(B) Tall is a dominant trait

Tallness is dominant over shortness. In F1, all plants are Tt (tall), as T masks t. In F2, self-pollination gives TT : Tt : tt = 3 tall : 1 short, showing the recessive allele reappears.

The examiner expects you to identify dominance and recessiveness as the explanation. The key NCERT concept is: "the trait that gets expressed is called the dominant trait and the other is called the recessive trait." Option A in the original choices matches this, but from the given OPTIONS list, (B) Tall is a dominant trait is the correct answer. Remember: the 3:1 ratio in F2 is the classic Mendelian result proving dominance, not blending.

Answer: B) 1:1

Cross: Tt × tt → Tt, Tt, tt, tt
Tall (Tt) : Short (tt) = 1:1

Source: Chapter 8, Section 8.2.2

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This is a test cross (not F2 self-pollination). When Tt × tt, the gametes are T/t from the tall parent and only t from the short parent, giving 50% Tt (tall) and 50% tt (short) — ratio 1:1. Students often confuse this with the F2 ratio (3:1, from Tt × Tt) or the genotypic ratio (1:2:1). Remember: 3:1 and 1:2:1 come from crossing two heterozygotes (Tt × Tt), not from a cross with a homozygous recessive (tt).

Answer: (B) Variations increase the chances that at least some individuals will survive drastic environmental changes.

The passage clearly states that if some variations are present in a population, there is a chance for some individuals to survive when a niche is drastically altered (e.g., heat-resistant bacteria surviving global warming). Options A, C, and D are factually incorrect as per the textbook — variations do not make offspring identical, they occur in both sexual and asexual reproduction, and they do not reduce offspring numbers.

Option B — The two traits are inherited independently of each other.

The 9:3:3:1 ratio in F2 of a dihybrid cross directly supports Mendel's Law of Independent Assortment, showing the two traits are inherited independently.

The question maps the 9:3:3:1 ratio to a Mendelian principle. The textbook explicitly states that in a dihybrid cross, F2 progeny show new combinations (tall+wrinkled, short+round), proving the traits are independently inherited — this is Independent Assortment. Note the options given in the MCQ body differ from the four lettered choices (A–D) in the question stem; among the actual answer options listed, B is correct. Examiners expect you to link the 9:3:3:1 ratio directly to independent assortment/independent inheritance.

(B) So that when two germ cells combine, the normal chromosome number is restored in the offspring.

Germ cells are formed by meiosis, carrying only half the chromosomes. At fertilisation, two germ cells fuse, restoring the full chromosome number in the offspring.

The passage (Chapter 7, Section 7.3.1) clearly states that germ cells have half the chromosomes (via meiosis), so that when they combine during sexual reproduction, the original chromosome number and DNA content are re-established. Option B directly reflects this. Options A, C, and D are either incorrect or unrelated to the reason given in the textbook. Examiners expect you to link meiosis → half chromosomes → fertilisation → full number restored.

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

Answer: B — Determined by environmental cues, not genes

Since crocodile sex depends on incubation temperature (an environmental factor), not chromosomes or genes, it is determined by environmental cues, not genetic inheritance.

Source: Chapter 8, Section 8.2.4 – Sex Determination

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• The passage explicitly states: "in some animals like a few reptiles, the temperature at which fertilised eggs are kept determines whether the animals developing in the eggs will be male or female."
• This is contrasted directly with human sex determination, which is genetically determined (XX/XY).
• Options A, C, and D all relate to genetic/chromosomal mechanisms — these apply to humans, not crocodiles.
• Key examiner tip: The word "environmental cues" from the textbook is the exact phrase to recall here.

Answer: (B)

The gene producing the efficient enzyme is dominant, because even a single copy produces enough hormone to make the plant tall, while two copies of the less efficient gene are needed for shortness.

Source: Heredity, Section 8.2.2

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The textbook states: "a single copy of 'T' is enough to make the plant tall… traits like 'T' are called dominant traits." Dominance means one copy is sufficient to express the trait. Here, one copy of the efficient-enzyme gene produces enough hormone for tallness — exactly the definition of a dominant allele. The less efficient enzyme gene requires two copies (homozygous recessive) to show shortness, making it recessive.

(A) Each gene controlling a trait is located on a separate chromosome — so the two traits assort independently during gamete formation (Mendel's Law of Independent Assortment).

The key concept is independent assortment: genes on different (non-homologous) chromosomes segregate independently during meiosis. The textbook states that factors controlling seed shape and seed colour "recombine" freely, producing new trait combinations in F2 — this only happens when the genes are on separate chromosomes. Options B, C, and D are incorrect biological statements not supported by the chapter.

(C) The trait present in 60%, because it has had more time to accumulate in the population through successive generations of asexual reproduction.

The textbook (Ch. 8, Q1 under Section 8.1) directly asks this: "If a trait A exists in 10% … and trait B in 60% … which arose earlier?" The expected reasoning is that in asexual reproduction, variations accumulate over generations; a higher frequency means more time has passed for it to spread — not that it is necessarily more beneficial. Avoid option D's reasoning ("always more beneficial") as the textbook does not say frequency equals benefit.

Answer: B

Sexual reproduction combines genetic material from two different individuals, generating new gene combinations, which produces more heritable variation than asexual reproduction.

Source: Chapter 7 – Sexual Reproduction; Chapter 8 – Introduction

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• The key concept is that sexual reproduction involves two parents, so their genetic material combines, creating new combinations of traits in offspring.
• The textbook explicitly states: "Modes of sexual reproduction allow for greater variation to be generated."
• Eliminate wrong options: Asexual reproduction does not always involve mutations (B wrong); sexual reproduction is not faster (C wrong); both parents' genes are inherited, not just the mother's (D wrong).
• Option A in the MCQ stem is the closest to correct but uses different wording — the mapped answer from the given options is B.

Answer: B — Tall with white flowers and short with violet flowers

In F2, independent assortment produces new combinations (TtWw × TtWw) not seen in either parent (TTWW or ttww).

• Parents are TTWW (tall, violet) and ttww (short, white); F1 is TtWw.
• F1 self-pollination gives 9:3:3:1 ratio — the 3:1 minority classes (tall+white and short+violet) are new recombinant phenotypes.
• The textbook explicitly states: "there would also be some F2 progeny that showed new combinations" when two traits are independently inherited — directly supporting option B.
• Options A and D describe only parental types; C describes only one parental type. Only B correctly identifies both new recombinant combinations.

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

The father is either XY; he passes X or Y to the child, while the mother (XX) always passes X, thus determining the child's sex.

The passage explicitly states: "the sex of the children will be determined by what they inherit from their father" — a child inheriting X from father is a girl, Y makes a boy. The mother always contributes X. So R directly and correctly explains A. Choose A.

Option C: A is true but R is false.

Assertion is correct — better-suited variants survive and reproduce more. Reason is false — variants do NOT have equal survival chances; environment selects favourably adapted ones.

The textbook explicitly states: "Do all these variations have equal chances of surviving? Obviously not." Environmental selection favours better-adapted variants (e.g., heat-resistant bacteria survive a heat wave). This directly contradicts Reason R, making C the correct choice.

Answer: A

Both A and R are true, and R is the correct explanation of A. In sexual reproduction, germ cells (egg and sperm) each contribute one copy of each chromosome, so body cells have two copies — one maternal, one paternal.

• A is true: Body cells of sexually reproducing organisms are diploid (two copies of each chromosome).
• R is true: One copy comes from the mother's egg cell and one from the father's sperm cell.
• R correctly explains A: The two-copy nature of body chromosomes is directly caused by the fusion of two germ cells during fertilisation, making R the correct explanation of A.
• Key phrase from the chapter: "Most human chromosomes have a maternal and a paternal copy" — this directly links R to A.

Blood group O is recessive.

Reasoning: The mother (blood group A) must have genotype I^A i, and the father (blood group B) must have genotype I^B i. The child inherited one 'i' allele from each parent, giving genotype 'ii' (blood group O). Since O only appears when both copies are 'i', it is recessive — a single copy of I^A or I^B is enough to suppress it.

• The key logic: if O were dominant, at least one parent would show blood group O — but neither does. So O must be recessive, hidden in both parents as a single copy (i).
• Examiners expect you to state the conclusion clearly and give the reasoning using the concept of dominant/recessive traits.
• You don't need to write full genotype notation, but it strengthens the answer.

Germ cells are produced by a special cell division called meiosis, which halves the number of chromosomes (and DNA content). This is necessary because when two germ cells combine during fertilisation, the normal chromosome number is restored in the new individual. If germ cells had two sets, the zygote would have double the DNA of normal body cells, disrupting cellular control.

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

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The key concept here is meiosis — the examiner expects you to name it and explain why it happens (to prevent doubling of chromosomes at fertilisation). Two marks = name the process + give the reason. Avoid writing a long paragraph; two crisp sentences covering both points earn full marks.

The genotype of the round-seeded parent is Rr (heterozygous).

Justification: A test cross is performed between the plant of unknown genotype and a homozygous recessive (rr) plant. The cross Rr × rr produces Rr (round) and rr (wrinkled) in a 1:1 ratio, which matches the equal proportion observed. If the parent were RR, all offspring would be round (Rr only).

Source: Heredity, Chapter 8, Section 8.2.2 & 8.2.3

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• Test cross always involves crossing with the homozygous recessive (rr). The 1:1 output of round : wrinkled is the classic result of a heterozygous × homozygous recessive cross.
• Examiners expect you to (1) state the genotype clearly, (2) show or describe the cross, and (3) explain why the other possibility (RR) is ruled out. All three points fit in the word budget for 2 marks.
• Write the cross explicitly (Rr × rr → Rr : rr) — it earns the second mark.

In snails, individuals can change their sex during their lifetime, which indicates that sex determination in snails is not genetically determined but depends on environmental or other non-genetic factors. In contrast, in human beings, sex is genetically determined — by the sex chromosomes (XX for female, XY for male) inherited from parents.

The key contrast examiners expect is: snails = non-genetic/environmental sex determination; humans = genetic sex determination via X and Y chromosomes. Mention both sides clearly for full 2 marks. The textbook explicitly states "individuals can change sex, indicating that sex is not genetically determined" — use this phrasing directly.

Source: Chapter 8, Section 8.2.4 Sex Determination

(i) The expected phenotypic ratio in F2 is 9 : 3 : 3 : 1
(Tall round : Tall wrinkled : Short round : Short wrinkled)

(ii) The two new phenotypic combinations appearing in F2 are:

• Tall plants with wrinkled seeds
• Short plants with round seeds

(iii) The appearance of new trait combinations in F2 indicates that the two traits (stem height and seed shape) are inherited independently of each other. The factors (genes) controlling different traits recombine independently during reproduction, giving rise to new combinations in offspring.

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

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• Part (i): A dihybrid cross always gives 9:3:3:1 in F2 — memorise this ratio.
• Part (ii): The two parental types were tall-round and short-wrinkled; so the new combinations are tall-wrinkled and short-round. Examiners specifically look for this distinction.
• Part (iii): The key phrase the examiner expects is "independently inherited" (or independent assortment). The textbook explicitly states: "the tall/short trait and the round seed/wrinkled seed trait are independently inherited." Always link the observation (new combinations) to the conclusion (independent inheritance of traits).

A gene is a segment of DNA that controls a specific characteristic by directing the synthesis of a particular enzyme.

For example, in pea plants, the gene for height (T/t) controls the production of enzymes that regulate the synthesis of plant hormones (such as gibberellins). These hormones promote cell elongation.

• The dominant gene T directs enzyme production that leads to higher hormone levels → tall plant.
• The recessive gene tt results in reduced enzyme activity → lower hormone levels → short plant.

Thus, the gene → enzyme → hormone → final trait (height) pathway determines the characteristic expressed.

Source: Chapter 8, Section 8.2.2; Chapter 6, Section 6.2

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• The examiner expects you to link the four terms in a logical chain: gene → enzyme → hormone → trait. Missing any link loses marks.
• The T/t example from Mendel (Chapter 8) is the prescribed context; gibberellin is the relevant plant hormone for height (Chapter 6 mentions plant hormones).
• Note: the source passages do not explicitly detail the enzyme-hormone chain for height, so use your textbook knowledge to bridge gene → enzyme → hormone, but keep the Mendelian T/t example as your anchor.
• Dominant (TT/Tt = tall) and recessive (tt = short) language from the passage should appear to show you understand the gene-to-trait connection.

(i) In the F1 generation, no medium-height plants appear because tallness is dominant over shortness (recessive). The F1 plants inherit one 'T' allele from the tall parent and one 't' allele from the short parent, making them Tt. A single copy of 'T' is sufficient to express tallness, so all F1 plants are tall. There is no blending of traits.

(ii) The reappearance of short plants in F2 proves that the F1 tall plants were heterozygous (Tt) — they carried both the tallness trait (T) and the shortness trait (t). The 't' allele was present but not expressed in F1 (being recessive). On self-pollination, tt combination appears in F2, expressing shortness. This shows that both traits were inherited but only tallness was expressed in F1.

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

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• For (i), the key point is dominance — examiners want you to state that T is dominant, Tt is the F1 genotype, and one copy of T is enough for tall phenotype. Never say traits "blend."
• For (ii), the key term examiners look for is heterozygous (Tt). You must explain that the recessive 't' was hidden in F1 but reappears when two Tt plants self-pollinate and produce tt offspring. The textbook phrase "both tallness and shortness traits were inherited in F1 plants, but only tallness was expressed" is the core idea to reproduce.

Cross Diagram:

| | X (from mother) | X (from mother) |
|---|---|---|
| X (from father) | XX (Girl) | XX (Girl) |
| Y (from father) | XY (Boy) | XY (Boy) |

Father determines sex: The mother always contributes an X chromosome to the child. The father can contribute either X or Y. If the child inherits X from the father → girl (XX); if Y → boy (XY). So the sex of the child depends on the father's contribution, not the mother's.

Equal probability: Since a father produces equal numbers of X-bearing and Y-bearing sperms, the chance of fertilisation by either type is equal (50:50), making the probability of a boy equal to the probability of a girl.

Source: Chapter 8, Section 8.2.4 – Sex Determination

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• The cross diagram (Punnett square) is essential — examiners expect it drawn/tabulated clearly.
• Key point: mother is always XX → always gives X; father is XY → gives X or Y randomly.
• "Father determines sex" must be explicitly stated with reasoning, not just implied.
• The 50:50 probability comes from equal production of X and Y sperms — mention this directly.
• All three parts carry marks, so answer each one clearly.

Asexual reproduction creates new individuals from a single parent through DNA copying. Although minor errors during DNA copying introduce some variation, the mechanism is largely identical each time, so very little variation is produced.

Sexual reproduction involves two individuals combining their DNA. This means traits from two parents are inherited separately and can appear in new combinations in the offspring. This mixing of genetic material allows considerably more variation to be generated, maximising the number of successful variations in the population.

Source: Chapter 7 (What you have learnt); Chapter 8 (Introduction)

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• The key contrast is: asexual = one parent, variation only from DNA copying errors (rare/minor); sexual = two parents, variation from combination/mixing of genes (much greater).
• Examiners expect you to name the source (one parent vs. two parents) and the mechanism (DNA copying errors vs. new combinations of traits).
• The phrase "new combinations of traits" from the Chapter 8 summary is a scoring keyword — use it.
• Do not write more than 3–4 sentences; this is a 3-mark answer.

(i) F₁ Genotype: RrYy; Phenotype: Round yellow seeds (R and Y are dominant).

(ii) Round green seeds = R\_yy genotype. From the F₂ Punnett square, this phenotype appears in 3/16 of the offspring.
Possible genotypes: RRyy and Rryy.

(iii) The F₂ genotype rryy breeds true for wrinkled yellow — wait, wrinkled yellow = rrY\_.
The true-breeding genotype is rrYY.
It is homozygous for both recessive (rr = wrinkled) and dominant (YY = yellow) alleles, so it will always produce wrinkled yellow offspring on self-pollination.

Source: Chapter 8, Section 8.2.3

• Part (i): Since RRYY × rryy, all F₁ get one R and one Y from each parent → RrYy. Round and yellow are dominant, so phenotype is round yellow.
• Part (ii): Use the 9:3:3:1 ratio from Fig. 8.5. Round green = 3/16. The two genotypes giving round green are RRyy and Rryy (R\_yy).
• Part (iii): "Breeds true" means homozygous. Wrinkled = rr, yellow = YY → rrYY. Examiners want you to state why it breeds true: because both gene pairs are homozygous, self-pollination always gives the same phenotype. Note: rrYy also gives wrinkled yellow but does NOT breed true.

(a) Cross Diagram – Tall × Short

| Generation | Cross |
|---|---|
| Parents (P) | TT (pure tall) × tt (pure short) |
| F₁ | All Tt (tall) |
| F₁ × F₁ | Tt × Tt |
| F₂ | TT : Tt : tt = 1 : 2 : 1 |

F₂ phenotype ratio → 3 tall : 1 short (TT and Tt are tall; tt is short).

(b) Fraction and Confirmatory Experiment

In F₂, the ratio of TT : Tt : tt is 1 : 2 : 1. Out of 3 tall plants, only 1/3 are pure-breeding (TT).

To confirm: allow each F₂ tall plant to self-pollinate (self-breed) and observe the progeny. A plant that produces only tall offspring in all subsequent generations is TT (pure-breeding), whereas one that gives 3 tall : 1 short is Tt.

(c) Advantage of Counting

Mendel was the first to count the number of individuals showing each trait in every generation rather than just noting traits qualitatively. This gave him exact ratios (3:1 in F₂), which revealed the mathematical pattern of inheritance and allowed him to deduce the laws governing heredity.

Source: Chapter 8 – Heredity, Section 8.2.2 Rules for the Inheritance of Traits – Mendel's Contributions

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• (a) Draw the cross as a table or Punnett-square-style layout. Label clearly: P generation genotypes (TT, tt), F₁ (Tt), F₁ × F₁, and F₂ (TT, Tt, Tt, tt). Examiners award marks for correct genotypes AND the ratio.
• (b) The key trap: students often say "1/4 of all F₂ are TT," which is correct, but the question asks what fraction of the tall plants are TT. Among 3 tall plants (TT + Tt + Tt), only 1 is TT → 1/3. The self-pollination experiment is directly from Activity 8.2 in the textbook.
• (c) The textbook explicitly states Mendel "was the first one to keep count of individuals exhibiting a particular trait in each generation. This helped him to arrive at the laws of inheritance." Use this language.

(a) Human females have two X chromosomes (XX) and males have one X and one Y chromosome (XY). A human cell has 23 pairs of chromosomes in total — 22 pairs are autosomes (perfectly matched pairs) and 1 pair is the sex chromosomes. Unlike the 22 autosomes, the sex chromosomes in males are mismatched: one normal-sized X and one shorter Y chromosome.

(b)

| Mother (XX) | × | Father (XY) |
|---|---|---|
| Eggs: all X | | Sperms: X or Y |

Offspring: XX (girl) or XY (boy) — 50% girls, 50% boys.

The father's contribution determines sex because the mother can only pass X chromosomes, while the father passes either X (→ girl) or Y (→ boy).

(c) In reptiles like crocodiles, sex is determined by incubation temperature, not genes. In snails, individuals can change sex, so sex is not genetically fixed. This shows that sex determination varies widely across species — it can be genetic (humans), environmental (reptiles), or flexible (snails).

Source: Chapter 8, Section 8.2.4 — Sex Determination

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• (a): Examiners expect the numbers 22 + 1 = 23 pairs, and the key distinction that autosomes are matched while the male sex chromosomes are mismatched (X and Y).
• (b): A simple cross table earns marks. The key conclusion — father's X or Y decides the child's sex — must be explicitly stated.
• (c): This 1-mark part only needs a brief, accurate contrast of the three examples and a one-line conclusion about diversity of sex-determination mechanisms. Don't over-write here.

(a) In asexually reproducing organisms, variations arise due to small inaccuracies (errors) in DNA copying during reproduction. For example, when one bacterium divides to produce four individuals, the resultant bacteria are very similar with only minor differences. Variations are limited because only one parent is involved — there is no mixing of genetic material from two individuals. In sexual reproduction, genetic material from two parents combines, generating far greater diversity.

(b) If bacteria living in temperate water face a rise in temperature due to global warming, most will die. However, variants that are heat-resistant will survive and reproduce. This environmental selection of variants is crucial for long-term survival because it ensures that at least some individuals can adapt when conditions change, preventing the species from being wiped out.

(c) Equal genetic contribution is ensured because each parent contributes one set of chromosomes through their gametes (sperm and egg). Since gametes are formed by meiosis (having half the chromosomes), when they fuse at fertilisation, the offspring receives practically equal amounts of DNA from each parent.

Source: Chapter 8, Section 8.1 — Accumulation of Variation During Reproduction; Chapter 7, Section 7.1.1 — The Importance of Variation

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• (a) The key phrase examiners look for is "inaccuracies in DNA copying" as the source of variation, and the point that only one parent means no mixing — hence limited variation.
• (b) Always use a specific example (bacteria + heat/global warming is directly from the textbook). Link selection → survival of species for full marks.
• (c) This is a 1-mark question — one crisp mechanistic point is enough: meiosis halves the chromosome number, fertilisation restores it, ensuring equal contribution. You don't need to write more than 2 lines.

(i) The yellow-seeded parent is likely heterozygous (Yy). The offspring ratio is approximately 48 yellow : 52 green ≈ 1:1, which indicates one parent contributed dominant (Y) and recessive (y) alleles, so the yellow parent must be Yy, not YY.

(ii)
Cross: Yy × yy
Gametes: (Y, y) × (y, y)
Offspring genotypes: Yy (yellow) and yy (green) in equal proportion (1:1).

(iii) If the yellow parent were YY, all gametes would be Y. Cross: YY × yy → all offspring Yy (yellow). Proportion of yellow-seeded offspring = 100%.

(iv) This cross is called a Test Cross. It is useful because the homozygous recessive parent (yy) reveals the unknown genotype — a 1:1 ratio indicates heterozygous (Yy), while all dominant offspring indicate homozygous dominant (YY).

Source: Chapter 8 – Heredity, Section 8.2.2

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• Key concept tested: Mendelian inheritance, dominant/recessive traits, test cross.
• The 1:1 ratio in offspring is the classic indicator of a test cross with a heterozygous parent.
• For (ii), always write gametes explicitly — examiners award marks for showing them.
• For (iii), remember YY × yy → all Yy, so 100% yellow; students often write ½ by mistake.
• Test cross is the exact NCERT term — use it, don't paraphrase it as "back cross" (which is slightly different).

(i) Sugarcane reproduces asexually (vegetative propagation), producing offspring from a single parent through DNA copying. Since no mixing of genetic material occurs, the plants are nearly identical to each other.

(ii) The primary source of variation in asexually reproducing organisms is errors/inaccuracies during DNA copying (replication).

(iii) In sexual reproduction, genetic material from two different individuals (parents) combines. This mixing produces new combinations of traits, generating greater variation. Asexual reproduction in sugarcane involves only one parent, so offspring inherit identical DNA with minimal variation.

(iv) Variation ensures that if the environment changes drastically (e.g., a new disease or climate change), at least some individuals with favourable traits will survive, preventing the entire species from being wiped out.

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

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• (i) & (iii): The key concept is one parent vs. two parents — asexual reproduction copies one genome; sexual reproduction mixes two, creating diversity.
• (ii): Many students write "mutation" — that is acceptable, but the textbook specifically says variation in asexual reproduction arises from DNA copying errors/inaccuracies.
• (iv): Always link variation → survival during environmental change → stability of the species. This is directly from Section 7.1.1 and is a favourite examiner point.
• Keep each answer to 1–2 lines since each carries only 1 mark.

(i) Both parents were RrYy (dihybrid). The 9:3:3:1 ratio is the classic F2 dihybrid ratio, which only appears when both parents are heterozygous for both traits (round = R dominant, yellow = Y dominant). Since both parents had identical phenotype (round yellow) yet produced all four phenotypic classes, each must have carried one dominant and one recessive allele for each gene.

(ii) Wrinkled green (rryy) = 1/16 of offspring.
Expected = (1/16) × 160 = 10 seeds

(iii) Round green and wrinkled yellow are new combinations not seen in either parent. Their appearance indicates that seed shape and seed colour are inherited independently of each other (independent assortment), allowing new combinations to arise in offspring.

(iv) If both traits were always inherited together, each parent would pass only one combined unit (e.g., RY or ry together) — effectively behaving as a single trait. The cross would reduce to Aa × Aa, giving only dominant : recessive = 3:1, with no new trait combinations possible.

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

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• (i) The 9:3:3:1 ratio is the definitive evidence for a dihybrid cross (RrYy × RrYy). Examiners want you to state the genotype AND justify it using the ratio.
• (ii) Straightforward calculation: wrinkled green = rryy = 1 part out of 16. Always show working.
• (iii) "New combinations" means recombinant phenotypes not present in parents. Link directly to independent inheritance/assortment — the key Mendelian principle the textbook states.
• (iv) Linked traits behave as one gene, so dihybrid becomes monohybrid → 3:1. The teacher's statement is testing whether you understand why independent assortment produces four classes instead of two.