INHERITANCE AND VARIATION
Question 1: Mention the advantages of selecting pea plant for experiment by Mendel.
Answer: Following are the advantages of selecting a pea plant for experiment by Mendel:
- It is a perennial plant, which means many generations of pea plant can be studied in a few years.
- Pea plant has many contrasting characters which are easily observable.
- Artificial cross-fertilisation can be easily carried out in pea plants.
Question 2: Differentiate between the following –
(a) Dominance and Recessive
Answer: A trait which prevents appearance of its contrasting trait is called a dominant trait. A trait which is unable to express in a particular generation because of its contrasting trait is called a recessive trait.
(b) Homozygous and Hetrozygous
Answer: When both the genes in a pair are similar, the genotype is called homozygous. When genes of a pair are different, the genotype is called heterozygous.
(c) Monohybrid and Dihybrid.
Answer: When one pair of contrasting characters is selected for study in a cross, the cross is called monohybrid cross. When two pairs of contrasting characters are selected for study in a cross, the cross is called dihybrid cross.
Question 3: A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer: This can be found by applying 2n formula; where n = number of loci. Hence, 2n = 24 = 16 types of gametes are produced.
Question 4: Explain the Law of Dominance using a monohybrid cross.
Answer: For monohybrid cross, let us take example of monohybrid cross for stem height, i.e. between tall plants and dwarf plants. During this experiment, Mendel made following observations:
- All plants in the F1 generation were tall plants.
- Some of the plants in the F2 generation were dwarf plants. The ratio of tall plants to dwarf plants was 3:1.
- The tall and dwarf traits did not show blending. In other words, the plants were either tall or dwarf, and there was no plant of medium height.
Based on his observations, Mendel proposed the law of dominance which is as follows: “In a dissimilar pair of factors, one factor dominates (dominant factor) over the other (recessive factor).”
Question 5: Define and design a test-cross.
Answer: Test Cross: When an organism showing a dominant phenotype (and whose genotype is to be determined) is crossed with the recessive parent (instead of self-crossing), the cross is called Test Cross. The progenies of a test cross can be easily analysed to predict the genotype of the test organism.
Following is an example of test cross when genotype needs to be known. Let us assume that all flowers are of violet colour and violet colour is the dominant trait. Let us cross them with homozygous recessive plants producing white flowers. Two situations may emerge after the test cross.
Situation 1: If plants with dominant trait are homozygous (WW) with violet flowers, then a cross with homozygous (ww) plants with white flowers will produce following result.
In this case, all plants in F2 generation will produce violet flowers because violet colour is the dominant trait. This will prove that genotype of F1 generation was homozygous (WW).
Situation 2: If plants with dominant trait are heterozygous (Ww) with violet flowers, then a cross with homozygous (ww) plants with white flowers will produce following result.
In this case, half of the plants in F2 generation will produce violet flowers (Ww) and remaining half will produce white flowers (ww). This will prove that genotype of F1 generation was heterozygous (Ww).
Question 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.
Answer: The Punette Square in previous question (Situation 2) shows a cross between a homozygous female and a heterozygous male. This shows that 50% plants of F1 generation would be heterozygous, while remaining 50% would be homozygous.
Question 7: When a cross in 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.
Answer: We get following Punette Square.
Tall and yellow (TTYy or TtYy) = 6 out of 16
Tall and green (TTyy or Ttyy) = 2 + 4 = 6 out of 16
Dwarf and green (ttyy) = 2 out of 16
Dwarf and yellow (ttYy) = 2 out of 16
Question 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?
Answer: In this case, the distribution of phenotypic feature would follow the same pattern as in dihybrid cross done by Mendel. It is important to recall the Law of Independent Assortment while answering this question. This law says that different traits segregate independent of each other; during gamete formation.
9. Briefly mention the contribution of T.H. Morgan in genetics.
Answer: Following are the contribution of T. H. Morgan in the field of genetics:
- His experiments on Drosophila resulted in discovery of production of variations during sexual reproduction.
- He gave the concept of linkage and recombination; which resulted in development of DNA mapping.
Question 10: What is pedigree analysis? Suggest how such an analysis, can be useful.
Answer: Study of the family history about inheritance of a particular trait over several generations of a family is called pedigree analysis. Pedigree analysis helps in understanding the likely prevalence of a genetic disease in future generations. It also helps in understand the cause of a particular sex-linked disease.
Question 11: How is sex determined in human beings?
Answer: XY type of sex determination is seen in many insects and in mammals (including man). In this case, both male and female have same number of chromosomes. The X-chromosome is larger, while the Y-chromosome is somewhat smaller. The somatic cell of a male contains an X and a Y chromosome in the last pair, while both chromosomes in the last pair of somatic cell of female are X-chromosomes. Thus, each egg has a single X-chromosome; as the sex-chromosome. 50% of sperms have X-chromosome, while the remaining 50% sperms have Y-chromosome as the sex-chromosome. If a sperm with X-chromosome fertilizes the egg, the zygote would develop into female. If a sperm with Y-chromosome fertilizes the egg, the zygote would develop into male. Examples: Human, Drosophila
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 offsprings.
Answer: The distribution of genotype can be given by following Punnette Square.
This shows that out of 4 children; 1 has AB blood group, 1 has A blood group, 1 has B blood group and 1 has ) blood group.
Question 13: Explain the following terms with example
Answer: To understand the concept of co-dominance, let us take example of ABO blood groupings in human beings. The plasma membrane of RBCs has sugar polymers that protrude from its surface and the kind of sugar is controlled by the gene. The gene (I) has three alleles; IA, IB and i. The alleles IA and IB produce a slightly different form of the sugar while the allele i does not produce a sugar. IA and IB are completely dominant over i. This means that when IA and i are present, only IA expresses. Similarly, when IB and i are present, only IB expresses. But when both IA and IB are present, both of them express their own types of sugars; because of co-dominance
(b) Incomplete dominance
Answer: Sometimes, a dominant trait may not completely dominate over a recessive trait. This condition is called incomplete dominance. We know that in a diploid organism, there are two copies of each gene, i.e. as a pair of alleles. One of the genes may be different due to some changes it has undergone
To understand this, let us take an example of a gene that contains the information for producing an enzyme. There are two copies of this gene. Let us assume that the normal allele produces the normal enzyme. The modified allele could be responsible for production of following:
- The normal or less efficient enzyme, or
- A non-functional enzyme, or
- No enzyme at all.
In the first case, the modified allele is same as the unmodified allele. It will produce the same phenotype, i.e. it will result in the transformation of substrate S. But, if the allele produces a non-functional enzyme or no enzyme, the phenotype may be affected. The unmodified allele is the dominant allele, while the modified allele is generally the recessive allele. The recessive trait is seen due to non-functional enzyme or because no enzyme is produced
Question 14: What is point mutation? Give one example.
Answer: When mutation arises due to change in a single base pair of DNA, it is called point mutation. Example; sickle-cell anaemia.
Question 15: Who had proposed the chromosomal theory of the inheritance?
Answer: Sutton and Boveri
Question 16: Mention any two autosomal genetic disorders with their symptoms.
Answer: Sickle Cell Anaemia: This is an autosome-linked recessive trait that can be transmitted from parents to the offspring when both the partners are carrier for the gene (or heterozygous). The disease is controlled by a single pair of allele; HbA and HbS. There are three possible combinations, i.e. HbAHbA, HbAHbS and HbSHbS. Out of them, only homozygous individuals with HbSHbS genotype are affected. Individuals with HbAHbS are carriers of this disease.
This defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the 6th position of the beta globin chain of the haemoglobin molecule. The mutant haemoglobin molecule undergoes polymerization under low oxygen tension causing the change in the shame of the RBC from biconcave disc to elongated sickle-like structure.
Phenylketonuria: This is an autosome-linked recessive trait. The affected individual lacks an enzyme that converts the amino acid phenylalanine into tyrosine. As a result, phenylalanine is accumulated and converted into phenylpyruvic acid and other derivatives. Accumulation of these substances in the brain results in mental retardation. These substances are also excreted through urine because of poor absorption by kidney.