NCERT Exemplar Class 12 Biology Ch 4 Inheritance & Variation PDF

Correct option: (b) Form one linkage group.

Concept used. Linkage is the tendency of genes that lie on the same chromosome to be inherited together, because they do not assort independently during meiosis. The full set of genes carried on one chromosome is called a linkage group. The number of linkage groups in an organism equals its haploid chromosome number (\(n\)).

1. Genes are physically located along a chromosome. When that chromosome moves to a gamete during meiosis, every gene on it moves together as one unit (unless crossing over separates them).
2. Therefore all genes on a single chromosome are inherited as a block. This whole block is, by definition, one linkage group: there is exactly one linkage group per chromosome.
3. Check the distractors. (a) is wrong: relative distance affects the recombination frequency between genes, not how many linkage groups they form. (c) is wrong: same-chromosome genes do form a linkage group. (d) confuses linkage with gene interaction (epistasis), which is unrelated.

Option (b): all genes on one chromosome form a single linkage group.

Correct option: (a) Aneuploidy.

Concept used. Aneuploidy is the gain or loss of one or a few individual chromosomes from the normal diploid set, caused by failure of chromosomes to separate (non-disjunction) during meiosis. It is written as \(2n\pm1\) (one chromosome extra/missing) or \(2n\pm2\). This is distinct from polyploidy, where the entire genome is multiplied (\(3n\), \(4n\), \(\ldots\)).

1. The notation \(2n+1\) means the diploid set plus one extra chromosome (trisomy); \(2n-1\) means one chromosome missing (monosomy); \(2n\pm2\) are tetrasomy/nullisomy-type changes.
2. Each of these changes the count by only one or two chromosomes, not by whole sets. By definition that is aneuploidy.
3. Eliminate distractors: (b) polyploidy and (c) allopolyploidy involve whole extra genomes (\(3n\), \(4n\)), not \(\pm1\) or \(\pm2\). (d) Monosomy is only the \(2n-1\) case, so it is too narrow to cover all four conditions listed.

Option (a): \(2n\pm1\) and \(2n\pm2\) are aneuploidy.

Correct option: (a) A direct relationship.

Concept used. Recombination frequency (RF) is the percentage of offspring that show new combinations of linked genes, produced by crossing over during meiosis. Sturtevant showed that the farther apart two genes lie on a chromosome, the more often a crossover occurs between them, so RF rises with distance. One map unit (centimorgan) \(=1\%\) recombination.

1. Crossing over happens at random points along a chromosome. The longer the stretch between two genes, the larger the chance that a crossover falls between them.
2. More crossovers between the genes means more recombinant gametes, so a higher recombination percentage.
3. Hence map distance and RF move in the same direction: as distance increases, RF increases. This is a direct relationship. (Inverse would mean RF falls as distance grows, which is the opposite of what is observed.)

Option (a): gene distance and recombination % are directly related.

Correct option: (d) Sex-linked recessive.

Concept used. An X-linked recessive disorder sits on the X chromosome and shows only when no normal allele is present. A female has two X chromosomes, so one normal allele masks the defective one and she is a healthy carrier (\(X^{A}X^{a}\)). A male has only one X (\(X^{a}Y\)), so a single defective allele is enough to make him affected.

1. The female is phenotypically normal but a carrier: genotype \(X^{A}X^{a}\). The normal father is \(X^{A}Y\).
2. Work the cross. Sons get their single X from the mother: half receive \(X^{A}\) (normal) and half receive \(X^{a}\) (affected). Daughters get \(X^{A}\) from the father, so all are at least carriers and none is affected. \[ X^{A}X^{a}\times X^{A}Y \;\Rightarrow\; \tfrac12\,X^{A}X^{A},\ \tfrac12\,X^{A}X^{a},\ \tfrac12\,X^{A}Y,\ \tfrac12\,X^{a}Y. \]
3. Only some sons (\(X^{a}Y\)) are affected and no daughters are. That pattern (carrier mother, some sons affected) is the signature of a sex-linked recessive trait, so (d) is correct. Autosomal traits would affect both sexes equally; sex-linked dominant would affect the carrier mother herself.

Option (d): Sex-linked (X-linked) recessive.

Correct option: (d) G U G.

Concept used. Sickle cell anaemia is caused by a point mutation in the gene for the beta chain of haemoglobin: the sixth codon GAG (glutamic acid) changes to GUG, so valine is inserted instead of glutamic acid. The codons given here are mRNA triplets; valine is coded by GUU, GUC, GUA and GUG.

1. The mutation is a single base change: the middle base \(A\) of GAG (glutamic acid) becomes \(U\), giving GUG.
2. GUG is one of the four valine codons, so it codes for valine. This is option (d).
3. Eliminate: GGG codes glycine, AAG codes lysine, GAA codes glutamic acid. None of these is valine.

Option (d): G U G codes for valine.

Correct option: (b) Co-dominance.

Concept used. Co-dominance is the condition where both alleles in a heterozygote are fully and independently expressed, so the phenotype shows the effect of both alleles, not a blend. In the ABO blood-group system the alleles \(I^{A}\) and \(I^{B}\) are co-dominant.

1. Allele \(I^{A}\) makes antigen A on red cells; allele \(I^{B}\) makes antigen B.
2. In genotype \(I^{A}I^{B}\) both alleles are expressed at the same time, so both antigen A and antigen B appear on the red cells. The blood group is AB, showing both alleles.
3. Because both alleles show fully and separately (no blending, no masking), this is co-dominance, option (b). It is not incomplete dominance: incomplete dominance would give an intermediate ``in-between'' antigen, which does not happen.

Option (b): Co-dominance (\(I^{A}\) and \(I^{B}\) both expressed).

Correct option: (d) Peacock.

Concept used. In the ZZ/ZW system of sex determination the female is heterogametic (ZW) and the male is homogametic (ZZ). This system is found in birds (and some reptiles, fishes, butterflies). A peacock is a bird.

1. Birds use ZW sex determination: male birds are ZZ, female birds are ZW. The peacock is a bird, so it follows ZZ/ZW.
2. Eliminate: cockroach uses the XX/XO system (insects); platypus has a complex multiple X-Y system; snails are largely hermaphrodite. None of these is the simple ZZ/ZW type.

Option (d): Peacock (a bird, ZZ male / ZW female).

Correct option: (b) Tt and Tt.

Concept used. Tallness (\(T\)) is dominant over dwarfness (\(t\)). A dwarf plant must be homozygous recessive (\(tt\)). For a \(tt\) offspring to appear, each parent must contribute a \(t\) allele, so both tall parents must carry a hidden \(t\), i.e. both are heterozygous \(Tt\).

1. A dwarf offspring is \(tt\). It needs one \(t\) from each parent.
2. Both parents are tall, so neither is \(tt\). The only tall genotype that can donate \(t\) is \(Tt\). Hence both parents are \(Tt\).
3. Verify with the Punnett square of \(Tt\times Tt\): \[ Tt\times Tt \;\Rightarrow\; \tfrac14\,TT,\ \tfrac12\,Tt,\ \tfrac14\,tt. \] This gives \(3\) tall \(:1\) dwarf, so ``few dwarf plants'' appear, matching the question. Option (b) is correct.
4. Eliminate: TT\(\times\)Tt and TT\(\times\)TT give no \(tt\) offspring (no dwarfs). Tt\(\times\)tt gives \(1{:}1\), which is a test cross with \(50\%\) dwarfs, not ``few''.

Option (b): both parents are \(Tt\) (heterozygous tall).

Correct option: (d) The alleles of two genes are segregating independently.

Concept used. Mendel's Law of Independent Assortment states that the alleles of two different genes segregate into gametes independently of each other. A dihybrid cross between two double heterozygotes (\(RrYy\times RrYy\)) then produces the \(9{:}3{:}3{:}1\) phenotypic ratio.

1. The \(9{:}3{:}3{:}1\) ratio is the direct mathematical result of combining two independent \(3{:}1\) ratios: \[ (3{:}1)\times(3{:}1)=9{:}3{:}3{:}1. \]
2. This multiplication is only valid if the two genes assort independently. So observing \(9{:}3{:}3{:}1\) tells us the genes are unlinked and their alleles segregate independently. That is option (d).
3. Eliminate: gene interaction (a) distorts the ratio (e.g. \(9{:}7\), \(12{:}3{:}1\)); multiple allelism (c) involves more than two alleles of one gene, not two genes; ``multigenic'' (b) describes polygenic traits with continuous variation, not a clean \(9{:}3{:}3{:}1\).

Option (d): the two genes' alleles segregate independently.

Correct option: (c) Linkage.

Concept used. Variation among siblings arises from processes that create new allele combinations: independent assortment, crossing over and mutation. Linkage does the opposite: it keeps genes on the same chromosome together, reducing the formation of new combinations.

1. Independent assortment shuffles non-homologous chromosomes, crossing over swaps segments between homologues, and mutation creates brand-new alleles. All three increase variation.
2. Linkage holds linked genes as one unit, so they pass together and do not generate new combinations. It restricts variation rather than producing it.
3. Therefore the process that does not create sibling variation is linkage, option (c).

Option (c): Linkage does not produce variation among siblings.

Correct option: (a) Non-homologous chromosomes.

Concept used. Independent assortment works only when two genes are on different (non-homologous) chromosomes, because separate chromosome pairs line up and separate independently at meiosis. Two genes on the same chromosome are linked and do not assort independently.

1. Independent assortment needs the two genes to be carried by two different chromosome pairs that orient randomly at metaphase I.
2. Genes on the same chromosome (option d) are linked, so they cannot assort independently. Extranuclear genes (option c) follow maternal, non-Mendelian inheritance.
3. Two genes on non-homologous chromosomes belong to different homologous pairs, so each pair segregates independently. This is the condition for independent assortment, option (a).

Option (a): non-homologous chromosomes (genetically exact). NCERT key states (b).

Correct option: (c) Pleiotropy.

Concept used. Pleiotropy is the phenomenon in which a single gene controls or influences several different phenotypic traits at once, usually because the gene's product takes part in many metabolic pathways.

1. ``One gene \(\to\) many effects'' is the literal definition of pleiotropy. Example: the phenylketonuria gene affects mental ability, hair colour and skin pigmentation together.
2. Eliminate: multiple allelism (a) is many alleles of one gene; polygeny (d) is many genes acting on one trait (the reverse); mosaicism (b) is two genetically different cell populations in one individual.

Option (c): Pleiotropy (one gene, many effects).

Correct option: (a) Males and females, respectively.

Concept used. Many insects use the XX/XO sex determination system. Females are XX (one extra sex chromosome) and have an even diploid number; males are XO (only one sex chromosome, no Y), so they have one chromosome fewer than females, giving an odd number.

1. Females XX: both sex chromosomes present, total \(=18\) (even).
2. Males XO: one sex chromosome missing, total \(=18-1=17\) (odd).
3. So the 17-chromosome insects are males and the 18-chromosome insects are females. That is option (a). The difference of exactly one chromosome is the hallmark of the XO system.

Option (a): 17 = males (XO), 18 = females (XX).

Correct option: (b) Mendelian trait.

Concept used. Pedigree analysis traces a single gene through a family tree to find its inheritance pattern (dominant/recessive, autosomal/sex-linked). It works only for traits controlled by one gene with clear-cut alternatives, i.e. Mendelian traits.

1. Pedigree symbols mark each person as affected or unaffected for one clearly inherited character. This needs a single-gene, discontinuous (qualitative) trait.
2. A single-gene, discrete trait that follows Mendel's rules is a Mendelian trait, so the character studied is equivalent to a Mendelian trait, option (b).
3. Eliminate: quantitative/polygenic traits (a, c) vary continuously and cannot be scored as simple affected vs unaffected; maternal trait (d) is extranuclear, not the usual pedigree subject.

Option (b): a Mendelian (single-gene) trait.

Correct option: (b) Observations that the F1 offspring of a cross between two contrasting parents shows only one character without any blending.

Concept used. Mendel concluded that hereditary factors (genes) are discrete particles, not blendable fluids, because in the F1 only one of the two contrasting parental characters appeared, fully and unblended (e.g. tall \(\times\) dwarf gave all tall, not medium-height plants).

1. If factors blended, a tall \(\times\) dwarf cross would give intermediate plants. Instead Mendel saw all tall F1 plants: one character, unblended.
2. The recessive character vanished in F1 but reappeared unchanged in F2. A factor that can hide and then reappear intact must be a discrete, independent particle.
3. Hence his proposition rests on the F1 non-blending observation, option (b). (a), (c), (d) describe later steps that confirmed, but did not originate, the discreteness idea.

Option (b): F1 shows one character with no blending.

Correct option: (d) \(1:1\).

Concept used. When two genes are completely linked, they travel together as one unit, so the dihybrid F1 (\(AB/ab\)) makes only the two parental gamete types (\(AB\) and \(ab\)) and no recombinant gametes. A test cross with \(aabb\) then reveals the gamete ratio directly.

1. For completely linked genes, F1 (\(AB/ab\)) produces only parental gametes: \(AB\) and \(ab\), in equal amounts \(1:1\).
2. Cross with \(aabb\) (which makes only \(ab\) gametes): \[ AB \times ab \to AaBb;\qquad ab\times ab \to aabb. \] Offspring are \(AaBb\) and \(aabb\) in a \(1:1\) ratio (two classes only, the two parental types).
3. Compare with unlinked genes: a normal dihybrid test cross gives \(1:1:1:1\) (four classes). Because linkage removes the recombinant classes, only the \(1:1\) parental ratio remains, option (d).

Option (d): \(1:1\) (only parental types, genes linked).

Correct option: (d) Phenotypes \(=4\); genotypes \(=9\).

Concept used. In a Mendelian dihybrid cross (\(RrYy\times RrYy\)), each gene independently gives \(3\) genotypes (\(RR,Rr,rr\)) and \(2\) phenotypes (dominant, recessive). For two independent genes the totals multiply.

1. Phenotypes: each gene has \(2\) phenotype classes, so two genes give \(2\times2=4\) phenotypes (the \(9{:}3{:}3{:}1\) classes).
2. Genotypes: each gene has \(3\) genotype classes, so two genes give \(3\times3=9\) genotypes.
3. Therefore F2 has \(4\) phenotypes and \(9\) genotypes, option (d). The \(16\) in option (a) is the size of the Punnett square (16 boxes), not the number of distinct genotypes.

Option (d): \(4\) phenotypes, \(9\) genotypes.

Correct option: (c) Both mother and father are heterozygous.

Concept used. In the ABO system, blood group O is genotype \(ii\) (homozygous recessive). For a child to be \(ii\), each parent must pass an \(i\) allele. So the group-A mother must carry \(i\) (\(I^{A}i\)) and the group-B father must carry \(i\) (\(I^{B}i\)): both are heterozygous.

1. Child is group O \(=ii\). One \(i\) comes from the mother, one from the father.
2. Mother is group A, so her genotype is \(I^{A}I^{A}\) or \(I^{A}i\). To donate \(i\) she must be \(I^{A}i\) (heterozygous).
3. Father is group B, so \(I^{B}I^{B}\) or \(I^{B}i\). To donate \(i\) he must be \(I^{B}i\) (heterozygous).
4. Cross \(I^{A}i\times I^{B}i\) gives \(I^{A}I^{B}\,(AB),\ I^{A}i\,(A),\ I^{B}i\,(B),\ ii\,(O)\) in \(1{:}1{:}1{:}1\), so an O child is possible. Both parents are heterozygous, option (c).

Option (c): mother \(I^{A}i\), father \(I^{B}i\) (both heterozygous).

Concept used. A test cross is a cross between an individual showing the dominant phenotype (unknown genotype) and a homozygous recessive individual. Because the recessive parent contributes only recessive alleles, the offspring phenotypes directly reveal the gametes, and hence the genotype, of the unknown parent.

1. The cross between an F1 (dominant phenotype) and the homozygous recessive parent is called a test cross (also a back cross when the recessive is a parent).
2. Usefulness: it tells whether the dominant individual is homozygous (\(TT\)) or heterozygous (\(Tt\)). If \(TT\times tt\), all offspring are tall. If \(Tt\times tt\), offspring are \(1\) tall \(:1\) dwarf. The appearance of any recessive offspring proves the test individual was heterozygous.

It is a test cross; it reveals an unknown dominant individual's genotype (homozygous vs heterozygous).

Concept used. Mendel's Law of Independent Assortment requires the two genes to be on different chromosomes. Genes on the same chromosome are linked and are inherited together, so they would not assort independently.

1. Yes, the results would have been different. If the seven chosen characters had been on the same chromosome, the genes would be linked.
2. Linked genes pass together into gametes (parental combinations dominate), so the dihybrid F2 ratio would deviate from \(9{:}3{:}3{:}1\), and the Law of Independent Assortment would not hold. (Mendel's Law of Segregation, which concerns a single gene, would still hold.)

Yes: same-chromosome genes are linked, so independent assortment (\(9{:}3{:}3{:}1\)) would fail.

Concept used. Controlled cross pollination is the deliberate transfer of pollen from a chosen male parent to the stigma of a chosen female parent, preventing self-pollination. Emasculation is the removal of anthers from a bisexual flower before they mature, so it cannot self-pollinate.

1. Steps: (i) Select the female parent and emasculate the bud (remove anthers before they dehisce). (ii) Bag the emasculated flower to stop unwanted pollen. (iii) Collect mature pollen from the chosen male parent. (iv) Dust this pollen on the mature stigma of the bagged female flower. (v) Re-bag and label/tag the flower until fruit sets.
2. Cucurbits (e.g. gourd, pumpkin) are unisexual: each flower is either male (staminate) or female (pistillate). A female flower has no anthers, so it cannot self-pollinate.
3. Therefore emasculation is not needed in cucurbits; only bagging of the female (pistillate) flower is required to keep out unwanted pollen.

Five steps (emasculate, bag, collect pollen, dust, re-bag/tag). Emasculation is not needed in cucurbits as flowers are unisexual.

Concept used. A good genetic-study organism must let many crosses be done quickly and scored unambiguously. Mendel's choice of the garden pea illustrates the ideal criteria.

1. The organism should: (i) have a short life cycle so many generations are obtained fast; (ii) produce a large number of offspring for reliable ratios; (iii) show clear, contrasting (qualitative) traits that are easy to score; (iv) be easy to grow/maintain and to cross and self at will (controlled mating possible); (v) have true-breeding (pure) lines available.
2. These criteria ensure statistically dependable ratios and unambiguous phenotype scoring, exactly why pea and Drosophila are classic choices.

Short life cycle, many offspring, clear contrasting traits, easy controlled mating, available pure lines.

Concept used. In a pedigree, squares are males, circles are females, and a filled (shaded) symbol means the trait is expressed. If two unaffected parents produce affected children, the trait must be recessive. If it appears equally in both sexes, the gene is autosomal, not sex-linked.

1. Both parents are unaffected (unshaded square and circle), yet some children are affected (shaded). A trait that skips the parents and appears in offspring must be recessive. The parents are unaffected carriers (\(Aa\times Aa\)).
2. The affected offspring include both a shaded square (male) and a shaded circle (female), so the trait appears in both sexes. A recessive trait affecting both sexes equally is autosomal recessive.
3. Conclusion: the trait is autosomal recessive; carrier \(\times\) carrier (\(Aa\times Aa\)) gives roughly one quarter affected children of either sex.

The trait is autosomal recessive; parents are heterozygous carriers (\(Aa\times Aa\)).

Concept used. The F1 comes from crossing two different pure parents. The F2 is obtained by selfing the F1, which lets each heterozygote's hidden recessive allele segregate and reappear.

1. For F1 Mendel needed two different pure parents (tall \(TT\) \(\times\) dwarf \(tt\)), so he had to cross-pollinate them. The F1 is all \(Tt\) (tall).
2. For F2, all F1 plants are genetically identical heterozygotes (\(Tt\)). Selfing such identical plants (\(Tt\times Tt\)) is enough to let segregation produce the \(3{:}1\) ratio; no second parent is needed. Pea flowers are bisexual and naturally self-pollinate, making selfing the simplest route.

F1 needs two different pure parents (cross); F2 only needs selfing identical \(Tt\) plants to reveal the \(3{:}1\) segregation.

Concept used. A gene is a segment of DNA that carries the coded instructions for making a specific polypeptide (protein) or RNA. The protein then performs a function that shows up as a visible trait.

1. The base sequence of a gene is a code. During expression it is transcribed to mRNA and translated to a specific protein (often an enzyme).
2. That protein carries out a biochemical reaction (e.g. an enzyme that makes a pigment). The product of this reaction is the trait we observe (e.g. flower colour, plant height).
3. So the information in the gene's sequence ultimately decides which protein is made and therefore which trait appears. A change in the gene (mutation) can change the protein and the trait.

A gene's DNA sequence codes for a protein; the protein produces the trait, so the gene holds the trait's information.

Concept used. Alleles are alternative forms of the same gene occupying the same locus on homologous chromosomes. They arise by mutation and differ in their DNA base sequence, which can change the protein and therefore the phenotype.

1. Alleles sit at the same position but have slightly different base sequences (e.g. allele \(T\) for tall and allele \(t\) for dwarf). The difference is a change in one or more nucleotides.
2. This sequence difference may give a fully functional protein (dominant allele) or an altered/non-functional protein (recessive allele), producing contrasting phenotypes.
3. Significance: allelic differences are the raw material of variation. They make Mendelian segregation observable and supply the variability on which natural selection acts.

Alleles differ in DNA sequence at the same locus; this difference creates contrasting phenotypes and is the basis of genetic variation.

Concept used. A pure-breeding plant is homozygous: it has two identical alleles. Capital (\(R\)) denotes the dominant allele (red), lower case (\(r\)) the recessive allele (white). Pure parents are written with two identical letters: \(RR\) and \(rr\).

1. The parents are pure-breeding, so each carries two identical alleles. The red parent shows the dominant trait and breeds true, so it is \(RR\). The white parent shows the recessive trait, so it is \(rr\).
2. Cross \(RR\times rr\): every F1 plant is \(Rr\). Since \(R\) (red) is dominant over \(r\) (white), all F1 are red, exactly as observed.
3. Selfing \(Rr\times Rr\) gives \(1\,RR:2\,Rr:1\,rr\), i.e. \(3\) red \(:1\) white. The reappearance of white in the \(3{:}1\) ratio confirms the parents were \(RR\) and \(rr\), justifying the symbols.

Pure parents are homozygous, so red \(=RR\) and white \(=rr\); this correctly predicts all-red F1 and the \(3{:}1\) F2.

Concept used. The phenotype is the result of genotype interacting with the environment. Genes set the potential range of a trait; the environment decides which value within that range is actually expressed.

1. The statement is true. Genes carry the coded potential for a trait (e.g. genes for plant height), but the trait will only develop fully if the environment is suitable.
2. Example: Himalayan rabbit has genes for dark fur, but the pigment-forming enzyme works only at low temperature, so dark fur appears only on cooler body extremities. The gene gives potential; temperature (environment) gives the chance to express it.
3. Hence both factors are needed: genotype \(+\) environment \(\to\) phenotype. The statement correctly describes this gene-environment interaction.

True: genes set the potential, the environment decides whether and how far that potential is realised (phenotype = genotype \(+\) environment).

Concept used. For independently assorting genes, work one gene at a time and then combine. Each parent's gamete for a gene is found from its genotype; the offspring genotype is the product across all three genes.

1. Gene \(A\): \(Aa\times aa\) gives gametes \(A\) or \(a\) (from first parent) with \(a\) (from second). Offspring: \(\tfrac12\,Aa\) and \(\tfrac12\,aa\).
2. Gene \(B\): \(bb\times bb\) gives only \(b\) gametes. All offspring are \(bb\).
3. Gene \(D\): \(DD\times dd\) gives \(D\) from one parent and \(d\) from the other. All offspring are \(Dd\).
4. Combine the three: offspring are \(Aa\,bb\,Dd\) and \(aa\,bb\,Dd\), each with probability \(\tfrac12\). So only two genotype types occur, in a \(1{:}1\) ratio.

Two genotypes: \(Aa\,bb\,Dd\) and \(aa\,bb\,Dd\), in a \(1{:}1\) ratio.

Concept used. In humans, females are homogametic (\(XX\)) and males are heterogametic (\(XY\)). The mother can give only an \(X\); the father gives either an \(X\) or a \(Y\). So the sex of the child is determined by the father's sperm.

1. The mother's eggs all carry \(X\). The father's sperm are of two kinds: \(X\)-bearing and \(Y\)-bearing, in equal numbers.
2. If an \(X\) sperm fertilises the egg, the child is \(XX\) (girl). If a \(Y\) sperm fertilises it, the child is \(XY\) (boy): \[ XX \times XY \;\Rightarrow\; XX\ (\text{girl}),\ XY\ (\text{boy}),\ \text{each } \tfrac12. \]
3. Since only the father supplies the \(Y\), the sex of the child is decided entirely by the father. Blaming the woman is therefore scientifically wrong and unjust.

No. The mother gives only \(X\); the father's \(X\) or \(Y\) sperm decides the child's sex, so blaming the woman is incorrect.

Concept used. The round/wrinkled seed trait depends on the starch-branching enzyme gene. The dominant allele makes a functional enzyme; a mutation (insertion) in the recessive allele makes a defective enzyme, so starch is not properly synthesised.

1. The round allele codes a functional starch-branching enzyme. It converts sugars into large branched starch, so seeds store more starch, hold water normally, and stay round on drying.
2. The wrinkled allele has a transposon insertion, giving a non-functional enzyme. Less starch is made and sugar accumulates, so the seed absorbs more water, then loses it on drying and collapses into a wrinkled shape.
3. Thus wrinkled is the recessive phenotype caused by a defective starch-branching enzyme; it appears only in homozygous recessive (\(rr\)) seeds.

Wrinkled seed is recessive: a mutant allele makes a non-functional starch-branching enzyme, so less starch and more sugar/water loss collapse the seed.

Concept used. Multiple allelism means a gene has more than two alleles in the population. But a gene sits at one locus, and a diploid individual has only two homologous chromosomes carrying that locus, hence only two alleles per individual.

1. A diploid organism has chromosomes in homologous pairs. The gene in question occupies one locus, present once on each chromosome of the pair, so two copies in total.
2. Even if the population has many alleles (e.g. ABO has \(I^{A}\), \(I^{B}\), \(i\)), any one individual can carry at most two of them, one on each homologue.
3. So multiple allelism is a population-level property; the two-allele limit per individual follows directly from diploidy.

Because a diploid individual has only two homologous chromosomes for that locus, it can carry only two of the many possible alleles.

Concept used. A mutagen is a physical or chemical agent that changes the DNA base sequence, producing a mutation. It does so by altering, adding, or deleting bases or by breaking the DNA.

1. Chemical mutagens act on bases. Example: nitrous acid deaminates cytosine to uracil, so a \(C{:}G\) pair is replaced by a \(T{:}A\) pair after replication, a point mutation.
2. Physical mutagens damage DNA directly. Example: UV light makes adjacent thymines join as a thymine dimer, distorting the helix and causing errors during repair or replication. Gamma/X-rays break the sugar-phosphate backbone.
3. Either way the changed sequence may alter the protein and the phenotype, which is the induced mutation.

A mutagen changes DNA (base alteration, addition/deletion, or strand break): e.g. UV forms thymine dimers; nitrous acid deaminates cytosine, causing point mutations.

Concept used. In a normal monohybrid cross the phenotypic ratio is \(3{:}1\) but the genotypic ratio is \(1{:}2{:}1\) (because the dominant allele masks the heterozygote). They become identical only when there is no dominance, i.e. incomplete dominance or co-dominance, where each genotype has its own phenotype.

1. For \(Rr\times Rr\) the genotypes are \(1\,RR:2\,Rr:1\,rr\). With complete dominance, \(RR\) and \(Rr\) look alike, so phenotype is \(3{:}1\) but genotype is \(1{:}2{:}1\). These differ.
2. If phenotype \(=\) genotype ratio (\(1{:}2{:}1\)), then the heterozygote \(Rr\) must have its own distinct phenotype. That happens only when the dominant allele does not fully mask the recessive.
3. Therefore the alleles show incomplete dominance (heterozygote intermediate) or co-dominance (both alleles expressed). Example: Mirabilis jalapa, red \(\times\) white gives \(1\) red \(:2\) pink \(:1\) white, ratio \(1{:}2{:}1\) for both genotype and phenotype.

The alleles are not completely dominant: incomplete dominance (or co-dominance), so each genotype has its own phenotype and both ratios are \(1{:}2{:}1\).

Concept used. ABO blood group is governed by three alleles: \(I^{A}\) and \(I^{B}\) (co-dominant) and \(i\) (recessive). Group O is genotype \(ii\). A group-A or group-B parent may carry a hidden \(i\).

1. Yes, it is possible. A group-A parent can be \(I^{A}I^{A}\) or \(I^{A}i\); a group-B parent can be \(I^{B}I^{B}\) or \(I^{B}i\).
2. If both parents are heterozygous (\(I^{A}i\times I^{B}i\)), cross them: \[ I^{A}i\times I^{B}i \;\Rightarrow\; I^{A}I^{B}\,(AB),\ I^{A}i\,(A),\ I^{B}i\,(B),\ ii\,(O), \] each with probability \(\tfrac14\).
3. The \(ii\) outcome is group O. So a child with blood group O is possible (probability \(\tfrac14\)) when both parents are heterozygous carriers of \(i\).

Yes: if parents are \(I^{A}i\) and \(I^{B}i\), a \(\tfrac14\) chance gives an \(ii\) (group O) child.

Concept used. Down's syndrome is a chromosomal disorder caused by trisomy of chromosome 21 (\(2n+1 = 47\)), arising from non-disjunction (failure of chromosome 21 to separate) during meiosis, usually in the egg.

1. Cause: an extra copy of chromosome 21, so the karyotype is \(47\) with three 21st chromosomes. It results from non-disjunction during egg formation.
2. Symptoms: short stature, broad flat face, slanting eyes with epicanthal fold, protruding furrowed tongue, partially open mouth, broad palm with a single crease, and delayed mental and physical development.
3. Age effect: in the mother, oocytes are arrested in meiosis for decades. With advancing age (over \(40\)) the spindle and chromosome-cohesion machinery weakens, so non-disjunction of chromosome 21 becomes more likely, raising the risk of a trisomic (Down's) child.

Down's syndrome \(=\) trisomy 21 (\(47\) chromosomes) from non-disjunction; symptoms include flat face, furrowed tongue, single palm crease, mental retardation; risk rises with maternal age (>40) due to ageing oocytes and faulty chromosome separation.

Concept used. The Chromosomal Theory of Inheritance (Sutton and Boveri) was concluded because the behaviour of chromosomes during meiosis exactly parallels the behaviour of Mendel's factors (genes).

1. Sutton and Boveri noticed that chromosomes occur in homologous pairs, separate during meiosis and reunite at fertilisation, just as Mendel's paired factors segregate and recombine.
2. This parallel suggested genes are carried on chromosomes. Morgan then proved it: studying eye colour in Drosophila, he found the gene was inherited along with the X chromosome (sex-linked inheritance).
3. Because a specific gene's inheritance tracked a specific chromosome, it was concluded that genes are physically located on chromosomes.

Sutton-Boveri saw chromosome behaviour parallels Mendelian factors; Morgan's sex-linked Drosophila eye-colour gene proved genes lie on chromosomes.

Concept used. Incomplete dominance is the condition where the heterozygote shows an intermediate phenotype because neither allele is fully dominant. An orange F1 from red \(\times\) yellow is the classic signature.

1. Let \(R^{1}R^{1}\) be red and \(R^{2}R^{2}\) be yellow. The cross \(R^{1}R^{1}\times R^{2}R^{2}\) gives all F1 \(=R^{1}R^{2}\).
2. The F1 is orange, an intermediate between red and yellow. Since the heterozygote is not red and not yellow but a blend, neither allele is completely dominant. This is incomplete dominance.
3. Predicting F2: selfing \(R^{1}R^{2}\times R^{1}R^{2}\) gives \(1\,R^{1}R^{1}\,(\text{red}):2\,R^{1}R^{2}\,(\text{orange}): 1\,R^{2}R^{2}\,(\text{yellow})\), i.e. phenotype ratio \(1{:}2{:}1\), equal to the genotype ratio.

Incomplete dominance: orange F1 (\(R^{1}R^{2}\)) is intermediate; F2 is \(1\) red \(:2\) orange \(:1\) yellow.

Concept used. A true-breeding (pure) line is one that is homozygous for the trait and, on self-pollination or inbreeding, produces offspring identical to the parent for that trait over many generations.

1. It is homozygous for the gene(s) concerned (e.g. \(TT\) or \(tt\)), so it carries no hidden contrasting allele.
2. On selfing/inbreeding for several generations it shows stable trait inheritance: all progeny are identical to the parent for that character, with no segregation.
3. Such lines are the essential starting material for controlled crosses (as in Mendel's experiments), because their genotype is known and constant.

A true-breeding line is homozygous and, on continued self-pollination, gives offspring identical to the parent (no segregation) for that trait.

Concept used. A dihybrid test cross crosses a plant of unknown genotype with a double homozygous recessive (\(tt\,rr\)). The four phenotype classes appearing in roughly equal numbers (\(\approx1{:}1{:}1{:}1\)) prove the tall-red parent was a double heterozygote.

1. Let \(T\) = tall (dominant), \(t\) = dwarf; \(R\) = red (dominant), \(r\) = white. The dwarf white parent shows both recessive traits, so its genotype is \(tt\,rr\).
2. The ratio of offspring is \(138:132:136:128\), which is very close to \(1{:}1{:}1{:}1\). A \(1{:}1{:}1{:}1\) test-cross ratio means the tall red parent produced four equally frequent gamete types (\(TR\), \(Tr\), \(tR\), \(tr\)), so it must be a double heterozygote \(Tt\,Rr\).
3. Cross \(Tt\,Rr \times tt\,rr\). The \(tt\,rr\) parent gives only \(tr\) gametes; the \(Tt\,Rr\) parent gives \(TR\), \(Tr\), \(tR\), \(tr\). Combining:
   • Tall, Red \(=Tt\,Rr\)
   • Tall, White \(=Tt\,rr\)
   • Dwarf, Red \(=tt\,Rr\)
   • Dwarf, White \(=tt\,rr\)
   each in equal proportion, matching the observed counts.

Parents: tall red \(=Tt\,Rr\), dwarf white \(=tt\,rr\). Offspring: \(Tt\,Rr\) (tall red), \(Tt\,rr\) (tall white), \(tt\,Rr\) (dwarf red), \(tt\,rr\) (dwarf white).

Concept used. Red-green colour blindness is an X-linked recessive trait. A male (\(XY\)) has only one X, so one defective allele makes him colour blind. A female (\(XX\)) needs the defective allele on both X chromosomes to be affected.

1. Male: genotype \(X^{c}Y\) is enough to be colour blind, because there is no second X to mask the defective allele. So any male carrying one \(X^{c}\) is affected.
2. Female: must be \(X^{c}X^{c}\) (defective allele on both X's) to be colour blind. A single \(X^{c}\) (i.e. \(X^{C}X^{c}\)) only makes her a normal carrier.
3. Since a female needs two copies of the rare allele while a male needs only one, affected males are far more frequent. (If allele frequency is \(q\), affected males \(\approx q\), affected females \(\approx q^{2}\), which is much smaller.)

Colour blindness is X-linked recessive; one \(X^{c}\) affects a male, but a female needs \(X^{c}X^{c}\), so frequency in males (\(q\)) far exceeds that in females (\(q^{2}\)).

Concept used. Colour blindness is X-linked. A son inherits his X chromosome from his mother and his Y from his father. So a father cannot pass an X-linked allele to his son.

1. The son's genotype for colour vision is \(X^{?}Y\). His Y came from the father; his single X came from the mother.
2. The colour-blindness gene is on the X. Since the son's X is maternal, his defective allele must have come from his mother, not his father.
3. Therefore it is not likely the son inherited it from his father. The mother must be at least a carrier (\(X^{C}X^{c}\)) and passed \(X^{c}\) to the son. The father being colour blind is coincidental for the son (though the father would pass \(X^{c}\) to all his daughters).

No: a son gets his X from his mother, so his colour blindness came from the (carrier) mother, not the father.

Concept used. An ideal genetic model organism allows many crosses, fast, with clear scoring. Drosophila melanogaster (fruit fly) meets every such criterion.

1. Short life cycle: about \(12\)–\(14\) days, so many generations are obtained quickly.
2. High fertility: a single mating yields a large number of offspring, giving statistically reliable ratios.
3. It is small, cheap and easy to culture on simple synthetic medium, and males and females are easily distinguished.
4. It shows many clear hereditary variations (eye colour, wing shape, body colour) and has only four pairs of chromosomes, with large polytene chromosomes in salivary glands that are easy to study.

Short life cycle, many offspring, cheap easy culture, distinct sexes, many visible mutants and only four chromosome pairs make Drosophila ideal for genetics.

Concept used. The Chromosomal Theory of Inheritance rests on the parallel between the behaviour of genes (factors) and chromosomes during meiosis and fertilisation.

1. Both occur in pairs: genes as paired alleles, chromosomes as homologous pairs.
2. Both segregate: paired alleles separate at gamete formation (Law of Segregation); homologous chromosomes separate at meiosis (anaphase I).
3. Both assort independently: non-allelic genes assort independently, just as non-homologous chromosomes orient and separate independently. Both are restored to the paired condition at fertilisation.

Genes and chromosomes both occur in pairs, segregate at gamete formation, assort independently, and are restored at fertilisation, the basis of the chromosomal theory.

Concept used. Recombination is the formation of new combinations of alleles/genes that differ from the parental combinations, naturally by crossing over during meiosis. In the lab, recombinant DNA technology deliberately joins DNA from different sources.

1. Natural recombination: during meiosis, crossing over between homologous chromosomes exchanges segments, producing gametes with new gene combinations (source of variation).
2. In genetic engineering, recombination is harnessed to make recombinant DNA: a desired gene is cut with restriction enzymes and joined into a vector (plasmid), then introduced into a host cell.
3. Applications: production of recombinant proteins like human insulin and growth hormone, gene therapy, disease-resistant and improved crops (e.g. Bt cotton), and vaccine production. All depend on artificially recombining DNA from different organisms.

Recombination \(=\) new allele combinations (naturally via crossing over). Genetic engineering uses recombinant DNA for insulin/hormone production, gene therapy, transgenic crops and vaccines.

Concept used. Artificial selection is the deliberate breeding of organisms by humans for chosen desirable traits. Natural selection is differential survival and reproduction driven by the environment.

1. In artificial selection, humans pick parents with desired traits (high yield, docile temperament) and breed only those, e.g. many dog breeds or high-milk cattle from wild ancestors.
2. Yes, it affects natural selection. Humans replace the natural environment as the selecting agent, so traits useful to humans are favoured even if they would be disadvantageous in the wild.
3. This can reduce genetic variability and produce organisms less fit for natural conditions (e.g. high-yield crops needing human care). So artificial selection redirects, and can oppose, the course natural selection would take.

Artificial selection is human-directed breeding for chosen traits; it overrides natural selection by making humans the selecting force, often favouring traits that are unfit in the wild.

Concept used. In incomplete dominance the heterozygote shows a blended, intermediate phenotype. In co-dominance the heterozygote shows both parental phenotypes fully and separately, with no blending.

1. Incomplete dominance: neither allele fully masks the other, so the heterozygote is intermediate. Example: Mirabilis jalapa, red (\(R^{1}R^{1}\)) \(\times\) white (\(R^{2}R^{2}\)) gives pink (\(R^{1}R^{2}\)) F1.
2. Co-dominance: both alleles are fully expressed at the same time, so the heterozygote shows both traits side by side. Example: ABO blood groups, \(I^{A}I^{B}\) produces blood group AB with both A and B antigens.
3. Key difference: incomplete dominance \(=\) one new intermediate phenotype (a blend); co-dominance \(=\) two distinct parental phenotypes appearing together (no blend).

Incomplete dominance: blended intermediate (pink in Mirabilis). Co-dominance: both traits expressed together (AB blood group).

Concept used. A harmful recessive allele can persist if the heterozygote has a survival advantage. For sickle cell, the heterozygote (\(Hb^{A}Hb^{S}\)) is resistant to malaria, called the heterozygote advantage.

1. The sickle allele is recessive for the disease. Homozygotes (\(Hb^{S}Hb^{S}\)) suffer severe anaemia and often die before reproducing, so selection acts against them.
2. However, heterozygotes (\(Hb^{A}Hb^{S}\)) are not anaemic and are resistant to malaria, so in malaria-endemic regions they survive better than normal homozygotes (\(Hb^{A}Hb^{A}\)).
3. Because heterozygotes are fitter where malaria is common, they keep transmitting the \(Hb^{S}\) allele to the next generation. This balancing selection keeps the harmful allele in the population instead of eliminating it.

The sickle allele persists because the heterozygote (\(Hb^{A}Hb^{S}\)) is malaria-resistant (heterozygote advantage), so balancing selection maintains it in malaria-endemic populations.

Concept used. A dihybrid cross follows two genes together. Mendel's Law of Independent Assortment says the two genes assort independently, giving a \(9{:}3{:}3{:}1\) F2 phenotypic ratio when the genes do not interact.

1. Symbols: \(T\) = tall (dominant), \(t\) = dwarf; \(R\) = red (dominant), \(r\) = white. Parents are pure: tall red \(TT\,RR\) \(\times\) dwarf white \(tt\,rr\).
2. F1: each parent gives one allele per gene, so all F1 are \(Tt\,Rr\), phenotype tall red.
3. F1 selfed: \(Tt\,Rr \times Tt\,Rr\). Each parent makes four gamete types in equal numbers: \(TR\), \(Tr\), \(tR\), \(tr\). A \(4\times4\) Punnett square gives the F2.
4. Counting phenotypes in the \(16\) boxes:
   • Tall Red \(=9\)
   • Tall White \(=3\)
   • Dwarf Red \(=3\)
   • Dwarf White \(=1\)
   So the standard dihybrid F2 ratio is \(\mathbf{9:3:3:1}\).
5. Gene interaction: yes, if the two genes interact (epistasis), one gene's product can mask or modify the other's effect. The \(9{:}3{:}3{:}1\) then collapses to modified ratios such as \(9{:}7\), \(9{:}3{:}4\), \(12{:}3{:}1\) or \(15{:}1\). So the values would deviate from \(9{:}3{:}3{:}1\) under gene interaction.

Standard dihybrid F2 ratio \(=\mathbf{9:3:3:1}\) (Tall Red : Tall White : Dwarf Red : Dwarf White). With gene interaction the ratio deviates (e.g. \(9{:}7\), \(12{:}3{:}1\), \(9{:}3{:}4\)).

Concept used. Homogametic means producing only one type of gamete with respect to sex chromosomes; heterogametic means producing two types. Sex is determined by which gamete combination forms the zygote; in some animals temperature (TSD) decides sex instead of chromosomes.

1. (a) Human female is \(XX\): all her eggs carry one \(X\), so she is homogametic (one gamete type). Human male is \(XY\): he makes two sperm types, \(X\)-bearing and \(Y\)-bearing, so he is heterogametic.
2. Reversed example: in birds, moths and butterflies the female is \(ZW\) (heterogametic, two egg types) and the male is \(ZZ\) (homogametic, one sperm type). So here males are homogametic and females heterogametic.
3. (b) The mother gives only \(X\). The father gives either \(X\) or \(Y\). An \(X\) sperm \(\to XX\) (girl); a \(Y\) sperm \(\to XY\) (boy). So the father determines the sex of the child.
4. Temperature: in humans temperature has no role. However, in some reptiles (many turtles, crocodiles) sex is decided by the incubation temperature of the eggs (temperature-dependent sex determination, TSD), not by sex chromosomes.

(a) Human female \(XX\) homogametic, male \(XY\) heterogametic; reversed in birds (female ZW, male ZZ). (b) The father's sperm (\(X\)/\(Y\)) determines sex; in humans temperature has no role, but reptiles like turtles/crocodiles show temperature-dependent sex determination.

Concept used. Colour blindness is an X-linked recessive trait. A woman whose father is colour blind (\(X^{c}Y\)) must have inherited his \(X^{c}\), so she is a carrier (\(X^{C}X^{c}\)). Sons get their single X from the mother; daughters get one X from each parent.

1. Genotypes. The woman's father is colour blind: \(X^{c}Y\). He passes \(X^{c}\) to his daughter, so the woman is \(X^{C}X^{c}\) (normal vision, carrier). Her husband is normal: \(X^{C}Y\).
2. Cross \(X^{C}X^{c} \times X^{C}Y\). Work out the four equally likely offspring: \[ \tfrac14\,X^{C}X^{C},\quad \tfrac14\,X^{C}X^{c},\quad \tfrac14\,X^{C}Y,\quad \tfrac14\,X^{c}Y. \]
3. Daughters (\(X^{C}X^{C}\) or \(X^{C}X^{c}\)): all have at least one \(X^{C}\) from the normal father, so none is colour blind. Probability of a colour-blind daughter \(=0\) (half are carriers).
4. Sons (\(X^{C}Y\) or \(X^{c}Y\)): half receive \(X^{C}\) (normal) and half receive \(X^{c}\) (colour blind). Probability of a colour-blind son \(=\tfrac12\) (\(50\%\)).

Daughters: probability of colour blindness \(=0\) (half are carriers). Sons: probability of colour blindness \(=\tfrac12\) (\(50\%\)).

Concept used. T. H. Morgan worked on Drosophila and established linkage, recombination and sex-linked inheritance. His student A. H. Sturtevant used recombination frequency to construct the first genetic map.

1. Morgan's contributions. (i) Discovered sex-linked inheritance: the white-eye gene in Drosophila is on the X chromosome, giving the chromosomal proof that genes lie on chromosomes. (ii) Discovered linkage: genes on the same chromosome are inherited together and do not assort independently. (iii) Showed recombination by crossing over, and that tightly linked genes recombine less often than loosely linked genes.
2. Sturtevant's contributions. He proposed that the recombination frequency between two genes measures the distance between them. Using Morgan's Drosophila data, he prepared the first genetic (linkage) map, arranging genes linearly with \(1\) map unit \(=1\%\) recombination (the centimorgan).
3. Together, Morgan's linkage and recombination work plus Sturtevant's mapping converted genetics from ratios into a physical, ordered picture of genes along a chromosome.

Morgan: sex-linked inheritance, linkage and recombination in Drosophila. Sturtevant: used recombination frequency to build the first linear genetic map (\(1\) map unit \(=1\%\) recombination).

Concept used. Aneuploidy is gain/loss of one or a few chromosomes (\(2n\pm1\), \(2n\pm2\)) due to non-disjunction. Polyploidy is the addition of one or more complete sets of chromosomes (\(3n\), \(4n\)).

1. Definition. Aneuploidy is the loss or gain of one or a few individual chromosomes from the normal \(2n\) set, caused by failure of chromosomes/chromatids to separate during meiosis (non-disjunction).
2. Difference from polyploidy. Aneuploidy changes the number by \(\pm1\) or \(\pm2\) chromosomes (e.g. \(2n+1\)); the genome is not a whole multiple. Polyploidy adds entire chromosome sets (\(3n\) triploid, \(4n\) tetraploid); common in plants, usually lethal in animals.
3. (a) Trisomy of chromosome 21 (\(2n+1=47\)): Down's syndrome. Features: short stature, flat broad face, slanting eyes, protruding furrowed tongue, broad palm with single crease, and mental retardation.
4. (b) XXY (\(44+XXY=47\)): Klinefelter's syndrome. The individual is male but has feminine features: gynaecomastia (breast development), sparse body hair, small testes, and is usually sterile.
5. (c) XO (\(44+XO=45\)): Turner's syndrome. The individual is female but sterile, with short stature, webbed neck, poorly developed ovaries (rudimentary) and lack of secondary sexual characters.

Aneuploidy \(=\) gain/loss of a few chromosomes (\(2n\pm1\), non-disjunction); polyploidy \(=\) extra whole sets (\(3n\), \(4n\)). (a) Trisomy 21 \(=\) Down's; (b) XXY \(=\) Klinefelter's; (c) XO \(=\) Turner's.