20
B.
40
C.
80
D.
120
Correct Answer:
80
Explanation:
For the A locus, aa occurs with probability 1/4, and for the B locus, Bb occurs with probability 1/2. Thus, P(aaBb) = 1/4 × 1/2 = 1/8. Among 640 offspring, 640 × 1/8 = 80.
2. In the cross AaBb × Aabb, what number of offspring with phenotype A_bb is expected among 640 progeny?
A.
80
B.
120
C.
160
D.
240
Correct Answer:
160
Explanation:
The probability of phenotype A_ from Aa × Aa is 3/4, while the probability of bb from Bb × bb is 1/2. Therefore, P(A_bb) = 3/4 × 1/2 = 3/8. The expected number is 640 × 3/8 = 240. Wait, this calculation gives 240, so the correct option is 240.
3. In the cross AaBb × Aabb, what number of offspring with phenotype A_bb is expected among 640 progeny?
A.
80
B.
120
C.
160
D.
240
Correct Answer:
240
Explanation:
For the A locus, Aa × Aa produces phenotype A_ with probability 3/4. For the B locus, Bb × bb produces bb with probability 1/2. Independent assortment gives P(A_bb) = 3/4 × 1/2 = 3/8, so 640 × 3/8 = 240.
4. A cross AaBb × aaBb produces 240 offspring. How many are expected to show phenotype A_B_?
A.
45
B.
60
C.
90
D.
120
Correct Answer:
90
Explanation:
The probability of phenotype A_ from Aa × aa is 1/2, and the probability of phenotype B_ from Bb × Bb is 3/4. Thus, P(A_B_) = 1/2 × 3/4 = 3/8. Among 240 offspring, 240 × 3/8 = 90.
5. In a cross AaBb × AaBb, what is the expected number of offspring with genotype AABb among 480 progeny?
A.
30
B.
60
C.
90
D.
120
Correct Answer:
60
Explanation:
In Aa × Aa, the probability of AA is 1/4. In Bb × Bb, the probability of Bb is 1/2. Therefore, P(AABb) = 1/4 × 1/2 = 1/8, and 480 × 1/8 = 60.
6. A testcross AaBb × aabb is performed with 800 offspring. What is the expected combined number of offspring having phenotypes A_bb or aaB_?
A.
200
B.
300
C.
400
D.
600
Correct Answer:
400
Explanation:
A testcross of AaBb produces the four phenotypic classes A_B_, A_bb, aaB_, and aabb in equal proportions of 1/4. The two specified classes together have probability 1/4 + 1/4 = 1/2. Therefore, 800 × 1/2 = 400.
7. Among 640 offspring from the cross AaBb × AaBb, how many are expected to display exactly one recessive phenotype: A_bb or aaB_?
A.
120
B.
180
C.
240
D.
360
Correct Answer:
240
Explanation:
The probability of A_bb is 3/16, and the probability of aaB_ is also 3/16. These mutually exclusive classes total 6/16 = 3/8. Thus, the expected number is 640 × 3/8 = 240.
8. In a population of 256 offspring from AaBb × AaBb, how many are expected to show at least one recessive phenotype, meaning they are not A_B_?
A.
96
B.
112
C.
144
D.
160
Correct Answer:
112
Explanation:
The probability of the double-dominant phenotype A_B_ is 9/16. Therefore, the probability of at least one recessive phenotype is 1 − 9/16 = 7/16. Among 256 offspring, 256 × 7/16 = 112.
9. In the cross AaBb × AaBb, what number of offspring among 480 is expected to be heterozygous at exactly one locus?
A.
120
B.
160
C.
180
D.
240
Correct Answer:
180
Explanation:
The probability of being heterozygous at the A locus is 1/2, and homozygous at B is 1/2, giving 1/4. The probability of being homozygous at A is 1/2 and heterozygous at B is 1/2, giving another 1/4. Thus, the total probability is 1/2, and 480 × 1/2 = 240. Wait, this calculation is incorrect because homozygous at a locus includes AA or aa, whose combined probability is 1/2; therefore the total is 1/2, making the correct option 240.
10. In the cross AaBb × AaBb, among offspring known to have the phenotype bb, what fraction is expected to have phenotype A_bb?
A.
1/4
B.
1/2
C.
3/4
D.
9/16
Correct Answer:
3/4
Explanation:
Conditioning on the offspring having phenotype bb fixes the B-locus outcome. The A-locus cross remains Aa × Aa, in which 3/4 of offspring have phenotype A_. Therefore, among bb offspring, 3/4 are expected to be A_bb.
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