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Principles of Inheritance and Variation · Question 88

AIPMT 2012 (Prelims)MCQMedium📈 Topic in 50% papers+4−1
1

A normal-visioned man whose father was colour-blind, marries a woman whose father was also colour-blind. They have their first child as a daughter. What are the chances that this child would be colour-blind ?

  1. 50%
  2. 25%
  3. 100%
  4. Zero percent

Answer

(4) Zero percent

Solution

Colour blindness is an X-linked recessive trait.

Man has normal vision ⇒ his genotype is XCY (his father's colour-blind X does not pass to a son).

Woman's father was colour-blind ⇒ she receives Xc from him, so she is a carrier XCXc.

Daughters = XC (from father) ( XC or Xc from mother) = XCXC or XCXc

∴ Every daughter receives the normal X from the father, so no daughter can be colour-blind (half may be carriers).

Answer: (4)

AIPMT 2012 (Prelims)

Keep practising — Principles of Inheritance and Variation

2

F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1. It represents a case of

  1. Monohybrid cross with incomplete dominance
  2. Co-dominance
  3. Monohybrid cross with complete dominance
  4. Dihybrid cross
AIPMT 2012 (Prelims)
3

Which one of the following conditions of the zygotic cell would lead to the birth of a normal human female child?

  1. only one Y chromosome
  2. only one X chromosome
  3. one X and one Y chromosome
  4. two X chromosomes
AIPMT 2011 (Mains)
4

Test cross in plants or in Drosophila involves crossing:

  1. the F1 hybrid with a double recessive genotype
  2. between two genotypes with dominant trait
  3. between two F1 hybrids
  4. between two genotypes with recessive trait
AIPMT 2011 (Mains)
5

Which one of the following conditions correctly describes the manner of determining the sex ?

  1. XO condition in humans as found in Turner Syndrome, determines female sex
  2. XO type of sex chromosomes determine male sex in grasshopper
  3. Homozygous sex chromosomes (ZZ) determine female sex in Birds
  4. Homozygous sex chromosomes (XX) produce male in Drosophila
AIPMT 2011 (Prelims)
6

Mutations can be induced with :

  1. Gamma radiation
  2. Infra Red radiations
  3. IAA
  4. Ethylene
AIPMT 2011 (Prelims)
7

ABO blood grouping is controlled by gene I which has three alleles and shows co-dominance. There are six genotypes. How many phenotypes in all are possible -

  1. five
  2. four
  3. three
  4. six
AIPMT 2010 (Mains)
8

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
AIPMT 2010 (Mains)
9

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)
10

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)
11

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)

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