31 Years NEET Previous Year Questions: Principles of Inheritance & Variation - 1


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QUESTION: 1

The production of gametes by the parents, formation of zygotes, the F1 and F2 plants, can be understood from a diagram called:   [2021]

Solution:

The production of gametes by the parents, the formation of the zygotes, the F1 and F2 plants can be understood from a diagram called Punnett Square. It helps in identifying the possible allelic combinations.


Fig: Punnett Square

QUESTION: 2

Experimental verification of the chromosomal theory of inheritance was done by:   [2020]

Solution:

Thomas Hunt Morgan, who studied fruit flies, provided the first strong confirmation of the chromosome theory. “Morgan discovered a mutation that affected fly eye color. The chromosome theory of inheritance states that genes are found at specific locations on chromosomes, and that the behavior of chromosomes during meiosis can explain Mendel's laws of inheritance.

QUESTION: 3

The frequency of recombination between gene present on the same chromosome as a measure of the distance between genes was explained by    [2019]

Solution:

Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes.

QUESTION: 4

A cell at telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase stage. There is no formation of cell plate and thus the cell is containing more number of chromosomes as compared to other dividing cells. This would result in:         [2016]

Solution:

This phenomenon is known as Polyploidy, wherein the cells contain more than two paired (homologous) sets of chromosomes, which is often seen in the case of plants.

The major cause of polyploidy is the non - disjunction of sister chromatids during meiotic recombination.


Fig: Polyploidy can make fruits bigger in size

QUESTION: 5

In a testcross involving F1 dihybrid flies, more parental-type offspring were produced than the recombinant-type offspring. This indicates ______.       [2016]

Solution:

When two genes in a dihybrid cross are situated on the same chromosome, the proportion of parental gene combinations are much higher than the non-parental or recombinant type. This is also called incomplete linkage.

QUESTION: 6

Pick out the correct statements :        [2016]

(a) Haemophilia is a sex-linked recessive disease.
(b) Down's syndrome is due to aneuploidy.
(c) Phenylketonuria is an autosomal recessive gene disorder.
(d) Sickle cell anaemia is a X-linked recessive gene disorder.

Solution:

Sickle cell anemia is not an X-linked recessive gene disorder. It is a point mutation caused by a point change in the sixth nitrogenous base pair.

Statements (a), (b) and (c)  are correct.

Hence, option D is correct.

QUESTION: 7

Match the terms in Column-I with their description in Column-II and choose the correct option : 
                  [2016] 

Solution:
  • Dominance means in a heterozygous pair of alleles, one allele suppresses the influence of another and dominates the trait.
    Example: Tt is tall because of the dominance of the T allele.
  • Codominance means when in a heterozygous pair both the alleles are equally dominant.
    Example: Blood group AB.
  • Pleiotropy means when a gene influences more than one character.
    Example: The appearance of a vestigial wing and eye colour in a fruit fly.
  • Polygenic inheritance is defined as quantitative inheritance, where multiple independent genes have an additive or similar effect on a single quantitative trait.

Hence, Option B is correct.

QUESTION: 8

A tall true breeding garden pea plant is crossed with a dwarf true breeding garden pea plant. When the F1 plants were selfed the resulting genotypes were in the ratio of:                [2016]

Solution:


Phenotypic ratio : 3: 1 (Tall :Dwarf) Genotypic Ratio: 1:2:1 (Homozygous Tall : Heterozygous Tall : Dwarf) 

QUESTION: 9

Which of the following most appropriately describes haemophilia ?                      [2016]

Solution:

Hemophilia A and hemophilia B are inherited in an X-linked recessive pattern. The genes associated with these conditions are located on the X chromosome, which is one of the two sex chromosomes.

  • In males (who have only one X chromosome), one altered copy of the gene in each cell is sufficient to cause the condition. A characteristic of X-linked inheritance is that fathers cannot pass X-linked traits to their sons.
  • In females (who have two X chromosomes), a mutation would have to occur in both copies of the gene to cause the disorder. Because it is unlikely that females will have two altered copies of this gene, it is very rare for females to have hemophilia.
QUESTION: 10

In the following human pedigree, the filled symbols represent the affected individuals. Identify the type of given pedigree.    [2015 RS]

Solution:

The given pedigree shows the autosomal recessive disorder. In this disorder, the individual inherits two mutated genes, one from each parent. This disorder is usually passed on by two earners.

The carriers have a 25% chance of having an unaffected child with normal genes, 50% chance of having an unaffected child who also is a carrier and a 25% chance of having an affected child with recessive genes.

QUESTION: 11

In his classic experiments on pea plants, Mendel did not use:                [2015 RS]

Solution:

Mendel used seven pea plant traits in his hybridization experiments:
(i) Flower colour (purple or white)
(ii) Flower position (axil or terminal)
(iii) Stem length (long or short)
(iv) Seed shape (round or wrinkled)
(v) Seed colour (yellow or green)
(vi) Pod shape (inflated or constricted)
(vii) Pod colour (yellow or green). 

Hence, Pod length is not included.

QUESTION: 12

 Alleles are :                             [2015 RS]

Solution:

Alleles are defined as alternative form of same gene. Humans are called diploid organisms because they have two alleles at each genetic locus, with one allele inherited from each parent.

Fig: Contrasting alleles for eye color on the same gene

QUESTION: 13

Which is the most common mechanism of genetic variation in the population of sexually reproducing organism?         [2015 RS]

Solution:
  • The most common cause of variations is recombination in the organism which are reproduced sexually. 
  • During meiosis in sexual organisms, two homologous chromosomes cross over and exchange genetic material. 
  • It is a process by which pieces of DNA are broken and recombined to produce new combinations of alleles.
  • This recombination process creates genetic diversity at the level of genes that reflects differences in the DNA sequences of different organisms.


Fig: Recombination in Homologous chromosomes

QUESTION: 14

How many pairs of contrasting characters in pea plants were studied by Mendel in his experiments ?         [2015 RS]

Solution:

Seven pairs of contrasting characters were selected in pea plant and studied by Mendel in his experiment. 

QUESTION: 15

 Fruit colour in squash is an example of: [2014]

Solution:
  • Epistasis is the phenomenon of suppression of phenotypic expression of a gene by a non-allelic gene which shows its own effect.
  • A dominant epistatic allele suppresses the expression of a non-allelic gene whether the latter is dominant or recessive.

Example: fruit colour of Summer Squash (Cucurbita pepo) is governed by a gene that produces yellow colour in the dominant state (Y-) and green colour in the recessive state (yy). 


Fig: Epistasis in Squash Fruit

QUESTION: 16

In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this data, the frequency of allele A in the population is:     [2014]

Solution:

According to Hardy-Weinberg equilibrium,

p2 + 2pq + q2  = 1

where p2 is the frequency of homozygous genotype AA
q2 is the frequency of genotype aa
pq is the frequency of genotype Aa

There are 1000 individuals, out of which 360 belong to genotype AA.

p2  = 360/1000 = 0.36

Then p = 0.6.

So, the correct answer is '0.6'.

QUESTION: 17

A man whose father was colour blind marries a woman who had a colour blind mother and normal father. What percentage of male children of this couple will be colour blind?      [2013]

Solution:

lt is given that the man had colour blind father, i.e., man's genotype would be XY. Now, the woman had a colourblind mother and normal father, thus her genotype would be XCX. A cross between them can be represented as below. Therefore, 50% of male children of this couple will be colourblind.

QUESTION: 18

If two pea plants having red (dominant) coloured flowers with unknown genotypes are crossed, 75% of the flowers are red and 25% are white. The genotypic constitution of the parents having red coloured flowers will be:  [2013]

Solution:

A cross is made between two pea plants in which one shows Red dominant flowers and the other genotype is not known. In the next generation, 75% of plants show red flowers and 25% of progeny are white flowers, i.e. the phenotype of parents is Rr and Rr.
Thus, both are heterozygous.

If two pea plants having red (dominant) coloured flowers wit
Fig: Cross between two heterozygous red dominant flowers

QUESTION: 19

A colour blind man marries a woman with normal sight who has no history of colour blindness in her family. What is the probability of their grandson being colour blind?   [2012]

Solution:
  • Colourblindness is an X-linked recessive disorder.
  • A man requires only one copy of the defective gene to be colourblind. Whereas a woman requires two copies of defective genes to be colourblind.
  • When a colourblind (XcY) man marries a normal woman (XX), all of their sons will be normal (XY) and all of their daughters will be carriers (XcX).
  • Hence, in the next generation, all the children of all of their sons will be normal. There will be a nil probability of being colourblind.
  • In the next generation of their daughter, all the daughters will be either normal or carrier. Whereas 50% of the sons will be colourblind acquiring defective gene (Xc) from their mother.
  • Hence, considering the grandsons of both son and daughter, there are 25% chances or 0.25 probability of grandson being colourblind.

A colourblind man marries a woman with normal sight who has no history of colour  blindness in her family. - Sarthaks eConnect | Largest Online Education  Community
Fig: Cross b/w normal woman and colorblind man

QUESTION: 20

Conditions of a karyotype 2n +1, 2n –1 and  2n + 2, 2n – 2 are called:   [2012]

Solution:

Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy. Aneuploidy arises due to non-disjunction of homologous chromosome.
Aneuploidy is of four types:

  • Monosomy = 2n – 1
  • Nullisomy = 2n – 2
  • Trisomy = 2n + 1
  • Tetrasomy = 2n + 2
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