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Principles of Inheritance and Variation · Question 95

AIPMT 2010 (Mains)MCQMedium📈 Topic in 50% papers+4−1
1

Study the pedigree chart of a certain family given below and select the correct conclusion which can be drawn for the character – [figure]

  1. The trait under study could not be colourblindness
  2. The female parent is heterozygous
  3. The parents could not have had a normal daughter for this character
  4. The male parent is homozygous dominant

Answer

(2) The female parent is heterozygous

Solution

Both parents are affected, but one son is normal.

⇒ Two affected parents have produced an unaffected child, so the trait is dominant and the normal son carries two recessive alleles (aa).

∴ Each parent must have given one recessive allele, i.e., both parents are heterozygous (Aa). Hence the female parent is heterozygous.

∵ A homozygous dominant (AA) male parent would give only affected children, option (4) is wrong.

∵ Aa × Aa gives 1/4 normal (aa) children of either sex, a normal daughter is also possible, so option (3) is wrong.

Answer: (2)

AIPMT 2010 (Mains)

Keep practising — Principles of Inheritance and Variation

2

A cross in which an organism showing a dominant phenotype is crossed with the recessive parent in order to know its genotype is called -

  1. Dihybrid cross
  2. Back cross
  3. Monohybrid cross
  4. Test cross
AIPMT 2010 (Mains)
3

In Antirrhinum two plants with pink flowers were hybridized. The F1 plants produced red, pink and white flowers in the proportion of 1 red, 2 pink and 1 white. What could be the genotype of the two plants used for hybridization? Red flower colour is determined by RR, and White by rr genes -

  1. rrrr
  2. rr
  3. Rr
  4. RR
AIPMT 2010 (Mains)
4

Which one of the following symbols and its representation, used in human pedigree analysis is correct –

  1. [figure] = mating between relatives
  2. [figure] = unaffected female
  3. [figure] = unaffected male
  4. [figure] = male affected
AIPMT 2010 (Prelims)
5

ABO blood groups in humans are controlled by the gene I. It has three alleles – IA, IB and i. Since there are three different alleles, six different genotypes are possible. How many phenotypes can occur –

  1. One
  2. Two
  3. Three
  4. Four
AIPMT 2010 (Prelims)
6

Select the correct statement from the ones given below with respect to dihybrid cross –

  1. Genes far apart on the same chromosome show very few recombinations
  2. Tightly linked genes on the same chromosome show very few recombinations
  3. Tightly linked genes on the same chromosome show higher recombinations
  4. Genes loosely linked on the same chromosome show similar recombinations as the tightly linked ones
AIPMT 2010 (Prelims)
7

The genotype of a plant showing the dominant phenotype can be determined by –

  1. Pedigree analysis
  2. Back cross
  3. Test cross
  4. Dihybrid cross
AIPMT 2010 (Prelims)
8

Which one of the following cannot be explained on the basis of Mendel's Law of Dominance?

  1. Out of one pair of factors one is dominant, and the other recessive
  2. Alleles do not show any blending and both the characters recover as such in F2 generation
  3. The discrete unit controlling a particular character is called a factor
  4. Factors occur in pairs
AIPMT 2010 (Prelims)
9

Select the incorrect statement from the following:

  1. Baldness is a sex limited trait
  2. Galactosemia is an inborn error of metabolism
  3. Linkage is an exception to the principle of independent assortment in heredity
  4. Small population size results in random genetic drift in a population
AIPMT 2009
10

Sickle-cell anaemia is:

  1. Caused by substitution of valine by glutamic acid in the beta globin chain of haemoglobin
  2. An autosomal linked dominant trait
  3. Caused by a change in a single base pair of DNA
  4. Characterised by elongated sickle like RBCs with a nucleus
AIPMT 2009
11

The most popularly known blood grouping is the ABO grouping. It is named ABO and not ABC, because "O" in it refers to having:

  1. Other antigens besides A and B on RBCs
  2. No antigens A and B on RBCs
  3. Overdominance of this type on the genes for A and B types
  4. One antibody only – either anti-A or anti-B on the RBCs
AIPMT 2009

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