Assertion A: A haemophilic female is extremely rare. Reason R: Her mother must be at least a carrier and her father must be haemophilic. Choose the correct answer.

Both A and R are correct and R is the correct explanation of A

A is correct but R is not correct

A is not correct but R is correct

Both A and R are correct but R is not the correct explanation of A

Principles of Inheritance and Variation

19 NEET-UG practice questions on Principles of Inheritance and Variation, part of Botany. The first 19 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Easy
How many phenotypes and genotypes does the ABO blood group system show in humans?
1. A3 phenotypes and 6 genotypes
2. B4 phenotypes and 3 genotypes
3. C6 phenotypes and 4 genotypes
4. D4 phenotypes and 6 genotypes
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Answer: D · 4 phenotypes and 6 genotypes
Three alleles, Iᴬ, Iᴮ and i, give six genotypes: IᴬIᴬ, Iᴬi, IᴮIᴮ, Iᴮi, IᴬIᴮ and ii. They show four blood groups: A, B, AB and O.
- A: Three is the number of alleles, not phenotypes.
- B: Three is the number of alleles, not genotypes.
- C: This swaps the two numbers.
Remember this
Three alleles give six genotypes and four phenotypes; Iᴬ and Iᴮ are codominant, and both dominate i.
Question 2 · Medium
Assertion A: A haemophilic female is extremely rare. Reason R: Her mother must be at least a carrier and her father must be haemophilic. Choose the correct answer.
1. AA is correct but R is not correct
2. BA is not correct but R is correct
3. CBoth A and R are correct and R is the correct explanation of A
4. DBoth A and R are correct but R is not the correct explanation of A
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Answer: C · Both A and R are correct and R is the correct explanation of A
Haemophilia is X-linked recessive, so an affected female needs a mutant X from each parent. That requires a carrier or affected mother and an affected father, a combination seldom found. So R explains A.
- A: R is correct, because a daughter needs a mutant X from each parent.
- B: A is correct, as NCERT states.
- D: R directly gives the reason for the rarity.
Remember this
X-linked recessive traits show in males readily, as hemizygotes, while females need two mutant alleles.
Question 3 · Medium
In a family, two unaffected parents have an affected daughter. Assuming control by a single gene and no new mutation, which mode of inheritance fits?
1. AAutosomal dominant
2. BAutosomal recessive
3. CX-linked recessive
4. DY-linked
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Answer: B · Autosomal recessive
Unaffected parents with an affected child point to a recessive allele hidden in both. An affected daughter of an X-linked recessive trait needs an affected father, which is not the case here. Hence the trait is autosomal recessive.
- A: A dominant trait would show in at least one parent.
- C: This requires the father to be affected.
- D: A Y-linked trait cannot occur in a daughter.
Remember this
Unaffected parents with an affected daughter indicate an autosomal recessive trait such as sickle-cell anaemia or phenylketonuria.
Question 4 · Easy
Which of these pea traits studied by Mendel is recessive?
1. ATerminal flower position
2. BInflated pod
3. CGreen pod
4. DYellow seed
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Answer: A · Terminal flower position
Axial flower position is dominant over terminal, so terminal is recessive. Inflated pod, green pod and yellow seed are the dominant forms.
- B: Inflated pod is dominant over constricted.
- C: Green pod is dominant over yellow pod.
- D: Yellow seed is dominant over green seed.
Remember this
Pod colour and seed colour differ: green pod is dominant, yet yellow seed is dominant over green seed.
Question 5 · Hard
Assume that skin colour depends on three unlinked genes, each dominant allele adding equally to colour. Two parents, both AaBbCc, have children. What fraction of the children will have the same skin colour as the parents?
1. A1/64
2. B1/8
3. C15/64
4. D5/16
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Answer: D · 5/16
The parents have three dominant alleles, so children must also carry exactly three out of six alleles. Number of such combinations is 6C3 = 20, out of 2⁶ = 64, so 20/64 = 5/16. Check by the full series 1, 6, 15, 20, 15, 6, 1, which adds to 64, with 20 at the centre.
- A: This is the extreme class with all or no dominant alleles.
- B: This counts only AaBbCc, leaving out others with three dominant alleles such as AABbcc.
- C: This is the class with two (or four) dominant alleles.
Remember this
Polygenic crosses follow the 1:6:15:20:15:6:1 distribution, and the middle class is the most common.
Question 6 · Easy
A breeder uses a test cross to find whether a tall pea plant is TT or Tt. Which plant is the partner in this cross?
1. AA heterozygous tall plant
2. BA homozygous dwarf plant
3. CA homozygous tall plant
4. DAny other tall plant
Answer and explanation in LearnFree account
Question 7 · Hard
A colour-blind man marries a woman who is a carrier of colour blindness. Which result is expected among their children?
1. ANo daughters and half the sons are colour blind
2. BHalf the daughters and half the sons are colour blind
3. CAll daughters and no sons are colour blind
4. DAll sons and half the daughters are colour blind
Answer and explanation in LearnFree account
Question 8 · Medium
Given below are two statements: Statement I: Genes that lie far apart on a chromosome show less recombination than genes lying close together. Statement II: In Drosophila, recombination between white and yellow genes was only 1.3%, while that between white and miniature was 37.2%. Choose the correct answer.
1. ABoth Statement I and Statement II are correct
2. BBoth Statement I and Statement II are incorrect
3. CStatement I is correct but Statement II is incorrect
4. DStatement I is incorrect but Statement II is correct
Answer and explanation in LearnFree account
Question 9 · Easy
In grasshoppers, the male differs from the female in sex chromosomes in that the male has
1. Aone Z and one W
2. Bone X and no Y
3. Ctwo X and no Y
4. Done X and one Y
Answer and explanation in LearnFree account
Question 10 · Medium
Match List-I with List-II. List-I (disorder) A. Turner's syndrome B. Klinefelter's syndrome C. Down's syndrome D. Sickle-cell anaemia List-II (basis) I. GAG changed to GUG in the β-globin gene II. 45, X0 III. Trisomy of chromosome 21 IV. 47, XXY
1. AA-IV, B-II, C-III, D-I
2. BA-II, B-IV, C-III, D-I
3. CA-II, B-III, C-IV, D-I
4. DA-III, B-IV, C-II, D-I
Answer and explanation in LearnFree account
Question 11 · Medium
Match List-I with List-II. List-I (scientist) A. Henking B. Morgan C. Sutton D. Sturtevant List-II (work) I. Used recombination frequencies to map genes II. United chromosome segregation with Mendelian principles III. Traced a nuclear structure named the X body in insect spermatogenesis IV. Showed linkage and recombination through Drosophila dihybrid crosses
1. AA-IV, B-III, C-II, D-I
2. BA-III, B-II, C-IV, D-I
3. CA-III, B-IV, C-I, D-II
4. DA-III, B-IV, C-II, D-I
Answer and explanation in LearnFree account
Question 12 · Hard
Read these statements on thalassaemia and sickle-cell anaemia. A. Thalassaemia is a quantitative defect in globin synthesis, whereas sickle-cell anaemia is a qualitative one. B. α-thalassaemia is controlled by the single gene HBB on chromosome 16. C. β-thalassaemia is controlled by the gene HBB on chromosome 11. D. Both disorders are X-linked. Which of these are correct?
1. AA, B and C only
2. BA, C and D only
3. CA and C only
4. DB and D only
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Question 13 · Medium
A dihybrid pea plant RrYy, with round yellow seeds, is selfed and gives 320 progeny. Roughly how many will have round green seeds?
1. A120
2. B20
3. C60
4. D180
Answer and explanation in LearnFree account
Question 14 · Medium
Arrange the steps of Mendel's monohybrid experiments in the correct order. A. Self-pollinate the F₁ plants B. Select true-breeding parental lines C. Self-pollinate the F₂ plants D. Cross-pollinate the contrasting parents
1. AD, B, A, C
2. BB, D, C, A
3. CB, A, D, C
4. DB, D, A, C
Answer and explanation in LearnFree account
Question 15 · Easy
Which chromosomal change causes Down's syndrome?
1. ATrisomy of chromosome 21
2. BMonosomy of chromosome 21
3. CAn extra X chromosome in a male
4. DAbsence of one X chromosome
Answer and explanation in LearnFree account
Question 16 · Medium
A pink-flowered F₁ snapdragon from a cross of red and white parents is selfed. Which result is expected in F₂?
1. APhenotypic ratio is 3:1 and genotypic ratio is 1:2:1
2. BPhenotypic ratio is 1:2:1 and genotypic ratio is 3:1
3. CPhenotypic and genotypic ratios are both 1:2:1
4. DPhenotypic and genotypic ratios are both 3:1
Answer and explanation in LearnFree account
Question 17 · Hard
An F₁ from AABB × aabb is test crossed. The progeny are 450 AB, 50 Ab, 50 aB and 450 ab in phenotype. What do these results show?
1. AThe genes are linked, with 90% recombination
2. BThe genes assort independently, with 10% recombination
3. CThe genes are linked, with 5% recombination
4. DThe genes are linked, with 10% recombination
Answer and explanation in LearnFree account
Question 18 · Medium
Assertion A: Among the four families in the table, family P is the only one whose child has a blood group that is genetically impossible. Reason R: Iᴬ and Iᴮ are each dominant over i, so a person of group A or group B may carry the allele i. Choose the correct answer.
Figure: MedExams
1. AA is not correct but R is correct
2. BBoth A and R are correct and R is the correct explanation of A
3. CA is correct but R is not correct
4. DBoth A and R are correct but R is not the correct explanation of A
Answer and explanation in LearnFree account
Question 19 · Medium
A trait controlled by a single gene with complete penetrance is traced in the pedigree. Filled symbols are affected. What is the probability that individual II-2 is a carrier of the allele for this trait?
Figure: MedExams
1. A1/4
2. B3/4
3. C2/3
4. D1/2
Answer and explanation in LearnFree account

Principles of Inheritance and Variation

How many phenotypes and genotypes does the ABO blood group system show in humans?

Assertion A: A haemophilic female is extremely rare. Reason R: Her mother must be at least a carrier and her father must be haemophilic. Choose the correct answer.

In a family, two unaffected parents have an affected daughter. Assuming control by a single gene and no new mutation, which mode of inheritance fits?

Which of these pea traits studied by Mendel is recessive?

Assume that skin colour depends on three unlinked genes, each dominant allele adding equally to colour. Two parents, both AaBbCc, have children. What fraction of the children will have the same skin colour as the parents?

A breeder uses a test cross to find whether a tall pea plant is TT or Tt. Which plant is the partner in this cross?

A colour-blind man marries a woman who is a carrier of colour blindness. Which result is expected among their children?

In grasshoppers, the male differs from the female in sex chromosomes in that the male has

Match List-I with List-II. List-I (disorder) A. Turner's syndrome B. Klinefelter's syndrome C. Down's syndrome D. Sickle-cell anaemia List-II (basis) I. GAG changed to GUG in the β-globin gene II. 45, X0 III. Trisomy of chromosome 21 IV. 47, XXY

A dihybrid pea plant RrYy, with round yellow seeds, is selfed and gives 320 progeny. Roughly how many will have round green seeds?

Arrange the steps of Mendel's monohybrid experiments in the correct order. A. Self-pollinate the F₁ plants B. Select true-breeding parental lines C. Self-pollinate the F₂ plants D. Cross-pollinate the contrasting parents

Which chromosomal change causes Down's syndrome?

A pink-flowered F₁ snapdragon from a cross of red and white parents is selfed. Which result is expected in F₂?

An F₁ from AABB × aabb is test crossed. The progeny are 450 AB, 50 Ab, 50 aB and 450 ab in phenotype. What do these results show?

Assertion A: Among the four families in the table, family P is the only one whose child has a blood group that is genetically impossible. Reason R: Iᴬ and Iᴮ are each dominant over i, so a person of group A or group B may carry the allele i. Choose the correct answer.

A trait controlled by a single gene with complete penetrance is traced in the pedigree. Filled symbols are affected. What is the probability that individual II-2 is a carrier of the allele for this trait?