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MCQ Practice Test & Solutions: Test: Heredity And Variation 4 - From Past 28 Years Questions (25 Questions)

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Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 25 minutes
  • - Number of Questions: 25

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Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 1

A and B genes are linked what shall be genotype of progeny in a cross between AB/ab and ab/ab?

[2001]

Detailed Solution: Question 1

Parents : AB/ab × ab/ab

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 2

In humans, male XXY and female XXXX occur due to

[2001]

Detailed Solution: Question 2

Aneuploidy is numerical change in the chromosome number in the genome. Euploidy is the phenomenon of having multiple or more than twice the number of genome.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 3

No. of barr body present in each somatic cell of a female is

[2001]

Detailed Solution: Question 3

One bar body is present in each somatic cell of female (XX).

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 4

Which of the following is correct match?

[2002]

Detailed Solution: Question 4

Sickle cell anaemia is a genetic disorder caused by a defective allele on chromosome 11. Hemophilia is a sex linked recessive disorder. Down's syndrome is caused by trisomy of the 21st chromosome.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 5

Which of the following is an example of sex linked disease?

[2002]

Detailed Solution: Question 5

AIDS is a viral disease. Syphilis and gonorrhoea are both sexually transmitted diseases. Colour blindness is a sex linked disease.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 6

Which of the following is an example of pleiotropic effect?

[2002]

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 7

A gene is said to be dominant if

[2002]

Detailed Solution: Question 7

A dominant gene would lead to the expression of its phenotype irrespective of the fact whether its allelic gene is dominant or recessive.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 8

A diseased man marries a normal woman. They have three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is

[2002]

Detailed Solution: Question 8

Sex linked disorders follow criss-cross inheritance pattern that affected father passes it on to their daughters. The daughters receive one of their X - chromosome from their fathers.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 9

On selfing a plant of F1 generation with genotype “AABbCC”,the genotypic ratio in F2 generation will be

[2002]

Detailed Solution: Question 9

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 10

There are three genes a, b, c. Percentage of crossing over between a and b is 20%, b and c is 28% an a and c is 8%. What is the sequence of genes on chromosome?

[2002]

Detailed Solution: Question 10

a − b = 20 %, b − c = 28 %, a − c = 8%

Percent crossing over between two genes is proportional to the distance between them. 

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 11

In Drosophila, the sex is determined by

[2003]

Detailed Solution: Question 11

Calvin Bridges demonstrated that in Drosophila, the sex is determined by ratio of the number of X chromosomes to the sets of autosomes.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 12

Genes for cytoplasmic male sterility in plants are generally located in

[2003, 05]

Detailed Solution: Question 12

Male sterility in plants can be controlled by nuclear genes or cytoplasm or by both. Three different mechanisms for control of male sterility in plants are :
(i) Genetic male sterility;
(ii) Cytoplasmic male sterility;
(iii) Cytoplasmic genetic male sterility. In Maize the genes for cytoplasmic control of male sterility are located in cytoplasm.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 13

Nicotiana sylvestris flowers only during long days and  N. tabacum flowers only during short days. If raised in the laboratory under different photoperiods, they can be induced to flower at the same time and can be cross fertilized  to produce self – fertile offspring. What is the best reason for considering N. sylvestris and N. tabacum to be separate species ?

[2003]

Detailed Solution: Question 13

Correct option is

C

They cannot interbreed in nature.

The term species evolves from word specific means only those group of plants or animals come under one species which can interbreed and produce fertile offsprings naturally. N. tobacum and N. sylvestris cannot produce flowers naturally have different photoperiods, hence they cannot belong to the same species. Thus the correct answer is 'They can not interbreed in nature.'

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 14

Which one of the following traits of garden pea studied by Mendel was a recessive feature ?

[2003]

Detailed Solution: Question 14

Yellow seed colour is dominant over green seed colour.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 15

The genes controlling the seven pea characters studied by Mendel are now known to be located on how many different chromosomes ?

[2003]

Detailed Solution: Question 15

The seventraits are now know to be present on 4 chromosome. But they do not show linkage, because of large distances between them on the chromosome.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 16

Two  crosses between the same pair of genotypes or phenotypes in which the source of the gametes are reversed in one cross, is known as

[2003]

Detailed Solution: Question 16

Dihybrid cross is made between individuals having contrasting traits in order to study the inheritance of two pairs of alleles. Test cross is a back cross between individual with a dominant trait and its recessive parent to know whether the individual is homozygous or heterozygous for the trait.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 17

The linkage map of X-chromosome of fruit fly has 66 units, with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) should be

[2003]

Detailed Solution: Question 17

Mapping of genes on chromosomes is based on the assumption that genes are arranged on the chromosome and that the percentage of crossing over (recombination frequency) between two genes is an index of their distance apart. Distances between genes is expressed as map units, where, one map unit 

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 18

Christmas disease is another name for

[2003]

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 19

Pattern baldness, moustaches and beard in human males are examples of

[2003]

Detailed Solution: Question 19

  Traits or characters which develop only in one sex are called sex-limited characters. They are produced and controlled by the genes which may be located on autosomes in only one sex, eg. Moustaches and beards in human males, breast in human females, milk secretion in human females.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 20

Down's syndrome is caused by an extra copy of chromosome number 21. What percentage of offspring produced by an affected mother and a normal father would be affected by this disorder?

[2003]

Detailed Solution: Question 20

50% of ova will have (n + 1) chromosome which would, on fertilisation, yield abnormal zygotes (n + 1) + (n) = 2n + 1.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 21

Lack of independent assortment of two genes A and B in fruit fly Drosophila is due to

[2004]

Detailed Solution: Question 21

The lack of inde pendent assor tment in sweet pea and Drosophila is due to linkage. 

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 22

The recessive genes located on X-chromosome in humans are always

[2004]

Detailed Solution: Question 22

Males have only one X - chromosome. Hence any gene present on the lone X - chromosome expresses itself  in males. Females have two X- chromosomes . The mutant  allele must be present on both the X-Chromosomes  to express itself phenotypically. If only one copy of allele present, then the female only becomes a carrier showing no external manifestation of the gene. Sublethal condition can never arise in males.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 23

A male human is heterozygous for autosomal genes A and B and is also hemizygous for hemophilic gene h. What proportion of his sperms will be abh?

[2004]

Detailed Solution: Question 23

Chance of getting a = 1/2
Chance of getting b = 1/2
Chance of getting h = 1/2
Chance of getting sperms with 

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 24

A normal woman, whose father was colourblind is married to a normal man. The sons would be

[2004]

Detailed Solution: Question 24

Normal woman with colour blind father would be a carrier

50% of the sons would be colour blind.

Test: Heredity And Variation 4 - From Past 28 Years Questions - Question 25

One of the parents of a cross has a mutation in its mitochondria. In that cross, that parent is taken as a male. During segregation of F2 progenies that mutation is found in

[2004]

Detailed Solution: Question 25

Mitochondria is an organelle present in the cytoplasm. A zygote receives its cytoplasm from the female parent gamete. Hence in the given question, the F2 progenies do not receive the mitochondrial genome from the male parent and mutation is not passed to progenies.

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