NEET Biology Principles of Inheritance and Variation Important Questions Set 01

These Class 12 NEET Biology important questions on Principles of Inheritance and Variation give 44 questions with step-by-step answers, arranged as very short, short and long answer types, as asked in school and board examinations. They help NEET aspirants revise the definitions, structures, differences and key terms of the chapter.

Very Short Answer Type Questions

Question 1. What is variation?
Answer: Degree by which progeny differ from parents is called variation.

Question 2. A tall garden pea plant is self-pollinated, one-fourth of the progeny produced is dwarf. What will be genotype(s) of remaining progeny?
Answer: Tt and TT

Question 3. Name the amino acid whose substitution at the 6th position of the beta chain of the haemoglobin molecule causes sickle-cell anaemia.
Answer: Glutamic acid (Glu)

Question 4. Name the chromosomal disorder in which an individual has overall masculine development, but feminine development is also expressed.
Answer: Klinefelter's syndrome

Question 5. What are mutagens? Give one example.
Answer: Chemical and physical factors that induce mutations are called mutagens, e.g., UV radiations.

Question 6. How many phenotypes and genotypes of blood groups are present in human population?
Answer: Four phenotypes and six genotypes.

Question 7. Name the Indian breed of cow that was developed through artificial selection and domestication of wild cows.
Answer: Sahiwal cows of Punjab.

Question 8. Name one autosomal dominant and autosomal recessive disorder.
Answer: (i) Autosomal dominant disorder = Myotonic dystrophy
(ii) Autosomal recessive disorder = Sickle-cell anaemia

Question 9. A diploid organism heterozygous for three loci can produced how many types of gametes?
Answer: Types of gametes = 2n n = 3 So, 8 types of gametes.

Question 10. What is a Punnett square?
Answer: It is a graphical representation to calculate the probability of all possible genotypes of offspring in a genetic cross.

Short Answer Type Questions

Question 11. What is pedigree analysis? How is it useful in human genetics?
Answer: (i) Analysis of inheritance of a particular trait represented in the family tree over generations is called pedigree analysis.
(ii) In human genetics, pedigree study provides a strong tool, which is utilised to trace the inheritance of a specific trait, abnormality or disease.

Question 12. Differentiate between co-dominance and incomplete dominance by giving examples.

Answer:

Question 13. (i) What are multiple alleles?
(ii) They can be found only when population studies are made. Explain it.
Answer: (i) If there are more than two alleles governing the same character, the condition is called multiple alleles e.g., ABO blood grouping.
(ii) Since only two alleles can be present in an individual, multiple alleles can be found only when population studies are made. 14 What is trisomy? Give one example. When an additional copy of a chromosome is included in an individual, the situation is known as trisomy. Example : Down's syndrome, trisomy of 21.

Question 15. Differentiate between male heterogamety and female heterogamety by giving example.
Answer: Male heterogamety Female heterogamety
1. If males produce two different types of 1. If two different types of gametes in gametes, the mechanism is terms of sex chromosomes are designated as male heterogamety. produced by females, the condition is called as female heterogamety.
2. e.g., Humans, Drosophila 2. e.g., Birds

Question 16. What is monosomy? How is it formed? Give one example.
Answer: (i) If an individual lacks one of any one pair of chromosomes the situation is known as monosomy.
(ii) It is formed due to failure of segregation of homologous pair of chromosomes during cell division cycle.
(iii) E.g., Turner's Syndrome

Question 17. (i) Why a cut results in non-stop bleeding in case of hemophilia?
(ii) The possibility of a female becoming a haemophilic is extremely rare. Why?
Answer: (i) In haemophilia, a single protein that is a part of the cascade of proteins involved in the clotting of blood is affected. Due to this, in an affected individual a simple cut will result in non-stop bleeding.
(ii) The possibility of a female becoming a haemophilic is extremely rare because mother of such a female has to be atleast carrier and the father should be haemophilic (unviable in the later stage of life).

Question 18. (i) What is point mutation? Give an example.
(ii) What is the cause of frame-shift mutations?

Answer:

(i) Mutation which arises due to change in a single base pair of DNA is known as point mutation e.g., sickle- cell anaemia.

(ii) Frame-shift mutations are caused by deletion and insertion of base pairs in DNA.

Question 19. What are Mendelian disorders? How can they be traced? Give examples.

Answer:

(i) Genetic disorders which are mainly determined by alteration or mutation in a single gene are called Mendelian disorders.

(ii) Mendelian disorders can be traced in a family by the pedigree analysis.

(iii) Example : Cystic fibrosis, Colour-blindness, Phenylketonuria etc.

Question 20. (i) What is the probability of a male or a female child in each pregnancy in human beings?
(ii) Genetic make up of which gamete determines the sex of the child in human beings?

Answer:

(i) In each pregnancy there is always 50 percent probability of either a male or a female child because females produce only one type of gametes and males produce two types of gametes.

(ii) Genetic makeup of the male gamete i.e., the sperm determines the sex of the child. If the sperm carrying X-chromosome fertilises the ovum it produces a female child and if the sperm with Y-chromosome fertilises the ovum it produces a male child.

Question 21. (i) What is the sex-determination mechanism in birds?
(ii) Which gamete is responsible for the sex of the chicks?

Answer:

(i) In birds the sex-determination mechanism is ZZ-ZW type (ZW type). Females are heterogametic i.e., they produce two different types of gametes in terms of the sex-chromosomes. Females have one Z and one W chromosomes, in addition to autosomes. Males have a pair of Z-chromosomes besides the autosomes. So, they produce one type of gamete.

(ii) Sex of the chicks depends on the type of female gametes.

Question 22. What are chromosomal disorders? Give two examples.
Answer: Disorders caused due to absence or excess or abnormal arrangement of one or more chromosomes are called chromosomal disorders. Examples : Down's syndrome, Turner's syndrome.

Question 23. (i) In what respect dihybrid crosses carried out by Morgan in Drosophila were different from Mendel's dihybrid crosses in pea?
(ii) What criteria was used by Alfred Sturtevant to map the position of genes on the chromosome?

Answer:

(i) Morgan's dihybrid crosses gave different results :

(a) He observed that the two genes did not segregate independently of each other.

(b) So, the F₂ ratio deviated very significantly from 9 : 3 : 3 : 1 ratio, which is expected when the two genes are independent.

(ii) Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their position on the chromosome.

Question 24. Write down the Mendelian monohybrid phenotypic and genotypic ratios of F₂ generation. How can it be mathematically condensed to form a binomial expression that has the gametes bearing genes T or t in equal frequency of 1/2 ?

Answer:

(i) In F₂ generation, Mendelian monohybrid Phenotypic ratio = 3 : 1 Genotypic ratio = 1 : 2 : 1
(ii) It can be mathematically condensed to the form of binomial expression (ax + by)², that has the gametes bearing genes T or t in equal frequency of 1/2. The expression is expanded as given below :

(½ T + ½ t)² = (½ T + ½ t) × (½ T + ½ t) = ¼ TT + ½ Tt + ¼ tt

Question 25. In a Mendelian cross involving true breeding tall pea plants having violet flowers with short pea plants having white flowers, derive the genotype and phenotype of the progeny in resulting generation.

Answer:

Question 26. Aman of blood group Amarries a woman with blood group B. The first child of the couple has O blood group. What other blood groups are expected in their future children?

Answer:

The first child has O blood group, so the genotype is ii (IOIO). Now, it is clear that both parents are heterozygous for A and B blood groups. Blood groups of future children :

Future children can have any blood group i.e., AB, A, B or O.

Question 27. Mention some symptoms shown by Klinefelter's syndrome sufferer. Explain the cause of the disorder.
Answer: (i) Symptoms of Klinefelter's syndrome :
(a) Overall masculine development.
(b) Feminine development like Gynaecomastia i.e., development of breast.
(c) Individuals are sterile.
(ii) It is caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.

Question 28. (i) State Mendel's Law of Independent Assortment.
(ii) Mendel made this generalisation based on what type of cross?
Answer: (i) The law 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.
(ii) It is based on the observations of dihybrid crosses i.e., crosses between plants differing in two traits.

Question 29. What is linkage? Explain tightly linked and loosely linked conditions of genes.
Answer: (i) Physical association of genes on a chromosome is called linkage.
(ii) Tightly linked genes show very low recombinations. Example : Genes for white-eye colour and yellow body colour are very tightly linked and showed only 1.3 percent recombination in Drosophila.
(iii) Loosely linked genes show higher recombinations. Example : Genes for white eye colour and miniature wing showed 37.2 percent recombination in Drosophila.

Question 30. In snapdragon (Antirrhinum) a cross between varieties with red and white flowers produces all pink F₁ progeny. Explain how it is a case of incomplete dominance and not of blending inheritance?
Answer: (i) In this case, F₁ had a phenotype that did not resemble either of the two parents and was in between the two.
(ii) When the F₁ was self-pollinated, the F₂ results in the ratio of 1 (Red) : 2 (Pink) : 1 (White).
(iii) It is not a case of blending as parental characters reappear in F₂ generation.
(iv) In this case, 'R' is not completely dominant over 'r' and this made it possible to distinguish Rr as pink from RR (red) and rr (white).

Long Answer Type Questions

Question 31. A round-seeded and violet-flowered (RrVv) pea plant is crossed with round-seeded and white-flowered (Rrvv) plant. Determine the phenotypic and genotypic ratios of the F₁ progeny.

Answer:

Round-seeded and Round-seeded and Parents violet-flowered plant white-flowered plant RrVv Rrvv

Gametes RV Rv rV rv Rv rv

RV Rv rV rv RRVv RRvv RrVv Rrvv Rv Round Round Round Round Violet White Violet White RrVv Rrvv rrVv rrvv rv Round Round Wrinkled Wrinkled Violet White Violet White

1. Phenotypic ratio = 3 : 3 : 1 : 1 Round Round Wrinkled Wrinkled Violet White Violet White
2. Genotypic ratio = RRVv : RrVv : RRvv : Rrvv : rrVv : rrvv 1 : 2 : 1 : 2 : 1 : 1

Question 32. Design a monohybrid cross and explain the Law of Dominance.

Answer:

Law of Dominance
(i) Characters are controlled by discrete units called factors.
(ii) Factors occur in pairs.
(iii) In a dissimilar pair of factors, one member of the pair dominates (dominant) the other (recessive). Law of dominance is used to explain the expression of only one of the parental traits in a monohybrid cross in the F₁ generation and the expression of both in the F₂ generation. It also explains the proportion of 3 : 1 obtained at the F₂. Parents TT tt (Tall) (Dwarf)

Gametes T t

F₁ Generation Tt Tt × Tt T t TT Tt T Tall Tall F₂ Generation Tt tt t Tall Dwarf Phenotypic ratio = 3 Tall : 1 Dwarf Genotypic ratio = 1 TT : 2 Tt : 1 tt

Question 33. (i) Using Punnett square, explain a dihybrid cross between homozygous round yellow and wrinkled green seed varieties of pea plant.
(ii) Give the F₂ phenotypic ratio.
(iii) Identify the Mendel's law which can be explained using this cross.

Answer:

(i) Round Yellow Green Wrinkled Parents RRYY rryy

Gametes RY ry

F₁ Generation RrYy (Round Yellow)

Selfing RrYy × RrYy RY Ry rY ry RRYY RRYy RrYY RrYy RY Round Round Round Round Yellow Yellow Yellow Yellow RRYy RRyy RrYy Rryy Ry Round Round Round Round Yellow Green Yellow Green F₂ Generation RrYY RrYy rrYY rrYy rY Round Round Wrinkled Wrinkled Yellow Yellow Yellow Yellow RrYy Rryy rrYy rryy ry Round Round Wrinkled Wrinkled Yellow Green Yellow Green
(ii) Phenotypic ratio of F₂ generation is :

Round yellow Round green Wrinkled yellow Wrinkled green 9 : 3 : 3 : 1

(iii) This cross explains Mendel's law of independent assortment.

Question 34. Explain the concept of dominance. Why are some alleles dominant and some recessive?
Answer: (i) Every gene contains information to express a particular trait.
(ii) In a diploid organism, there are two copies of each gene, i.e., as a pair of alleles.
(iii) The two alleles may be identical (homozygous) or different (heterozygous).
(iv) One of them may be different due to some changes that it has undergone which modifies the information that particular allele contains.
(v) If the modified allele produces a non-functional enzyme or no enzyme the phenotype will be affected.
(vi) The unmodified (functioning) allele, which represents the original phenotype is dominant allele.
(vii) The modified allele is generally recessive allele.
(viii) The recessive trait is seen due to non-functional enzyme or because no enzyme is produced.

Question 35. What is pedigree analysis? What are the symbols used in such an analysis?

Answer:

(i) Study of the family history about inheritance of a particular trait in several generations is called pedigree analysis.
(ii) It is a strong tool, which is utilised to trace the inheritance of a specific trait, abnormality or disease.
(iii) Standard symbols used in pedigree analysis are as follows :

(a) Squares are used to denote males i.e.,

(b) Circle for females i.e.,

(c) Mating is shown by horizontal line connecting male and female i.e.,

(d) Offsprings are arranged from left to right in order of birth below the parents i.e.,

(e) Clear symbols represent absence of trait.

(f) Parents with male child affected with disease

(g) Sex unspecified

(h) Mating between relatives (consanguineous mating)

5
(i) Five unaffected offspring

(j) Affected individuals

Question 36. (i) Name the gene which controls ABO blood groups in humans.
(ii) How many alleles exists for this gene? What is the difference in their functioning?
(iii) Why AB blood group shows co-dominance?
(iv) Write down the possible genotypes for blood group B.
Answer: (i) The gene I controls ABO blood groups.
(ii) The gene I has three alleles IA, IB and i. Alleles IA and IB produce slightly different form of the sugar while allele i does not produce any sugar.
(iii) Organism with AB blood group possess both IA and IB and they express their own types of sugars so this is co-dominance.
(iv) IBIB, IBi (IBIO)

Question 37. (i) Describe an example of a condition where a single gene produces more than one effect.
(ii) What does this explain about the property of the gene?
Answer: (i) (a) Starch synthesis in pea seeds is controlled by one gene. It has two alleles B and b.
(b) Starch is synthesised effectively by BB homozygotes and therefore, large starch grains are produced.
(c) 'bb' homozygotes have lesser efficiency in starch synthesis and produce smaller starch grains.
(d) After maturation of the seeds, BB seeds are round and bb seeds are wrinkled. Heterozygotes produce round seeds and so, B seems to be the dominant allele.
(e) But, the starch grains produced are of intermediate size in Bb seeds.
(f) So, if starch grain size is considered as the phenotype then from this angle, the alleles show incomplete dominance.
(ii) This explains that dominance is not an autonomous feature of a gene or the product that it has information for. It depends as much on the gene product and the production of a particular phenotype from this product as it does on the particular phenotype that we choose to examine, when more than one phenotype is influenced by the same gene.

Question 38. (i) Define the terms genotype, dominant allele and recessive allele.
(ii) In what way the inheritance of blood group (ABO system) in man and height of stem in Pisum sativum are different?

Answer:

(i) (a) Genotype is the genetic constitution of an organism e.g., TT and tt are called the genotype.
(b) Dominant allele : The allele which express in heterozygous (F₁) condition is dominant.
(c) Recessive allele : The allele which does not express in heterozygous (F₁) condition is recessive. e.g., in Tt = T (for tallness) is dominant over t (for dwarfness). (ii) Blood Group in Man Height of stem in Pisum sativum
(a) The gene for this trait shows multiple The gene for this shows two alleles. alleles.
(b) IAand IB together show co-dominance. Tis dominant and t is recessive.

(c) Four phenotypes and six genotypes Two phenotypes and three appear in population. genotypes appear in population.

Question 39. (i) Study the given pedigree chart and answer the questions:
(a) Is the trait dominant or recessive?
(b) Is the trait autosomal or sex-linked?
(c) Give one example of the disease.
(ii) Give examples of some most common and prevalent Mendelian disorders.

Answer:

(i) (a) It is a recessive trait.

(b) It is an autosomal disorder.

(c) Sickle-cell anaemia.

(ii) Most common and prevalent Mendelian disorders are haemophilia, cystic fibrosis, sickle-cell anaemia, colour- blindness, phenylketonuria, thalassemia etc.

Question 40. Explain the symptoms as well as chromosome complement of two sex-linked chromosomal disorders.

Answer:

Sex-linked chromosomal disorders are:

(i) Klinefelter's syndrome

(ii) Turner's syndrome

(i) Klinefelter's syndrome :

This genetic disorder is caused due to the presence of an additional copy of X-chromosome resulting into a karyotype of 47, XXY.

Symptoms : 1. Overall masculine development

2. Tall stature

3. Feminine development, i.e., development of breast (Gynaecomastia).

4. Individuals are sterile.

(ii) Turner's syndrome :

Such a disorder is caused due to the absence of one of the X-chromosomes, i.e., 45 with XO.

Symptoms : 1. Such females are sterile.

2. Short stature and underdeveloped feminine character.

3. Ovaries are rudimentary.

4. Lack of secondary sexual characters.

Question 41. (i) Study the given pedigree chart and answer the questions:
(a) What type of trait is seen?
(b) Give an example of the genetic disorder showing the given pedigree.
(ii) Mention the major symptoms associated with Down's syndrome.
(iii) Why the family pedigree of Queen Victoria shows a number of haemophilic descendents?
Answer: (i) (a) Autosomal dominant or X-linked dominant
(b) Myotonic dystrophy
(ii) Major symptoms of Down's syndrome are :
(a) Broad flat face
(b) Flat back of head
(c) Palm is broad with characteristic palm crease
(d) Many loops on fingertips
(e) Big and furrowed tongue
(f) Congenital heart disease
(g) Physical, psychomotor and mental development is retarded.
(h) Short statured with small round head
(i) Partially open mouth
(iii) Family pedigree of Queen Victoria shows a number of haemophilic descendents as she was a carrier of the disease.

Question 42. On the basis of Mendel's Laws of Inheritance, find out the result of following crosses on Pisum sativum.
(i) A tall homozygous and dominant variety is crossed with a dwarf variety.
(ii) Progeny of (i) is selfed.
(iii) Progeny of (i) is crossed with tall parent.
(iv) Progeny form (i) is crossed with dwarf parent.
(v) What type of cross is attempted in (iv)?

Answer:

(i) Parents TT tt (Tall) (Dwarf)

Gametes T t

Tt Tall All the offsprings are tall. (ii) Tt Tt

Gametes T t T t

TT Tt Tt tt Tall Dwarf 3 : 1 3/4th are tall and 1/4th are dwarf. Tt TT (iii)

Gametes T t T

TT Tt Tall All offsprings are tall.
(iv) Tt tt

Gametes T t t

Tt tt

1 : 1 50% of progenies are tall and 50% dwarf.
(v) It is a test cross.

Question 43. (i) When both parents are normal but exhibiting sickle-cell trait, what is the probability of their offspring having:
(a) Sickle-cell trait but apparently unaffected
(b) Normal and homozygous condition
(c) Diseased phenotype
(ii) If a cell has more than two sets of chromosomes then what is the condition? Parents HbAHbS HbAHbS

Answer:

(i)

Gametes HbA HbS HbA HbS

Offsprings HbAHbA : HbAHbS : HbAHbS : HbS HbS 1 : 2 : 1 Normal Sickle-cell Sickle-cell trait anaemia
(a) 50% of the offsprings (HbAHbS) have sickle-cell trait.
(b) 25% of the offsprings (HbAHbA) are normal and homozygous.
(c) 25% will have the diseased phenotype (HbSHbS).
(ii) The condition is polyploidy.

Question 44. How many F₁ gametes (types), F₂ genotypes and F₂ phenotypes would be expected from the following crosses:
(i) AA × aa
(ii) AABB × aaBB

Answer:

(i) Parents AA aa

Gametes A a F₁ Generation Aa Aa Aa A a A a

F₂ Generation AA Aa Aa aa
1. F₁ Gametes = A and a
2. F₂ Genotypes = AA : Aa : Aa : aa i.e., three genotypes 1 : 2 : 1
3. F₂ Phenotypes = 3 : 1 i.e., two phenotypes
(ii) Parents AABB aaBB Gametes AB aB F₁ Generation AaBB AaBB AaBB Gametes AB aB AB aB

F₂ Generation AABB AaBB AaBB aaBB 1 : 2 : 1
(a) F₁ Gametes = AB and aB
(b) F₂ Genotypes = 3 (1 : 2 : 1)
(c) F₂ Phenotypes = 2 (3 : 1)

Question 45. Discuss the mechanism of sex determination in humans.
Answer: (i) Sex determining mechanism in case of humans is XY type.
(ii) Out of 23 pairs of chromosomes present, 22 pairs are exactly same in both males and females, these are autosomes.
(iii) A pair of X-chromosomes are present in the female.
(iv) Presence of X and Y chromosomes are determinant of male characteristic.
(v) During spermatogenesis among males, two types of gametes are produced. 50% carry X-chromosome and 50% has Y-chromosome besides autosomes.
(vi) Females produce only one type of ovum with an X-chromosome.
(vii) When ovum fertilises with a sperm carrying X-chromosome the zygote develops into a female (XX) and fertilisation of ovum with Y-chromosome carrying sperm results into male offspring.
(viii) Genetic make-up of the sperm determines the sex of the child.