NCERT Solutions Class 12 Biology Chapter 5

Q.1 Mention the advantages of selecting pea plant for experiment by Mendel.

Ans-

Mendel chose the pea plants to carry out his genetics experiment because of the following features:

1. Pea plants have naturally occurring visible contrasting traits such as purple/white flowers, tall/dwarf plants, yellow/green pods etc.
2. Peas are self-pollinating plants i.e. pollen from a flower fertilises the eggs of the same flower giving rise to purebred lines having the same trait generations after generations.
3. Pea plants can be easily cross-pollinated by emasculation whereby, the stamen of the plant is removed and the pistil is dusted with the stamen of the desired parent.
4. The life cycle of pea plants is short and they produce many seeds in one generation enabling the statistical analysis of the result.

Q.2 Differentiate between the following –
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid

Ans-

(a) A dominant factor or allele expresses itself in the presence or absence of the recessive allele. For example in a pea plant tall, violet flowers, green pod etc. are dominant factors. A recessive factor or allele cannot express itself in the presence of the dominant character. For example in a pea plant dwarf, white flowers, yellow pods are recessive factors.

(b) A homozygous genotype is the one in which both the alleles for a trait are similar. The genotype can be homozygous for a dominant (TT) or a recessive (tt) trait. All gametes produced by the homozygous organism will carry the same allele. A heterozygous genotype is one in which the two alleles for a trait are different. This type of genotype will have both dominant and recessive allele for example Tt. The gametes produced will be of two kinds one will only carry the dominant allele and one will carry the recessive allele.

(c) A monohybrid cross is a cross between two pure breed parents that differ in only one pair of contrasting character. For example a cross between tall (TT) and dwarf (tt) pea plants. A dihybrid cross is a cross between two parents that differ in two pairs of contrasting characters. For example a cross between pea plants having round yellow (RRYY) and wrinkled green (rryy) seeds.

Q.3 A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Ans-

Loci means the position on the chromosome at which a gene lies. If an organism is heterozygous for 4 loci it means that for four genes the organism has dissimilar alleles. During meiosis the alleles will segregate in 16 possible ways to form 16 different types of gametes given that the genes are not linked. If the genes are linked then the outcome could vary as two or more alleles could co-segregate.

Explanation: To calculate the number of possible gametes when an organism is heterozygous for a locus the formula is as follows:

*The above holds true for genes that are not linked and segregate independently.

Q.4 Explain the Law of Dominance using a monohybrid cross.

Ans-

According to Mendel’s Law of Dominance, a physical trait is controlled by a pair of factors (also called alleles). When the factors are dissimilar one factor dominates (dominant factor) the other (recessive factor). But the recessive factor is not lost and appears in the second generation.

This law is explained by a monohybrid cross which is a cross of two parents that are pure for one contrasting trait, for example, tall (TT) and dwarf (tt). The F₁ generation yields all the progeny with a heterozygous genotype Tt that express the dominant trait – Tall. The F₂ generation which is a self cross (TtxTt) yields progeny with a phenotypic ratio of 3:1 ::: tall:short and a genotypic ratio of TT:Tt:tt::1:2:1. The recessive trait (dwarf plant) appears in the F₂ generation which was masked by the dominant trait (tall plant) in the F₁ generation.

Parents: TT (tall) X tt (dwarf)

F1 generation

Genotype – Tt

Phenotype – Tall

Parents: Tt (tall) X Tt (tall)

F2 Generation

Genotype – TT:Tt:tt::1:2:1

Phenotype – 3:1:: tall:short

Q.5 Define and design a test-cross.

Ans-

A test cross is when an organism exhibiting the dominant phenotype is crossed with the homozygous recessive parent to determine the genotype of the former by analyzing proportions of offspring displaying the recessive phenotype. If all offspring display the dominant phenotype, the individual in question is homozygous dominant; if the offspring displays both dominant and recessive phenotypes, then the individual is heterozygous.

Test cross design:

Plants with violet flowers (dominant trait; unknown genotype WW/Ww) were crossed with a purebred line yielding white flowers. The outcome of the experiment determines the genotype of the parent with the dominant trait. If all offspring yield violet flowers the parent with the dominant trait would be homozygous (WW) whereas, if the offspring yield both violet and white flowers 50% of the times the parent with the dominant trait would be heterozygous (Ww).

Parents: Violet (unknown parent could be WW/Ww) X White (ww)

Phenotype – 100% flowers Purple

Inferred parental genotype – WW

Offspring Genotype – Ww

Parents: Violet (unknown parent could be WW/Ww) X White (ww)

Phenotype – 50% flowers purple, 50% flowers white

Inferred parental genotype – Ww

Offspring Genotype – 50% Ww, 50% ww

Q.6 Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Ans-

Let us assume that the man is heterozygous for the blood group antigen AB and the female is homozygous BB. The first filial generation of the two parents will have the following distribution:

Parents: Female (BB) X Male (AB)

Progeny Genotype – AB:BB::1:1

Progeny Phenotype – AB:B::1:1

Therefore in this situation, the phenotypic and genotypic ratio of the progeny will be the same. Also, the genotype of the female and male will be equally expressed.

Q.7 When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be
(a) Tall and green
(b) Dwarf and green

Ans-

A dihybrid cross between two parents differing in two pairs of contrasting traits: Plant height and seed colour was made using a Punnet square.

Parent: TtYy X Ttyy

Number of offspring with phenotype

(a) Tall and Green – 3

(b) Dwarf and Green – 1

Explanation: This phenomenon is based on Law of Independent assortment coined by Mendel in which he states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

Q.8 Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dibybrid cross?

Ans-

When two genes are linked they do not follow the expected ratio for a dihybrid cross between heterozygous parents as seen in Mendel’s crosses (9:3:3:1). Instead, the phenotype ratio will be like that of a monohybrid cross if the two genes are very tightly linked because they will be inherited together. Recombinant phenotypes may also appear in low numbers or varying numbers depending on the distance/extent of linkage between the two loci.

Q.9 Briefly mention the contribution of T.H. Morgan in genetics.

Ans-

Morgan formulated the theory of inheritance of chromosomes. He performed several dihybrid crosses in Drosophila to demonstrate linkage and recombination (the two terms coined by him) of genes on the X chromosome. He saw that some pair of contrasting characters did not segregate in the ratio of 9:3:3:1 (the expected outcome when the two genes are independent). Morgan attributed this to the genes being physically linked in such a way that they didn’t segregate independently of each other. Recombination was the event that was used to describe the generation of non-parental gene combinations. He demonstrated that tightly linked genes showed very low recombination while others that were loosely linked showed a higher rate of recombination. Morgan won the Nobel Prize in Physiology or Medicine in 1933 for his contribution to genetics.

Q.10 What is pedigree analysis? Suggest how such an analysis can be useful.

Ans-

Pedigree analysis is a type of genetic analysis that analyses the pattern of inheritance of a particular trait, disease or abnormality. This type of analysis is made in several generations of a family using special symbols and lines.

Uses of pedigree analysis:

1. Can help to determine if the trait is dominant or recessive
2. Whether the trait is linked to the sex chromosome

Predict and trace the pattern of inheritance of single mutation disorder like Haemophilia, sickle cell anaemia etc so that parents who are carrying the mutated genes or belong to families afflicted by these disorders can make an informed decision about their progeny.

Q.11 How is sex determined in human beings?

Ans-

Sex determination in humans is associated with sex chromosomes that are different between male and female individuals. Normal human females have two sex chromosomes – XX. The normal human male has 2 sex chromosomes – XY. The males produce two different types of gametes one having the X chromosome and one with the Y chromosome, while all-female gametes have an X chromosome. The sex of the baby is determined by whether the fertilising sperm contains the X or the Y chromosome. There is an equal probability of having a girl or a boy for each fertilisation event.

Parent: Male (XY) X Female (XX)

Progeny Genotype – XX:XY::1:1

Q.12 A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offspring.

Ans-

The child of a set of parents that have blood group A and B can only be of blood group O if the genotype of the parents is heterozygous having one allele as the gene “i” that expresses neither antigen A or B. Therefore the genotypes of the parents and the offspring will be:

The following Punnet cross explains how the various genotypes and phenotypes are derived:

Parent: Father (I^Ai) X Mother (I^Bi)

Progeny Genotype – I^A I^B: I^A i: I^B i:ii::1:1:1:1

Progeny Phenotype – AB:A:B:O:: 1:1:1:1

Explanation:

Human blood type is determined by co-dominant alleles. It means that in a heterozygous situation, both alleles are expressed. In the case of blood type, there are three different alleles that result in variable expression of antigens on the RBCs.

Each of us has two ABO alleles, one from each parent. A pair of allele is called the genotype for that trait. Since there are three alleles 6 different genotypes are possible at the ABO genetic locus.

*Two recombinations result in AB genotype resulting in 6 different genotypes.

Q.13 Explain the following terms with example
(a) Co-dominance
(b) Incomplete dominance

Ans-

(a) Co-dominance is a genetic phenomenon where the F1 generation resembles both parents i.e. the two alleles of a gene are equally dominant and both are expressed in a heterozygous condition. ABO blood group is an example of co-dominance. The ABO blood group is controlled by gene I which has three alleles IA, IB and i. IA and IB produce two different types of antigens on the RBC while the I allele doesn’t produce any antigen. In an individual, with the genotype IAIB both the A and B antigens are expressed on the RBC because of co-dominance.

(b) Incomplete dominance refers to a phenomenon whereby one allele does not completely dominate another allele, and therefore the progeny resembles neither of the parents resulting in a new phenotype which is a mixture of parental phenotypes. For example, a cross between the purebred red (RR) and white (rr) flowered plants of snapdragon (Antirrhinum sp.). The F1 generation (Rr) of such a cross yields 100% pink flowers as opposed to the expected red colour. This is because the RR genotype is partially dominant in the recessive trait of white flowers. Therefore, the white pigment of the flowers is also expressed resulting in pink colour flowers.

Parents: Red (RR) X White (rr)

Phenotype – 100% flowers Pink

Genotype – 100% Rr (Red trait shows incomplete dominance)

Parents: Pink (Rr) X Pink (Rr)

Phenotype – 1:2:1::Red:Pink:White

Genotype – 1:2:1::RR:Rr:rr

Q.14 What is point mutation? Give one example.

Ans-

A gene mutation involving the substitution, addition, or deletion of a single nucleotide base is called a point mutation.

Sickle-cell anaemia is caused by the substitution of glutamic acid by valine at the sixth position of the beta-globin chain of the haemoglobin molecule. This substitution is caused due to a point mutation that changes the nucleotide A to T in the coding strand of the DNA.

Sequence for Normal Haemoglobin (HbA gene)

Sequence for Sickle Cell Haemoglobin (HbS gene)

Explanation:

Point mutations can have one of three outcomes.

Point mutations may arise from spontaneous mutations that occur during DNA replication. The rate of mutation may be increased by mutagens which are physical, chemical or biological agents that increase the frequency of mutations above the normal level. Mutagens associated with cancers are often studied to learn about cancer and its prevention.

Sickle-cell anaemia is an autosome linked recessive trait that is transmitted from parents to offspring when both parents are carriers of the point mutated gene. The disease is controlled by a single pair of allele, HbA and HbS. Double recessive individuals of genotype HbSHbS show the disease phenotype. The heterozygous individuals are only carriers of the disease and are unaffected.

Q.15 Who had proposed the chromosomal theory of the inheritance?

Ans-

The chromosomal theory of inheritance was proposed by Walter Sutton and Theodore Bovery in 1902. The theory linked the inheritance of traits described in Mendelian Laws to the inheritance of chromosomes.

Explanation: The theory explains that chromosomes carried the factors described in Mendelian inheritance. It also stated that the chromosomes are linear structures with genes located at specific sites called loci along their length.

Q.16 Mention any two autosomal genetic disorders with their symptoms.

Ans-

1. Sickle cell anaemia: This autosomal recessive genetic disorder is caused by the substitution of glutamic acid by valine at the sixth position of the beta-globin chain of the haemoglobin molecule. This point mutation causes a physiological change in the haemoglobin, which in turn changes the shape of the RBC from biconcave disc to sickle-like structure.

Symptoms: Shortness of breath, dizziness, headaches, cold hands and feet and pale jaundiced skin.

2. Phenylketonurea: This autosomal recessive genetic disorder leads to the absence of an enzyme that converts the amino acid phenylalanine to tyrosine. This causes the accumulation of phenylalanine, which in turn leads to the associated symptoms.

Symptoms: Mental retardation, seizures, delayed development, behavioural problems, and psychiatric disorders.