biology : class12 Product Rule in Inheritance : Dihybrid Cross, Independent Assortment : Law of Independent Assortment MCQ Questions and Answers
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Product Rule in Inheritance
MCQ questions and answers.
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MCQ Questions
1. In a dihybrid cross AaBb × AaBb, what is the probability that an offspring will have genotype Aabb?
A.
1/16
B.
1/8
C.
3/16
D.
1/4
Correct Answer:
1/8
Explanation:
The probability of genotype Aa at the first locus is 1/2, and the probability of genotype bb at the second locus is 1/4. By the product rule, (1/2) × (1/4) = 1/8.
2. For the cross AaBb × Aabb, what is the probability that an offspring will have genotype aaBb?
A.
1/16
B.
1/8
C.
1/4
D.
3/8
Correct Answer:
1/8
Explanation:
At the A locus, Aa × Aa gives aa with probability 1/4. At the B locus, Bb × bb gives Bb with probability 1/2. Independent probabilities are multiplied: (1/4) × (1/2) = 1/8.
3. In the testcross AaBb × aabb, what is the probability that an offspring will show the dominant phenotype for gene A and the recessive phenotype for gene B?
A.
1/2
B.
1/8
C.
1/4
D.
3/4
Correct Answer:
1/4
Explanation:
The probability of showing the dominant A phenotype is 1/2, while the probability of showing the recessive bb phenotype is 1/2. Their joint probability is (1/2) × (1/2) = 1/4.
4. In the self-cross AaBb × AaBb, what is the probability that an offspring will show a dominant phenotype for one gene and a recessive phenotype for the other gene?
A.
1/4
B.
3/16
C.
1/2
D.
3/8
Correct Answer:
3/8
Explanation:
The probability of A_bb is (3/4) × (1/4) = 3/16. The probability of aaB_ is (1/4) × (3/4) = 3/16. Adding these mutually exclusive outcomes gives 3/16 + 3/16 = 3/8.
5. Two offspring are produced from the cross AaBb × AaBb. What is the probability that both offspring will have genotype aabb?
A.
1/64
B.
1/128
C.
1/256
D.
1/512
Correct Answer:
1/256
Explanation:
For each offspring, the probability of aabb is (1/4) × (1/4) = 1/16. For two independent offspring, the product rule gives (1/16) × (1/16) = 1/256.
6. In the cross AaBb × AaBb, what is the probability that the first offspring will be Aabb and the second offspring will be aaBb?
A.
1/32
B.
1/64
C.
1/128
D.
1/256
Correct Answer:
1/64
Explanation:
The probability of Aabb in one offspring is 1/8, and the probability of aaBb in another offspring is also 1/8. For the specified order, multiply them: (1/8) × (1/8) = 1/64.
7. For the cross AaBb × AABb, what is the probability that an offspring will have genotype AABb?
A.
1/8
B.
1/4
C.
3/8
D.
1/2
Correct Answer:
1/4
Explanation:
At the A locus, Aa × AA produces AA with probability 1/2. At the B locus, Bb × Bb produces Bb with probability 1/2. By the product rule, (1/2) × (1/2) = 1/4.
8. In the cross AaBb × aaBb, what is the probability that an offspring will have genotype Aa and show the dominant phenotype for gene B?
A.
1/4
B.
3/8
C.
1/2
D.
5/8
Correct Answer:
3/8
Explanation:
The probability of Aa from Aa × aa is 1/2. The probability of the dominant B phenotype from Bb × Bb is 3/4. Since the loci assort independently, the joint probability is (1/2) × (3/4) = 3/8.
9. Four offspring are produced from the cross AaBb × AaBb. What is the probability that all four will have genotype Aabb?
A.
1/256
B.
1/512
C.
1/1024
D.
1/4096
Correct Answer:
1/4096
Explanation:
For one offspring, the probability of Aabb is (1/2) × (1/4) = 1/8. The probability that four independent offspring all have this genotype is (1/8)^4 = 1/4096.
10. In the cross AaBb × AaBb, what is the probability that an offspring will be heterozygous at both loci or homozygous recessive at both loci?
A.
1/8
B.
5/32
C.
3/16
D.
1/4
Correct Answer:
5/32
Explanation:
The probability of AaBb is (1/2) × (1/2) = 1/4. The probability of aabb is (1/4) × (1/4) = 1/16. These outcomes are mutually exclusive, so their combined probability is 1/4 + 1/16 = 5/16. Therefore, the correct value is 5/16.
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