ScienceCBSE Class 10 · Chapter 8

Heredity Class 10: Mendel's Experiments and Sex Determination

Updated 14 October 2026 · Fact-checked

Heredity is the passing of traits from parents to offspring through genes. This Class 10 chapter covers how variation builds up during reproduction, Mendel's pea experiments, dominant and recessive traits, independent inheritance, how genes control traits and sex determination in humans. Board questions ask you to work out crosses, ratios and genotypes, and to explain XX and XY inheritance.

What Heredity is about

This chapter answers a simple question: why do children look like their parents, but never exactly like them? You start with how small differences, called variations, build up from one generation to the next. Then you follow Mendel's pea plant experiments, which showed that traits are controlled by pairs of factors, now called genes. You also see how a gene actually controls a trait, and how the sex of a child is decided in humans.

It builds directly on the previous chapter, How do Organisms Reproduce?. That chapter explained DNA copying, germ cells and fertilisation. This chapter uses those ideas to explain why both parents contribute equally to a child, and why germ cells carry only one set of chromosomes. Evolution, which once followed in the same chapter, is now school-assessed only and is not part of the board exam.

Board questions come in several shapes. MCQs ask you to read a genotype from a ratio of offspring. Assertion-reason items test why variation helps survival or how genes control height. Short answers ask you to work out a cross step by step with ratios. Case-based questions give you a Mendel-style experiment or a passage on chromosomes to interpret. Long answers ask for a full cross with explanation, or for sex determination with reasoning.

Weightage: This chapter is part of Unit II, World of Living, which carries 25 of the 80 theory marks together with Life Processes, Control and Coordination and How do Organisms Reproduce?.Source: CBSE Science syllabus 2026-27, Unit II World of Living

In real board papers: asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.8 marks per paper. See the questions

Key concepts

Variation During Reproduction

NCERT section 8.1 Accumulation of Variation During Reproduction

Every new generation inherits two things from the one before it. It gets a common basic body design and small changes in that design. When this generation reproduces, its offspring carry the old differences plus fresh new ones. So variation keeps building up over generations.

How much variation appears depends on how the organism reproduces. In asexual reproduction, one bacterium divides into two, and those two divide into four. The four are very similar. The only differences come from small inaccuracies when DNA is copied. In sexual reproduction, DNA from two different parents comes together, so much greater diversity is produced. This is why a field of sugarcane shows very little variation, while humans show clearly visible differences.

Why does this matter? Not all variations have an equal chance of surviving. Suppose a heat wave strikes. Bacteria that happen to withstand heat will survive better than the rest. Variation therefore protects a species from being wiped out by one change in its surroundings.

Worked example

(a) One bacterium divides twice to give four bacteria. Separately, four children are born to the same two parents. Which group will show more variation? Give a reason for each group.
(b) How can variation in a population of bacteria help it during a heat wave? (3 marks)

Try it first, then show the solutionSolution

The four bacteria show very little variation. Their only differences come from small inaccuracies in DNA copying during division. (1 mark)
The four children show much more variation. Sexual reproduction combines DNA from two different parents, so each child gets a new combination. (1 mark)
In a heat wave, the variants that can withstand heat survive better. So the population is not wiped out completely. (1 mark)

Answer: The children show more variation, because sexual reproduction mixes DNA from two parents, while the bacteria differ only by DNA copying errors. Heat-tolerant variants survive a heat wave, so variation helps the species survive.

Inherited Traits and Activity 8.1

NCERT section 8.2.1 Inherited Traits

Heredity means the rules by which traits and characteristics are reliably passed from parents to offspring. A child always has all the basic features of a human being. But the child does not look exactly like its parents, and human populations show a lot of variation.

A clear example is the earlobe, the lowest part of the ear. In some people it hangs free. In others it is closely attached to the side of the head. Free and attached earlobes are two variants of the same trait.

In Activity 8.1 you survey your class. You count how many students have free and how many have attached earlobes, and you work out the percentage of each. Then you find out the earlobe type of each student's parents and compare. The comparison is the important part. If a child shows a type that neither parent shows, that type must have been carried by the parents without showing. Thinking like this is how you suggest a rule of inheritance from family evidence, just as Mendel did with peas.

Worked example

In a class of 40 students doing Activity 8.1, 28 have free earlobes and 12 have attached earlobes.
(a) Calculate the percentage of students with each type.
(b) One student, Ravi, has attached earlobes, but both his parents have free earlobes. What does this suggest about the inheritance of earlobe type? (3 marks)

Try it first, then show the solutionSolution

Free earlobes: $\frac{28}{40} \times 100 = 70%$ . (½ mark)
Attached earlobes: $\frac{12}{40} \times 100 = 30%$ . (½ mark)
Both parents show free earlobes, yet the child shows attached earlobes. So the attached trait was present in the parents but was not expressed. (1 mark)
This suggests that attached earlobe behaves as a recessive trait and free earlobe as dominant. Each parent passed on one hidden copy for attached earlobes. (1 mark)

Answer: 70% free and 30% attached. The evidence suggests that attached earlobe is recessive: it can stay hidden in parents and show up in a child who receives it from both.

Mendel's Monohybrid Cross (Tall × Short)

NCERT section 8.2.2 Rules for the Inheritance of Traits – Mendel's Contributions

Mendel used garden peas with clear contrasting characters, such as tall/short plants, round/wrinkled seeds and violet/white flowers. He crossed a tall plant with a short plant. Then he counted how many offspring showed each trait. Counting was the key to his success.

All the first-generation (F1) plants were tall. There were no medium-height plants, so the traits did not blend. Only one parental trait showed.

Next, he let the F1 tall plants self-pollinate. Their offspring, the F2, were not all tall: one quarter were short. So the F1 plants must have carried the shortness trait without showing it.

Mendel proposed that each sexually reproducing organism has two copies of a factor (now called a gene) for each trait. Pure tall is $T T$ and pure short is $tt$ . F1 is $T t$ . Selfing $T t$ gives $T T$ , $T t$ and $tt$ in a $1 : 2 : 1$ ratio. Because $T T$ and $T t$ both look tall, the plants appear in a $3 : 1$ ratio of tall to short.

To confirm the $1 : 2 : 1$ ratio (Activity 8.2), you self-pollinate each F2 tall plant. A $T T$ plant gives only tall offspring. A $T t$ plant gives both tall and short offspring.

Worked example

A pure tall pea plant ( $T T$ ) is crossed with a pure short pea plant ( $tt$ ).
(a) Give the genotype and phenotype of the F1 plants.
(b) The F1 plants are self-pollinated. Show the F2 genotypes and give the genotype and phenotype ratios.
(c) If 800 F2 plants are raised, how many are expected to be short, and how many tall plants would have the genotype $T t$ ? (3 marks)

Try it first, then show the solutionSolution

Gametes: the $T T$ parent gives only $T$ and the $tt$ parent gives only $t$ . So all F1 plants are $T t$ , and all are tall. (1 mark)
F1 selfing: $T t \times T t$ . Each parent gives $T$ or $t$ , so the F2 is $T T$ , $T t$ , $T t$ , $tt$ . The genotype ratio is $1 : 2 : 1$ and the phenotype ratio is 3 tall : 1 short. (1 mark)
Short plants $= \frac{1}{4} \times 800 = 200$ . Plants with $T t$ $= \frac{2}{4} \times 800 = 400$ . (1 mark)

Answer: F1: all $T t$ , tall. F2: $T T : T t : tt = 1 : 2 : 1$ , tall : short $= 3 : 1$ . Out of 800 F2 plants, 200 are short and 400 are $T t$ (tall).

Dominant and Recessive Traits

NCERT section 8.2.2 Rules for the Inheritance of Traits – Mendel's Contributions

In Mendel's explanation, both $T T$ and $T t$ plants are tall. Only $tt$ is short. A single copy of $T$ is enough to make the plant tall. The plant is short only when both copies are $t$ .

When the two copies of a gene are different, the trait that is expressed is called dominant. The trait that stays hidden is called recessive. By convention, the dominant form is written with a capital letter and the recessive form with the small letter of the same letter.

Note that the recessive trait is not lost or destroyed in a $T t$ plant. It is only masked. That is why it reappears in one quarter of the F2. A plant showing a recessive trait must have two recessive copies. A plant showing a dominant trait could have one or two dominant copies.

This gives you a useful tool. Cross a plant showing the dominant trait with a plant showing the recessive trait. If any offspring show the recessive trait, the dominant-looking parent must have been carrying a hidden recessive copy.

Worked example

In pea plants, violet flower colour ( $W$ ) is dominant over white ( $w$ ). A violet-flowered plant is crossed with a white-flowered plant. The progeny are 52 violet and 48 white.
(a) What is the genotype of the violet parent? Give a reason.
(b) What result would you expect if the violet parent had been $W W$ ? (3 marks)

Try it first, then show the solutionSolution

The white parent shows the recessive trait, so it is $w w$ and can pass on only $w$ . (1 mark)
About half the progeny are white ( $w w$ ), so the violet parent must have passed on $w$ to half of them. Its genotype is $W w$ , and $W w \times w w$ gives $W w : w w = 1 : 1$ . (1 mark)
If the violet parent were $W W$ , it would give only $W$ . All progeny would then be $W w$ and all would have violet flowers. (1 mark)

Answer: The violet parent is $W w$ , because half the progeny are white. A $W W$ parent would have given all violet progeny.

Dihybrid Cross and Independent Inheritance

NCERT section 8.2.2 Rules for the Inheritance of Traits – Mendel's Contributions

Mendel also crossed plants that differed in two characters at once. A tall plant with round seeds crossed with a short plant with wrinkled seeds gave F1 plants that were all tall with round seeds. So tallness and round seeds are dominant. When the F1 self-pollinated, the F2 had the two parental types and also new combinations: tall with wrinkled seeds, and short with round seeds.

The seed shape and colour cross works the same way. Round green ( $R R y y$ ) crossed with wrinkled yellow ( $r r Y Y$ ) gives F1 $R r Y y$ , which is round and yellow. Each F1 plant forms four kinds of germ cells: $R Y$ , $R y$ , $r Y$ and $r y$ . These combine in 16 ways.

NCERT's F2 counts are 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green, out of 556 seeds. This is close to $9 : 3 : 3 : 1$ .

New combinations could only appear if the gene for shape and the gene for colour go into germ cells separately. So different traits are inherited independently of each other.

Worked example

A pea plant with round green seeds ( $R R y y$ ) is crossed with one with wrinkled yellow seeds ( $r r Y Y$ ). The F1 is self-pollinated, and the F2 gives 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green seeds.
(a) Which traits are dominant? How do you know?
(b) Write the F1 genotype and the types of gametes it forms.
(c) Which F2 classes are new combinations, not seen in either parent?
(d) What conclusion does this cross support? (4 marks)

Try it first, then show the solutionSolution

Round and yellow are dominant, because the F1 from round green × wrinkled yellow is round and yellow. (1 mark)
The F1 is $R r Y y$ . It forms four types of gametes: $R Y$ , $R y$ , $r Y$ and $r y$ . (1 mark)
The parents were round green and wrinkled yellow. So the new combinations are round yellow and wrinkled green. (1 mark)
The ratio is close to $9 : 3 : 3 : 1$ . This shows that seed shape and seed colour are inherited independently of each other. (1 mark)

Answer: Round and yellow are dominant. F1 is $R r Y y$ , with gametes $R Y$ , $R y$ , $r Y$ and $r y$ . The new combinations are round yellow and wrinkled green. The cross shows independent inheritance of the two traits.

How Genes Control Traits

NCERT section 8.2.3 How do these Traits get Expressed?

Mendel called them factors; today we call them genes. But what does a gene actually do inside a plant? The DNA in a cell holds the instructions for every protein the cell builds. One stretch of DNA that codes for a single protein is the gene for that protein.

Now think about height. Growth in a plant is switched on by hormones. So a plant grows taller when it has more of a certain growth hormone, and stays shorter when it has less. How much hormone the plant ends up with depends on how well it can produce it, and a protein called an enzyme plays a key part in that production.

Suppose the gene for this enzyme is in its normal form. The enzyme does its job well, plenty of hormone is produced, and the plant grows tall. Now suppose the gene is changed so that the enzyme it codes for works poorly. Hormone production drops, and the plant stays short.

So the chain runs: gene → protein (enzyme) → amount of hormone → trait. This ties Mendel's tall and short versions of one gene to a real difference inside the cell, and shows that genes control traits through the proteins they make.

Worked example

Explain, with the example of plant height, how a change in a single gene can make a pea plant short instead of tall. (3 marks)

Try it first, then show the solutionSolution

A gene is a stretch of DNA that holds the instructions for making one protein, such as an enzyme. (1 mark)
Plant height depends on the amount of a growth hormone. An enzyme is important for making this hormone, so an efficient enzyme means more hormone and a tall plant. (1 mark)
If the gene for this enzyme is altered, the enzyme becomes less efficient. Less hormone is made and the plant stays short. So the gene controls the trait through the protein it makes. (1 mark)

Answer: An altered gene makes a less efficient enzyme, so less growth hormone is made and the plant is short. Genes control traits through the proteins they code for.

Chromosomes and Equal Genetic Contribution

NCERT section 8.2.3 How do these Traits get Expressed?

Mendel's results only make sense if both parents contribute equally to the DNA of the progeny. So each pea plant must carry two sets of all genes, one from each parent. For this to keep working, each germ cell must carry only one gene set.

How does a germ cell end up with one set? A gene set is not one long thread of DNA. It is present as separate, independent pieces called chromosomes. Each body cell has two copies of each chromosome, one from the male parent and one from the female parent. Every germ cell takes one chromosome from each pair, and that chromosome may be of maternal or paternal origin.

When two germ cells fuse, the normal number of chromosomes is restored. This keeps the DNA of the species stable from generation to generation.

This also explains independent inheritance. If a whole gene set passed on as one piece, the $R$ and $y$ of a parent would always travel together, and new combinations could never appear. All sexually reproducing organisms use this mechanism.

Worked example

A human body cell has 22 pairs of chromosomes plus one pair of sex chromosomes.
(a) How many chromosomes will a human germ cell have?
(b) Suppose germ cells carried the full body-cell number. What would happen to the chromosome number of the offspring over generations?
(c) How do chromosomes being separate pieces explain why seed shape and seed colour are inherited independently? (3 marks)

Try it first, then show the solutionSolution

A body cell has 23 pairs, which is 46 chromosomes. A germ cell takes one from each pair, so it has 23. (1 mark)
If each germ cell had 46, the zygote would have 92, and the number would double every generation. The DNA of the species would not stay stable. (1 mark)
Each germ cell takes one chromosome from each pair, and each may be maternal or paternal. Genes on different chromosomes go into germ cells separately, so $R$ and $y$ are not tied together and new combinations form. (1 mark)

Answer: (a) 23. (b) The chromosome number would double each generation, so species DNA would be unstable. (c) Separate chromosomes are sorted independently into germ cells, so traits on them are inherited independently.

Sex Determination in Humans

NCERT section 8.2.4 Sex Determination

Different species decide the sex of a newborn in different ways. In a few reptiles, the temperature at which the fertilised eggs are kept decides whether each animal becomes male or female. In snails, individuals can change sex, so sex is not genetically fixed there. In human beings, sex is largely decided by genes.

How can genes do this if both parents give similar gene sets? Humans have 22 pairs of chromosomes with matching maternal and paternal copies. The remaining pair, the sex chromosomes, is not always a perfect pair. Women have two X chromosomes (XX). Men have one normal-sized X and one short Y chromosome (XY).

A mother can only pass on an X. A father passes on either an X or a Y. If the sperm carries X, the child is XX, a girl. If it carries Y, the child is XY, a boy. On average, half the children are boys and half are girls. The sex of the child is therefore decided by the chromosome inherited from the father.

Worked example

(a) Write the sex chromosome carried by every human egg and by each of the two kinds of human sperm.
(b) A zygote has the sex chromosomes XY. Which parent supplied each of these chromosomes? Give a reason.
(c) Explain why the mother's egg can never decide the sex of the child. (3 marks)

Try it first, then show the solutionSolution

(a) The mother is XX, so every egg carries one X. The father is XY, so half his sperms carry X and half carry Y. (1 mark)
(b) The X came from the mother's egg, because eggs carry only X. The Y must have come from the father's sperm, the only source of a Y chromosome. (1 mark)
(c) Every egg carries the same sex chromosome, X, whether the child will be a son or a daughter. Only the sperm differs, so the father's X or Y decides whether the zygote is XX (girl) or XY (boy). (1 mark)

Answer: (a) Egg: X only; sperm: X or Y. (b) X from the mother, Y from the father. (c) All eggs carry X, so only the sperm's X or Y can decide the sex.

Science in real life: Heredity

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Variation During Reproduction

Imagine your school biology lab keeps a bacterial culture that grew from a single bacterium. One afternoon the incubator overheats. When the teacher checks, most bacteria have died, but a few have survived.

All these bacteria came from one parent by division. Why were any of them different enough to survive? What does this tell you about the value of variation?

Try it first, then show the solutionSolution

The bacteria were produced by repeated division, which is asexual reproduction.
During each division the DNA is copied, and small inaccuracies in copying create minor differences between the bacteria.
Some of these variants happened to be able to withstand heat.
When the temperature rose, only these heat-tolerant variants survived.
If all the bacteria had been exactly identical, a single change like this could have killed the whole population.

Answer: Small DNA copying errors created a few heat-tolerant variants, and these survived the overheating. Variation lets at least some individuals survive when the environment changes.

Memory hook: An overheated incubator shows why even small variations can save a population.

Inherited Traits and Activity 8.1

Suppose at a family gathering you check everyone's earlobes. Your grandparents on one side both have free earlobes. Their son (your uncle) has attached earlobes, and your father has free earlobes.

How can two parents with free earlobes have a son with attached earlobes? What can you suggest about which earlobe type behaves as recessive?

Try it first, then show the solutionSolution

Each child receives one copy of the earlobe gene from each parent.
The uncle shows attached earlobes even though neither of his parents does.
So both grandparents must have carried a hidden copy for attached earlobes.
A trait that can stay hidden in a parent and show up in a child behaves as a recessive trait.
The uncle must have received the attached copy from both parents, while your father received at least one free copy.

Answer: Both grandparents carry a hidden copy for attached earlobes. Attached earlobe behaves as recessive and free earlobe as dominant.

Memory hook: A trait that skips the parents and shows in a child is the recessive one.

Mendel's Monohybrid Cross (Tall × Short)

Imagine your school eco club repeats Mendel's work in the school garden. They cross pure tall and pure short pea plants, grow the F1, and let the F1 plants self-pollinate. From the seeds collected, 240 F2 plants grow.

(a) How many F2 plants are expected to be short? (b) How many are expected to be $T T$ , and how many $T t$ ? (c) A club member says every tall F2 plant is just like the F1 plants. Is she right?

Try it first, then show the solutionSolution

The F1 plants are $T t$ . Selfing $T t \times T t$ gives $T T : T t : tt = 1 : 2 : 1$ .
Short ( $tt$ ) $= \frac{1}{4} \times 240 = 60$ .
$T T = \frac{1}{4} \times 240 = 60$ and $T t = \frac{2}{4} \times 240 = 120$ .
The tall plants number $60 + 120 = 180$ , of which 120 are $T t$ like the F1 and 60 are $T T$ .
So one-third of the tall F2 plants are pure $T T$ , unlike the F1.

Answer: (a) 60 short. (b) 60 $T T$ and 120 $T t$ . (c) No: 60 of the 180 tall plants are $T T$ , so not every tall plant is like the F1.

Memory hook: In the garden, one in four plants is short, but not every tall plant is the same inside.

Dominant and Recessive Traits

Suppose the school gardener has a bed of violet-flowered pea plants. She wants to save seeds only from plants that will always give violet flowers. Some of the plants may secretly carry the white-flower copy. Violet ( $W$ ) is dominant over white ( $w$ ).

Suggest a cross that will tell her which violet plants are $W W$ , and explain how she should read the results.

Try it first, then show the solutionSolution

Cross each violet plant with a white-flowered plant ( $w w$ ), which can give only $w$ .
If the violet plant is $W W$ , all the progeny are $W w$ and all have violet flowers.
If the violet plant is $W w$ , the progeny are $W w : w w = 1 : 1$ , so about half have white flowers.
She should keep seeds only from the violet plants whose cross gave no white-flowered progeny.

Answer: Cross each violet plant with a white one. Plants giving all-violet progeny are $W W$ and are the ones to save. Plants giving about half white progeny are $W w$ .

Memory hook: A white partner exposes any hidden white copy in a violet plant.

Sex Determination in Humans

Imagine a family where a couple has two daughters. An elderly relative says the mother is responsible for there being no son. You have just studied sex determination.

Use the inheritance of sex chromosomes to explain whether the relative is right, and give the chance of a boy at the next birth.

Try it first, then show the solutionSolution

The mother is XX, so every egg she makes carries an X chromosome.
The father is XY, so half his sperm carry X and half carry Y.
If a sperm with X fertilises the egg, the child is XX, a girl. If a sperm with Y fertilises it, the child is XY, a boy.
So the sex of the child depends on which chromosome comes from the father, not the mother.
Each birth is a fresh event with an equal chance of X or Y from the father, so the chance of a boy is one half.

Answer: The relative is wrong. The mother always gives X, and the father's X or Y decides the sex. The chance of a boy at the next birth is one half (50%).

Memory hook: Mum always gives X; Dad's X or Y decides girl or boy.

Topics in this chapter

Heredity and VariationComing soon
Mendel's Experiments and Monohybrid Cross
Dihybrid Cross and Laws of Inheritance
How Traits Get ExpressedComing soon
Sex Determination in Humans
Evolution

Heredity: formulas and key facts

Monohybrid cross: F1: $T T \times tt \to T t (all tall)$; Use this to show that F1 shows only the dominant trait and that there is no blending.
Monohybrid F2 genotype ratio: $T T : T t : tt = 1 : 2 : 1$; Use it when a question asks for genotypes or genetic make-up of the F2.
Monohybrid F2 phenotype ratio: $Tall : Short = 3 : 1$; Use it when a question asks what the F2 plants look like. One quarter of the F2 is short.
Cross with a recessive parent: $T t \times tt \to T t : tt = 1 : 1$; If about half the progeny show the recessive trait, the dominant-looking parent is a hybrid.
Dihybrid cross: F1: $R R y y \times r r Y Y \to R r Y y (round, yellow)$; Round and yellow are dominant. Learn this as the cross shown in the NCERT figure.
Gametes of a dihybrid F1: $R r Y y \to R Y, R y, r Y, r y$; There are four types of gametes, each with one gene for shape and one for colour. They make 16 boxes in the checkerboard.
Dihybrid F2 phenotype ratio: $9 : 3 : 3 : 1$; The order is: both dominant, one dominant and one recessive, the reverse, then both recessive. NCERT's counts of 315 : 108 : 101 : 32 out of 556 are close to this.
Dominant trait: The trait expressed when the two copies of a gene differ; one copy (e.g. $T$ ) is enough; Use this in definitions and in reasoning from F1 results.
Recessive trait: The trait that shows only when both copies are recessive (e.g. $tt$ ); A recessive trait is hidden, not lost, in a hybrid. It reappears in the F2.
Gene: A section of DNA that provides the information for one protein; Use this in questions on how traits get expressed.
Gene → trait chain: Gene $\to$ enzyme $\to$ amount of hormone $\to$ height; This is the NCERT example for how a gene controls a trait.
Germ cells: Each germ cell carries one chromosome from each pair, of maternal or paternal origin; This explains equal contribution by both parents, stable chromosome number and independent inheritance.
Human chromosomes: $22 pairs + 1 pair of sex chromosomes = 23 pairs (46)$; A body cell has 23 pairs. A germ cell has 23 single chromosomes.
Sex chromosomes: $Female = X X, Male = X Y$; Y is the short chromosome. The mother's eggs always carry X.
Sex determination cross: $X X \times X Y \to X X : X Y = 1 : 1$; Half girls and half boys. The father's sperm (X or Y) decides the sex.
Source of variation: Asexual: small DNA copying inaccuracies; sexual: DNA from two parents combined; Sexual reproduction produces much greater diversity. Use this in variation and survival questions.

Heredity: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Monohybrid cross: tall × short pea plants over two generations

3- or 5-mark questions asking you to show Mendel's cross with a flow chart or checkerboard, or to work out F2 ratios.

Draw

Top line: Parents, tall ( $T T$ ) × short ( $tt$ ).
Below: gametes $T$ (from tall) and $t$ (from short).
Arrow down to F1: $T t$ , all tall.
F1 self-pollinated: $T t$ × $T t$ , gametes $T$ , $t$ from each.
A 2 × 2 checkerboard: male gametes across the top, female gametes down the side, four boxes filled $T T$ , $T t$ , $T t$ , $tt$ .
Below the board: phenotype ratio and genotype ratio.

Label

Parents (P)
Gametes
F1 generation – all tall ( $T t$ )
Self-pollination
F2 generation
Phenotype ratio 3 tall : 1 short
Genotype ratio 1 $T T$ : 2 $T t$ : 1 $tt$

Watch out: Gametes carry only one letter ( $T$ or $t$ ), never $T t$ . Do not call $T t$ plants medium-height: they are tall. Write both ratios, phenotype and genotype, as separate lines.

Dihybrid cross checkerboard: seed shape and seed colour (or height and seed shape)

5-mark or case-based questions on independent inheritance and new combinations in the F2.

Draw

Parents: round green ( $R R y y$ ) × wrinkled yellow ( $r r Y Y$ ).
Gametes: $R y$ and $r Y$ .
F1: $R r Y y$ , all round and yellow.
F1 gametes: four kinds, $R Y$ , $R y$ , $r Y$ , $r y$ .
A 4 × 4 checkerboard with these four gametes across the top and down the side; fill all 16 boxes by combining row and column letters.
Group the 16 boxes into four phenotype classes.

Label

Parents
Gametes
F1 – round, yellow ( $R r Y y$ )
F1 gametes $R Y$ , $R y$ , $r Y$ , $r y$
F2 (16 combinations)
9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green

Watch out: Each gamete carries one letter for each trait (two letters in all). The common slip is writing F1 gametes as only $R y$ and $r Y$ ; the F1 makes all four kinds, which is why new combinations appear.

Sex determination in human beings (flow chart)

3- or 5-mark questions: 'Explain with a flow chart how the sex of a child is determined.'

Draw

Top: Father $X Y$ on one side, Mother $X X$ on the other.
Below the father: two kinds of sperm, $X$ and $Y$ . Below the mother: eggs, all $X$ .
Arrows from the gametes meeting at fertilisation.
Two zygote boxes: $X$ egg + $X$ sperm gives $X X$ ; $X$ egg + $Y$ sperm gives $X Y$ .
Under them: $X X$ = girl, $X Y$ = boy; write 'half girls, half boys'.

Label

Father (male) $X Y$
Mother (female) $X X$
Sperms: $X$ and $Y$
Eggs: $X$ only
Fertilisation
Zygote $X X$ – girl
Zygote $X Y$ – boy
Ratio 1 : 1

Watch out: Show the mother giving only $X$ eggs and the father giving two kinds of sperm. Never write an egg with $Y$ or an $X X$ gamete: a gamete carries one sex chromosome.

How the board asks Heredity

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
MCQ: find the genotype from the progeny	A cross is described with the traits of the progeny (for example, all violet but about half short), and you choose the parent's genotype from four options.	Take one trait at a time. If all the progeny show the dominant trait, that parent is pure dominant for it. If about half show the recessive trait, that parent is a hybrid. Combine the two answers and only then look at the options.
Assertion-reason on variation or gene expression	The assertion is about something like heat-tolerant bacteria surviving or an enzyme controlling height. The reason gives a general rule, such as variation helping survival or genes making proteins.	Judge A and R separately as true or false first. Then ask whether R actually explains A. Watch for a reason that is true but about a different link in the chain.
Very short answer (2 marks): observations and definitions	You may be asked to write two observations Mendel made about F1 progeny, define dominant and recessive traits, or say why variation helps survival with an example.	Give two crisp, separate points. For F1: only one parental trait appears, and there are no halfway or medium forms. Use the NCERT examples (tall/short peas, heat-tolerant bacteria).
Short answer (3 marks): work out a cross	You are asked to show a monohybrid or dihybrid cross from parents to F2, and give ratios or the number of plants expected out of a given total.	Write the parents, then the gametes, then the F1, then the F2 combinations, then the ratio. Marks are given at each stage. For numbers, multiply the fraction (for example ¼) by the total.
Case-based (4 marks): Mendel's experiments or chromosomes	A passage describes a Mendel-style experiment or human sex chromosomes, followed by sub-questions on ratios, dominance, new combinations or which parent decides sex.	Underline the data in the passage. Answer each sub-part with a reason taken from that data. In a choice between sub-parts, pick the one where you can show working.
Long answer (5 marks): full explanation	You may be asked to explain how Mendel's experiments show dominance and independent inheritance, or to explain sex determination in humans with a cross and say why the father decides the sex.	Write in steps: experiment, result, interpretation, cross or checkerboard in words, conclusion. Always end with the rule that the experiment proves.
Data and percentage questions	You are given counts (an earlobe survey, F2 seed numbers) or a cross, and asked for percentages or what the data suggests.	Work out each percentage as part ÷ total × 100. Then state the inference, such as which trait can hide in parents. Give a reason, because the marking scheme gives a mark for it.

Tricks to remember Heredity

Short memory aids. Say each one aloud once, then use it on a question the same day.

Capital wins
A capital letter (dominant) shows whenever it is present. A small letter (recessive) shows only when it is doubled: $T T$ and $T t$ are tall, only $tt$ is short.
Use it when: Use it to read any genotype into a phenotype in a second.
3-1 looks, 1-2-1 genes
Say "three-one looks, one-two-one genes". Phenotype F2 = 3 : 1, genotype F2 = 1 : 2 : 1.
Use it when: Use it to avoid mixing up the two monohybrid ratios.
9-3-3-1 adds to 16
The dihybrid ratio always adds up to 16, matching the 4 × 4 checkerboard. Both dominant = 9, both recessive = 1.
Use it when: Use it to check your dihybrid answer and to turn ratios into fractions (for example, 3 out of 16).
Half or none
A dominant-looking parent crossed with a recessive one: no recessive progeny means pure; about half recessive progeny means hybrid.
Use it when: Use it for MCQs that ask for a parent's genotype.
Dad decides
Mum is XX and gives only X. Dad is XY and gives X or Y. Dad's chromosome decides girl or boy.
Use it when: Use it for sex determination MCQs and long answers.
Two make one, one plus one make two
Body cells have chromosomes in pairs. Germ cells take one from each pair. Fusion of two germ cells restores the pairs.
Use it when: Use it for questions on equal contribution by both parents and stability of the species' DNA.

Heredity: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Ways in which sex is determined · 1 mark · MCQBoard level

Which one of the following statements about sex determination is correct?
(a) In a few reptiles, the temperature at which fertilised eggs are kept decides whether the young will be male or female.
(b) In snails, the sex of an individual is fixed by its chromosomes and can never change.
(c) In human beings, the chromosome inherited from the mother decides the sex of the child.
(d) Human females have a mismatched pair of sex chromosomes, $X$ and $Y$ .

Try it first, then show the solutionSolution

(b) is wrong: snails can change sex, which shows sex is not genetically fixed in them.
(c) is wrong: every child gets an $X$ from the mother, so the father's chromosome decides sex.
(d) is wrong: females are $X X$ ; males have the mismatched $X Y$ pair.
(a) matches the environmental cue described for some reptiles.

Answer: (a) In a few reptiles, the temperature at which fertilised eggs are kept decides the sex of the young.

Example 2: Why the F1 is all tall · 1 mark · Assertion-reasonBoard level

Assertion (A): When a pure tall pea plant is crossed with a pure short pea plant, all the F1 plants are tall.
Reason (R): A single copy of the factor for tallness is enough to make a pea plant tall.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

A is true: Mendel found only tall plants in the F1, with no medium-height plants.
R is true: both $T T$ and $T t$ are tall, so one $T$ is enough.
Every F1 plant is $T t$ ; it shows tallness because one $T$ is enough, so R explains A.

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 3: Counting F2 classes in a two-trait cross · 3 marks · Short answerHarder board level

A pure tall pea plant with round seeds ( $T T R R$ ) is crossed with a pure short plant with wrinkled seeds ( $tt r r$ ). The F1 plants are self-pollinated and 160 F2 plants are raised.
(a) Write the genotype of the F1 and the gametes it forms.
(b) How many F2 plants are expected to be tall with wrinkled seeds, and how many short with round seeds?
(c) What do these two classes show? (3 marks)

Try it first, then show the solutionSolution

(a) F1 genotype: $T tR r$ (tall, round). It forms four kinds of gametes: $T R$ , $T r$ , $tR$ , $t r$ .
(b) In the 16-box checkerboard the F2 classes come out as 9 tall round : 3 tall wrinkled : 3 short round : 1 short wrinkled.
Tall wrinkled = $\frac{3}{16} \times 160 = 30$ plants. Short round = $\frac{3}{16} \times 160 = 30$ plants.
(c) These are new combinations not present in either parent. They show that the height factor and the seed-shape factor are inherited independently.

Answer: (a) $T tR r$ ; gametes $T R$ , $T r$ , $tR$ , $t r$ . (b) 30 tall wrinkled and 30 short round. (c) Height and seed shape are inherited independently.

Example 4: Parents' genotypes from a 3 : 3 : 1 : 1 ratio · 3 marks · Short answerHarder board level

A tall pea plant with round seeds is crossed with a tall pea plant with wrinkled seeds. Tall ( $T$ ) and round ( $R$ ) are dominant. The 160 progeny are: 60 tall round, 60 tall wrinkled, 20 short round and 20 short wrinkled.
(a) Work out the genotypes of both parents, taking one trait at a time.
(b) Check your answer by finding the expected number of short wrinkled progeny.
(c) Which idea from Mendel's work allows you to treat the two traits separately? (3 marks)

Try it first, then show the solutionSolution

(a) Height: tall = 60 + 60 = 120 and short = 20 + 20 = 40, so tall : short = 3 : 1. This ratio comes from $T t \times T t$ , so both parents are $T t$ . (1 mark)
Seed shape: round = 60 + 20 = 80 and wrinkled = 60 + 20 = 80, so round : wrinkled = 1 : 1. This comes from $R r \times r r$ , so the round parent is $R r$ and the wrinkled parent is $r r$ . Parents: $T tR r$ (tall round) and $T t r r$ (tall wrinkled). (1 mark)
(b) Short wrinkled needs $tt$ (one quarter of the progeny) and $r r$ (one half): $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$ , and $\frac{1}{8} \times 160 = 20$ . This matches the data. (½ mark)
(c) Height and seed shape are inherited independently, so the ratio for each trait can be worked out on its own and then combined. (½ mark)

Answer: (a) $T tR r$ (tall round) and $T t r r$ (tall wrinkled). (b) 20 short wrinkled progeny are expected, which matches the data. (c) Independent inheritance of the two traits.

Example 5: Percentages and pure plants in a dihybrid F2 · 4 marks · Short answerHarder board level

The F1 of round green ( $R R y y$ ) × wrinkled yellow ( $r r Y Y$ ) is self-pollinated and 640 F2 seeds are obtained.
(a) What percentage of the F2 seeds are expected to be round and green?
(b) How many round green seeds are expected, and how many of them will be pure-breeding ( $R R y y$ )?
(c) What fraction of the round yellow seeds are expected to be $R R Y Y$ ?
(d) A student says every round green F2 seed will give only round green plants on self-pollination. Is she right? (4 marks)

Try it first, then show the solutionSolution

(a) The F1 is $R r Y y$ , and its F2 is 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. Round green $= \frac{3}{16} \times 100 = 18.75%$ . (1 mark)
(b) Round green seeds $= \frac{3}{16} \times 640 = 120$ . In the checkerboard the three round green boxes are $R R y y$ , $R r y y$ and $R r y y$ , so one-third are pure: $\frac{1}{3} \times 120 = 40$ seeds. (1 mark)
(c) Of the 9 round yellow boxes, only one is $R R Y Y$ , so the fraction is $\frac{1}{9}$ , which is 40 of the 360 round yellow seeds. (1 mark)
(d) No. Only the $R R y y$ seeds breed true. The $R r y y$ seeds (two-thirds, that is 80 seeds) carry a hidden $r$ and will give some wrinkled green plants on self-pollination. (1 mark)

Answer: (a) 18.75%. (b) 120 round green seeds, of which 40 are pure $R R y y$ . (c) $\frac{1}{9}$ . (d) No: two-thirds of the round green seeds are $R r y y$ and can give wrinkled green plants.

Example 6: Sex determination with a flow chart · 5 marks · Long answerHarder board level

(a) Draw a labelled flow chart to show how the sex of a child is determined in human beings.
(b) Why are the chances of having a boy or a girl equal?
(c) A relative blames a mother for having only daughters. Use your flow chart to explain why this is wrong.
(d) Name one animal in which sex is not genetically determined and say what decides it. (5 marks)

Try it first, then show the solutionSolution

(a) Draw and label: Father $X Y$ and Mother $X X$ at the top; under the father, two kinds of sperm, $X$ and $Y$ ; under the mother, eggs, all $X$ ; arrows meeting at 'Fertilisation'; two zygotes, $X X$ labelled 'Girl' and $X Y$ labelled 'Boy'; write 'ratio 1 : 1'.
Slip to avoid: never show a $Y$ egg or an $X X$ sperm; each gamete carries one sex chromosome.
(b) Humans have 22 pairs of matched chromosomes and one pair of sex chromosomes. The father makes $X$ sperms and $Y$ sperms in equal numbers, so an egg is equally likely to be fertilised by either. Half the children are expected to be girls and half boys.
(c) Every egg from the mother carries an $X$ , so she gives the same chromosome to sons and daughters. A child is a girl if the father's sperm carries $X$ , and a boy if it carries $Y$ . So the father's contribution decides the sex, and the mother cannot be blamed.
(d) In a few reptiles, the temperature at which the fertilised eggs are kept decides whether the young become male or female. (Snails, which can change sex, are also accepted.)

Answer: Flow chart: $X Y$ father gives $X$ or $Y$ sperm, $X X$ mother gives only $X$ eggs; $X + X = X X$ (girl), $X + Y = X Y$ (boy) in a 1 : 1 ratio. The sex is decided by the sperm, so blaming the mother is wrong. In some reptiles, egg temperature decides sex.

Example 7: Activity 8.2: testing the 1 : 2 : 1 ratio · 3 marks · Short answerBoard level

In Mendel's tall × short cross, the F2 is said to contain $T T$ , $T t$ and $tt$ plants in the ratio 1 : 2 : 1.
(a) Why can you not confirm this just by looking at the F2 plants?
(b) Describe an experiment to confirm it.
(c) State the result that would confirm the ratio. (3 marks)

Try it first, then show the solutionSolution

(a) $T T$ and $T t$ plants are both tall, because $T$ is dominant. They look the same, so only a 3 tall : 1 short ratio can be seen directly.
(b) Allow every F2 plant to self-pollinate. Collect and grow the seeds of each plant separately, and record whether its progeny are all tall, all short, or a mixture.
(c) Plants giving only tall progeny are $T T$ ; plants giving both tall and short (about 3 : 1) are $T t$ ; plants giving only short progeny are $tt$ .
If the numbers of these three kinds of F2 plants are in the ratio 1 : 2 : 1, the genotype ratio is confirmed.

Answer: Self-pollinate each F2 plant and grow its seeds separately. About one quarter give only tall progeny ( $T T$ ), half give tall and short ( $T t$ ) and one quarter give only short ( $tt$ ), confirming 1 : 2 : 1.

Example 8: Dominance, germ cells and independent inheritance · 5 marks · Long answerHarder board level

(a) How did Mendel's tall × short cross show that one trait can be dominant and the other recessive?
(b) Each pea plant has two sets of genes. Why must each germ cell carry only one set?
(c) How does the fact that genes are carried on separate chromosomes explain independent inheritance of two traits? (5 marks)

Try it first, then show the solutionSolution

(a) All F1 plants were tall, with no medium plants: only one parental trait appeared.
When F1 plants were self-pollinated, one quarter of the F2 was short. So the short trait had been present in the F1 but not expressed. The expressed trait (tall) is dominant; the hidden one (short) is recessive.
(b) Each parent contributes one copy of every gene to the progeny. If germ cells carried both sets, the zygote would have four sets, and the amount of DNA would double every generation.
With one set in each germ cell, fusion of two germ cells restores the normal two sets, keeping the DNA of the species stable.
(c) The gene set is not one long thread but separate pieces called chromosomes. Each germ cell takes one chromosome from each pair, maternal or paternal, independently of the other pairs. So genes on different chromosomes, such as those for height and seed shape, can sort into new combinations instead of staying linked.

Answer: (a) F1 all tall, F2 one quarter short: tall is dominant, short recessive. (b) One set per germ cell keeps the two-set number constant after fertilisation. (c) Each chromosome pair sorts independently into germ cells, so traits on different chromosomes are inherited independently.

Example 9: Find the errors in a student's cross · 3 marks · Short answerHOTS / competency

A student wrote this answer:
"The F1 plant from tall ( $T T$ ) × short ( $tt$ ) is $T t$ . Since it has one tall and one short factor, it is a medium-height plant. When the F1 is self-pollinated, the F2 is 1 tall : 2 medium : 1 short."
Identify the errors and write the correct conclusion. (3 marks)

Try it first, then show the solutionSolution

Error 1: $T t$ is not medium-height. Mendel found no halfway characteristics; $T$ is dominant, so a single $T$ makes the plant tall.
Error 2: the F2 ratio confuses genotype with phenotype. The genotypes are 1 $T T$ : 2 $T t$ : 1 $tt$ .
Since both $T T$ and $T t$ are tall, the correct phenotype ratio is 3 tall : 1 short.

Answer: The F1 ( $T t$ ) is tall, not medium, because tallness is dominant. The F2 genotype ratio is 1 $T T$ : 2 $T t$ : 1 $tt$ and the phenotype ratio is 3 tall : 1 short.

Example 10: Two tall parents with some short progeny · 3 marks · Short answerHOTS / competency

Two tall pea plants, P and Q, are crossed. Of 120 progeny, 30 are short.
(a) Work out the genotypes of P and Q, with a reason.
(b) How many of the tall progeny are expected to be pure tall?
(c) How could you find out which tall progeny are pure? (3 marks)

Try it first, then show the solutionSolution

(a) Short plants are $tt$ , so each one got a $t$ from both parents. Both tall parents must therefore carry $t$ : P = $T t$ , Q = $T t$ . Check: $\frac{30}{120} = \frac{1}{4}$ short, as expected from $T t \times T t$ .
(b) Tall progeny = 90, in the ratio 1 $T T$ : 2 $T t$ . Pure tall = $\frac{1}{3} \times 90 = 30$ .
(c) Self-pollinate each tall plant and grow its seeds separately. A plant whose progeny are all tall is pure ( $T T$ ); one that gives some short progeny is $T t$ .

Answer: (a) Both P and Q are $T t$ . (b) About 30 of the 90 tall progeny are pure ( $T T$ ). (c) Self-pollinate each tall plant: only pure tall plants give all-tall progeny.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Flower colour and height in the school garden

Suppose the eco club of a school in Pune repeats a Mendelian experiment in the school garden. They cross a pure tall pea plant bearing violet flowers with a pure short plant bearing white flowers. All the F1 plants are tall with violet flowers. The F1 plants are allowed to self-pollinate, and the students grow 320 F2 plants. Their record reads:

Tall, violet: 180
Tall, white: 60
Short, violet: 60
Short, white: 20

The teacher asks them to use symbols $T$ / $t$ for height and $W$ / $w$ for flower colour and explain the result.

(i) · 1 mark
Which traits are dominant? Give the evidence from the F1.
Answer
Answer: Tallness and violet flower colour are dominant.
The F1 plants got one factor for each trait from each parent, yet all of them were tall with violet flowers. The traits that show in the F1 are the dominant ones.
(ii) · 1 mark
Name the F2 classes that are new combinations, and write the genotype of the F1.
Answer
Answer: Tall white and short violet are new combinations. F1 genotype: $T t W w$ .
The parents were tall violet ( $T T W W$ ) and short white ( $ttw w$ ). Tall white (60) and short violet (60) were not seen in either parent. The F1 received $T W$ from one parent and $tw$ from the other, so it is $T t W w$ .
(iii) · 2 marks
Show that the data fit a 9 : 3 : 3 : 1 ratio, and state the conclusion this supports.
Answer
Answer: 180 : 60 : 60 : 20 = 9 : 3 : 3 : 1. Height and flower colour are inherited independently.
Divide each number by 20: 180 ÷ 20 = 9, 60 ÷ 20 = 3, 60 ÷ 20 = 3, 20 ÷ 20 = 1. This is the ratio expected when the F1 forms four kinds of gametes ( $T W$ , $T w$ , $t W$ , $tw$ ) in equal numbers. Because new combinations appear, the factor for height and the factor for flower colour must separate and recombine independently.

Counting chromosomes in gametes and zygotes

Suppose your biology teacher puts up a chart in the lab. It says that a human body cell has 22 pairs of chromosomes, each pair with a maternal and a paternal copy, plus one pair of sex chromosomes. Women have $X X$ and men have $X Y$ , where $Y$ is the shorter chromosome. Each germ cell takes one chromosome from every pair, and fertilisation joins one sperm with one egg. The teacher asks you to use only the chart to answer the questions below.

(i) · 1 mark
How many chromosomes are present in a human sperm and in a human egg? How many of these are sex chromosomes?
Answer
Answer: 23 in each; one of them is a sex chromosome.
A body cell has 22 + 1 = 23 pairs, that is 46 chromosomes. A germ cell takes one chromosome from each pair, so it has 23: 22 others and 1 sex chromosome.
(ii) · 1 mark
Write the chromosome make-up of the two kinds of sperm and of an egg, in the form '22 + sex chromosome'.
Answer
Answer: Sperm: 22 + $X$ or 22 + $Y$ . Egg: 22 + $X$ only.
The father is $X Y$ , so a sperm gets either his $X$ or his $Y$ along with one of each of the other 22 pairs. The mother is $X X$ , so every egg gets an $X$ .
(iii) · 2 marks
A zygote has 44 + $X X$ . (i) Will the child be a boy or a girl, and which parent's germ cell decided this? (ii) Explain how fertilisation keeps the chromosome number the same from parents to child.
Answer
Answer: (i) A girl; the father's sperm decided it by carrying $X$ . (ii) Each germ cell carries 23 chromosomes, and fusion of a sperm and an egg gives 46, the same number as in the parents' body cells.
(i) The egg always carries $X$ , so the second $X$ must have come from the sperm. Since a sperm could have carried $Y$ instead, the father's germ cell decided that the child is a girl. (1 mark) (ii) Because each germ cell takes only one chromosome from each pair, it carries half the body-cell number. Fertilisation adds 23 + 23 = 46 and restores the pairs, so the DNA of the species stays stable from one generation to the next. (1 mark)

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

In Mendel's cross between a tall and a short pea plant, no medium-height plants appeared in the F1 generation. Write two conclusions that can be drawn from this observation. (2 marks) (2 marks)

Step	Marks
Only one of the two parental traits (tallness) is seen in the F1. The traits do not blend or mix to give a halfway character.	1
Tallness is the dominant trait, because one copy of the tall factor ( $T$ ) is enough to make the plant tall. Shortness is recessive, as it stays hidden in the F1 ( $T t$ ) plants.	1

How do Mendel's experiments show that traits are inherited independently? Explain with a cross between a tall plant with round seeds and a short plant with wrinkled seeds. (3 marks) (3 marks)

Step	Marks
Parents: tall, round ( $T T R R$ ) $\times$ short, wrinkled ( $tt r r$ ). All F1 plants are tall with round seeds ( $T tR r$ ), so tallness and roundness are dominant.	1
When the F1 is self-pollinated, the F2 has the two parental types (tall round, short wrinkled) and also two new combinations: tall with wrinkled seeds and short with round seeds.	1
New combinations can only form if the height factor and the seed-shape factor separate and recombine independently in the gametes. So the two traits are inherited independently of each other.	1

(a) Why does sexual reproduction produce more variation than asexual reproduction?
(b) How can variations help a species survive? Give an example.
(c) Explain, using plant height as an example, how a gene controls a trait. (5 marks) (5 marks)

Step	Marks
(a) In asexual reproduction, variation comes only from small inaccuracies in DNA copying, so offspring are very similar.	1
(a) In sexual reproduction, the DNA of two different parents combines in the offspring, so much greater diversity is produced.	1
(b) Different variants have different advantages in an environment. Example: if a heat wave occurs, bacteria that can withstand heat survive better, so the species is not wiped out.	1
(c) A gene is a section of DNA that gives information for making one protein, such as an enzyme. Plant height depends on the amount of a growth hormone, which an enzyme helps to make.	1
(c) If the gene gives an efficient enzyme, much hormone is made and the plant is tall. If the gene is altered so the enzyme is less efficient, less hormone is made and the plant is short. So genes control traits.	1

Where students lose marks in Heredity

The slips examiners see most often, and what to write instead.

Mistake: Mixing up the genotype and phenotype ratios in F2
Both ratios come from the same cross, and students write whichever they remember first.
Wrong working
The F2 of $T t \times T t$ shows the phenotype ratio 1 : 2 : 1.
Right working
Genotype ratio $T T : T t : tt = 1 : 2 : 1$ . Phenotype ratio tall : short $= 3 : 1$ , because $T T$ and $T t$ both look tall.
Do this: Always label the ratio as "genotype" or "phenotype", and check which one the question asks for.
Usually costs: ½ to 1 mark
Mistake: Writing a pair of letters as a gamete
Students forget that a germ cell carries only one copy of each gene.
Wrong working
Gametes of the F1: $T t$ and $T t$ .
Right working
Each F1 plant ( $T t$ ) forms two kinds of gametes, $T$ and $t$ . A dihybrid $R r Y y$ forms $R Y$ , $R y$ , $r Y$ and $r y$ .
Do this: Write the gametes on a separate line, with one letter per gene, before combining them.
Usually costs: ½ to 1 mark in a cross question
Mistake: Saying the mother decides the sex of the child
This is a common belief, and students do not reason from XX and XY.
Wrong working
The sex of the child depends on the mother's egg.
Right working
The mother's eggs all carry X. The father's sperm carry X or Y. An X sperm gives a girl (XX) and a Y sperm gives a boy (XY), so the father's chromosome decides the sex.
Do this: Always write the parents as XX and XY and show the gametes. The answer then follows.
Usually costs: 1 to 2 marks
Mistake: Naming the wrong classes as new combinations in a dihybrid cross
Students assume round yellow is always a parental type, as it is in many textbook diagrams.
Wrong working
For $R R y y \times r r Y Y$ , the new F2 combinations are round green and wrinkled yellow.
Right working
The parents here are round green and wrinkled yellow. So the new combinations are round yellow and wrinkled green.
Do this: Before naming new combinations, write down exactly what the two original parents looked like.
Usually costs: 1 mark
Mistake: Describing F1 as a blend of the two parents
Students expect a mixture of tall and short to give medium plants.
Wrong working
The F1 of tall × short are medium-height plants.
Right working
All F1 plants are tall. There are no halfway characteristics, so only one parental trait (the dominant one) is expressed.
Do this: Remember Mendel's first observation: no medium plants in F1.
Usually costs: 1 mark
Mistake: Deciding dominance from too little data
Students conclude a rule from one family or from a trait being common.
Wrong working
Many light-eyed children have light-eyed parents, so light eye colour is dominant.
Right working
Such data alone cannot show dominance. You need crosses where the two forms differ and you count how many progeny show each form.
Do this: Ask yourself: does a hidden trait reappear in the offspring? Are there enough counts? If not, say the data is not enough and explain why.
Usually costs: 1 to 2 marks

NCERT exercise guide: Heredity

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

In-text questions (first box, on sections 8.1 and 8.2.1)

What it tests: How variation builds up during reproduction, why a trait that is more common in an asexual population is likely to be older, and how variation helps survival.

How to approach it: Link each answer to time and survival. In asexual reproduction, a new variation is passed on to all descendants, so the longer it has existed, the more individuals carry it. For survival questions, name an environmental change and say which variant survives it, using the heat-tolerant bacteria idea.

Mirrors the in-text question on which trait arose earlier in an asexually reproducing population

In a population of an asexually reproducing organism, trait P is found in 5% of individuals and trait Q in 45%. Which trait probably arose earlier? Give a reason.

Show the solutionSolution

In asexual reproduction, a variation that arises in one individual is copied into all of its descendants.
The longer a variation has existed, the more generations have reproduced with it, so the more individuals carry it.
Trait Q is found in far more individuals (45%) than trait P (5%).

Answer: Trait Q probably arose earlier, because it has had more generations to be passed on and spread through the population.

In-text questions (second box, on sections 8.2.2 to 8.2.4)

What it tests: How Mendel's crosses show dominant and recessive traits and independent inheritance, whether a single family's data is enough to decide dominance, and how sex is determined in humans.

How to approach it: For dominance questions, quote the F1 (only one parental trait seen) and the F2 (the hidden trait reappears in one quarter). For independence, mention the new combinations in the F2 of a two-trait cross. For 'is this enough to decide' questions, test both possibilities: if both fit the data, the answer is no.

Mirrors the in-text question on deciding dominance from one family

A man with free earlobes marries a woman with attached earlobes. Their son has attached earlobes. Is this enough to tell which earlobe type is dominant? Why or why not?

Show the solutionSolution

Case 1: suppose attached is dominant. The son can get the attached copy from his mother and show it. This fits the data.
Case 2: suppose free is dominant. The father could carry one hidden attached copy. If the son gets an attached copy from each parent, he has attached earlobes. This also fits.
Both cases explain the result, and only one child is observed.

Answer: No. The information fits both possibilities, so we cannot decide which type is dominant. Data from many families, including crosses where both parents show the same type, would be needed.

End-of-chapter exercises

What it tests: Working out a parent's genotype from the progeny of a two-trait cross, judging dominance from survey data, planning a project to find a dominant trait, and how both parents contribute equally to the progeny.

How to approach it: For genotype questions, take one trait at a time: if all progeny show the dominant trait, that parent is pure ( $X X$ ) for it; if about half show the recessive trait, that parent is mixed ( $X x$ ). Then join the two answers. For project and 'equal contribution' questions, write the steps in order: crossing, counting, and the role of germ cells with one chromosome from each pair.

Mirrors the end-of-chapter MCQ on finding the genotype of a parent

Pea plants with yellow, round seeds were crossed with plants having green, wrinkled seeds. All the progeny had yellow seeds, but about half had wrinkled seeds. Yellow ( $Y$ ) and round ( $R$ ) are dominant. The genotype of the yellow, round parent is:
(a) $Y Y R R$
(b) $Y y R R$
(c) $Y Y R r$
(d) $Y y R r$

Show the solutionSolution

The green, wrinkled parent is $y y r r$ and gives only $y$ and $r$ to every offspring.
Colour: all progeny are yellow, so the other parent gives $Y$ every time. It must be $Y Y$ .
Shape: about half the progeny are wrinkled ( $r r$ ), so the other parent gives $r$ half the time. It must be $R r$ .
Combine: $Y Y R r$ .

Answer: (c) $Y Y R r$

Activity 8.1

What it tests: Observing free and attached earlobes in the class, calculating percentages, comparing with parents and suggesting a rule of inheritance.

How to approach it: Count carefully, then use percentage = (number with the trait ÷ total) × 100. When you compare with parents, look especially for children who show a trait neither parent shows: that trait must have been hidden in the parents, which points to it being recessive.

Mirrors the data-handling part of Activity 8.1

In a class of 50 students, 35 have free earlobes and the rest have attached earlobes. (a) Find the percentage of each type. (b) A student with attached earlobes has two parents with free earlobes. What rule does this suggest?

Show the solutionSolution

Attached earlobes: 50 − 35 = 15 students.
Free: $\frac{35}{50} \times 100 = 70%$ . Attached: $\frac{15}{50} \times 100 = 30%$ .
Neither parent shows attached earlobes, yet the child does, so each parent must carry a hidden copy for attached earlobes.
A trait that stays hidden when paired with another copy behaves as recessive.

Answer: (a) Free 70%, attached 30%. (b) It suggests that attached earlobe is a recessive trait and free earlobe is dominant.

Activity 8.2

What it tests: Designing a test to confirm that the F2 of a tall × short cross really has $T T$ , $T t$ and $tt$ in the ratio 1 : 2 : 1.

How to approach it: Tall plants $T T$ and $T t$ look alike, so you cannot tell them apart by looking. Let each F2 plant self-pollinate and study its progeny: pure plants give only their own type, mixed plants give both tall and short. Count how many F2 plants fall in each group.

Mirrors the reasoning asked in Activity 8.2

From an F2 of a tall × short cross, 300 tall plants are self-pollinated one by one. How many would you expect to give only tall progeny, and how many to give both tall and short? Explain.

Show the solutionSolution

Tall F2 plants are $T T$ and $T t$ in the ratio 1 : 2.
$T T$ plants, on self-pollination, give only tall progeny: $\frac{1}{3} \times 300 = 100$ plants.
$T t$ plants give tall and short progeny (about 3 : 1): $\frac{2}{3} \times 300 = 200$ plants.
Short F2 plants ( $tt$ ) would give only short progeny.

Answer: About 100 tall plants breed true (all tall progeny) and about 200 give both tall and short progeny. With the short plants breeding true, this confirms 1 $T T$ : 2 $T t$ : 1 $tt$ .

Heredity: previous year questions

Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.8 marks per paper; 55 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Heredity

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 31/1/1, Q6 · 1 mark · MCQ

Sex is determined by different factors in various species. However, in human beings, it is determined genetically. Which amongst the following option(s) is/are correct for human beings?
(i) Gamete carrying X chromosome from female parent.
(ii) Gamete carrying X chromosome from male parent.
(iii) Gamete carrying Y chromosome from male parent.

(A)(ii) and (iii)
(B)(i) only
(C)(i) and (iii)
(D)(iii) only

Try it first, then show our solutionSolution

A woman is XX, so every egg carries an X chromosome. A man is XY, so half his sperms carry X and half carry Y.
Sex of the child depends on the sperm. An X-carrying sperm from the father gives a girl (XX). A Y-carrying sperm from the father gives a boy (XY). So the deciding gametes are (ii) and (iii). Correct option: (A).
1 mark

Answer: (A) (ii) and (iii)

Where the marks are

Chooses (A) (1 mark)

See it in the full 2026 board paper, set 31/1/1

2026, set 31/1/1, Q15(a) · 4 marks · Short answer

Mendel took garden pea plants with different characteristics, such as height to study the inheritance pattern of factors (genes). He crossed tall pea plant with short pea plant and obtained all the tall plants in the F₁ generation. Answer the following questions:

(a)
Why only tall pea plants were observed in F₁ progeny?
1 mark
(b)
By which method did Mendel obtain F₂ progeny?
1 mark
(c)(i)
Write one difference between dominant and recessive trait.
2 marks

Try it first, then show our solutionSolution

(a) The tall parent (TT) gives T and the short parent (tt) gives t, so every F₁ plant is Tt. Tallness (T) is dominant over shortness (t), so all F₁ plants appear tall.
1 mark
(b) Mendel obtained the F₂ progeny by self-pollination of the F₁ tall plants.
1 mark
(c)(i) A dominant trait expresses itself even in the presence of the recessive factor, so it shows in both TT and Tt. A recessive trait is masked when the dominant factor is present; it shows only when both factors are recessive (tt).
2 marks

Answer: (a) F₁ is Tt and tall is dominant over short; (b) self-pollination of F₁; (c)(i) dominant trait shows in TT and Tt, recessive shows only in tt.

Where the marks are

F₁ is Tt; tall is dominant (1 mark)
Self-pollination (1 mark)
Correct difference between dominant and recessive traits (2 marks)

See it in the full 2026 board paper, set 31/1/1

2026, set 31/1/1, Q15(c)(ii) · 2 marks · Very short answer

Write two observations made by Mendel about F₁ progeny.

Try it first, then show our solutionSolution

Observation 1: all plants of the F₁ progeny were tall.
1 mark
Observation 2: no short or medium-height plants appeared in F₁; the F₁ plants resembled only one parent (the tall one).
1 mark

Answer: All F₁ plants were tall; no medium or short plants appeared (F₁ resembled only one parent).

Where the marks are

All F₁ plants tall (1 mark)
No medium/short plants, or F₁ resembled one parent only (1 mark)

See it in the full 2026 board paper, set 31/1/1

Every Heredity question, by year

For 2027

Sample paper 2026-27: Q15 (4 marks, solved)

2026

Set 31/1/1: Q6 (1 mark, solved), Q15(a) (4 marks, solved), Q15(c)(ii) (2 marks, solved)

Set 31/2/1: Q1 (1 mark, solved), Q11 (2 marks, solved), Q14 (3 marks, solved)

Set 31/3/1: Q14 (3 marks, solved)

Set 31/4/1: Q4 (1 mark, solved), Q6 (1 mark, solved), Q14 (3 marks, solved)

Set 31/5/1: Q5 (1 mark, solved), Q8 (1 mark, solved), Q13 (3 marks, solved)

Sample paper 2025-26: Q5 (1 mark, solved), Q8 (1 mark, solved), Q14 (3 marks, solved)

2025

Set 31/1/1: Q11 (1 mark, solved), Q18 (1 mark, solved), Q30 (3 marks, solved)

Set 31/2/1: Q18 (1 mark, solved), Q30 (3 marks, solved)

Set 31/3/1: Q24 (2 marks, solved), Q30 (3 marks, solved)

Set 31/4/1: Q9 (1 mark, solved), Q19 (1 mark, solved), Q24 (2 marks, solved)

Set 31/5/1: Q38 (4 marks, solved)

Set 31/6/1: Q8 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q30 (3 marks, solved)

Sample paper 2024-25: Q12 (1 mark, solved), Q30 (3 marks, solved)

2024

Set 31/1/1: Q8 (1 mark, solved), Q9 (1 mark), Q38 (4 marks, solved)

Set 31/2/1: Q12 (1 mark, solved), Q18 (1 mark, solved), Q23 (2 marks, solved)

Set 31/3/1: Q10 (1 mark, solved), Q12 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved)

Set 31/4/1: Q24 (2 marks, solved), Q30 (3 marks, solved)

Set 31/5/1: Q18 (1 mark, solved), Q30 (3 marks, solved)

Sample paper 2023-24: Q11 (1 mark, solved), Q30 (3 marks, solved), Q38 (4 marks, solved)

2023

Set 31/1/1: Q10 (1 mark, solved), Q18 (1 mark, solved), Q38 (4 marks, solved)

Set 31/2/1: Q12 (1 mark, solved), Q23 (2 marks, solved)

Set 31/4/1: Q10 (1 mark, solved), Q18 (1 mark, solved), Q38 (4 marks, solved)

Set 31/5/1: Q10 (1 mark, solved), Q18 (1 mark, solved), Q38 (4 marks, solved)

Set 31/6/1: Q9 (1 mark, solved), Q38 (4 marks, solved)

Practice questions on Heredity

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Define dominant trait and recessive trait. Give one example of each from Mendel's work with pea plants.
Hint
Think of which trait shows in a $T t$ plant and which stays hidden.
Answer
When the two copies of a gene are different, the trait that is expressed is the dominant trait and the one that is not expressed is the recessive trait. Example: tallness ( $T$ ) is dominant and shortness ( $t$ ) is recessive; a $T t$ plant is tall.
Q2 · 1 mark · MCQHarder board level
Consider these statements about human chromosomes:
(i) Every human egg carries 22 chromosomes other than the sex chromosome, plus one $X$ .
(ii) Every human sperm carries 22 chromosomes other than the sex chromosome, plus either an $X$ or a $Y$ .
(iii) A zygote that develops into a boy has 44 chromosomes other than the sex chromosomes, plus one $X$ and one $Y$ .
(iv) A human egg may carry either an $X$ or a $Y$ chromosome.
Which statements are correct?
(a) (i) and (ii) only
(b) (i), (ii) and (iii)
(c) (ii) and (iv)
(d) (i), (iii) and (iv)
Hint
A germ cell takes one chromosome from each of the 23 pairs; a zygote gets one set from each parent.
Answer
(b) (i), (ii) and (iii). Each gamete has one chromosome from each pair: 22 others plus one sex chromosome. The mother is $X X$ , so eggs carry only $X$ , which makes (iv) false. A boy's zygote has 22 + 22 = 44 others, plus $X$ from the egg and $Y$ from the sperm.
Q3 · 2 marks · Very short answerBoard level
Sex is not decided by genes in every species. Give two examples from other animals and say what decides sex in each.
Hint
One depends on an environmental cue, the other on a change during life.
Answer
In a few reptiles, the temperature at which the fertilised eggs are kept decides whether the young will be male or female. In snails, individuals can change sex, showing that sex is not genetically fixed.
Q4 · 3 marks · Short answerHarder board level
A tall pea plant with violet flowers is crossed with a short plant with white flowers. The progeny are tall violet, tall white, short violet and short white in roughly equal numbers. Tall ( $T$ ) and violet ( $W$ ) are dominant. Find the genotype of the tall violet parent, with reasons.
Hint
The short white parent is $ttw w$ . Take one trait at a time.
Answer
Height: about half the progeny are short ( $tt$ ), so the tall parent must give $t$ half the time: $T t$ . Colour: about half are white ( $w w$ ), so it must give $w$ half the time: $W w$ . Genotype of the tall violet parent: $T t W w$ .
Q5 · 3 marks · Short answerBoard level
How is the equal genetic contribution of the male and female parent ensured in the progeny? How does this keep the DNA of the species stable?
Hint
Think about what each germ cell takes from every chromosome pair.
Answer
Each body cell has two copies of each chromosome, one from each parent. A germ cell takes only one chromosome from each pair, so it has a single gene set. Male and female germ cells each give one set at fertilisation, so both parents contribute equally, and the zygote gets back the normal two sets. The chromosome number stays the same every generation, keeping the species' DNA stable.
Q6 · 5 marks · Long answerBoard level
A pure violet-flowered pea plant ( $W W$ ) is crossed with a pure white-flowered plant ( $w w$ ).
(a) Give the F1 genotype and phenotype.
(b) Show the F2 using a checkerboard.
(c) Write the phenotype and genotype ratios.
(d) If 400 F2 plants are grown, how many are white, and how many violet plants are pure?
Hint
Gametes carry one letter each; violet is dominant.
Answer
(a) F1: $W w$ , all violet. (b) Gametes $W$ , $w$ from each F1 parent; boxes $W W$ , $W w$ , $W w$ , $w w$ . (c) Phenotype 3 violet : 1 white; genotype 1 $W W$ : 2 $W w$ : 1 $w w$ . (d) White = $\frac{1}{4} \times 400 = 100$ ; pure violet ( $W W$ ) = $\frac{1}{4} \times 400 = 100$ .
Q7 · 3 marks · Short answerHarder board level
A pea plant with round yellow seeds is crossed with a plant with round green seeds. Round ( $R$ ) and yellow ( $Y$ ) are dominant. The 240 progeny are: 90 round yellow, 90 round green, 30 wrinkled yellow and 30 wrinkled green.
(a) Find the genotypes of both parents, taking one trait at a time.
(b) Check your answer by working out the expected number of wrinkled green progeny.
(c) How many of the round green progeny are expected to be pure-breeding ( $R R y y$ )?
Hint
Add up the round and wrinkled classes, then the yellow and green classes, and compare each pair as a ratio.
Answer
(a) Shape: round 180 : wrinkled 60 = 3 : 1, so both parents are $R r$ . Colour: yellow 120 : green 120 = 1 : 1, so the yellow parent is $Y y$ and the green parent is $y y$ . Parents: $R r Y y$ (round yellow) and $R r y y$ (round green). (b) Wrinkled green needs $r r$ (one quarter) and $y y$ (one half): $\frac{1}{4} \times \frac{1}{2} \times 240 = 30$ , which matches. (c) Among round progeny of $R r \times R r$ , one-third are $R R$ , so pure round green = $\frac{1}{3} \times 90 = 30$ .
Q8 · 3 marks · Short answerHarder board level
A pure tall violet pea plant ( $T T W W$ ) is crossed with a pure short white plant ( $ttw w$ ), and the F1 is self-pollinated to give 800 F2 plants.
(a) What percentage of the F2 plants are expected to be short with violet flowers?
(b) What fraction of the tall violet F2 plants are expected to be $T T W W$ ?
(c) How many tall white F2 plants are expected, and how many of them will breed true?
Hint
The F1 is $T t W w$ , so the F2 follows 9 : 3 : 3 : 1. Look inside the checkerboard for the pure boxes.
Answer
(a) F2 ratio: 9 tall violet : 3 tall white : 3 short violet : 1 short white. Short violet = $\frac{3}{16} \times 100 = 18.75%$ . (b) Only one of the 9 tall violet boxes is $T T W W$ , so the fraction is $\frac{1}{9}$ . (c) Tall white = $\frac{3}{16} \times 800 = 150$ . Of the three tall white boxes ( $T T w w$ , $T tw w$ , $T tw w$ ) only $T T w w$ breeds true, so $\frac{1}{3} \times 150 = 50$ plants.
Q9 · 1 mark · Assertion-reasonHarder board level
Assertion (A): When tall F1 pea plants ( $T t$ ) are self-pollinated, about one quarter of the progeny are short.
Reason (R): The height of a pea plant depends on how much growth hormone it makes, and this depends on how efficiently an enzyme works.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
R is a correct fact. Does it explain why short plants reappear in the F2?
Answer
(b) Both A and R are true but R is not the correct explanation of A. A is true: one quarter of the F2 is short. R is true: it describes how a gene controls height. But A is explained by each F1 parent passing the hidden $t$ into half its gametes, so one quarter of the offspring get $tt$ . R explains how the gene acts, not why short plants reappear.
Q10 · 3 marks · Short answerHarder board level
A couple already has two sons. What is the chance that their third child will be a daughter? Explain using the sex chromosomes, and say why the earlier births do not change this chance.
Hint
Each fertilisation is a fresh event.
Answer
The mother's eggs all carry $X$ ; the father's sperms carry $X$ or $Y$ in equal numbers. An $X$ sperm gives $X X$ (girl), a $Y$ sperm gives $X Y$ (boy), so the chance of a girl is $\frac{1}{2}$ (50%). Each fertilisation happens independently, so earlier sons do not affect which sperm fertilises the next egg. The chance stays $\frac{1}{2}$ .
Q11 · 3 marks · Short answerHOTS / competency
A survey found that most children with attached earlobes have at least one parent with attached earlobes. A student concludes that attached earlobe must be the dominant trait. Do you agree? Justify your answer and suggest what data would settle the question.
Hint
Could the same pattern arise if attached earlobe were recessive?
Answer
No, the conclusion does not follow. Children resemble parents whether a trait is dominant or recessive, because both parents pass on copies of the gene. Better data: look at families where both parents show the same type. If two free-earlobed parents have an attached-earlobed child, attached must be hidden in them and so recessive; if two attached-earlobed parents have a free-earlobed child, attached would be dominant.
Q12 · 3 marks · Short answerHOTS / competency
A plant species has red or white flowers. Outline a project to find out which flower colour is dominant.
Hint
Follow Mendel: pure lines, cross, look at F1, then self-pollinate F1.
Answer
1. Raise pure-breeding red and white plants (plants that give only their own colour when self-pollinated for a few generations). 2. Cross red with white and record F1 flower colour; the colour shown by all F1 plants is dominant. 3. Self-pollinate the F1 and count the F2: the dominant colour should appear in about three quarters and the other colour reappear in about one quarter, confirming it is recessive. Use a large number of plants for reliable counts.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What fraction of the F2 is short in Mendel's tall × short cross?Show answer
One quarter (tall : short = 3 : 1).
2.What is the F2 genotype ratio of a monohybrid cross?Show answer
$T T : T t : tt = 1 : 2 : 1$ .
3.Define a recessive trait.Show answer
A trait that is expressed only when both copies of the gene are recessive. In a hybrid it is hidden by the dominant trait.
4.What is a gene?Show answer
A section of DNA that provides the information for one protein.
5.What four gametes does $R r Y y$ form?Show answer
$R Y$ , $R y$ , $r Y$ and $r y$ .
6.What does the dihybrid cross prove?Show answer
That different traits, such as seed shape and seed colour, are inherited independently, giving new combinations in F2.
7.How many chromosomes does a human germ cell carry?Show answer
23: one from each of the 23 pairs (22 pairs plus the sex chromosome pair).
8.Give one non-genetic way sex is decided in animals.Show answer
In a few reptiles, the temperature at which the eggs are kept decides the sex. Snails can change sex.

Best order to study it

Start with variation (8.1): how differences build up, why sexual reproduction creates more diversity, and why variation helps survival.
Learn inherited traits with the earlobe example and Activity 8.1, so the idea of a hidden trait makes sense.
Master the monohybrid cross step by step (parents, gametes, F1, F2, both ratios), along with dominant and recessive traits and Activity 8.2.
Move to the dihybrid cross: the F1 genotype, the four gametes, the 16-box checkerboard, the 9 : 3 : 3 : 1 ratio and new combinations.
Then study how genes control traits (gene → enzyme → hormone → height) and how chromosomes ensure equal contribution and independent inheritance.
Finish with sex determination in humans (XX, XY, the father's chromosome) and the non-genetic examples, then practise mixed crosses under timed conditions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you list Mendel's two observations on the F1 of tall × short?
Can you write the full monohybrid cross from $T T \times tt$ to F2 with both ratios?
Can you say what experiment confirms the 1 : 2 : 1 ratio (Activity 8.2)?
Can you find a parent's genotype when about half the progeny show the recessive trait?
Can you write the gametes of $R r Y y$ and name the new combinations from $R R y y \times r r Y Y$ ?
Can you explain the gene → enzyme → hormone → height chain in three lines?
Can you explain why germ cells carry one chromosome from each pair, and what happens at fertilisation?
Can you explain why chromosomes being separate pieces allows independent inheritance?
Can you show XX × XY, give the 1 : 1 result and say who decides the sex?
Can you give one reason variation helps a species survive, using the bacteria and heat wave example?

Heredity: frequently asked questions

Is evolution part of the Class 10 Heredity chapter for the board exam?

No. Evolution, speciation, fossils and human evolution have been removed from the NCERT chapter, which is now called Heredity. In the 2026-27 syllabus they are school-assessed only and do not come in the board exam.

What is the difference between the 3:1 and 9:3:3:1 ratios?

The 3:1 ratio is the F2 phenotype ratio when you follow one trait, such as tall and short. The 9:3:3:1 ratio is the F2 phenotype ratio when you follow two traits together, such as seed shape and seed colour. The second ratio adds up to 16 because a dihybrid forms four kinds of gametes.

Why is it the father who decides the sex of the child?

The mother is XX, so all her eggs carry an X chromosome. The father is XY, so his sperm carry either X or Y. Whichever sperm fertilises the egg decides whether the child is a girl (XX) or a boy (XY).

Do I need to draw the full checkerboard for a dihybrid cross?

It is the clearest way to show your working and earns step marks. At the very least, write the parents, the F1 genotype, the four gametes and the final ratio with the names of each class. Show the working clearly, because the marking scheme gives marks at each step.

How much detail do I need on sex determination?

The syllabus asks only for a brief introduction. Know that some reptiles depend on egg temperature and snails can change sex, and that humans have 22 pairs of chromosomes plus the XX or XY pair. Also know why the father's chromosome decides the sex, and that half the children are expected to be boys and half girls.

Based on: CBSE Science (086) syllabus 2026-27; NCERT Science, Textbook for Class X (Reprint 2026-27)