CBSE Class 10 Science Heredity and Evolution MCQs Set 10

Answer: (d) Natural selection can alter frequency of an inherited trait.

Explanation:
1. Mendel's results show that only one parental trait appears in F₁, that each trait is inherited as two copies, and that a recessive trait needs two identical copies.
2. Natural selection acting on traits is not a conclusion from his crosses, so (d) is not a direct conclusion.

Teacher's Note:
1. Mendel did not study natural selection.
2. Read the word NOT before choosing.

Answer: (d) all of these.

Explanation:
1. Mendel studied one character at a time, the pea plant he used was true-breeding, and the characters he chose were not affected by linkage.
2. All these factors contributed, so the answer is all of these.

Teacher's Note:
1. He also kept careful numerical records of his crosses.
2. Counting large numbers of progeny helped him find the ratios.

Answer: (b) Pea plants can be self-fertilised.

Explanation:
1. Pea plants can be self-fertilised, so pure lines are easy to obtain and the number of progeny is large.
2. Dogs are unisexual and cannot be self-fertilised, so self-fertilisation made peas more suitable.

Teacher's Note:
1. Peas also have a short life cycle and clearly contrasting traits.
2. Dogs produce few offspring in a long time.

Answer: (a) Trichomes – Glandular or non-glandular

Explanation:
1. Mendel studied seed colour, pod shape and stem height among his seven characters.
2. Glandular or non-glandular trichomes were not studied.

Teacher's Note:
1. Mendel's seven characters are visible and have two clear forms.
2. Learn the list in the NCERT table.

Answer: (b) They were having seven pairs of contrasting characters

Explanation:
1. Pea plants had seven pairs of clearly contrasting characters that were easy to observe.
2. They also had short life cycle and could be self- or cross-pollinated.

Teacher's Note:
1. Mendel's choice was scientific, not economic.
2. Clear contrasting characters made the results easy to interpret.

Answer: (d) Law of segregation

Explanation:
1. Each gamete gets only one allele of a gene, so it is pure for that character.
2. This is Mendel's law of segregation.

Teacher's Note:
1. Independent assortment is Mendel's second law.
2. There is no law of variation among Mendel's laws.

Answer: (d) All of the above

Explanation:
1. The law of segregation is also called the law of purity of gametes.
2. Alleles separate from each other at gamete formation, and this happens because the chromosomes of a pair separate during meiosis.
3. All the statements are true.

Teacher's Note:
1. Alleles and chromosomes behave in a parallel way.
2. Segregation is Mendel's first law.

Answer: (c) Hybrid plant

Explanation:
1. Cross-pollination between two different varieties combines their genes.
2. The offspring plant formed from such a cross is called a hybrid.

Teacher's Note:
1. Mendel's F₁ plants were hybrids.
2. Do not confuse a hybrid with a mutant, which arises from a change in DNA.

Answer: (b) Heredity

Explanation:
1. Transmission of characters from parents to offspring is called heredity (inheritance).
2. Variation is a difference, gene and allele are units of heredity, so the answer is heredity.

Teacher's Note:
1. Heredity ensures similarity between generations.
2. Learn the three basic terms: heredity, variation and genetics.

Answer: (c) Cross fertilisation

Explanation:
1. Two different plants bearing pink flowers were crossed with each other.
2. Fusion of gametes from two different plants is cross fertilisation.

Teacher's Note:
1. The 1 : 2 : 1 result shows that pink is a hybrid (heterozygous) colour.
2. Self-pollination would be within the same flower or plant.

Answer: (a) Tallness is the dominant trait

Explanation:
1. All F₁ plants are Tt, with one tall and one short allele.
2. They are all tall, so the tall trait is expressed over the short trait.
3. Hence tallness is dominant.

Teacher's Note:
1. The trait that appears in F₁ is dominant.
2. The trait that is hidden in F₁ and returns in F₂ is recessive.

Answer: (c) 1:1

Explanation:
1. F₁ plants are all Tt. On selfing, F₂ has TT : Tt : tt in the ratio 1 : 2 : 1.
2. Pure tall plants (TT) are 1 part and pure short plants (tt) are 1 part.
3. The ratio is 1 : 1.

Teacher's Note:
1. Do not confuse the genotypic ratio 1 : 2 : 1 with the phenotypic ratio 3 : 1.
2. Pure means homozygous.

Answer: (a) 1:2:1 (homozygous tall: heterozygous tall: dwarf)

Explanation:
1. F₁ plants are Tt. On selfing, the F₂ genotypes are TT : Tt : tt in the ratio 1 : 2 : 1.
2. TT is homozygous tall, Tt is heterozygous tall and tt is dwarf.
3. So the ratio is 1 : 2 : 1 (homozygous tall : heterozygous tall : dwarf).

Teacher's Note:
1. The ratio 3 : 1 is the phenotypic ratio, not the genotypic ratio.
2. Write the Punnett square to avoid mistakes.

Answer: (b) the genotype of F₂ plant

Explanation:
1. In a test cross an individual showing the dominant trait is crossed with a homozygous recessive individual.
2. The offspring show whether it is homozygous or heterozygous, so the test cross determines the genotype of the tested plant (for example an F₂ plant).

Teacher's Note:
1. If all offspring show the dominant trait, the tested plant is homozygous.
2. If about half show the recessive trait, it is heterozygous.

Answer: (d) A are dwarf and B are tall

Explanation:
1. A trait that disappears in F₁ and reappears in F₂ is the recessive trait.
2. The recessive trait of height is dwarfness, so plants A are dwarf and plants B (whose trait shows in F₁) are tall.

Teacher's Note:
1. The trait that shows in F₁ is dominant.
2. Recessive traits reappear in one-fourth of F₂.

Answer: (b) recessive

Explanation:
1. Only the dominant trait appears in F₁.
2. The trait that does not appear in F₁ is the recessive trait.

Teacher's Note:
1. The recessive trait reappears in F₂.
2. It is hidden, not lost, in F₁.

Answer: (b) law of segregation

Explanation:
1. During formation of gametes the two alleles of a gene separate so that each gamete receives only one allele.
2. This is Mendel's law of segregation.

Teacher's Note:
1. Segregation means separation of alleles at gamete formation.
2. Independent assortment deals with two pairs of traits together.

Answer: (b) A-p, B-q, C-r, D-s

Explanation:
1. BB × bb gives all Bb, so A matches (p).
2. Bb × Bb gives 1 BB : 2 Bb : 1 bb, so B matches (q).
3. BB × BB gives only BB, so C matches (r).
4. bb × bb gives only bb, so D matches (s).
5. Hence the matching is A-p, B-q, C-r, D-s.

Teacher's Note:
1. Write the gametes of each parent and combine them.
2. Crosses between two homozygous parents of the same type give offspring identical to them.

Answer: (b) Tt × tt

Explanation:
1. A test cross is made between an individual showing the dominant trait but of unknown genotype and a homozygous recessive individual.
2. Of the given crosses, Tt × tt fits this description.

Teacher's Note:
1. Test cross shows whether the dominant-looking plant is TT or Tt.
2. The recessive parent is always tt.

Answer: (c) All red-flowered

Explanation:
1. Taking red as dominant, the cross is RR × rr and all offspring are Rr.
2. The dominant red colour is expressed in every offspring, so all are red-flowered.

Teacher's Note:
1. This is the F₁ generation of a monohybrid cross.
2. The white trait is hidden but reappears in F₂.

Answer: (b) 3 smooth : 1 sugary

Explanation:
1. The F₁ seeds are smooth, so smooth is dominant. Let the F₁ genotype be Ss.
2. Ss × Ss gives SS, Ss, Ss and ss.
3. Three plants have smooth seeds and one has sugary seeds, a ratio of 3 smooth : 1 sugary.

Teacher's Note:
1. A monohybrid F₂ always shows 3 : 1 of dominant to recessive phenotype.
2. Draw a Punnett square to check.

Answer: (b) Tt and tt

Explanation:
1. Tt produces gametes T and t, and tt produces only t.
2. The offspring are Tt and tt in equal numbers, i.e. Tt and tt.

Teacher's Note:
1. This is a test cross.
2. A 1 : 1 result shows that the tested plant is heterozygous.

Answer: (d) None of these

Explanation:
1. AA produces only A gametes and aa produces only a gametes.
2. All F₁ offspring are Aa, so the probability of aa is 0%, which is none of these.

Teacher's Note:
1. The recessive aa genotype appears only in F₂ when Aa is selfed.
2. Do not confuse F₁ with F₂.

Answer: (d) Green coloured pod and rounded seed

Explanation:
1. In pea plants the dominant traits are tall plant, round seed, green pod, inflated pod, axial flower and violet flower.
2. Only option (d), green coloured pod and rounded seed, lists two dominant characters.

Teacher's Note:
1. Dwarf, wrinkled, terminal and yellow pod are recessive traits.
2. Learn Mendel's seven pairs of traits.

Answer: (d) 135

Explanation:
1. All F₁ are tall and white, so tall and white are dominant.
2. In F₂ the double recessive (short, blue) is 1/16 of the plants.
3. 2160 ÷ 16 = 135.

Teacher's Note:
1. Dihybrid F₂: 9 : 3 : 3 : 1.
2. The recessive-recessive class is always 1 out of 16.

Answer: (a) 0

Explanation:
1. I^AI^O gives I^A or I^O gametes and I^AI^B gives I^A or I^B gametes.
2. The possible children are I^AI^A, I^AI^B, I^AI^O and I^BI^O.
3. Group O needs I^OI^O, which is not possible, so the probability is 0.

Teacher's Note:
1. A child needs one allele from each parent.
2. The second parent has no I^O allele.

Answer: (d) 100%

Explanation:
1. Attached earlobes are recessive, so each parent with attached earlobes has the genotype ff.
2. ff × ff gives only ff children.
3. So the chance of a child with attached earlobes is 100%.

Teacher's Note:
1. A recessive trait is expressed only in the homozygous recessive condition.
2. Two parents showing a recessive trait always have children showing it.

Answer: (a) 75

Explanation:
1. Tt × Tt gives TT, Tt, Tt and tt, so one-fourth of the progeny are short (tt).
2. One-fourth of 300 is 75.

Teacher's Note:
1. Short plants are homozygous recessive.
2. In a monohybrid F₂ the recessive phenotype forms 25%.

Answer: (d) 1:1 and 1:1

Explanation:
1. Rr gives R and r gametes and rr gives only r gametes.
2. The offspring are Rr and rr in equal numbers, so the genotypic ratio is 1 : 1.
3. Rr shows the dominant trait and rr the recessive one, so the phenotypic ratio is also 1 : 1.

Teacher's Note:
1. A cross with a homozygous recessive parent is a test cross.
2. Here genotypic and phenotypic ratios are the same.

Answer: (a) 25%

Explanation:
1. BB × bb gives all Bb in F₁.
2. Bb × Bb gives 1 BB : 2 Bb : 1 bb, so one-fourth of the F₂ are bb and have white fur.
3. The answer is 25%.

Teacher's Note:
1. White fur is recessive and needs bb.
2. The F₂ phenotypic ratio is 3 black : 1 white.

Answer: (d) One parent is heterozygous.

Explanation:
1. Cattle with horns are hh.
2. A 1 : 1 ratio of horned (hh) to hornless offspring needs the hornless parent to give both H and h, so it is Hh.
3. Hence one parent is heterozygous (Hh × hh).

Teacher's Note:
1. A 1 : 1 ratio is the sign of a test cross.
2. If the hornless parent were HH, all offspring would be hornless.

Answer: (c) 1:2:1

Explanation:
1. F₁ plants are Tt.
2. Tt × Tt gives TT, Tt, Tt, tt, which is a genotypic ratio of 1 : 2 : 1.

Teacher's Note:
1. Phenotypic ratio is 3 : 1, but genotypic ratio is 1 : 2 : 1.
2. Do not mix up the two.

Answer: (d) One parent is heterozygous.

Explanation:
1. The white-furred pig is bb.
2. A 1 : 1 ratio of black to white offspring means the black parent can give both B and b, so it is Bb.
3. Hence one parent is heterozygous (Bb × bb).

Teacher's Note:
1. If the black parent were BB, all offspring would be black.
2. Both parents cannot be homozygous because bb × BB gives only Bb.

Answer: (a) three

Explanation:
1. The F₁ are all heterozygous, and an F₁ × F₁ cross gives red and white in the ratio 3 : 1.
2. White-eyed individuals are 1/4 of 12 = 3.

Teacher's Note:
1. Recessive trait forms 25% of the F₂.
2. Multiply the fraction by the total number.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Genotype is the complete genetic make-up of an organism, that is, all its dominant and recessive genes.
2. So the reason correctly explains the assertion.

Teacher's Note:
1. Genotype is the gene combination, for example Tt.
2. Phenotype is the visible trait.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. A dominant allele shows its effect in TT and Tt, and a recessive allele shows it only in tt, so A is true.
2. R states why: one dominant copy is enough while two recessive copies are needed.
3. So R explains A.

Teacher's Note:
1. The source printed 'heterozygous trait' in the assertion, which has been corrected to 'recessive trait'.
2. Dominant masks recessive in heterozygotes.