(a) In sexual reproduction, DNA from two different individuals (mother and father) combines. During this process, variations arise from DNA copying errors and the mixing of parental genetic material. This produces offspring with new combinations of traits, generating far greater variation than asexual reproduction, where a single parent copies its DNA to produce nearly identical offspring.

(b) Variation is useful because if the environment changes drastically, at least some individuals with favourable variations will survive. For example, if water temperature rises due to global warming, heat-resistant bacteria in a population will survive and reproduce, keeping the species from being wiped out.

Source: Chapter 7, Section 7.1.1 — The Importance of Variation; Chapter 8, Introduction

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• Part (a) is worth ~1.5 marks: mention two parents, combining of DNA, and greater variation. The contrast with asexual (single parent, near-identical copies) strengthens the answer.
• Part (b) is worth ~1.5 marks: state the principle (survival when environment changes) and give the exact bacteria/global warming example from the textbook — examiners look for this specific example.
• Do not write vague points like "variation helps evolution"; stick to what the textbook says — survival of species over time.

Trait B (present in 60%) most likely arose earlier.

In an asexually reproducing population, variations arise due to minor inaccuracies in DNA copying over successive generations. A trait that appeared earlier would have had more generations to be inherited and passed on, thus accumulating in a greater proportion of the population over time.

Source: Chapter 8, Section 8.1 – Accumulation of Variation During Reproduction

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• The key logic: older variation = more time to accumulate = higher frequency in the population.
• The textbook question (Q1 under Section 8.1) directly asks this — examiners expect you to link frequency of a trait to how long it has been present.
• Avoid over-explaining; just state which trait, then give the one-line reason about inheritance over generations.
• Don't confuse this with dominance/recessiveness — that applies to sexual reproduction, not this concept.

Population Y (sexually reproducing) would show greater variation.

Reasons:

1. In asexual reproduction, new individuals arise from a single parent through DNA copying. Only minor differences occur due to small inaccuracies in DNA copying, so Population X shows very little variation.

1. In sexual reproduction, two individuals contribute DNA, leading to new combinations of genes in each offspring. This generates far greater diversity among individuals in Population Y with every generation.

As stated in the textbook, variations are maximised by the process of sexual reproduction.

Source: Chapter 8, Section 8.1 — Accumulation of Variation During Reproduction

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• The question is worth 3 marks: 1 mark for identifying Population Y, 2 marks for two distinct reasons.
• Examiners expect you to contrast asexual (DNA copying errors only → minor variation) with sexual reproduction (gene combinations → greater variation). Use textbook language like "inaccuracies in DNA copying" and "new combinations of genes."
• Avoid vague statements like "sexual reproduction is better" — be specific about the mechanism causing variation.

Answer: B — (ii) only

Pre-existing genetic variations in the population gave some individuals better heat tolerance, allowing them to survive and reproduce when environmental temperature rose suddenly.

The passage (Ch. 7, Sec. 7.1.1) explicitly states that if some variations were present in a few individuals, those variants would survive a drastic niche change — e.g., bacteria resistant to heat survive a heat wave. This is natural selection acting on pre-existing variation, not acquired adaptation (i) or effort-based survival (iii). Option (iv) is wrong because all individuals belong to the same species. Examiners look for the key phrase "pre-existing variation."

(i) In the F1 plant (RrYy), during gamete formation, the factors (genes) for seed shape (R/r) and seed colour (Y/y) separate independently and recombine randomly to form zygotes. This produces new combinations like Rryy (round green) and rrYy (wrinkled yellow) that were absent in the parents. These new trait combinations arise because factors controlling seed shape and seed colour recombine freely during fertilisation to form F2 offspring.

(ii) The appearance of these new combinations shows that seed shape and seed colour are independently inherited — the inheritance of one trait does not influence the other.

Source: Chapter 8, Section 8.2.2

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• (i) The key phrase from the textbook is: "factors controlling for seed shape and seed colour recombine to form zygote leading to form F2 offspring." Examiners want: independent separation of factors during gamete formation + random recombination at fertilisation.
• (ii) The exact conclusion from the textbook is: "the tall/short trait and the round seed/wrinkled seed trait are independently inherited." Use this language directly — it scores full marks.
• Avoid over-explaining meiosis; at Class 10 level, "factors separate and recombine independently" is sufficient.

Both parents have attached earlobes, yet the student has free earlobes. This tells us that earlobe type is inherited — it follows Mendel's rules of inheritance, where each child receives genetic material from both parents.

Since both parents have attached earlobes but still produced a child with free earlobes, the free earlobe trait must be recessive. The parents each carried one copy of the free earlobe gene (recessive) without expressing it, as attached earlobe is the dominant trait. When the child inherited one recessive copy from each parent, she expressed the free earlobe trait.

Source: Chapter 8, Section 8.2.2 — Rules for the Inheritance of Traits

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• The key logic: if a trait "skips" a generation (parents don't show it but child does), the trait is recessive.
• Parents with attached earlobes must be heterozygous (Aa) — carrying one dominant (A) and one recessive (a) allele. The student received aa, expressing the recessive free earlobe.
• Examiners expect you to use the terms dominant and recessive correctly and justify with Mendelian reasoning. Avoid vague answers like "it skipped generations" without explaining why.

The reappearance of short plants in F2 proves that the F1 tall plants were not pure for tallness — they were carrying both the tallness factor (T) and the shortness factor (t), i.e., they were Tt (heterozygous). The 't' factor was present but not expressed because 'T' is dominant over 't'. Only when two copies of 't' came together (tt) in F2 did the short trait reappear. This proves that both traits were inherited by F1 plants, but only the dominant trait (tallness) was expressed.

Source: Chapter 8, Section 8.2.2 – Rules for the Inheritance of Traits

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• The examiner wants three key ideas: (1) F1 plants were heterozygous (Tt), not pure tall (TT); (2) the recessive 't' factor was hidden/not expressed in F1; (3) F2 self-pollination produced tt offspring, revealing the hidden recessive trait.
• Use terms: dominant, recessive, heterozygous, factors/genes — these fetch marks.
• The textbook explicitly states: "both the tallness and shortness traits were inherited in the F1 plants, but only the tallness trait was expressed." Quote or paraphrase this logic directly.

(B) The T allele is dominant and its single copy is sufficient to express the tall trait, masking the effect of t.

The textbook states clearly: "a single copy of 'T' is enough to make the plant tall." Both TT and Tt are tall; only tt is short. Examiners expect students to use the terms dominant and recessive and the idea that one copy of a dominant allele is sufficient for expression.

The absence of medium-height plants in the F1 generation shows that traits do not blend. If blending occurred, all F1 offspring would be medium-height (a mix of tall and short). Instead, all offspring were fully tall, meaning one parental trait was expressed completely while the other was hidden, not mixed.

Source: Chapter 8, Section 8.2.2

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The examiner wants students to directly link the observation (no medium-height plants, all tall) to why blending inheritance is disproved. Key point: blending would predict intermediate offspring; Mendel's result showed one trait was completely expressed and the other was hidden (recessive), not diluted. Mention both the expected outcome under blending AND the actual outcome for full marks.

In a dihybrid cross, RrYy (round, yellow) parent produces gametes by independent segregation of alleles. The R allele (round) and y allele (green) can combine in a gamete as Ry, and fertilisation can produce Rryy offspring. Although neither parent had round + green together, the alleles for each trait were present separately and can recombine independently during gamete formation.

This demonstrates Mendel's Law of Independent Assortment — factors controlling different traits (seed shape and seed colour) are inherited independently of each other, allowing new combinations to appear in F2 progeny.

Source: Chapter 8, Section 8.2.2

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• Examiners expect you to mention gamete formation (specifically that alleles segregate independently) and name the law explicitly.
• The key phrase from the textbook is: "factors controlling for seed shape and seed colour recombine to form zygote leading to F2 offspring" — use this idea.
• Do NOT confuse this with the Law of Segregation (which is about two alleles of the same gene separating). Independent Assortment is about different genes behaving independently.
• For 3 marks, cover: (1) how Ry gametes are formed, (2) recombination at fertilisation gives Rryy, (3) name of the principle.

Test Cross (Cross with homozygous recessive):

To confirm whether the tall F₂ plants are TT or Tt, each tall F₂ plant is crossed individually with a true-breeding short plant (tt) — this is called a test cross.

Expected results:

• If the tall F₂ plant is TT: all offspring will be tall (Tt).
• If the tall F₂ plant is Tt: offspring will be tall and short in 1:1 ratio (Tt : tt).

By observing the progeny of each cross, we can identify which tall F₂ plants are TT and which are Tt, confirming the expected 1:2 ratio of TT:Tt among the tall plants.

Source: Chapter 8, Heredity

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• Examiners expect the term test cross to be named.
• Key point: the tall F₂ plant is crossed with tt (not another tall plant), because tt produces only 't' gametes, so any variation in offspring reveals the unknown genotype.
• State both outcomes clearly — all tall = TT; 1:1 tall:short = Tt. This is where most marks lie.
• No need for a full Punnett square unless specifically asked; stating the ratios is sufficient for 3 marks.

In sexually reproducing organisms, a child receives one copy of each gene from the father and one from the mother, making two copies per trait in the offspring. For this to work correctly, each germ cell (sperm or egg) must carry only one copy of each gene. If germ cells carried two copies, the offspring would receive four copies per trait, violating Mendel's principle that sexually reproducing organisms have exactly two copies of genes controlling each trait.

Source: Chapter 8, Section 8.2.2

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• Examiners want you to link germ cell formation → fertilisation → two copies in offspring as a logical chain.
• The key Mendelian idea: sexually reproducing organisms have two copies of each gene (one paternal, one maternal). This only holds if each gamete contributes exactly one copy.
• Avoid over-explaining; two tight sentences covering "why one" and "what goes wrong with two" is enough for 2 marks.

P cross: RRyy × rrYY → F₁: all RrYy (round, yellow)

F₁ self-pollination: RrYy × RrYy

For seed shape: Rr × Rr → 1 RR : 2 Rr : 1 rr → wrinkled (rr) = 1/4

For seed colour: Yy × Yy → 1 YY : 2 Yy : 1 yy → yellow (Y_) = 3/4

Since the genes are on separate chromosomes, they assort independently:

Fraction of wrinkled and yellow = 1/4 × 3/4 = 3/16

This matches the 3 wrinkled-yellow seeds seen in the 9:3:3:1 F₂ ratio.

Source: Chapter 8, Section 8.2.3; Figure 8.5

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• Examiners expect you to show each trait analysed separately, then multiply the fractions — this demonstrates understanding of the Law of Independent Assortment.
• The Punnett square in Fig. 8.5 confirms the 9:3:3:1 ratio; wrinkled-yellow is the second 3, i.e., 3/16.
• Always state the final fraction clearly; writing only "3 out of 16" without working loses a mark.
• The key phrase to include: "genes on separate chromosomes assort independently."

The trait reappeared in F2 because F1 plants carried two copies of the gene (one dominant T, one recessive t); only organisms with two copies can hide one trait while passing it on, causing its reappearance in tt offspring.

Source: Chapter 8, Section 8.2.2

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The examiner wants you to link the disappearance in F1 and reappearance in F2 directly to the idea of two copies per organism. Key point: if there were only one copy, a recessive trait could never be "hidden" and then recovered. Mention T and t (or dominant/recessive) to show you understand the mechanism. Keep it to one crisp line for 1 mark.

A gene contains instructions for making a specific protein (enzyme). If a gene undergoes a change (mutation), the enzyme it codes for may become non-functional or altered. In pea plants, if the gene controlling the enzyme needed to produce a growth hormone is changed, that enzyme may not form correctly. As a result, the growth hormone is not produced or is produced in insufficient amounts. Without the hormone, the plant cannot grow to its normal height, and a dwarf (short) phenotype is observed instead of the tall phenotype — showing how a gene change directly alters physical characteristics.

Source: Chapter 8, Heredity (Mendel's Contributions / gene–trait relationship)

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• Examiners expect a gene → enzyme → hormone → phenotype chain of reasoning — all four links must be present for full marks.
• Use the words gene, enzyme, hormone, and phenotype explicitly.
• The example from the question (growth hormone in pea plants) should be referenced — don't give a generic answer.
• A common mistake is stopping at "the enzyme doesn't work" without connecting it to the visible change in the plant.

Mendel's Law of Independent Assortment and Linked Genes

Mendel's law of independent assortment states that two different traits are inherited independently of each other. This holds true only when the two genes are located on different (non-homologous) chromosomes, so they segregate independently during gamete formation.

If both genes are on the same chromosome (linked genes):

• The two genes tend to be inherited together as a unit, since they are physically attached on the same chromosome.
• During gamete formation, linked genes do not assort independently; they travel together to the same gamete.
• New combinations of traits (recombinants) would appear far less frequently than expected, unlike the free mixing seen in a standard dihybrid cross.

Effect on F2 phenotypic ratio:

• In a normal dihybrid cross (genes on different chromosomes), the F2 ratio is 9:3:3:1.
• If both genes are linked, this ratio would not be obtained. Instead, parental combinations (e.g., tall + round seeds; short + wrinkled seeds) would appear in a much higher proportion, and recombinant types (tall + wrinkled; short + round) would be very rare.
• The ratio would deviate significantly from 9:3:3:1, approximating a 3:1 ratio (like a monohybrid cross) if linkage is complete.

Thus, linkage prevents independent assortment, reducing genetic variation in offspring.

Source: Chapter 8, Section 8.2.2 – Rules for the Inheritance of Traits

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• The examiner expects you to first state what independent assortment means and its condition (genes on non-homologous chromosomes).
• Then explain the concept of linkage — genes on the same chromosome travel together.
• Finally, clearly state the change in F2 ratio: 9:3:3:1 breaks down; parental types dominate; recombinants are rare; ratio approaches 3:1 for complete linkage.
• The NCERT text explicitly shows that new combinations (tall+wrinkled, short+round) arise in F2 because the traits are independently inherited — use this as your contrast point.
• Avoid writing a long essay; keep points crisp and logically ordered.

The DNA of a tall pea plant contains the gene T (dominant allele). This gene is transcribed into mRNA, which is then translated by ribosomes to produce a specific enzyme (protein). This enzyme catalyses the synthesis of a gibberellin-type growth hormone in large amounts. The hormone promotes elongation of stem cells, resulting in the tall phenotype. In short plants (genotype tt), the non-functional enzyme produces little or no hormone, so stems remain short.

Chain: DNA (gene T) → mRNA → Enzyme (protein) → Growth hormone → Tall plant.

Source: Chapter 8 (Heredity), Chapter 6, Section 6.2.2

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• The examiner expects the DNA → RNA → Protein → Trait chain explicitly stated.
• Mention that the protein acts as an enzyme that controls hormone production — this links molecular biology to the observable trait.
• The textbook mentions gibberellins help in stem growth (Ch. 6.2.2); connecting that to the T gene is the key bridge the question demands.
• Keep the chain logical and sequential — examiners award one mark per link in the chain (gene → enzyme → hormone → trait).

The mutant plant will most likely appear shorter than normal (dwarf-like).

Justification: Genes control traits by producing specific enzymes. Here, the gene for growth-hormone synthesis produces an enzyme working at only 20% efficiency, meaning far less growth hormone is synthesised. Since adequate growth hormone is not available, the plant cannot achieve normal height. This demonstrates the direct gene → enzyme → trait relationship: a defective gene leads to a less functional enzyme, which alters the final observable trait (phenotype).

• The key concept tested is gene → enzyme → trait: genes code for enzymes; enzymes drive biochemical reactions; reactions determine the trait.
• The examiner wants you to link reduced enzyme efficiency → reduced hormone production → altered phenotype in a logical chain.
• At 20% efficiency, growth hormone is severely limited, so the most predictable effect is stunted/reduced growth — draw this conclusion clearly.
• This question is application-based; the source passages establish that genes control traits (Chapter 8), so ground your answer in that principle even though the exact enzyme scenario isn't explicitly described.

Body cells carry two copies of each gene (one from each parent). When two gametes fuse during fertilisation, the offspring must receive the correct total — two copies. To ensure this, gametes are formed by meiosis, carrying only one copy (half the chromosomes).

If gametes were not formed this way, each fertilisation would double the chromosome number in every generation, disrupting the cell's genetic control and making normal development impossible.

Source: Chapter 7, Section 7.3.1 — Why the Sexual Mode of Reproduction?

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• The key idea is meiosis halving the chromosome number so fertilisation restores it.
• Examiners expect two clear points: (1) why gametes have half — so fusion gives the correct number; (2) consequence if they didn't — chromosome number doubles each generation.
• Quote or paraphrase the textbook phrase: "re-establishment of the number of chromosomes and DNA content in the new generation."
• Avoid over-explaining; two focused sentences covering both marks is sufficient.

In Mendel's dihybrid cross, factors (genes) controlling seed shape and seed colour recombine independently when forming a zygote, producing new combinations like tall with wrinkled seeds or short with round seeds in F2 offspring.

If each parent contributed a single, unbroken set of all genes, the traits would always stay linked together and pass as one unit. Independent recombination of individual factors would be impossible — new combinations of traits could never arise in offspring. Mendel's results are only explained by each trait being inherited separately, not as part of an intact, inseparable package.

Source: Chapter 8, Section 8.2.2

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• The examiner wants you to connect the dihybrid cross outcome (new trait combinations in F2) to the mechanism — independent inheritance of separate gene pairs.
• Key idea: if genes were inherited as a single unbroken set, all traits from one parent would always travel together; reshuffling would be impossible.
• Use the textbook phrase: "factors controlling for seed shape and seed colour recombine to form zygote."
• Avoid over-explaining Mendel's ratios — that is not asked. Focus on why independent inheritance requires genes to be separable units.

I agree with the student's argument.

Mendel's Law of Independent Assortment states that genes for different traits are inherited independently of each other. This is possible because different traits are located on different chromosomes. During gamete formation, chromosomes separate and assort independently into sex cells.

If all genes were on a single continuous DNA strand (one chromosome), all traits would always be inherited together — they would be linked. New combinations like "tall + wrinkled seeds" observed in F2 progeny (as in Mendel's dihybrid cross) would never arise. The existence of separate chromosomes ensures that traits like height and seed shape assort independently, producing new combinations in offspring.

Source: Chapter 8, Section 8.2.2

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• The examiner wants you to connect chromosomes to independent assortment — that's the core biological reasoning.
• Key terms to use: linked genes, independent assortment, gamete formation, chromosomes.
• Refer to the dihybrid cross result (tall + wrinkled / short + round new combinations) as evidence — it shows traits assort independently because they are on different chromosomes.
• Don't just define the law; apply it to the hypothetical single-strand scenario to show why it would fail.

During sexual reproduction, each parent produces gametes (sperm or egg) through a process where chromosome pairs separate, so each gamete carries only one chromosome from each pair. When two gametes fuse at fertilisation, the resulting offspring receives one chromosome from each parent, restoring the full chromosome number (46 in humans = 22 pairs + sex chromosomes). This ensures the chromosome number remains constant across every generation.

Genetic variation arises because:

• Each offspring receives a unique combination of chromosomes from both parents.
• Sexually reproducing individuals have two copies of each gene; these may differ (dominant/recessive), producing varied traits.
• Traits can be inherited independently, giving rise to new combinations in offspring.

Thus, sexual reproduction both maintains chromosomal stability and maximises variation, which may improve survival of the species.

Source: Chapter 8 — Heredity; Chapter 7 — What you have learnt

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What examiners look for (5 marks):

1. (~1 mark) Gametes carry one chromosome from each pair (halving).
2. (~1 mark) Fertilisation restores the full chromosome number — this is the key mechanism for preservation.
3. (~1 mark) Unique chromosome combinations from two parents → variation.
4. (~1 mark) Dominant/recessive gene pairs contribute to variation.
5. (~1 mark) Independent inheritance of traits → new trait combinations.

Key tip: Always explicitly link halving in gametes + fusion at fertilisation = constant chromosome number. Many students explain variation well but forget to clearly state the restoration mechanism — that's where marks are lost.

The mother has only X chromosomes (XX), so she can only pass an X chromosome to her child. The father has XY chromosomes, so he can pass either X or Y. If the father passes X, the child is a girl (XX); if he passes Y, the child is a boy (XY). Thus, the sex of the child is determined by the father.

Source: Chapter 8, Section 8.2.4 – Sex Determination

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• The key logic: mother always contributes X, so she cannot determine sex. Only the father's contribution (X or Y) decides the outcome.
• Examiners look for: (1) mother's limitation (only X), (2) father's two possibilities (X or Y), (3) clear conclusion linking Y → boy, X → girl.
• Avoid vague statements — be specific about which chromosome does what.

No, the father is not correct.

In human beings, sex is determined by the father's contribution, not the mother's.

• All eggs produced by the mother (XX) carry only the X chromosome.
• The father (XY) produces two types of sperms — 50% carry X and 50% carry Y.
• If a sperm carrying X fertilises the egg → girl (XX)
• If a sperm carrying Y fertilises the egg → boy (XY)

Since the mother can only contribute an X chromosome, the sex of the child depends entirely on which type of sperm from the father fertilises the egg. The father is responsible, not the mother.

Source: Chapter 8, Section 8.2.4 — Sex Determination

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What examiners look for:

1. Clear statement that the father is wrong (1 mark).
2. Explanation that mother is XX → only X-bearing eggs (1 mark).
3. Father is XY → produces X and Y sperms; Y sperm gives a boy, X sperm gives a girl — sex determined by father's sperm (1 mark).

Key tip: Always state the conclusion first, then justify. Don't forget to mention both types of sperms from the father — that's the heart of the answer.

Both examples show that characteristics of an organism are determined by an interplay of genes and environment.

In human beings, sex is genetically determined — a child inheriting X from the father is a girl (XX), and one inheriting Y is a boy (XY). The genes alone decide the sex.

In contrast, in reptiles, environmental factors (like temperature) determine sex, showing that genes are not always in control.

This tells us that some traits are controlled primarily by genes, some primarily by the environment, and some by both together. Neither genes nor environment alone can explain all characteristics; the balance varies from organism to organism and trait to trait.

Source: Chapter 8, Section 8.2.4 – Sex Determination

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• The examiner wants you to use both examples — human sex determination (genetic) and reptile sex determination (environmental) — to draw a general conclusion.
• The key phrase to write is that genes and environment both play a role; which factor dominates depends on the organism and the trait.
• Avoid writing only about humans or only about environment — the question specifically tests whether you can synthesise both sides.
• "Genes vs environment" is a classic CBSE concept; stating that it is "not always one or the other" earns the concluding mark.

Probability of the fourth child being a boy = 1/2 (50%)

Chromosomal Diagram at Fertilisation:

```
Mother (XX) Father (XY)
| |
X X X Y
\ / \ /
egg sperm
X X or Y (equal chance)
```

| Sperm from Father | Egg from Mother | Child | Sex |
|---|---|---|---|
| X | X | XX | Girl |
| Y | X | XY | Boy |

Explanation:

Women are XX and men are XY. During gamete formation, the mother always produces eggs carrying one X chromosome. The father produces two types of sperm — half carrying X and half carrying Y — in equal numbers.

The sex of the child depends entirely on which sperm fertilises the egg. An X-bearing sperm produces a girl (XX); a Y-bearing sperm produces a boy (XY).

Each pregnancy is an independent event. The previous three births being girls do not affect the sperm available in the next fertilisation. Therefore, the probability of a boy remains 1/2 every time.

Source: Sex Determination, Chapter 8, Section 8.2.4

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• Examiners expect you to state the probability clearly at the start.
• The diagram (Punnett square or cross diagram) is expected — it earns dedicated marks.
• Key point to stress: each pregnancy is independent; past outcomes don't change probabilities.
• Quote the textbook idea directly: "sex of the children is determined by what they inherit from their father."
• Avoid writing more than needed — the table + two short paragraphs is the ideal length for 5 marks.

The population of pea plants (reproducing sexually) is more likely to have survivors.

Sexual reproduction generates greater variation among offspring due to the combination of genetic material from two parents. When environmental temperature changes suddenly, some individuals with favourable variations may survive and reproduce.

Asexual reproduction produces offspring that are nearly identical to the parent, with only minor variations from DNA copying errors. So if the temperature change is lethal to the parent type, most or all bacteria are likely to die.

Thus, greater variation in sexually reproducing populations increases the chance of some individuals surviving environmental changes.

Source: Chapter 7, Section 7.1.1; Chapter 8, Section 8.1

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• The key concept being tested is variation and survival — more variation = better chance of some individuals surviving a change.
• Examiners expect you to: (1) name the correct population, (2) link sexual reproduction to greater variation, (3) link asexual reproduction to near-identical offspring with little variation.
• The textbook example of heat-resistant bacteria dying in a heat wave directly supports this answer — use it as your evidence for asexual reproduction.
• Don't write about the mechanism of reproduction in detail; focus on variation as the reason for survival.

In F1 plants, the gene for tallness (T) comes from one parent and the gene for shortness (t) comes from the other, giving genotype Tt. Because a single copy of T is sufficient to produce a tall plant, T is dominant and t is recessive. The t allele is present but its effect is completely masked by T, so F1 plants appear tall.

During self-pollination of F1 (Tt × Tt), the two alleles separate and recombine randomly, producing offspring in the ratio TT : Tt : tt = 1 : 2 : 1. Only the tt plants (one-quarter) lack any T allele, so the recessive shortness trait is expressed again in F2.

Source: Chapter 8, Section 8.2.2

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• Key terms examiners expect: dominant, recessive, genotype (Tt, TT, tt), self-pollination, 1:2:1 ratio.
• The core idea is masking of recessive allele by dominant allele in heterozygous condition, and segregation during gamete formation restoring the recessive phenotype.
• Avoid vague phrases like "the gene disappeared" — always say it was present but not expressed.
• A Punnett square is not required for 3 marks but mentioning the 1:2:1 genotypic ratio earns credit.

Flaw: The argument is incorrect. The mother cannot determine the sex of the child.

Correct Mechanism: All women are XX, so a mother can only pass an X chromosome to her child. Men are XY, so a father can pass either an X or a Y chromosome. If the child inherits X from the father → girl (XX); if the child inherits Y from the father → boy (XY). Since the father's contribution determines whether X or Y is passed, it is the father, not the mother, who determines the sex of the child.

Source: Chapter 8, Section 8.2.4 – Sex Determination

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• The key examiner expectation is: (1) identify the flaw clearly — mother only has X chromosomes to give, so she cannot determine sex; (2) state the correct mechanism — father has both X and Y, making him the deciding factor.
• Use the terms XX (female) and XY (male) — examiners specifically look for these.
• A common mistake is writing a vague answer without explaining why the mother cannot determine sex. Always link it back to chromosome composition.

The alleles for seed shape and plant height are inherited independently because the two genes are located on different (non-homologous) chromosomes. During gamete formation, chromosomes of each pair separate independently of other pairs (Mendel's Law of Independent Assortment), so the allele inherited for one trait does not affect the allele inherited for the other.

Cross: RrTt × RrTt (self-pollination)

Gametes produced: RT, Rt, rT, rt (each combination equally likely)

The F2 progeny show new combinations such as tall-wrinkled (R\_tt) and short-round (rrT\_), which could not arise if the traits were always inherited together. This confirms independent assortment.

Source: Chapter 8 – Heredity, Section 8.2.2

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• The key concept is independent assortment, which applies only when genes are on different chromosomes (as stated in the question and confirmed in the passage: "the tall/short trait and the round seed/wrinkled seed trait are independently inherited").
• Examiners expect: (1) a clear statement that inheritance is independent, (2) the reason (different chromosomes → independent separation during gamete formation), and (3) evidence from the cross — specifically the appearance of new trait combinations in F2.
• Do not draw a full 4×4 Punnett square for a 3-mark question; just list gametes and mention the new combinations as proof.

Germ cells are formed by meiosis, which halves the chromosome number. When two germ cells fuse during fertilisation, the normal number is restored, giving offspring one copy of each gene from each parent. Without this reduction, the chromosome number would double with every generation.

This question is conceptual (not a standard Assertion-Reason format despite the options listed). The key ideas from Chapter 7 (section 7.3.1) are: meiosis halves chromosomes in germ cells; fertilisation restores the full number; without meiosis, each generation would have twice the chromosomes of the previous one. Examiners expect you to mention meiosis, halving, fertilisation, and the doubling consequence clearly.

(i) Gene T encodes an enzyme that produces a plant hormone. In genotype TT, two functional copies produce a large amount of hormone. In Tt, one functional copy (T) produces a moderate amount — which is still enough to make the plant tall. Since even one copy of T produces sufficient hormone for full height, T is dominant. In tt, no functional enzyme is produced, so hormone level is too low and the plant is short. Thus, dominance of T is due to the sufficiency of a single functional allele to produce the required hormone.

(ii) Cross: Tt × Tt

| | T | t |
|---|----|----|
| T | TT | Tt |
| t | Tt | tt |

Offspring ratio: 1 TT : 2 Tt : 1 tt
Short plants = tt → 1/4 of the offspring would be short.

(iii) In asexually reproducing plants, variations cannot arise from genetic recombination (no sexual reproduction). The likely source of variation would be mutations — random changes in DNA during cell division (DNA copying errors).

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• Part (i) tests molecular understanding of dominance: one functional allele of T produces enough enzyme/hormone for full height — this is the key molecular reason. Don't just say "T is dominant because it is expressed."
• Part (ii) always show the Punnett square; the answer 1/4 (25%) must follow logically from it.
• Part (iii) is directly from the chapter concept: asexual reproduction produces very little variation; what variation does arise comes from DNA copying errors/mutations, not recombination.

(a) Gene T is located on a chromosome. During germ cell (gamete) formation, the process of meiosis occurs, in which the chromosome pair carrying gene T separates. Each germ cell receives only one chromosome from the pair — so it carries either T or t, not both. This ensures each gamete has a single copy of the gene.

(b) In the F1 plant (Tt), only the tall trait is expressed because T is dominant over t. A single copy of T is sufficient to produce the tall phenotype. The trait 't' (shortness) is recessive — it is hidden but not lost. Whether a trait is expressed or hidden depends on whether it is dominant or recessive; recessive traits are expressed only when both copies are identical (tt).

(c) When genes for height (T/t) and seed colour (Y/y) are on different (non-homologous) chromosomes, they assort independently during meiosis. The chromosome pairs separate independently of each other, so all combinations — TY, Ty, tY, ty — are equally possible in gametes. This independent assortment produces new trait combinations in F2 offspring, as Mendel demonstrated.

Source: Chapter 8, Section 8.2.2

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• (a) Examiners expect the term "meiosis" and the idea of separation of chromosome pairs so each gamete gets one allele.
• (b) Key terms: dominant, recessive. Quote the textbook logic: "a single copy of T is enough to make the plant tall; both copies must be t for the plant to be short."
• (c) The phrase "independent assortment" (or independent inheritance) and the reason — genes on different chromosomes separate independently — are the scoring points. Mention new combinations in F2 for full marks.