Flow chart: Sex Determination in Human Beings

```
Parents: Father (44 + XY) × Mother (44 + XX)
↓ ↓
Gametes: 50% sperm with X All eggs with X
50% sperm with Y
↓
Fertilisation
↓
┌────────────────┴────────────────┐
Egg (X) + Sperm (X) Egg (X) + Sperm (Y)
↓ ↓
Child = XX Child = XY
(Girl) (Boy)
```

• All children inherit X chromosome from mother.
• Sex is determined by the chromosome inherited from the father.
• Father's X → Girl; Father's Y → Boy.
• Half the children will be boys and half will be girls.

Source: Chapter 8, Section 8.2.4 – Sex Determination

---

• Examiners expect a clearly labelled flow chart showing parents → gametes → fertilisation → offspring. Draw it neatly in the exam.
• Key point to state: mother always contributes X; the father's contribution (X or Y) determines the sex.
• Mention the 44 autosomes + sex chromosomes notation for full marks.
• The conclusion — 50% boys, 50% girls — must appear either in the chart or as a point below it.

(a) The 'Y' chromosome is shorter/smaller in size compared to the 'X' chromosome. The mismatched pair of sex chromosomes in humans is XY (found in males).

(b) There is 1 pair of sex chromosomes in human beings. In the female (XX), a perfect/matched pair of sex chromosomes is present.

(c) Two examples showing sex is not always genetically determined:

1. Snails – some individuals can change their sex during their lifetime depending on environmental or social conditions.
2. Reptiles (e.g., lizards/turtles) – the temperature of the environment during egg incubation determines the sex of the offspring, not genes.

Source: Heredity and Evolution, Sex Determination

---

• (a) Examiners expect mention of relative size difference (Y smaller than X) and identification of XY as the mismatched pair.
• (b) Key facts: only 1 pair of sex chromosomes; females have XX (perfect/matched pair), males have XY.
• (c) Two distinct examples are required. The passage directly states that "some animals rely on environmental cues" and "some can change their sex during lifetime" — link your examples clearly to these two points. Snails (sex change) and reptiles (temperature-based determination) are standard NCERT examples. Award is typically 1 mark per valid example with brief justification.

Cross between Red (RR) × White (rr) flowers:

$$RR \times rr$$
$$\downarrow$$
$$\text{F}_1: \text{All } Rr \text{ (Red flowers)}$$
$$\downarrow \text{(Self-pollination)}$$
$$\text{F}_2: RR : Rr : rr = 1:2:1$$

Ratio: 3 Red (RR + Rr) : 1 White (rr) → 3:1

Red (R) is the dominant trait; White (r) is the recessive trait.

Source: Chapter 8 – Heredity, Section 8.2.2

---

• The examiner wants a flow chart format showing P, F₁, and F₂ generations — not a paragraph. Draw the cross clearly with arrows.
• Write the genotypes (RR, Rr, rr) alongside phenotypes.
• Always state the 3:1 phenotypic ratio and identify which trait is dominant/recessive — these are the likely marking points for 2 marks.
• Since red flowers appear in F₁, red is dominant; white reappears in F₂, confirming it is recessive.

In pea plants, height is controlled by a gene with two forms (alleles): 'T' (tallness) and 't' (dwarfness). When a tall plant (TT) is crossed with a short plant (tt), all F₁ plants are tall (Tt). When F₁ plants self-pollinate, F₂ plants are tall (TT/Tt) or short (tt) in a 3:1 ratio. This shows that the gene 'T' or 't' directly determines whether the plant is tall or dwarf — proving genes control traits.

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

---

• Examiners want you to mention: alleles (T and t), F₁ result (all tall), F₂ result (3 tall : 1 short), and the conclusion that genes control traits.
• State that 'T' is dominant (one copy sufficient) and 'tt' gives short plant (recessive — both copies needed).
• Avoid lengthy diagrams unless asked; a brief cross notation (TT × tt → Tt) is enough to support the explanation within the word limit.

(c)

In F₂ progeny from a dihybrid cross, the phenotypic ratio is 9 : 3 : 3 : 1.

Total plants = 1600

(i) Tall with round seeds (9 parts):
$$\frac{9}{16} \times 1600 = \textbf{900 plants}$$

(ii) Short with wrinkled seeds (1 part):
$$\frac{1}{16} \times 1600 = \textbf{100 plants}$$

Conclusion:
Mendel concluded that two traits (height and seed shape) are inherited independently of each other. The F₂ generation shows a 9:3:3:1 ratio, which proves that factors (genes) for different characters assort independently during gamete formation. This is known as the Law of Independent Assortment.

Source: Heredity and Evolution, Mendel's Laws of Inheritance

---

• The dihybrid cross ratio 9:3:3:1 must be stated clearly; examiners expect students to know it by heart.
• Show the calculation (fraction × 1600) for full method marks.
• The conclusion must name the Law of Independent Assortment — this is the key scoring point for the last part.
• Do not mix up 9:3:3:1 with the 3:1 monohybrid ratio.

(a) The F1 progeny will all be tall plants with round seeds, as tall and round are the dominant traits over short and wrinkled respectively.

(b) The recessive traits in the above cross are:

• Short height (dwarf)
• Wrinkled seeds

(c) When F1 progeny (tall, round) is allowed to self-pollinate, the F2 generation shows two new combinations along with parental types in the ratio 9:3:3:1:

| Phenotype | Ratio |
|---|---|
| Tall Round | 9 |
| Tall Wrinkled (new) | 3 |
| Short Round (new) | 3 |
| Short Wrinkled | 1 |

The two new combinations are: Tall Wrinkled and Short Round.

Source: Chapter – Heredity and Evolution, Dihybrid Cross / Laws of Inheritance

---

• (a) Always state that F1 shows only dominant traits — the answer must name both dominant characters (tall + round).
• (b) Examiners want both recessive traits named clearly; one mark each, so don't miss either.
• (c) This is a 2-mark part — you must name the two new combinations AND give the 9:3:3:1 ratio. Missing either loses a mark. The table format is neat but a list is equally acceptable.

(a) Men produce two types of gametes (sperms):

• Sperms carrying X chromosome
• Sperms carrying Y chromosome

(b) No, a male child does not inherit the X chromosome from his father.
Justification: A male child has XY sex chromosomes. He inherits the Y chromosome from his father and the X chromosome from his mother. The father passes the Y chromosome to his son, not the X chromosome.

(c) A human female produces only one type of gamete — the egg (ovum) — all carrying the X chromosome.

---

• This question is from Chapter 8 (How do Organisms Reproduce?) — specifically the section on sex determination in humans.
• Key point: Father determines the sex of the child (XY father passes either X or Y); mother always contributes X.
• The ½ mark sub-questions need single-line answers; the 1-mark justification needs 2–3 clear lines.
• Examiners look for the correct chromosome names (X and Y), not just "two types."

(a) A zygote has 46 chromosomes (44 autosomes + 2 sex chromosomes). Each gamete (sperm or egg) has 23 chromosomes.

(b) In some reptiles like turtles, the temperature at which eggs are incubated determines whether the offspring will be male or female. There are no sex chromosomes involved; sex is determined by environmental factors.

(c)

$$
\underbrace{\text{Mother (44+XX)}}_{\text{Female}} \quad \underbrace{\text{Father (44+XY)}}_{\text{Male}}
$$

$$\downarrow \quad\quad\quad\quad \downarrow$$

$$\text{Eggs: 22+X} \quad \text{Sperms: 22+X \ or \ 22+Y}$$

$$\downarrow$$

$$\text{22+X + 22+X} = \textbf{Girl} \quad \text{22+X + 22+Y} = \textbf{Boy}$$

Since the mother always contributes X, the sex of the child depends on whether the sperm carries X or Y — which is a matter of chance.

Source: Heredity and Evolution, Sex Determination

---

• (a) Remember: gametes are haploid (23), zygote is diploid (46).
• (b) This is a direct factbook point — temperature-dependent sex determination in reptiles.
• (c) The examiner wants a flowchart, not a paragraph. Show both parents, their gametes, and the two possible combinations. The key conclusion is that the father's sperm (X or Y) determines the child's sex, not the mother, so no parent is "responsible."

(c) 1 : 1

In F₂, the genotypes are TT : Tt : tt = 1 : 2 : 1. Pure tall (TT) : pure short (tt) = 1 : 1.

Students often confuse the phenotypic ratio (3 tall : 1 short) with the genotypic ratio. The question asks for pure plants only — TT and tt — both present in 1 out of 4 offspring each, giving 1 : 1. Hybrid tall (Tt) plants are tall in appearance but not pure, so they are excluded from the count.

(a) Two pairs of contrasting characters used by Mendel (other than seed shape and colour):

• Tall plant vs. Dwarf plant
• Inflated pod vs. Constricted pod

(b) A dominant trait is the one that expresses itself in the F₁ generation (e.g., round, yellow seeds). A recessive trait is the one that remains hidden in F₁ but reappears in F₂ (e.g., wrinkled, green seeds).

(c) The F₂ generation shows seeds in the ratio 9 : 3 : 3 : 1
(9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green).

Interpretation: The traits of seed shape and seed colour are independently inherited; they segregate independently during gamete formation and recombine freely. This is Mendel's Law of Independent Assortment.

Source: Heredity and Evolution, Mendel's Contribution

---

• (a) Any two of the seven pairs Mendel used are acceptable; tall/dwarf and inflated/constricted pod are safe choices.
• (b) The key distinction is expression in F₁ vs. hidden in F₁ but visible in F₂. Use the words "expresses" and "remains hidden/suppressed."
• (c) The 9:3:3:1 ratio is the standard dihybrid ratio — must be stated with the full description of each class. The interpretation must mention independent assortment; examiners look for this conclusion explicitly.

(c) (A) is true, but (R) is false.

Assertion is correct — sex is genetically determined by chromosomes inherited from parents. But the Reason is false — a male child inherits his X chromosome from his mother, not his father; he inherits the Y chromosome from his father.

The key fact: men are XY, women are XX. Every child gets X from the mother. The father contributes either X (girl) or Y (boy). So the X in a son comes from the mother, making (R) false. Assertion (A) is true as the textbook clearly states sex is determined by genes (chromosomes) inherited from parents.

(a) In individuals of a given species, a specific gene is located on a particular chromosome.

Source: Chapter 8, Heredity

• Option (b) is wrong — DNA is the information source for making proteins (Chapter 7).
• Option (c) is wrong — each chromosome carries many genes, not just one.
• Option (d) is wrong — in humans, inherited traits are controlled by genes.
• Option (a) is correct: genes have fixed loci on specific chromosomes within a species, which is a fundamental characteristic of genes covered in the heredity chapter.

Pair 1: Plant Height

• Dominant character: Tall plant
• Recessive character: Short plant

Pair 2: Seed Shape

• Dominant character: Round seeds
• Recessive character: Wrinkled seeds

(Any other valid pair such as Flower Colour — Violet (dominant) / White (recessive) is also acceptable.)

Source: Chapter 8, Section 8.2.2

---

• The question asks for two pairs, each with the dominant and recessive character clearly named — that's 1 mark per pair.
• Mendel's seven pairs include: tall/short, round/wrinkled seeds, violet/white flowers, yellow/green seeds, etc. State any two.
• Always write which is dominant and which is recessive explicitly — don't just list them, or you may lose marks.

Option A: Tt and Tt

When two tall plants with genotype Tt are crossed, offspring are TT : Tt : tt in ratio 1:2:1, producing 25% dwarf (tt) plants along with tall plants.

Key logic: dwarf plants appear in offspring only if both parents carry the recessive allele 't'. A Tt × Tt cross gives ¼ tt (dwarf). Option B (Tt × tt) gives 50% dwarf — too many. Options C and D cannot produce dwarf offspring at all (TT is present). Examiners expect you to recall the Tt × Tt cross from Mendel's F1 self-pollination experiment.

Option C: 9 : 3 : 3 : 1

When two traits (shape and colour) are independently inherited, the F₂ progeny shows four combinations in the ratio 9 : 3 : 3 : 1 (round-yellow : round-green : wrinkled-yellow : wrinkled-green).

Source: Chapter 8, Section 8.2.2

• This is Mendel's Law of Independent Assortment applied to a dihybrid cross.
• The 9:3:3:1 ratio appears in F₂ when two independently assorting traits are considered together.
• Examiners expect you to identify option C and briefly justify it — just one line of reasoning is enough for 1 mark.
• Do not confuse with monohybrid ratios: 3:1 (phenotype) or 1:2:1 (genotype).

(D) (A) is false, but (R) is true.

Human female has a perfect pair of sex chromosomes (XX) — this is true, so (A) is actually true. Wait — re-reading: (A) states the female has a perfect pair, which is correct. The sex chromosome contributed by the father (X or Y) determines the sex of the child — (R) is also true. However, (R) does not explain (A); both are independent facts.

Answer: (B) — Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Source: Chapter 8, Section 8.2.4 – Sex Determination

---

• (A) is TRUE: The textbook clearly states "Women have a perfect pair of sex chromosomes, both called X."
• (R) is TRUE: "The sex of the children will be determined by what they inherit from their father."
• But (R) explains sex determination, not why females have a perfect pair — so (R) does not explain (A). Hence (B) is correct.
• Common mistake: Students may choose (A), confusing "correct explanation" with just "both being true."

(C) Tt and tt

When Tt × tt cross is performed, offspring are 50% Tt (tall) and 50% tt (short), matching the given result.

This is a test cross (tall heterozygous × short homozygous recessive). Option A (Tt × Tt) gives 75% tall : 25% short; Option B gives all tall or 3:1; Option D gives all tall. Only Tt × tt produces the 1:1 ratio (50:50) described. Examiners expect you to recall basic monohybrid cross ratios.

(i) In F₁ plants, Mendel observed:

• All plants were tall (no short plants appeared).
• Only one trait (tallness) was expressed; shortness was hidden.

(ii) A dominant trait is the one that expresses itself in F₁ generation (e.g., tallness), whereas a recessive trait is the one that remains hidden in F₁ but reappears in F₂ generation (e.g., shortness).

(iii) Mendel obtained F₂ generation by self-pollination of F₁ plants (RrYy × RrYy).

The ratio of parental combinations (Round Yellow : Wrinkled Green) in F₂ = 9 : 3 : 3 : 1 overall; parental types (RRYY and rryy types) appeared in ratio 9 : 1 among combinations.

Conclusion: The two pairs of traits assorted independently of each other — this is the Law of Independent Assortment.

Source: Heredity and Evolution, Mendel's Laws of Inheritance

---

• For (i), examiners expect two distinct observations about F₁ — all tall, and shortness hidden. One mark requires both points briefly stated.
• For (ii), a clear one-line contrast with examples scores full marks. Avoid long definitions.
• For (iii), three things are tested: method (self-pollination/selfing of F₁), the ratio (9:3:3:1 and parental combinations 9 round yellow : 3 wrinkled yellow : 3 round green : 1 wrinkled green, parental combos ratio 9:1), and the conclusion (Law of Independent Assortment). Cover all three for full 2 marks.

(c) (i), (ii) and (iv)

Statements (i), (ii), and (iv) are correct. Statement (iii) is wrong because human reproductive cells (gametes) are haploid — they contain only one set of chromosomes, not pairs.

• The key trap here is statement (iii): somatic (body) cells have chromosomes in pairs (diploid), but reproductive cells (gametes) have half the number — they are haploid. So (iii) is false.
• Statements (i), (ii), and (iv) are standard textbook facts about chromosomes: they carry hereditary/DNA information, are thread-like structures in the nucleus, and are involved in cell division (DNA copies separate during division).
• The source passage confirms chromosomes carry information for inheritance and are involved in DNA copying/cell division.

(b) (ii) and (iv)

Statement (ii) is correct — sex depends on what the child inherits from the father (X or Y chromosome). Statement (iv) is correct — sex is determined at fertilisation when gametes fuse.

Source: Chapter 8, Section 8.2.4 Sex Determination

---

• Statement (i) is wrong: all children inherit X from the mother, so the mother does not determine sex.
• Statement (iii) is wrong: the passage clearly states half the children will be boys and half will be girls — equal probability.
• Statements (ii) and (iv) are directly supported by the text: the father's gamete (X or Y) decides sex, and this is fixed at fertilisation.

(a) Two observations in F₁ generation:

1. All plants had purple flowers — only one parental trait was expressed.
2. There were no intermediate forms (no "light purple" flowers).

(b)
(i) 25% white-flowered plants in F₂ generation.
(ii) Ratio of WW : Ww : ww = 1 : 2 : 1

(c)

| Dominant Trait | Recessive Trait |
|---|---|
| Expressed even when a single copy of the gene is present (e.g., WW or Ww) | Expressed only when both copies are identical recessive genes (ww) |

Source: Heredity, Section 8.2.2

---

• (a) Examiners expect: (1) only one trait expressed / no blending, and (2) all F₁ plants look alike (all purple). Both are directly from Mendel's observations.
• (b) The cross Ww × Ww gives 1WW : 2Ww : 1ww. Only ww = white = 1/4 = 25%. State both the percentage and the ratio clearly.
• (c) The key distinction: dominant needs only one copy; recessive needs two copies (homozygous). A one-line tabular format is clean and scores full marks for a 1-mark difference question.

Proteins are responsible for expressing traits in an organism. For example, tallness in plants is controlled by a specific protein — the enzyme that produces the plant hormone gibberellin. If the gene for this enzyme is present and functional, gibberellin is produced, the plant grows tall. If the gene is absent or non-functional, the enzyme (protein) is not made, gibberellin is not produced, and the plant remains short (dwarf).

Thus, genes control traits by directing the synthesis of specific proteins.

This question links genetics (Chapter 8) to biochemistry. The key idea is the gene → protein → trait pathway. Examiners expect students to:

1. State that genes control protein synthesis.
2. Give the specific example: gene → enzyme → gibberellin → tallness.
3. Show the consequence when the gene/protein is absent (dwarf plant).

Avoid vague statements like "genes determine height" without explaining the protein link.

(D) 100% tall with round seeds

When TTRR × ttrr are crossed, all F₁ offspring receive one T and one R allele from each parent, giving genotype TtRr — tall (T dominant over t) and round (R dominant over r).

In a dihybrid cross between two pure-breeding parents, all F₁ offspring are heterozygous (TtRr). Since tall (T) and round (R) are dominant traits, 100% of F₁ plants show both dominant phenotypes. Examiners expect you to identify dominance and apply it directly — no Punnett square needed for this straightforward cross.

(a) Chromosomes are thread-like structures present in the nucleus of a cell that carry genetic information (genes) from parents to offspring.

(b) In human males, the sex chromosomes are X and Y. Since X and Y chromosomes are of different sizes and shapes (not perfectly paired), they are called a mismatched pair.

(c) Flow chart showing 50:50 probability:

```
Parents: Mother (XX) × Father (XY)
↓ ↓
Gametes: X X X Y
↓
Zygotes: XX XX XY XY
(Girl)(Girl)(Boy)(Boy)
```

∴ Probability of girl : boy = 50% : 50%

Source: Heredity and Evolution, Sex Determination

---

• (a) is 1 mark — one line definition is enough.
• (b) is 1 mark — key term is "mismatched" because X and Y differ in size/shape.
• (c) is 2 marks — examiners expect a proper cross/flow chart showing gametes from both parents and resulting combinations. Writing all four zygote possibilities (XX, XX, XY, XY) and stating the 50:50 ratio is essential for full marks.

Flow Chart:

$$\text{Father (XY)} \xrightarrow{\text{produces}} \begin{cases} \text{Sperm with X} \\ \text{Sperm with Y} \end{cases}$$

$$\text{Mother (XX)} \xrightarrow{\text{produces}} \text{Egg with X only}$$

• X sperm + X egg → XX → Girl
• Y sperm + X egg → XY → Boy

Since the mother always contributes X, the sex of the child depends on whether the father contributes X or Y chromosome. Thus, the father is responsible for determining the sex of the child.

Source: Chapter 8, Section 8.2.4 – Sex Determination

---

• The examiner wants a flow chart (arrows showing the cross) plus a one-line conclusion — both are needed for full 2 marks.
• Key point: mother is always XX, so only the paternal chromosome (X or Y) decides the sex.
• Avoid writing long paragraphs; the flow chart itself carries most marks here.

(a) Both, Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

In reptiles, temperature determines sex of offspring because sex is not genetically determined in such animals — R correctly explains A.

Both statements are factually correct as per the textbook: in some reptiles, incubation temperature determines sex, and in animals like snails, sex is not genetically determined. R directly explains why A is true, making option (a) the correct choice. Students must check whether R correctly explains A, not just whether both are true.

(c) 100% ; 75%

When pure-tall (TT) is crossed with pure-dwarf (tt), all F₁ plants are Tt (tall) = 100% tall. In F₂, ratio is 1TT : 2Tt : 1tt, giving 3 tall : 1 dwarf = 75% tall.

Remember: tall is dominant over dwarf. F₁ gives 100% tall (all Tt). In F₂, self-crossing Tt × Tt gives the classic 3:1 Mendelian ratio — 75% tall, 25% dwarf. This is the foundational result of Mendel's monohybrid cross.

Mendel conducted experiments on pea plants using two traits simultaneously (dihybrid cross). He crossed tall plants with round seeds with short plants with wrinkled seeds. The F1 progeny were all tall with round seeds. When F1 plants were self-pollinated, the F2 generation showed new combinations — tall with wrinkled seeds and short with round seeds — in addition to the original combinations.

This proved that the tall/short trait and round/wrinkled seed trait were inherited independently of each other, as genes controlling different traits recombine independently during reproduction to form new combinations in offspring.

Source: Chapter 8, Section 8.2.2

---

• The key concept examiners look for is the dihybrid cross and the appearance of new trait combinations in F2.
• Use the exact phrase "independently inherited" as it appears in the textbook.
• Mention F1 and F2 generations clearly — this shows you understand the experimental design.
• Do not confuse this with the monohybrid cross (tall/short only); the question specifically asks about independent inheritance, which requires two traits.

In humans, females have two X chromosomes (XX) and males have one X and one Y chromosome (XY). The mother can only pass on an X chromosome to every child. The father, however, can pass either X or Y. A child inheriting X from the father is a girl; inheriting Y makes a boy. Hence, the sex of the child is determined by the father.

Source: Chapter 8, Section 8.2.4 – Sex Determination

---

The key logic examiners want: mother is always XX → only X possible from her → father's contribution (X or Y) is the deciding factor. State this clearly with the XX/XY notation. Two marks = two key points: (1) mother contributes only X, (2) father contributes either X or Y, determining the sex. Don't forget to mention which combination gives a girl (XX) and which gives a boy (XY).

In Mendel's cross, each trait is controlled by two factors (genes). One factor came from each parent. Since only one parental trait (tall) appeared in F1, the trait from one parent was dominant over the other (recessive). The dominant trait completely masked the recessive one, so no intermediate or halfway characteristics were observed.

Source: Chapter 8, Section 8.2.2

---

• The examiner expects you to use the terms dominant and recessive correctly.
• The key idea is that one allele completely masks the other — there is no blending — which is why F1 shows only the dominant trait, not a mix.
• Avoid vague phrases like "stronger gene"; say "dominant trait masks the recessive trait."

(a) In males, the pair of sex chromosomes consists of one X chromosome and one Y chromosome, which differ in both type and size (Y is much smaller than X). Hence, it is called a mismatched pair.

(b) Female has a perfect pair of sex chromosomes (XX). Since both chromosomes are identical, all eggs produced will carry one X chromosome — so gametes will be of the same kind.

(c) (Note: The alternatives for part (c) are not printed in the question. Based on the passage, the standard answer is:)
The sex of a child is determined by the father, not the mother. All eggs from the mother carry X chromosome. The father produces two types of sperms — 50% carrying X and 50% carrying Y. If a sperm with X fertilises the egg → girl (XX); if a sperm with Y fertilises the egg → boy (XY). Thus, the father's sperm determines the sex of the child.

Source: Chapter 8 – How do Organisms Reproduce?, Sex Determination

---

• For (a), the key point is that X and Y differ in both type and size — examiners expect both words.
• For (b), remember females are XX (homogametic) → one type of gamete; males are XY (heterogametic) → two types. This directly follows from the passage.
• For (c), since the alternatives were missing, the standard NCERT-based answer about father determining sex is used. If alternatives are printed, choose the one stating the father/male parent is responsible for the child's sex.

(i) In F₁ progeny, all plants will have violet flowers (like the dominant parent). The white flower trait (recessive) will not appear; all offspring show only the dominant character.

(ii) In F₂ progeny, both violet-flowered and white-flowered plants will appear in the ratio 3:1 (3 violet : 1 white). The recessive trait (white flower) reappears.

(iii) In a 3:1 ratio, 25 plants out of 100 will show the recessive trait (white flowers).

• This is a standard Mendelian monohybrid cross question. F₁ shows only the dominant trait; F₂ gives a 3:1 phenotypic ratio.
• Examiners expect you to clearly state "all violet" for F₁, the 3:1 ratio for F₂, and the calculation (100 × 1/4 = 25) for part (iii).
• Key terms: dominant, recessive, F₁, F₂, 3:1 ratio — use them explicitly.

Variation is the difference in characteristics among individuals of the same species arising during reproduction.

Two main reasons for variation:

1. Inaccurate DNA copying — Small errors occur during DNA replication, producing subtle differences in the new individual.
2. Sexual reproduction — Combining DNA from two different individuals creates new combinations of variations in the offspring.

Source: Chapter 7, Section 7.1.1 & 7.3.1; Chapter 8, Section 8.1

---

• The definition should mention "differences among individuals of the same species" linked to reproduction — examiners expect this context.
• For the two reasons, the textbook specifically highlights DNA copying errors and sexual reproduction (combining DNA from two parents). Naming both scores full marks.
• Do not confuse "variation" with "mutation" — use the textbook terminology.
• At 2 marks, one line for the definition and two labelled points is the ideal format.

(a) The sex of a newborn is determined by the type of sex chromosome inherited from the father. If the child inherits an X chromosome from the father, it is a girl (XX); if it inherits a Y chromosome, it is a boy (XY).

(b) In males, the two sex chromosomes are X and Y, which are not perfectly homologous — they differ in size and genetic content. Hence, they are called a mismatched pair.

(c) During gamete formation (meiosis), the chromosome number is halved — each gamete carries only one set (haploid, n). When the male gamete (n) fuses with the female gamete (n) during fertilisation, the original diploid number (2n) is restored in the progeny.

Source: Heredity and Evolution, Sex Determination / Chromosomes and Sex

---

• (a) Examiners want mention of the father's contribution (X or Y) and the resulting XX/XY combination.
• (b) Key phrase is "mismatched" — X and Y differ in size and gene content; one mark for stating this clearly.
• (c) This is a 2-mark answer: name meiosis (halving) + fertilisation (restoration). Both steps must be mentioned for full credit.

(c) Assertion (A) is true, but Reason (R) is false.

A is true — humans show great variation. R is false — not all variations have equal survival chances; environment selects certain variants over others.

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

The textbook explicitly states: "Do all these variations in a species have equal chances of surviving in the environment in which they find themselves? Obviously not." Environmental selection favours certain variants (e.g., heat-resistant bacteria survive a heat wave). So the Assertion is correct, but the Reason directly contradicts the textbook — making option (c) the right choice.

(c) Dominant trait, Inherited trait

Statement (i) describes the trait expressed in F1 generation — this is the dominant trait. Statement (ii) describes a trait passed from generation to generation — this is an inherited trait.

• "Dominant trait" is the one expressed when two different copies of a gene are present (as seen in F1).
• "Recessive trait" is hidden in F1, so option (a) and (b) are wrong.
• The trait in statement (ii) keeps passing across generations, which fits "inherited trait," not specifically "recessive." This rules out option (b).
• Hence (c) is correct.

(i) In sexual reproduction, offspring receive traits from both parents. Since each parent contributes different combinations of genes/traits, the offspring inherits a mix of characteristics from both, making it not an exact copy of either parent.

(ii) The F1 generation (first filial) shows only the dominant trait — all offspring look alike. The F2 generation (second filial), obtained by self-pollinating F1 plants, shows both dominant and recessive traits in a 3:1 ratio.

(iii) (Sub-question (A) or (B) text was not provided in the question. Please include the full text of option A or B to receive an answer.)

Source: Chapter — Heredity and Evolution, Section — Heredity

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• (i) Key point: sexual reproduction involves two parents → genetic mixing → variations. Mention "genes" or "traits from both parents."
• (ii) Examiners expect: F1 = only dominant trait visible; F2 = both traits appear in 3:1 ratio. Use correct terminology (dominant/recessive).
• (iii) Since sub-question (iii) options (A)/(B) were not included in the question text, they cannot be answered. Always copy the full question when seeking model answers.

(d) (A) is false but (R) is true.

Assertion is false because progeny with bb (homozygous recessive) will have brown eyes, not black. Reason is true — black eye colour (B) is the dominant trait.

• The key error in Assertion (A) is including bb: since 'b' is recessive, bb gives brown eyes, not black.
• BB and Bb both give black eyes (dominant trait expressed), but bb gives brown eyes.
• Reason (R) is correctly stated — dominance of B explains why BB and Bb show black eyes, but it does not save the Assertion from being false.
• So the correct option is (d).

(c) 3 : 1

Cross: vv × VV → F1 all Vv (violet). F1 × F1: VV : Vv : vv = 1:2:1, so violet (VV + Vv) : white (vv) = 3 : 1.

The cross vv × VV gives all Vv in F1. F2 from Vv × Vv gives 1VV : 2Vv : 1vv. Since V is dominant, both VV and Vv show violet flowers — so violet : white = 3 : 1. This is the classic Mendelian monohybrid ratio. Examiners expect you to pick option (c) and ideally know why (F1 × F1 logic).

(a) The gene combination in F1 plants is Tt (hybrid tall).

(b) 'T' (tallness) is a dominant trait. A single copy of 'T' is enough to make the plant tall, so 'tt' (shortness) being recessive is not expressed in F1.

(c) The ratio of plants in F2 generation is 3 tall : 1 short (i.e., 1 TT : 2 Tt : 1 tt).

Source: Chapter 8, Section 8.2.2

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• Examiners expect the genotype Tt for part (a) — not just "hybrid."
• For part (b), use the key terms dominant and recessive; state that one copy of T is sufficient.
• For part (c), the phenotypic ratio 3:1 is essential. Writing the genotypic ratio 1:2:1 alongside shows deeper understanding but the phenotypic ratio is the primary answer expected here.
• All three parts together carry 2 marks, so keep each answer to one line.

(b) 44 + XY

When a Y-carrying sperm fertilises an egg (which carries X), the zygote has 44 autosomes + XY sex chromosomes.

Source: What you have learnt, Chapter 8

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• Every human egg carries 22 autosomes + X; every sperm carries 22 autosomes + X or Y.
• At fertilisation, both sets combine: 22 + 22 = 44 autosomes, and X (egg) + Y (sperm) = XY.
• Total = 44 + XY → male offspring.
• Common mistake: choosing 22 + XY (forgetting that fertilisation doubles the chromosome number from each gamete).

(a) The F₁ progeny plants all looked tall. Their gene combination was Tt (hybrid tall).

(b) In F₁ plants, the gene for shortness (t) could not be expressed because the gene for tallness (T) is dominant over the gene for shortness (t), which is recessive. A single copy of 'T' is sufficient to make the plant tall.

(c) In F₂ progeny, the ratio of tall to short plants was 3 : 1.
Conclusion: Both tallness and shortness traits were inherited by F₁ plants, but only tallness was expressed. Recessive traits are not lost; they reappear in later generations.

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

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• (a) Cross TT × tt → all offspring are Tt. Since T is dominant, all appear tall. Examiners expect both the phenotype and genotype.
• (b) The key terms examiners look for are dominant (T) and recessive (t). The textbook explicitly states "a single copy of T is enough to make the plant tall."
• (c) The 3:1 phenotypic ratio (3 tall : 1 short) is a standard fact. The conclusion must mention that the recessive trait is not lost but reappears — this is the heart of Mendel's finding and is directly from the textbook passage.

(D) Assertion (A) is false, but Reason (R) is true.

A is false because sex is determined by what children inherit from their father (X gives girl, Y gives boy), not the mother. R is true — women are XX.

The passage clearly states: "the sex of the children will be determined by what they inherit from their father." Since mothers are always XX, they can only pass an X chromosome — so the mother's contribution does not determine sex. The father's X or Y does. Reason (R) is a true fact but it actually explains why the Assertion is false (mother only contributes X, so she cannot determine sex).

Differences between Dominant and Recessive Traits:

| Dominant Trait | Recessive Trait |
|---|---|
| Expressed when even one copy of the gene is present (e.g., Tt or TT) | Expressed only when both copies are the same recessive gene (e.g., tt) |
| Masks the recessive trait | Gets masked/suppressed in the presence of the dominant gene |

Percentage of yellow-seeded plants in F2 generation:

Cross: YY × yy → F1: Yy (all yellow)

F1 × F1: Yy × Yy → F2: YY : Yy : yy = 1 : 2 : 1

Plants with yellow seeds (YY + Yy) = 75%

Source: Heredity, Section 8.2.2

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• Give exactly two differences — one mark each.
• The cross/calculation carries one mark; show the F2 ratio clearly and state 75%. Examiners look for the Punnett logic (F1 = Yy, F2 ratio) and the final percentage. Writing just "75%" without the cross loses marks.
• Remember: yellow (Y) is dominant over green (y), so YY and Yy both appear yellow.

(a) The seeds of F₁ generation will be round and yellow, as roundness (R) and yellowness (Y) are dominant traits.

(b) The four possible combinations in F₂ generation are:

• Round, Yellow
• Round, Green
• Wrinkled, Yellow
• Wrinkled, Green

Their ratio is 9 : 3 : 3 : 1.

(c) The new combinations (Round Green and Wrinkled Yellow) appear because the factors (genes) controlling seed shape and seed colour assort independently during gamete formation and recombine randomly during fertilisation to form the zygote.

Source: Chapter 8, Section 8.2.2

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• (a) is straightforward — F₁ of a dihybrid cross always shows both dominant traits (RrYy phenotype).
• (b) Examiners expect all four phenotypes listed clearly with the 9:3:3:1 ratio. Missing the ratio or any combination loses the mark.
• (c) The key phrase the examiner looks for is independent assortment of traits/genes. The textbook specifically states traits are "independently inherited" and new combinations arise when "factors controlling seed shape and seed colour recombine to form the zygote." Use this language.

(a) Proteins control characteristics by acting as enzymes or structural components. For example, in pea plants, genes produce a specific enzyme that synthesises plant hormones (like gibberellin). More hormone → taller plant; less hormone → dwarf plant.

(b) The section of DNA that controls characteristics in an organism is called a gene.

• For (a), the key idea is: gene → protein (enzyme) → characteristic. Use the tallness example: enzyme controls gibberellin production, which determines plant height.
• For (b), the one-word answer "gene" is sufficient. Examiners want the exact term — do not write "DNA segment" without naming it as a gene.
• These are standard NCERT Chapter 9 (Heredity and Evolution) concepts. Memorise the gene–protein–trait link.

(a) Since free earlobe (F) is dominant over attached earlobe (f), the mother with free earlobes must be heterozygous (Ff) and the father with attached earlobes is homozygous recessive (ff).

Cross: Ff × ff

| | F | f |
|---|---|---|
| f | Ff | ff |

Progeny: 50% Ff (free earlobe) and 50% ff (attached earlobe) — this explains the 1:1 ratio observed.

(b) Father's gene combination: ff (attached earlobes)
Mother's gene combination: Ff (free earlobes)

Source: Chapter 8, Section 8.2.1 – Inherited Traits

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• The 50:50 ratio is a classic test cross result (Ff × ff), which tells the examiner the mother must be heterozygous, not homozygous dominant (FF × ff would give 100% free earlobes).
• Always show a Punnett square for 2-mark genetics questions — it earns method marks.
• For part (b), write gene combinations clearly using standard notation (Ff, ff); do not just write "dominant/recessive."

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

A mango seed inherits genetic material from parent plant, and heredity ensures traits are reliably passed to offspring, so it germinates into a mango tree only.

Source: Chapter 8, Section 8.2 (Heredity)

• The assertion is true: a mango seed carries the genetic blueprint of the mango plant and will always grow into a mango tree.
• The reason is also true and directly explains the assertion: heredity (rules of inheritance) ensures that offspring reliably resemble their parents.
• Since R correctly explains why A is true, option (A) is the right choice.
• Key line from textbook: "The rules of heredity determine the process by which traits and characteristics are reliably inherited."

(a) Chromosomes are thread-like structures found in the nucleus of a cell, made up of DNA and proteins. They carry genes which transmit hereditary information from parents to offspring.

(b) In sexually reproducing organisms, gametes (sperm and egg) are formed by meiosis, which halves the chromosome number. When two gametes fuse during fertilisation, the original chromosome number is restored. This ensures that the DNA content remains constant in every generation of a species.

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• (a) One mark — state what chromosomes are (location + composition). Mentioning DNA and genes earns the mark.
• (b) Two marks — the key concept is the meiosis–fertilisation cycle: meiosis halves the chromosome number (1 mark), fertilisation restores it (1 mark). Use correct terms: meiosis, gametes, fertilisation. No diagram is needed for 2 marks but mentioning haploid/diploid can strengthen the answer.

(C) Assertion (A) is true, but Reason (R) is false.

A child bears all basic human features, but it does not look exactly like its parents — human populations show a great deal of variation.

The textbook (Chapter 8, Section 8.2.1) explicitly states: "a child bears all the basic features of a human being. However, it does not look exactly like its parents, and human populations show a great deal of variation." So A is correct, but R contradicts the textbook — making R false. Option (C) is the only fit.

(C) 100% round and yellow

When RRyy × rrYY are crossed, all F₁ offspring are RrYy — round (R dominant over r) and yellow (Y dominant over y).

From the textbook (Fig. 8.4, chapter 8): RRyy (round, green) × rrYY (wrinkled, yellow) → all F₁ = RrYy. Round (R) is dominant over wrinkled (r), and yellow (Y) is dominant over green (y), so 100% of F₁ seeds are round and yellow. Examiners expect you to recall dominant-recessive relationships and the F₁ genotype directly.

Cross:

P generation: TT (Pure Tall) × tt (Pure Short)

F₁ progeny: All Tt (Tall)

$$TT \times tt \rightarrow Tt, \ Tt, \ Tt, \ Tt$$

Observations:

1. All F₁ plants were tall — no medium-height plants were observed.
2. Only one parental trait (tallness) was expressed in F₁; shortness was not visible.
3. This showed that tallness is dominant over shortness (recessive).
4. The F₁ plants (Tt) inherited both factors but expressed only the dominant one.

Source: Chapter 8, Section 8.2.2

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• The cross diagram (P → F₁) with genotypes earns 1 mark; observations earn the remaining 2 marks — so write at least 3 clear observations.
• Examiners expect the terms dominant and recessive to appear.
• State explicitly that no intermediate (medium) height plants appeared — this is a key Mendelian observation.
• Use genotype notation (TT, tt, Tt) to show you understand the factor/gene concept.

(a) Sex of offspring is not always genetically determined in all animals. In some animals, environmental factors decide sex:

• Snails – social factors can determine sex.
• Lizards/Turtles – temperature during incubation of eggs determines the sex of offspring.

Thus, factors other than sex chromosomes can determine sex in certain animals.

(b)

| Feature | Female | Male |
|---|---|---|
| Sex chromosomes | XX (two X chromosomes) | XY (one X, one Y chromosome) |
| Gametes produced | All eggs carry X | 50% sperms carry X, 50% carry Y |

The Y chromosome in males is smaller and carries fewer genes than the X chromosome.

• Part (a) is 2 marks — give two distinct examples with a brief explanation each. Snails (social/environmental), lizards/turtles (temperature) are standard NCERT examples. The key point is that environmental factors, not genes, determine sex here.
• Part (b) is 1 mark — a simple comparison: Female = XX, Male = XY. One crisp differentiating point is sufficient. Don't over-explain.
• The source passages provided are unrelated (Chemistry/Physics), so this answer is based on NCERT Class 10 Biology (Chapter: Heredity and Evolution), as required by the question.

Variation means subtle differences among individuals of a species. When the environment changes drastically, most individuals may die, but variants better suited to the new conditions survive and reproduce.

Example: A population of bacteria lives in temperate water. Due to global warming, water temperature rises. Most bacteria die, but heat-resistant variants survive and multiply, ensuring the species continues.

Thus, variation is useful for survival of a species over time.

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

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• Examiners expect: a definition/explanation of variation + a clear example — the bacteria/heat example from the textbook is ideal and safest to use.
• The key idea is that variation helps the species (not every individual) survive when the environment changes.
• Avoid writing a long essay; two crisp points + one example is sufficient for 2 marks.
• Always name the organism, the environmental change, and the outcome in your example.

(D) 1:2:1

F1 hybrid is Gg. On self-pollination: GG : Gg : gg = 1 : 2 : 1.

F1 cross is Gg × Gg. The Punnett square gives GG, Gg, Gg, gg — ratio 1:2:1. This is the standard monohybrid F2 genotypic ratio examiners expect students to recall directly.

Cross: RRYY × rryy

F₁ Generation:
All F₁ plants are RrYy (round, yellow) — dominant traits appear.

F₂ Generation (F₁ × F₁: RrYy × RrYy):

| | RY | Ry | rY | ry |
|---|---|---|---|---|
| RY | RRYY | RRYy | RrYY | RrYy |
| Ry | RRYy | RRyy | RrYy | Rryy |
| rY | RrYY | RrYy | rrYY | rrYy |
| ry | RrYy | Rryy | rrYy | rryy |

F₂ phenotypic ratio = 9 : 3 : 3 : 1 (out of 16 combinations)

(i) RRYY — 1 plant out of 16

(ii) rryy — 1 plant out of 16

Source: Chapter 8 – Heredity, Section 8.2.3 (Figure 8.5)

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• The question asks for a dihybrid cross up to F₂; draw the Punnett square clearly.
• Note: The textbook (Fig. 8.5) shows RRyy × rrYY, but the question uses RRYY × rryy. The F₁ is still RrYy and the F₂ Punnett square is identical.
• Both RRYY and rryy appear once each in the 4×4 Punnett square (1/16 each). Examiners award marks for the correct Punnett square AND the specific counts.
• Always state F₁ genotype before drawing F₂.

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

The textbook (Ch. 8) explicitly states that bacteria able to withstand heat survive better in a heat wave — confirming A. It also states that accumulation of variations gives species better chances of survival in a changed environment — confirming R. R directly explains A because heat-resistant bacteria survive due to the presence of accumulated variations in the population. Option (A) is therefore correct.

(C) 44 + XY

When the egg (22 + X) is fertilized by a sperm carrying Y chromosome (22 + Y), the zygote formed has 44 + XY chromosomes, developing into a male child.

The human egg always carries an X chromosome along with 22 autosomes (total 23). A sperm carrying Y has 22 + Y. On fertilization: (22 + X) + (22 + Y) = 44 + XY. Remember, sex is determined by the father's sperm, not the mother's egg.

(a) In the F₁ generation, only tall plants were observed because tallness is the dominant trait. When a tall plant (TT) is crossed with a short plant (tt), all F₁ offspring are Tt. Since a single copy of 'T' is enough to make the plant tall, all plants appeared tall.

(b) Mendel obtained F₂ progeny by self-pollination of the F₁ tall plants.

(c)

| Dominant Trait | Recessive Trait |
|---|---|
| The trait that expresses itself in the presence of even one copy of the factor (e.g., tallness – T). | The trait that expresses itself only when both copies of the factor are the same recessive type (e.g., shortness – tt). |

Source: Chapter 8, Section 8.2.2

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• (a) Always mention the genotypes (TT × tt → Tt) and state that one copy of 'T' is sufficient for expression — this earns full marks.
• (b) The key term is self-pollination — do not write "cross-pollination."
• (c) Two marks means two distinct points — one for dominant, one for recessive. Using a table format is clear and saves time. The examiner looks for the idea that dominant is expressed with one copy, recessive needs two copies of the recessive allele.

Answer: A — (ii) and (iii)

In human beings, females are XX and males are XY. The female parent always gives an X chromosome to all children. The male parent gives either X (producing a girl) or Y (producing a boy) through his gametes. So statements (ii) and (iii) are correct — it is the father's gamete (carrying X or Y) that determines sex, not the mother's.

Source: Chapter 8, Section 8.2.4 – Sex Determination

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• Statement (i) is incorrect because the female (mother) always contributes an X chromosome — this does not determine sex, as it is constant.
• Statements (ii) and (iii) are correct because the male contributes either X or Y, making sex determination entirely dependent on the paternal gamete.
• Key exam point: Sex is determined by the father, not the mother in human beings.

(a) Cross:

Let Red (R) be dominant over white (r).

P: RR (Red) × rr (White)

F₁: All Rr (Red)

F₁ × F₁: Rr × Rr

F₂:

| | R | r |
|--|--|--|
| R | RR | Rr |
| r | Rr | rr |

F₂: RR : Rr : rr → 1 : 2 : 1

(b) F₂ Ratios:

• Phenotypic ratio (external appearance): Red : White = 3 : 1 (33 red : 11 white, approximately 3:1)

• Genotypic ratio (genetic makeup): RR : Rr : rr = 1 : 2 : 1

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• (a) CBSE expects the actual Punnett square or gamete cross notation. Show P, F₁, and F₁×F₁ steps clearly using dominant (capital) and recessive (small) letters.
• (b) Two separate ratios must be stated: phenotypic (3:1) and genotypic (1:2:1). Examiners deduct marks if either is missing. The 33:11 from the table confirms the 3:1 ratio.

(D) (i) and (iii)

The parental combinations were round green (RRyy) and wrinkled yellow (rrYY). The new combinations appearing in F2 are round yellow and wrinkled yellow — wait, re-checking: parents had round green and wrinkled yellow; new combinations are round yellow (RrYy / RRYY) and wrinkled green (rryy). So the answer is (B) (i) and (iv).

(B) (ii) and (iii)

The correct answer is (D) (ii) and (iii) — No.

Correct Answer: (B)

The parents were round green and wrinkled yellow. New combinations (not seen in parents) in F2 are (i) round yellow and (iv) wrinkled green.

The parents are RRyy (round, green) and rrYY (wrinkled, yellow). So the parental phenotypes are "round green" and "wrinkled yellow." In F2 (from Fig. 8.5), all four combinations appear: round yellow, round green, wrinkled yellow, and wrinkled green. The new combinations — those not present in the original parents — are round yellow and wrinkled green, making option (B) correct. Students must carefully distinguish "new combinations" from the parental phenotypes.

(B) (i) and (iv)

Swimming technique and dancing skills are acquired traits (learned through practice) and cannot be transmitted genetically to offspring.

The key concept here is the difference between inherited traits (determined by genes, e.g., blood group, eye colour) and acquired traits (skills learned during one's lifetime, e.g., swimming, dancing). Acquired traits do not cause changes in DNA/genes and therefore cannot be passed on to offspring. Blood group and eye colour are genetically determined and are heritable.