Cytogenetics Questions and Answers – Monohybrid Cross

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This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Monohybrid Cross”.

1. What is the maximum number of allele that monohybrid cross can consider?
a) 1
b) 2
c) 4
d) 8
View Answer

Answer: b
Explanation: If a gene is heterozygous it will have two different alleles for the same gene, thus in case of monohybrid cross where only a single gene is considered maximum two alleles are under consideration. For dihybrid cross the number will be 4.
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2. The monohybrid phenotypic and genotypic ratio is same in the case of ________________
a) Multiple allele
b) Codominance
c) Incomplete dominance
d) Normal dominance recessive relation
View Answer

Answer: c
Explanation: In the case of incomplete dominance the phenotype of the heterozygote is intermediate to that of the homozygote in both recessive and dominant cases. Then the ratio of both genotype and phenotype is 1:2:1.

3. If a true breeding tall pea plant is crossed with a true breeding short pea plant, what will be the phenotype of the F1 generation?
a) All short
b) All tall
c) 3:1 short: tall
d) 1:3 short: tall
View Answer

Answer: b
Explanation: In this case tall is the dominant trait. Then all the F1 progeny will be heterozygotes where the dominant phenotype tall is expressed.
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4. If an H is for tall trait which is dominant and h is the recessive trait for short, which of the following cross will result in 1:1 tall: short progeny?
a) HH X hh
b) Hh X Hh
c) Hh X hh
d) hh X hh
View Answer

Answer: c
Explanation: The Punett square shows that the cross between Hh and hh produces Hh and hh in the ratio of 1:1. Here Hh will be tall and hh will be short, thus it produces them in 1:1 ratio as well.

5. The cross between a black and white fowl will give _______
a) Gray
b) Andulasian
c) White
d) Black
View Answer

Answer: b
Explanation: Feather colour in fowls show an incomplete dominance relation where the wild types are white and black and heterozygotes generated by their mating are Andulasian.
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6. Which of the following is a test cross?
a) Hh X Hh
b) Hh X HH
c) HH X HH
d) Hh X hh
View Answer

Answer: d
Explanation: While back cross is crossing the progeny with any of the parents, test cross is performed by crossing the F1 progeny with the double recessive parent. The others are examples of back cross except option a.

7. The back cross can distinguish heterozygotes from homozygotes as they give _________ phenotypic ratio respectively.
a) 1:1 dominant: recessive ; all recessive
b) All dominant; all dominant
c) 1:1 recessive; dominant; all dominant
d) All dominant; 1:1 recessive: dominant
View Answer

Answer: c
Explanation: Test cross with a double recessive parent and a homozygous dominant will give all heterozygotes, which will express dominant trait. On the other hand heterozygotes will give 1:1 heterozygote: homozygote which will express dominant: recessive.
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8. The F1 generation is determined by crossing P/P with p/p. Then the progeny obtained from them were intercrossed. What will be the ratio of pure breeding flowers to not pure breeding flowers in F2?
a) 3:1 pure: non- pure
b) 1:3 pure: non- pure
c) 1:1 pure: non- pure
d) 1:2 pure: non- pure
View Answer

Answer: c
Explanation: Both homozygous dominant and homozygous recessive will be pure breeding. Thus the probability of pure: heterozygous will be 1:1.

9. In a cross between wild type pure breeding recessive and heterozygous dominant trait, if 4 progeny are found to be expressing the dominant phenotype, what will be the expected number of recessive?
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: d
Explanation: In this case, the ratio of recessive to dominant will be 1:1, so if there are 4 dominant then around 4 recessive progeny should be produced.
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10. Crossing 2 Andulasian fowls you get 10 white fowl. What will be the expected number of Andulasian fowls as progeny?
a) 10
b) 20
c) 30
d) 40
View Answer

Answer: b
Explanation: Here the parents are heterozygotes so the ratio of the progeny will be 1:2:1 for white: Andulasian: black. If there are 10 white then according to the ratio there should be 20 Andulasian fowls produced. If we consider the number of black fowls that would be 10.

Sanfoundry Global Education & Learning Series – Cytogenetics.

To practice all areas of Cytogenetics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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