Biology

Genetics and Plant Breeding

1,240 Questions

Genetics and Plant Breeding explores the principles of heredity, chromosomal inheritance, and hybridization techniques. It includes key concepts like Mendelian genetics, gene linkage, and polygenic inheritance. This topic is essential for students tackling advanced biology or botany sections in competitive examinations.

Mendelian InheritanceChromosome TheoryGene LinkagePolygenic InheritancePlant HybridizationPopulation Genetics Equilibrium

Genetics and Plant Breeding Questions

Multiple choice
  1. 0.24

  2. 0.16

  3. 0.48

  4. 0.36

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

  The genotype proportions p2, 2pq, and q2 are called the Hardy-Weinberg proportions.Two alleles denoted A and a with frequencies f(A) = p and f(a) = q, respectively,Expected genotype frequencies are-  f(AA) = p2 for the AA homozygotes,  f(aa) = q2 for the aa homozygotes, and f(Aa) = 2pq for the heterozygotes. So, the frequency of heterozygotes = 2pq= 2 X 0.6 X 0.4 = 0.48 

Multiple choice
  1. AaBb

  2. aabb

  3. AABB

  4. aaBB

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 In order to find out the different types of gametes produced by a pea plant having the genotype AaBb, it should be crossed to a plant with the homozygous recessive genotype.It is called Test Cross.So, in above options, it should be crossed with aabb double recessive plant.

Multiple choice
  1. must have normal colour vision.

  2. may be colour blind or may be of normal vision.

  3. will be partially colour blind since he is heterozygous for the colour blind mutant allele.

  4. must be colour blind.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of this couple was a boy. This boy may be colour blind or may be of normal vision.

Multiple choice
  1. He is either homozygous dominant (DD) or heterozygous (Dd) for this trait.

  2. He will be homozygous dominant (DD).

  3. He will be homozygous recessive (dd).

  4. He may be either homozygous recessive (dd) or heterozygous (Dd) for this trait.

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Rh blood factor is a dominant trait. Therefore, both homozygous dominant (DD) and heterozygous (Dd) people have Rh antigens on the surface of their red cells, which make them Rh positive.

Multiple choice
  1. genome imprinting

  2. intermediate expression

  3. pleiotropy

  4. codominance

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This hypothetical example would be an example of the intermediate expression of a trait. Apparently, this is rare. For most genes, it is expected that individuals with heterozygous genotypes will have phenotypes like homozygous dominant individuals.

Multiple choice
  1. The trait has no genetic component.

  2. The trait is completely determined by genetics.

  3. Genes contribute to the trait, but are not deterministic.

  4. The trait must be due to the biology of twinning.

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

There is an increased twin concordance, suggestive of a genetic contribution, but since twins are not fully concordant, other non-genetic factors must also be involved.

Multiple choice
  1. The child will be a carrier.

  2. The child will be affected if recombination takes place in the father only.

  3. The child will be homozygous unaffected if recombination takes place in the mother.

  4. All of the above

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Yes, all the statements are correct.

Multiple choice
  1. ½ sons and ½ daughters

  2. all daughters and no sons

  3. all sons and no daughters

  4. ¼ daughters and ¼ sons

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an X-linked dominant disorder: affected female (X^A X^a) × normal male (X^a Y). Each child gets one X chromosome from the mother (either X^A or X^a, 50% probability) and either X^a or Y from the father. Daughters: 50% will be X^A X^a (affected), 50% will be X^a X^a (normal). Sons: 50% will be X^A Y (affected), 50% will be X^a Y (normal). Therefore, 50% of sons and 50% of daughters will manifest the disease. This matches option A. Option B would describe X-linked dominant inheritance from an affected father, while option C would describe Y-linked inheritance (which doesn't apply here).