Chapter Exam

Chapter 2: Classical Genetics — Online MCQ Test

Q: What is the expected phenotypic ratio of a standard monohybrid test cross in F2 generation?

Answer: B. 1:1 — A monohybrid test cross involves crossing an F1 heterozygote (Tt) with a homozygous recessive parent (tt), resulting in a phenotypic ratio of 1:1.

Q: Carl Correns discovered the phenomenon of incomplete dominance using which of the following plants?

Answer: B. Mirabilis jalapa — Carl Correns studied incomplete dominance using Mirabilis jalapa (Four o'clock plant), where crossing homozygous red and white flowers produces pink-flowered F1 heterozygotes.

Q: In Lablab purpureus, the gene interaction that alters the standard Mendelian dihybrid ratio to 9:3:4 is an example of:

Answer: B. Recessive epistasis — Recessive epistasis (like in Lablab seed coat color) occurs when the recessive homozygote at one locus suppresses the phenotypic expression of genes at another locus, resulting in a 9:3:4 ratio.

Q: An organism with the genotype AaBbCc is self-pollinated. Assuming all three genes assort independently, what is the probability of obtaining offspring with the completely homozygous recessive genotype (aabbcc)?

Answer: B. 1/64 — For each heterozygous locus, the chance of a homozygous recessive offspring is 1/4. For three independent loci, the combined probability is (1/4) * (1/4) * (1/4) = 1/64.

Q: In Nicotiana, self-incompatibility is governed by multiple alleles (S1, S2, S3, etc.). If a female plant of genotype S1S2 is pollinated by a male plant of genotype S2S3, which of the following represents the genotypes of the viable progeny?

Answer: B. S1S3 and S2S3 — Pollen grains carrying an allele present in the diploid pistil cannot germinate. Thus, S2 pollen from the male is rejected by the S1S2 pistil, while S3 pollen is compatible and fertilizes S1 and S2 eggs, producing S1S3 and S2S3 plants.

BOTANY · CLASS 12th · Tamil Nadu State Board

Practice Chapter 2: Classical Genetics with a free chapter-wise online MCQ test for Tamil Nadu State Board CLASS 12th BOTANY. This chapter covers: Exploring Mendelian genetics this chapter details monohybrid dihybrid and trihybrid inheritance patterns. Students examine Mendelian laws gene interactions like epistasis and compl.... AI-generated questions from basic to board-exam level, with instant results and explanations.

Questions

20m

Time Limit

Attempts Left

10 random questions from this chapter (mixed difficulty)
Questions you've seen before won't repeat until the pool resets
You have 20 minutes — exam auto-submits when time is up
Maximum 3 attempts per chapter
Results and explanations shown immediately after submission

New here? Register free — includes 3 free chapter exams.

Chapter 2: Classical Genetics — Important Questions & Answers (FAQ)

Frequently asked questions from Tamil Nadu State Board CLASS 12th BOTANY — Chapter 2: Classical Genetics, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is the expected phenotypic ratio of a standard monohybrid test cross in F2 generation?

A. 3:1
B. 1:1 ✓
C. 9:3:3:1
D. 1:2:1

Answer: B. 1:1
A monohybrid test cross involves crossing an F1 heterozygote (Tt) with a homozygous recessive parent (tt), resulting in a phenotypic ratio of 1:1.

Carl Correns discovered the phenomenon of incomplete dominance using which of the following plants?

A. Lathyrus odoratus
B. Mirabilis jalapa ✓
C. Antirrhinum majus
D. Pisum sativum

Answer: B. Mirabilis jalapa
Carl Correns studied incomplete dominance using Mirabilis jalapa (Four o'clock plant), where crossing homozygous red and white flowers produces pink-flowered F1 heterozygotes.

In Lablab purpureus, the gene interaction that alters the standard Mendelian dihybrid ratio to 9:3:4 is an example of:

A. Dominant epistasis
B. Recessive epistasis ✓
C. Complementary genes
D. Duplicate genes

Answer: B. Recessive epistasis
Recessive epistasis (like in Lablab seed coat color) occurs when the recessive homozygote at one locus suppresses the phenotypic expression of genes at another locus, resulting in a 9:3:4 ratio.

An organism with the genotype AaBbCc is self-pollinated. Assuming all three genes assort independently, what is the probability of obtaining offspring with the completely homozygous recessive genotype (aabbcc)?

A. 1/16
B. 1/64 ✓
C. 1/8
D. 27/64

Answer: B. 1/64
For each heterozygous locus, the chance of a homozygous recessive offspring is 1/4. For three independent loci, the combined probability is (1/4) * (1/4) * (1/4) = 1/64.

In Nicotiana, self-incompatibility is governed by multiple alleles (S1, S2, S3, etc.). If a female plant of genotype S1S2 is pollinated by a male plant of genotype S2S3, which of the following represents the genotypes of the viable progeny?

A. S1S2 and S2S3
B. S1S3 and S2S3 ✓
C. S1S2, S1S3, S2S3, S2S2
D. S3S3 only

Answer: B. S1S3 and S2S3
Pollen grains carrying an allele present in the diploid pistil cannot germinate. Thus, S2 pollen from the male is rejected by the S1S2 pistil, while S3 pollen is compatible and fertilizes S1 and S2 eggs, producing S1S3 and S2S3 plants.