Principle of Inheritance and Variation - NEET Questions

Mendel’s laws and other concepts of genetics and inheritance are significant in terms of NEET exam perspective. Here are some important questions from the Chapter: Principles of Inheritance and Variation. It is curated based on Biology NEET syllabus.

1. How many pairs of contrasting traits were studied by Mendel in the pea plant?

1. 6
2. 7
3. 8
4. 10

Answer : b. 7

Discussion: Mendel selected seven characters. Each character had a pair of contrasting traits. The 7 pairs of contrasting traits are mentioned in the below diagram.

2. In hybridisation experiments conducted by Mendel, what is the correct sequence of steps he followed?

1. Bagging
2. Pollination
3. Emasculation
4. Rebagging
5. Selection of parents
6. Collection of seeds

1. 1, 3, 4, 2, 6, 5
2. 5, 3, 1, 2, 4, 6
3. 2, 3, 5, 6, 4, 1
4. 1, 2, 3, 6, 5, 4

Answer: b. 5, 3, 1, 2, 4, 6

Discussion: The correct sequence of steps followed by Mendel is –

1. Selection of parents with the desired character.
2. Emasculation – removal of anthers from a bisexual flower which was selected as a female parent to prevent self-pollination.
3. Bagging – covering the emasculated flowers with a bag to prevent contamination with unwanted pollen.
4. Cross-pollination – transfer of desired pollen to the stigma of the emasculated flower.
5. Rebagging – again covering the emasculated flowers to avoid contamination.
6. Finally, collection of seeds.

3. Mendel proposed which of the following terms for hereditary units?

1. Factor
2. Allele
3. Gene
4. Chromosome

Answer: a. Factor

Discussion: Based on hybridisation experiments conducted by Mendel, he proposed that something is being passed on from one generation to the next unchanged via gametes. He termed them ‘factors’. Factors are currently called genes and alleles are the alternative forms of a gene.

Genes are units of inheritance and they contain the information required to express a particular trait.

4. 1:1 phenotypic ratio is obtained in the offspring when ⸻ for the character considered in the cross.

1. Both parents carry dissimilar alleles.
2. One parent carries both the dominant alleles and the other parent carries both the recessive alleles.
3. One parent carries both the dominant alleles and the other parent carries dissimilar alleles.
4. One parent carries both the recessive alleles and the other parent carries dissimilar alleles.

Answer: d. One parent carries both the recessive alleles and the other parent carries dissimilar alleles.

Discussion: A cross between a homozygous recessive parent (tt) and a dominant F1 phenotype (Tt/TT) is called a test cross. The test determines whether the plant is homozygous or heterozygous dominant.

If a 1:1 phenotypic ratio is obtained in the offspring, then one parent carries both the recessive alleles (tt) and the other parent carries dissimilar alleles (Tt) for the character considered in the cross (test cross).

Also Check: MCQs on Mendel’s Laws of Inheritance

5. The allele which gets expressed in both heterozygous and homozygous conditions is known as the ⸻ allele.

1. Recessive
2. Dominant
3. Dependent
4. Independent

Answer: b. Dominant

Discussion: Alleles are of two types – dominant and recessive alleles. Dominant alleles get expressed in both heterozygous and homozygous conditions. Whereas recessive alleles are expressed only in homozygous conditions.

6. How can you find out the genotype of a pea plant that produces violet flowers?

1. By crossing it with a true-breeding plant for violet flowers.
2. By crossing it with a plant showing a dominant phenotype for flower colour.
3. By crossing it with a plant showing recessive phenotype for flower colour.
4. By performing a reciprocal cross.

Answer: c. By crossing it with a plant showing recessive phenotype for flower colour

Discussion: In a test cross, an organism showing a dominant phenotype for a particular character is crossed with the plant showing a recessive phenotype for the same character. Hence, the plant producing violet flowers (WW, Ww) is crossed with a plant producing white flowers (ww) to find out the genotype of violet flowered plant.

7. A pea plant producing round seeds is crossed with another pea plant producing wrinkled seeds. In the F1 generation, 50% of the offspring produced wrinkled seeds. Predict the genotype of the parents.

1. RR ✕ rr
2. Rr ✕ Rr
3. RR ✕ Rr
4. Rr ✕ rr

Answer: Rr ✕ rr

Discussion: Consider the cross, Rr ✕ rr. 50% of the offspring will be having round seeds (Rr) while 50% of the offspring will be having wrinkled seeds (rr).

8. The total number of progenies obtained through the dihybrid cross of Mendel is 1280 in F2 generation. How many of these are recombinants?

1. 240
2. 360
3. 480
4. 720

Answer: c. 480

Discussion: In Mendel’s dihybrid cross, the phenotypic ratio in F2 generation is 9:3:3:1. In this ratio, out of 16 progenies –

• 10 progenies (9+1) show parental type
• 6 progenies (3+3) show recombinant type

Therefore, the required number of recombinants out of 1280 progenies = 6×1280÷16 = 480

9. Which one of the following genotypes can form only two different types of gametes?

1. AaBbccDd
2. AaBbccDDEe
3. AABbCCddEE
4. AaBbCcDDEe

Answer: c. AABbCCddEE

Discussion: Formula to calculate how many different types of gametes can be formed is = 2ⁿ where n is the number of heterozygous allele pairs. The AABbCCddEE genotype has one heterozygous gene pair of Bb. Hence, it will form 2 types of gametes – ABCdE and AbCdE.

Applying the formula,

• AaBbccDd forms 2³ = 8 gametes
• AaBbccDDEe forms 2³ = 8 gametes
• AaBbCcDDEe forms 2⁴ = 16 gametes

10. How many of the following statements are correct?

Statement 1: Recombination of genes is directly proportional to the strength of linkage.

Statement 2: The more the distance between the 2 genes on a chromosome, the less the strength of linkage.

Statement 3: Recombination frequency is directly proportional to the distance between the genes present on a chromosome.

1. Three
2. Two
3. Four
4. One

Answer: b. Two

Discussion: If genes are placed very close to each other on the same chromosome, the chances of crossing over are very low. This lowers the chances of recombination. Hence, the strength of linkage would be more.

Recombination ∝ 1/Linkage

If the distance between genes on the same chromosome is more, the chances of crossing over between them are high. This increases the chances of recombination. Hence, the strength of linkage would be less.

Strength of linkage ∝ 1/Distance between the genes

11. The phenotypic and genotypic ratios of the F2 generation are similar in which of the following cases?

1. Incomplete dominance in snapdragon
2. Mendelian monohybrid cross
3. Mendelian dihybrid cross
4. All of the above

Answer: a. Incomplete dominance in snapdragon

Discussion:

Selfing of the F1 generation will result in 1:2:1 phenotypic and genotypic ratios.

Selfing of F1: Rr (Pink) ✕ Rr (Pink)

12. If a man with blood group A marries a woman with blood group O, the possible blood groups of their children would be

1. A only
2. A and O
3. O only
4. A, AB and O

Answer: b. A and O

Discussion: First possibility is A (I^AI^A) ✕ O (ii)

Second possibility is A (I^Ai) ✕ O (ii)

In both cases, A and O are the possible blood types that can be seen in the progenies.

13. How many types of gametes can be produced by a diploid organism that is heterozygous for 4 loci?

1. 4
2. 8
3. 16
4. 32

Answer: c. 16

Discussion: The number of different types of gametes produced by a diploid organism can be calculated by formula 2ⁿ. Where n is the number of heterozygous loci. In the given case, the organism is heterozygous for 4 loci. Hence, the number of different types of gametes produced by the organism would be 2⁴ = 16.

14. According to the chromosomal theory of inheritance, the behaviour of ⸻ (A) was parallel to the behaviour of ⸻ (B)

1. (A) – chromosomes, (B) – genes
2. (A) – genes, (B) – proteins
3. (A) – chromosomes, (B) – proteins
4. (A) – genes, (B) – cell organelles

Answer: a. (A) – chromosomes, (B) – genes

Discussion: According to the chromosomal theory of inheritance, the behaviour of chromosomes was parallel to the behaviour of genes.

Do Check: MCQs on Chromosome Structure

15. Which among the following is an incorrect statement?

1. A polygenic character is controlled by multiple genes.
2. Numerous intermediate phenotypes are found in between the two extremes in polygenic inheritance.
3. In humans, height and skin colour are polygenic traits.
4. Polygenic inheritance is controlled by multiple alleles of a single gene.

Answer: d. Polygenic inheritance is controlled by multiple alleles of a single gene.

Discussion: The inheritance pattern when the complete expression of a trait is controlled by two or more genes in which the dominant allele of each gene contributes only a fraction of the trait is called qualitative or polygenic inheritance. Here, the total phenotypic expression is the sum total or cumulative effect of all dominant alleles of the respective genes (polygenes).

There are several intermediate phenotypes between the two extremes. For example, in humans, skin colour, height, hair and eye colour are polygenic traits.

16. Point mutation may occur due to

1. Alteration in DNA sequence
2. Change in a single base pair of DNA
3. Deletion of a segment of DNA
4. Gain of a segment in DNA

Answer: b. Change in a single base pair of DNA

Discussion: Point mutation is caused by a change in a single base pair of DNA. Such a change can be caused due to –

• Replacement of a nucleotide with another one.
• Deletion or addition of a single nucleotide pair.

For example – Sickle cell anaemia is caused due to the substitution of thymine by adenine in the β globin gene.

17. If a colour blind man marries the daughter (who is a carrier with a colour blind gene) of a colour blind father, then the next generation will be

1. None of their daughters will be colour blind
2. All the sons will be colour blind
3. All the daughters will be colour blind
4. Half their sons will be colour blind

Answer: d. Half their sons will be colour blind

Discussion: Carrier mother (X^CX) ✕ affected father (X^CY)

18. Haemophilia is rare in women because

1. It is a recessive autosomal disorder
2. Women are homozygous
3. They have only one X chromosome
4. They are more resistant to this

Answer: b. Women are homozygous

Discussion: Haemophilia is a sex-linked recessive trait and the trait is associated with the X chromosome.

19. Which of the following features characterise chromosomal disorders?

1. They are caused by mutations in single nucleotides
2. Their transmission to subsequent generations follows Mendelian principles
3. They are caused due to abnormal copies of chromosomes
4. They are caused by mutations in single gene

Answer: c. They are caused due to abnormal copies of chromosomes

Discussion: Genetic disorders of 2 types –

1. Mendelian disorders are caused due to mutation in the single gene.
2. Chromosomal disorders are caused due to the presence of an extra copy or absence of one or few chromosomes.

20. Identify the mismatched pairs:

1. 1,3
2. 2,4
3. 1,2
4. 3,4

Answer: d. 3,4

Discussion:

1. Haemophilia is a sex-linked recessive genetic disorder (x-linked) which is characterised by impaired blood clotting.
2. Down’s syndrome is a trisomy of the 21st chromosome. It occurs both in males and females.
3. Turner syndrome occurs in females. It is due to the removal of one X chromosome from the sex chromosome – 44A + X0
4. Klinefelter’s syndrome is due to extra X chromosome in sex chromosome. It occurs only in males. The chromosomal composition is – 44A + XXY

Explore similar important NEET questions from the other chapters, only at BYJU’S. Also check NEET notes for Principles of Inheritance and Variations.

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