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NEET Biology: Genetics and Evolution · Chapter 1

Principles of Inheritance and Variation

What to remember

  • Mendel's laws: dominance, segregation (alleles separate in gametes, monohybrid F2 = 3:1 phenotype and 1:2:1 genotype) and independent assortment (dihybrid F2 = 9:3:3:1).
  • Incomplete dominance (snapdragon, F2 1:2:1 in both phenotype and genotype) and co-dominance (ABO blood groups, alleles IA, IB and i) are exceptions to complete dominance.
  • Sex is decided by chromosomes: XX-XY in humans (male is heterogametic), ZZ-ZW in birds (female is heterogametic), XX-XO in grasshopper and haplodiploidy in honey bee.
  • Mendelian disorders follow gene rules (haemophilia and colour blindness are X-linked recessive; sickle-cell anaemia, thalassaemia and phenylketonuria are autosomal recessive); chromosomal disorders arise from a wrong number of chromosomes (Down 47 with trisomy 21, Klinefelter 47 XXY, Turner 45 XO).

Mendel's experiments on pea

Gregor Mendel worked on the garden pea, Pisum sativum, for about seven years (1856 to 1863). Pea was a good choice. It has many clear contrasting traits, it is self-pollinating, it can be cross-pollinated by hand, it grows easily and it has a short life cycle. Mendel used pure lines (true-breeding plants that gave the same trait after many generations of self-pollination). He studied seven pairs of contrasting traits.

CharacterDominant traitRecessive trait
Stem heightTallDwarf
Flower colourVioletWhite
Flower positionAxialTerminal
Pod shapeInflatedConstricted
Pod colourGreenYellow
Seed shapeRoundWrinkled
Seed colourYellowGreen

Mendel counted the plants of each type in every generation. He used mathematics to analyse the results, and this was new at that time.

Inheritance of one gene (monohybrid cross)

A gene has different forms called alleles. A cross between a tall (TT) and a dwarf (tt) plant gives F1 plants that are all tall (Tt). When F1 plants are self-pollinated, the F2 generation has tall and dwarf plants in the ratio 3:1. The dwarf trait, which is hidden in F1, comes back in F2.

  • Phenotype is the outward appearance. Genotype is the genetic make-up.
  • F2 phenotypic ratio is 3 tall : 1 dwarf. F2 genotypic ratio is 1 TT : 2 Tt : 1 tt.
  • Homozygous means two identical alleles (TT or tt). Heterozygous means two different alleles (Tt).
  • A Punnett square (devised by Reginald Punnett) is a grid that shows all possible gamete combinations. A cross of Tt × Tt gives four boxes, with TT, Tt, Tt and tt.

Law of dominance: Characters are controlled by pairs of factors (alleles). When two different alleles are present, one (the dominant) expresses and the other (the recessive) does not.

Law of segregation (law of purity of gametes): The two alleles of a pair separate during gamete formation. Each gamete receives only one allele. This is the only law that is universally true, because it has no exceptions.

Test cross: An organism with a dominant phenotype is crossed with the homozygous recessive parent to find its genotype. If all offspring show the dominant trait, the unknown parent is homozygous (TT). If the offspring are 1:1 dominant to recessive, the unknown parent is heterozygous (Tt). A dihybrid test cross of AaBb gives 1:1:1:1.

Deviations from Mendelian ratios

Incomplete dominance: The F1 hybrid is intermediate between the parents. In snapdragon (Antirrhinum), a red (RR) × white (rr) cross gives pink (Rr) F1. In F2 the ratio is 1 red : 2 pink : 1 white. The phenotypic ratio and the genotypic ratio are the same (1:2:1). The gene R makes a functional product, but one copy does not give enough of it for full red colour.

Co-dominance: Both alleles express fully in the heterozygote. ABO blood groups are the standard example.

Multiple alleles: A gene has more than two alleles in the population, but any one individual has only two of them. The ABO gene (gene I) has three alleles: IA, IB and i. IA makes sugar type A on the red blood cell surface. IB makes sugar type B. The allele i makes no sugar. IA and IB are both dominant over i. IA and IB are co-dominant with each other.

Blood group (phenotype)Genotype(s)
AIAIA or IAi
BIBIB or IBi
ABIAIB
Oii

Three alleles give six genotypes and four phenotypes. A person with IAIB shows both A and B antigens, which is co-dominance.

Inheritance of two genes (dihybrid cross)

Mendel crossed round yellow seeds (RRYY) with wrinkled green seeds (rryy). F1 was all round yellow (RrYy). Selfing of F1 gives a 16-box Punnett square. The F2 ratio is 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green (9:3:3:1). There are 9 genotypes and 4 phenotypes. Each gene alone still gives a 3:1 ratio.

Law of independent assortment: When two pairs of traits are considered, the alleles of one pair separate independently of the alleles of the other pair. F1 (RrYy) makes four types of gametes (RY, Ry, rY, ry) in equal numbers. This law holds for genes on different chromosomes. It does not hold for genes that are linked.

Chromosomal theory of inheritance

Walter Sutton and Theodor Boveri (1902) noticed that chromosomes behave like Mendel's factors. Both occur in pairs, both separate at gamete formation, and one of each pair goes to a gamete. Sutton used grasshopper cells to study this. They joined these two ideas to form the chromosomal theory of inheritance. Thomas Hunt Morgan and his group later gave experimental proof using the fruit fly, Drosophila melanogaster.

Linkage and recombination

Morgan crossed yellow-bodied, white-eyed female Drosophila with brown-bodied, red-eyed males. The two recessive genes (on the X chromosome) stayed together in the F2 more often than expected. Dihybrid ratio 9:3:3:1 was not obtained. Morgan named this physical association of genes on one chromosome linkage. The new combinations of genes that arise by crossing over are recombination.

  • Genes that are very close on the chromosome are tightly linked, so they show a very low recombination frequency.
  • Genes that are far apart are loosely linked and show a higher recombination frequency.
  • Alfred Sturtevant, a student of Morgan, used the frequency of recombination between gene pairs as a measure of the distance between them. He drew the first genetic maps of chromosomes.

Polygenic inheritance and pleiotropy

Polygenic inheritance: A trait is controlled by three or more genes. Each dominant allele adds to the effect, so the effects are cumulative. Human skin colour and height are examples. With three genes (AaBbCc), a cross gives seven phenotype classes. Their numbers form a bell-shaped curve. The extreme classes (all dominant or all recessive) are the rarest. In AaBbCc × AaBbCc, the all-dominant genotype AABBCC occurs in 1/64 of the offspring.

Pleiotropy: One gene affects more than one trait. Phenylketonuria (PKU) is an example. The gene that makes the enzyme phenylalanine hydroxylase is faulty. Phenylalanine builds up and is changed to other compounds. The effects are mental retardation, and reduced hair and skin pigmentation.

Sex determination

TypeMaleFemaleExamples
XX-XOXO (one X, 2n odd)XXGrasshopper
XX-XYXY (heterogametic)XX (homogametic)Humans, Drosophila
ZZ-ZWZZ (homogametic)ZW (heterogametic)Birds
HaplodiploidyHaploid (n = 16), from unfertilised eggDiploid (2n = 32), from fertilised eggHoney bee
  • In humans 22 pairs are autosomes and one pair is the sex chromosomes. Half of the sperm carry X and half carry Y. All eggs carry X. The father's sperm decides the sex of the child.
  • In the honey bee, the queen and the workers are diploid females. Drones are haploid males formed by parthenogenesis. Drones make sperm by mitosis. This system is called haplodiploidy.

Mutation

A mutation is a sudden, heritable change in DNA.

  • Point mutation: change in a single base pair. Sickle-cell anaemia is an example, where one base substitution changes one amino acid.
  • Frameshift mutation: deletion or insertion of one or two base pairs changes the reading frame of the code.
  • Chromosomal aberrations: changes in chromosome structure (deletion, duplication, inversion, translocation) or in chromosome number. Gain or loss of one or a few chromosomes is aneuploidy, caused by failure of separation (non-disjunction) in meiosis. Gain of whole sets of chromosomes is polyploidy. Polyploidy occurs when the cell does not divide after the chromosomes have doubled, and it is common in plants.
  • Mutagens are agents that cause mutation, such as UV radiation and some chemicals.

Genetic disorders and pedigree analysis

A pedigree is a family tree for one trait. Symbols: a square is a male, a circle is a female, a horizontal line joining them shows mating, a vertical line shows offspring, and shaded symbols show affected persons. A double line between mates shows a mating between relatives (consanguineous). Generations are marked with Roman numerals.

DisorderType of inheritanceKey facts
HaemophiliaX-linked recessiveBlood does not clot well because a clotting protein is missing. A heterozygous female is a carrier. A female is affected only if her mother is a carrier or affected and her father is affected.
Colour blindnessX-linked recessiveDefect in red or green cone cells. It is common in males (about 8%) and rare in females (about 0.4%).
Sickle-cell anaemiaAutosomal recessiveSubstitution of valine for glutamic acid at position 6 of the β chain of haemoglobin (HbS). Red cells become sickle-shaped.
ThalassaemiaAutosomal recessiveReduced rate of making α or β globin chains. The result is anaemia.
PhenylketonuriaAutosomal recessiveLack of phenylalanine hydroxylase.

In sex-linked recessive traits, the disease passes from an affected father to his grandson through his carrier daughter. A son gets his X only from his mother, so a carrier mother passes the disease to half of her sons.

Chromosomal disorderKaryotypeFeatures
Down syndrome47, trisomy of chromosome 21Short height, small round head, furrowed tongue, partly open mouth, broad palm with a characteristic crease, delayed physical and mental development.
Klinefelter syndrome47, XXYMale with overall masculine build, but breast development (gynaecomastia) and other feminine features. Sterile.
Turner syndrome45, XOFemale, short, rudimentary ovaries, lacks secondary sexual characters. Sterile.

The risk of Down syndrome rises with the mother's age. The cause of all three is non-disjunction.

NEET traps

  • Law of dominance is not universal (incomplete and co-dominance exist), but segregation always holds.
  • Incomplete dominance gives 1:2:1 phenotypes. Complete dominance gives 3:1 phenotypes. In both cases the genotype ratio is 1:2:1.
  • Co-dominance (both alleles show, as in AB blood group) differs from incomplete dominance (blended, intermediate trait as in pink flower).
  • A person has only two ABO alleles, though the population has three.
  • Male is heterogametic in humans, but female is heterogametic in birds.
  • Pleiotropy is one gene with many effects. Polygenic is many genes with one trait.
  • Test cross uses the homozygous recessive parent. A back cross uses either parent.
  • Tightly linked genes have low recombination. Far apart genes have high recombination.
  • Haemophilia and colour blindness are X-linked. Sickle-cell, thalassaemia and PKU are autosomal.
  • Drone honey bees are haploid and have no father. Queen and workers are diploid.

One-liners

  • 1. Mendel worked on Pisum sativum and studied seven pairs of contrasting traits.
  • 2. Dominant traits in pea: tall, violet, axial, inflated, green pod, round and yellow seed.
  • 3. Monohybrid F2 phenotypic ratio is 3:1 and genotypic ratio is 1:2:1.
  • 4. Test cross uses the homozygous recessive parent.
  • 5. Snapdragon red × white gives pink F1 and 1:2:1 F2.
  • 6. IA and IB are co-dominant, and both are dominant over i.
  • 7. Blood group O has genotype ii.
  • 8. Dihybrid F2 ratio is 9:3:3:1, with 9 genotypes and 4 phenotypes.
  • 9. Sutton and Boveri proposed the chromosomal theory of inheritance.
  • 10. Morgan used Drosophila and discovered linkage.
  • 11. Sturtevant made the first genetic maps using recombination frequency.
  • 12. Human skin colour is polygenic.
  • 13. PKU is pleiotropic and autosomal recessive.
  • 14. In birds the female is ZW and the male is ZZ.
  • 15. Down syndrome is trisomy 21, Klinefelter is XXY and Turner is XO.

Practice questions

  1. In Mendel's pea plant, the dominant form of the pod colour trait is

    1. Constricted
    2. Yellow
    3. Green
    4. Violet
    Answer

    C. Green

    Green pod is dominant over yellow pod. Violet is a flower colour and constricted is a pod shape.

  2. How many phenotypic classes appear in the F2 generation of a Mendelian dihybrid cross with complete dominance?

    1. 2
    2. 4
    3. 8
    4. 16
    Answer

    B. 4

    The 9:3:3:1 ratio has four phenotypic classes. The 16 is the number of boxes in the Punnett square.

  3. A test cross is made by crossing an individual that shows the dominant phenotype with

    1. an F1 hybrid
    2. a homozygous dominant individual
    3. a heterozygous individual
    4. a homozygous recessive individual
    Answer

    D. a homozygous recessive individual

    The homozygous recessive parent gives only recessive gametes, so the offspring reveal the genotype of the test individual.

  4. Selfing of a pink-flowered snapdragon plant gives red, pink and white flowers in the ratio

    1. 1 : 2 : 1
    2. 1 : 1
    3. 9 : 3 : 3 : 1
    4. 3 : 1
    Answer

    A. 1 : 2 : 1

    In incomplete dominance, Rr x Rr gives 1 RR red : 2 Rr pink : 1 rr white.

  5. How many different genotypes are possible for the human ABO blood group gene?

    1. 3
    2. 6
    3. 4
    4. 9
    Answer

    B. 6

    IAIA, IAi, IBIB, IBi, IAIB and ii are the six genotypes. They give four phenotypes.

  6. Who used the frequency of recombination between gene pairs as a measure of the distance between them on a chromosome?

    1. Sutton
    2. Boveri
    3. Sturtevant
    4. Morgan
    Answer

    C. Sturtevant

    Alfred Sturtevant, a student of Morgan, used recombination frequency to make the first genetic maps.

  7. Genes that are tightly linked on a chromosome show

    1. independent assortment
    2. a 9 : 3 : 3 : 1 ratio in F2
    3. a high frequency of recombination
    4. a very low frequency of recombination
    Answer

    D. a very low frequency of recombination

    Close genes are rarely separated by crossing over, so recombinants are few.

  8. Human skin colour is an example of

    1. polygenic inheritance
    2. co-dominance
    3. incomplete dominance of a single gene
    4. pleiotropy
    Answer

    A. polygenic inheritance

    Skin colour is controlled by several genes whose effects add up, which gives a continuous range of shades.

  9. The enzyme that is defective in phenylketonuria is

    1. amylase
    2. tyrosinase
    3. phenylalanine hydroxylase
    4. DNA ligase
    Answer

    C. phenylalanine hydroxylase

    Without phenylalanine hydroxylase, phenylalanine is not changed to tyrosine and it builds up.

  10. In honey bees, drones develop from

    1. fertilised eggs
    2. unfertilised eggs by parthenogenesis
    3. fertilised eggs of workers only
    4. zygotes with ZW chromosomes
    Answer

    B. unfertilised eggs by parthenogenesis

    Drones are haploid males (n = 16) that come from unfertilised eggs.

  11. A human female with a 45, XO chromosome complement has

    1. Turner syndrome
    2. Down syndrome
    3. Patau syndrome
    4. Klinefelter syndrome
    Answer

    A. Turner syndrome

    Turner syndrome is monosomy of the sex chromosome (45, XO). It gives a sterile, short female.

  12. In sickle-cell anaemia the change in the haemoglobin β chain is

    1. valine replaced by glutamic acid at position 6
    2. glutamic acid deleted at position 6
    3. glycine replaced by alanine at position 6
    4. glutamic acid replaced by valine at position 6
    Answer

    D. glutamic acid replaced by valine at position 6

    One base substitution changes the sixth amino acid from glutamic acid to valine, which makes HbS.

  13. A frameshift mutation is caused by

    1. deletion or insertion of one or two base pairs
    2. deletion of exactly three base pairs
    3. substitution of a single base by another
    4. doubling of the chromosome number
    Answer

    A. deletion or insertion of one or two base pairs

    Adding or removing one or two bases shifts the reading frame of every codon after that point.

  14. In birds, the heterogametic sex is

    1. the male, with ZZ chromosomes
    2. the female, with ZW chromosomes
    3. both male and female
    4. neither, because birds lack sex chromosomes
    Answer

    B. the female, with ZW chromosomes

    In the ZZ-ZW system the female makes two types of gametes (Z and W).

  15. The approximate frequency of red-green colour blindness among human males is

    1. about 25 per cent
    2. about 0.4 per cent
    3. about 8 per cent
    4. about 50 per cent
    Answer

    C. about 8 per cent

    Males have one X, so one recessive allele is enough. About 8% of males and about 0.4% of females are affected.

  16. Which of the following disorders is autosomal recessive?

    1. Turner syndrome
    2. Haemophilia
    3. Colour blindness
    4. Sickle-cell anaemia
    Answer

    D. Sickle-cell anaemia

    Sickle-cell anaemia is controlled by an autosomal recessive gene. Haemophilia and colour blindness are X-linked, and Turner is chromosomal.

  17. Given below are two statements: Statement I: The law of segregation holds true even when a gene shows incomplete dominance. Statement II: In incomplete dominance, the F2 phenotypic ratio is 3 : 1.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Statement I is correct because the two alleles still separate in gametes. Statement II is incorrect because the F2 ratio is 1 : 2 : 1.

  18. Given below are two statements: Statement I: The sex of a human child is decided by the sperm. Statement II: All normal human eggs carry an X chromosome.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Sperm carry X or Y while all eggs carry X, so the sperm decides the sex of the child.

  19. Given below are two statements: Statement I: Haemophilia is an X-linked dominant disorder. Statement II: A heterozygous female for the haemophilia allele is a carrier.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Haemophilia is X-linked recessive, so Statement I is incorrect. A heterozygous female has one normal allele and is a carrier, so Statement II is correct.

  20. Given below are two statements: Statement I: In the ZZ-ZW system the female is homozygous for the sex chromosomes. Statement II: Drone honey bees are diploid.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    The ZW female is heterozygous, and drones are haploid. Both statements are incorrect.

  21. Which of the following statements are correct? 1. Polygenic traits show a bell-shaped distribution in a population. 2. A pleiotropic gene affects more than one trait. 3. Sutton used Drosophila to study the chromosomal theory.

    1. 1 and 2 only
    2. 2 and 3 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    Statements 1 and 2 are correct. Sutton used grasshopper cells, and Morgan used Drosophila.

  22. Given below are two statements: Statement I: Genes that are far apart on a chromosome show a higher recombination frequency than genes that are close together. Statement II: The law of independent assortment holds for genes that are linked.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    Crossing over separates distant genes more often, so Statement I is correct. Linked genes do not assort independently, so Statement II is incorrect.

  23. Given below are two statements: Statement I: Thalassaemia is a sex-linked disorder. Statement II: In thalassaemia the rate of synthesis of one or more globin chains is reduced.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    Thalassaemia is autosomal recessive, so Statement I is incorrect. Statement II is correct.

  24. Which of the following statements about ABO blood groups are correct? 1. The I gene has three alleles. 2. A person carries only two of these alleles. 3. IA and i are co-dominant.

    1. 2 and 3 only
    2. 1 and 2 only
    3. 1 and 3 only
    4. 1, 2 and 3
    Answer

    B. 1 and 2 only

    IA and IB are co-dominant, while IA is dominant over i. So statement 3 is wrong.

  25. Given below are two statements: Statement I: A couple with blood groups A and B can have an AB child only if both are heterozygous. Statement II: A couple with blood groups A and B can have an O child if both are heterozygous.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    IAIA x IBIB gives only AB children, so Statement I is incorrect. IAi x IBi can give an ii (O) child, so Statement II is correct.

  26. Given below are two statements: Statement I: In a dihybrid cross, each gene alone gives a 3 : 1 ratio in F2. Statement II: The F1 hybrid RrYy makes four types of gametes in equal numbers.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both statements are correct. The 9:3:3:1 ratio is made of two independent 3:1 ratios.

  27. Assertion (A): A test cross of a heterozygous (Tt) plant gives a 1 : 1 ratio of tall to dwarf offspring. Reason (R): A heterozygous plant makes two types of gametes in equal numbers.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

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

    Tt makes T and t gametes equally. The tt parent gives only t, so the offspring are 1 Tt : 1 tt.

  28. Assertion (A): The heterozygous snapdragon plant has pink flowers. Reason (R): In snapdragon the red allele is completely dominant over the white allele.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    Pink shows incomplete dominance, so R is false. One copy of the red allele does not give a full red colour.

  29. Assertion (A): Genes that are linked do not show independent assortment. Reason (R): Sturtevant used recombination frequency to map the positions of genes.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

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

    Both are true, but the mapping method does not explain why linked genes fail to assort independently. Physical closeness on the chromosome is the cause.

  30. Assertion (A): The human male is homogametic. Reason (R): The human male makes two types of sperm, carrying X or Y.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    The male (XY) is heterogametic, so A is false. R is true.

  31. Assertion (A): Individuals with Down syndrome show delayed mental development. Reason (R): They have an extra copy of chromosome 21 caused by non-disjunction.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

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

    Trisomy 21 arises from failure of chromosome 21 to separate in meiosis, and the extra gene dosage causes the features.

  32. Assertion (A): A haemophilic female is rarely seen in a population. Reason (R): A female becomes haemophilic only if both of her X chromosomes carry the recessive allele.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

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

    Both are true and R explains A. The father must be haemophilic and the mother a carrier or affected, so the chance is low.

  33. Match List I with List II. List I: a. A gene with more than two alleles b. One gene with many effects c. Many genes controlling one trait d. Both alleles expressing in a heterozygote. List II: i. Pleiotropy ii. Polygenic inheritance iii. Co-dominance iv. Multiple allelism.

    1. a-iii, b-i, c-ii, d-iv
    2. a-iv, b-ii, c-i, d-iii
    3. a-iv, b-i, c-ii, d-iii
    4. a-i, b-iv, c-ii, d-iii
    Answer

    C. a-iv, b-i, c-ii, d-iii

    a is multiple allelism (iv), b is pleiotropy (i), c is polygenic (ii) and d is co-dominance (iii).

  34. Match List I with List II. List I: a. Grasshopper b. Birds c. Honey bee d. Humans. List II: i. ZZ-ZW ii. XX-XY iii. Haplodiploidy iv. XX-XO.

    1. a-iii, b-i, c-iv, d-ii
    2. a-iv, b-i, c-iii, d-ii
    3. a-i, b-iv, c-iii, d-ii
    4. a-iv, b-ii, c-iii, d-i
    Answer

    B. a-iv, b-i, c-iii, d-ii

    Grasshopper is XX-XO (iv), birds are ZZ-ZW (i), honey bee is haplodiploid (iii) and humans are XX-XY (ii).

  35. Match List I with List II. List I: a. Down syndrome b. Klinefelter syndrome c. Turner syndrome d. Sickle-cell anaemia. List II: i. 45, XO ii. 47, XXY iii. Trisomy of chromosome 21 iv. Autosomal recessive disorder.

    1. a-ii, b-iii, c-i, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-iii, b-i, c-ii, d-iv
    4. a-iv, b-ii, c-i, d-iii
    Answer

    B. a-iii, b-ii, c-i, d-iv

    Down is trisomy 21 (iii), Klinefelter is 47 XXY (ii), Turner is 45 XO (i) and sickle-cell anaemia is autosomal recessive (iv).

  36. Match List I with List II. List I: a. Sutton and Boveri b. Morgan c. Sturtevant d. Mendel. List II: i. First genetic maps ii. Chromosomal theory of inheritance iii. Linkage in Drosophila iv. Law of segregation from pea experiments.

    1. a-ii, b-i, c-iii, d-iv
    2. a-iii, b-ii, c-i, d-iv
    3. a-iv, b-iii, c-i, d-ii
    4. a-ii, b-iii, c-i, d-iv
    Answer

    D. a-ii, b-iii, c-i, d-iv

    Sutton and Boveri proposed the chromosomal theory (ii), Morgan found linkage (iii), Sturtevant made maps (i) and Mendel did the pea work (iv).

  37. Match List I with List II. List I: a. Haemophilia b. Thalassaemia c. Phenylketonuria d. Colour blindness. List II: i. X-linked recessive, failure of blood clotting ii. Reduced synthesis of globin chains iii. Lack of phenylalanine hydroxylase iv. X-linked recessive, defect in cone cells.

    1. a-ii, b-i, c-iii, d-iv
    2. a-i, b-iii, c-ii, d-iv
    3. a-i, b-ii, c-iii, d-iv
    4. a-iv, b-ii, c-iii, d-i
    Answer

    C. a-i, b-ii, c-iii, d-iv

    Haemophilia affects clotting (i), thalassaemia affects globin chain synthesis (ii), PKU lacks phenylalanine hydroxylase (iii) and colour blindness is a cone defect (iv).

  38. Match List I with List II. List I: a. Monohybrid test cross b. Dihybrid F2 c. Incomplete dominance F2 d. Dihybrid test cross. List II: i. 1 : 2 : 1 ii. 9 : 3 : 3 : 1 iii. 1 : 1 iv. 1 : 1 : 1 : 1.

    1. a-iii, b-i, c-ii, d-iv
    2. a-iv, b-ii, c-i, d-iii
    3. a-iii, b-ii, c-i, d-iv
    4. a-i, b-ii, c-iii, d-iv
    Answer

    C. a-iii, b-ii, c-i, d-iv

    Tt x tt gives 1:1 (iii), the dihybrid F2 is 9:3:3:1 (ii), incomplete dominance F2 is 1:2:1 (i) and AaBb x aabb gives 1:1:1:1 (iv).

  39. A heterozygous tall pea plant (Tt) is crossed with a dwarf plant (tt). What percentage of the offspring is expected to be dwarf?

    1. 75 per cent
    2. 0 per cent
    3. 25 per cent
    4. 50 per cent
    Answer

    D. 50 per cent

    Tt x tt gives 1 Tt : 1 tt, so half of the offspring are dwarf.

  40. A man with blood group A, whose mother had group O, marries a woman with blood group AB. Which blood groups are possible among their children?

    1. A and AB only
    2. A, B and AB
    3. A, B, AB and O
    4. B and O only
    Answer

    B. A, B and AB

    The man is IAi and the woman is IAIB. Children can be IAIA, IAi (A), IAIB (AB) and IBi (B). No child can be O.

  41. A woman who is a carrier of haemophilia marries a normal man. What fraction of their sons is expected to be haemophilic?

    1. None
    2. One half
    3. All
    4. One fourth
    Answer

    B. One half

    Sons receive the mother's X. Half get the X with the recessive allele, so half are affected. Daughters get the normal X from the father.

  42. Two plants of genotype AaBb are crossed (independent assortment, complete dominance). What is the fraction of offspring that are aaB_ (aa and showing the dominant B trait)?

    1. 1/16
    2. 9/16
    3. 1/4
    4. 3/16
    Answer

    D. 3/16

    aa is 1/4 and B_ is 3/4. Their product is 3/16.

  43. A colour-blind man marries a homozygous normal woman. What is expected in their children?

    1. All daughters are carriers and all sons are normal
    2. All children are normal and none is a carrier
    3. Half of the daughters are colour-blind
    4. All sons are colour-blind
    Answer

    A. All daughters are carriers and all sons are normal

    The father gives his X (with the allele) to daughters and Y to sons. Daughters are carriers. Sons get the mother's normal X.

  44. Height in a plant is controlled by three genes (AaBbCc), each dominant allele adding equally. In AaBbCc x AaBbCc, what fraction of offspring shows the extreme phenotype with all six alleles dominant (AABBCC)?

    1. 1/16
    2. 1/8
    3. 1/64
    4. 1/32
    Answer

    C. 1/64

    Each gene gives 1/4 of the dominant homozygote, so (1/4)^3 = 1/64.

  45. A couple, one with blood group A and the other with group B, has a child with blood group O. The genotypes of the parents must be

    1. IAIA and IBi
    2. IAi and IBi
    3. IAIA and IBIB
    4. IAi and IBIB
    Answer

    B. IAi and IBi

    The child ii must receive an i from each parent, so both parents are heterozygous.

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