A man and a woman, who do not show any apparent signs of a certai

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 Multiple Choice QuestionsMultiple Choice Questions

241.

The trisomy for 21st chromosome is called

  • Down's syndrome

  • Turner's syndrome

  • Sickle cell anaemia

  • Klinfelter's syndrome


242.

Chromosome complement with 2n - 1 is called

  • monosomy

  • nullisomy

  • trisomy

  • tetrasomy


243.

L- shaped chromosomes are also called

  • acrocentric

  • telocentric

  • sub- metacentric

  • None of these


244.

If root of a flowering plant has 24 chromosomes, then its gamete has how many chromosomes?

  • 24

  • 12

  • 4

  • 8


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245.

Balbiani rings are the structural features of

  • allosomes

  • polytene chromosomes

  • autosomes

  • Lampbrush chromosomes


246.

X- linked recessive gene is

  • always expressed in male

  • always expressed in femlae

  • lethal

  • sub- lethal


247.

Which of the following is a test cross?

  • Tt ×tt

  • TT ×tt

  • Tt ×Tt

  • tt ×tt


248.

1 : 2 : 1 phenotypic and genotypic ratio is found in

  • complementary genes

  • blending inheritance

  • multiple alleles

  • pseudo alleles


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249.

A man and a woman, who do not show any apparent signs of a certain inherited disease, have seven children (2 daughters and 5 sons). Three of the sons suffer from the given disease but none of the daughters are affected. Which of the following mode of inheritance do you suggest for this disease?

  • Autosomal dominant

  • Sex-linked dominant

  • Sex-limited recessive

  • Sex-linked recessive


D.

Sex-linked recessive

Sex linked characteristic is determined by the genes located on the sex chromosome ie X or Y. Sex limited characteristic is the condition where expression of sex linked genes is limited to one sex.

In the given problem, neither man nor woman are deceased by the disease. Therefore, women must be the carrier of the disease causing gene. Also, none of the daughters are suffereing from the disease except the sons, ie, daughters are also carrier.

Suppose, XY = genotype of man

XdX = genotype of woman (d refers to disease causing gene)

 

The probability for each combination is 25%. Therefore, among seven children, 2 are normal daughter, 3 diseased sons and 2 normal sons are possible.


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250.

A haemophilic woman marries a normal man then

  • all the children will be normal

  • all the sons will be haemophilic

  • all the girls will be haemophilic

  • half girls will be haemophilic


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