Biology Question Bank on “Inheritance of Two Genes-2”.
1. What is the phenotypic ratio of F2 generation in a dihybrid cross?
a) 3:1
b) 1:2:1
c) 9:3:1
d) 9:3:3:1
Answer: d
Clarification: The F2 generation in a dihybrid cross produces all possible phenotypes. The ratio obtained for it is 9:3:3:1.
2. In Mendel’s experiments on garden pea plants, he performed a dihybrid cross of round yellow and green wrinkled seed plants. In the F2 generation, he sampled 1600 plants. Which of the following represents the correct number of plants of each phenotype?
a) 900 round yellow, 300 round green, 300 wrinkled yellow and 100 wrinkled green
b) 900 round green, 300 round yellow, 300 wrinkled yellow and 100 wrinkled green
c) 900 wrinkled green, 300 round yellow, 300 wrinkled yellow and 100 round green
d) 900 round green, 300 wrinkled green, 300 wrinkled yellow and 100 round yellow
Answer: a
Clarification: The F2 progeny of a dihybrid Mendelian cross has a phenotypic ratio of 9:3:3:1. Thus of the 1600 plants sampled, 900 will be round yellow, 300 round green, 300 wrinkled yellow and 100 wrinkled green
3. Which law was proposed by Mendel based on his dihybrid cross studies?
a) Law of Dominance
b) Law of Recessiveness
c) Law of Segregation
d) Law of Independent Assortment
Answer: d
Clarification: The first two laws of Mendel were based on his studies on monohybrid crosses. These were the law of dominance and the law of segregation. The dihybrid cross studies supported them, but also led to the formulation of a third law: the law of independent assortment.
4. What law talks about the segregation of two or more traits independent of one another?
a) Law of Dominance
b) Law of Recessiveness
c) Law of Segregation
d) Law of Independent Assortment
Answer: d
Clarification: The Law of Independent Assortment states the independent segregation of traits that are unrelated. Mendel drew this conclusion based on his studies on garden pea plants.
5. Which of the following is a correct interpretation of the law of independent assortment?
a) Factors exist in pairs
b) Factors segregate such that each gamete gets a single copy
c) For multiple traits under consideration, each segregate independently of one another
d) For multiple traits under consideration, they segregate in a specific pattern
Answer: c
Clarification: The law of independent assortment was the third law of Mendel. It states that when multiple traits are under consideration, each trait segregates independently of the other.
6. If x is the phenotypic ratio of monohybrid cross for trait A and y is the phenotypic ratio of monohybrid cross for trait B, what would be the phenotypic ratio of a dihybrid cross involving traits A and B?
a) xy
b) x + y
c) x2 + y2
d) (xy)2
Answer: a
Clarification: For a dihybrid cross involving two traits, the independent law of assortment states that the traits would segregate independently of each other. Thus, the phenotypic ratio would be xy.
7. The law of independent assortment states that a dihybrid heterozygote would produce four types of gametes.
a) True
b) False
Answer: a
Clarification: The law of independent assortment states that the segregation of traits is independent of one another in a dihybrid cross. Thus, a heterozygote AaBb will produce four types of gametes: AB, Ab, aB, and ab.
8. Which of the following are the correct gametes produced by TtYy?
a) Tt, Yy, TY, ty
b) TY, ty, Tt, Yt
c) Yt, yt, YT, yT
d) YT, yt, yY, tT
Answer: c
Clarification: TtYy is a dihybrid heterozygote. The independent law of assortment applies to it. Thus, it will produce four types of gametes. These will be TY, Ty, tY, and ty.
9. For a dihybrid cross, the number of squares is 16 in a Punnett square of F2 generation.
a) True
b) False
Answer: a
Clarification: For a dihybrid cross, four types of gametes are produced by the heterozygotic generation. This gives rise to 4*4=16 combinations. Hence 16 squares are required in the Punnett square.
10. For a poly hybrid cross concerning N trait, what is the number of squares required in the Punnett square to check for the F2 generation?
a) 2N
b) N2
c) 2(2N)
d) (2N)2
Answer: c
Clarification: For a poly hybrid cross involving N traits, according to the law of independent assortment, 2N types of gametes will be produced by each parent. Therefore, the cross of these progeny will give (2N)*(2N) = 2(2N).
11. For a poly hybrid cross concerning N traits, what is the number of types of gametes produced by the F1 generation?
a) 2N
b) N2
c) 2(2N)
d) (2N)2
Answer: a
Clarification: In a poly hybrid cross, each F1 is a heterozygote. According to the independent law of assortment, each gamete will get one copy of all N factors. Thus, there are 2N types of gametes possible.
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