A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be probabilit

NCERT Exemplar · Q51

Q. A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.

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Concept understanding — Pedigree Analysis

Pedigree Analysis: A Family Tree for Traits

Imagine you're looking at a family photograph album. You notice that your grandfather, your uncle, and your cousin all have the same unusual hairline or the same colour blindness. You start wondering: Is this just a coincidence, or is it something that runs in the family? How does this pattern pass from one generation to the next?

That curiosity is exactly where pedigree analysis begins.

What is a Pedigree?

A pedigree is simply a family tree that records the presence or absence of a particular trait — like a disease, a physical feature, or a genetic condition — across several generations. Instead of writing names and stories, biologists use a standard set of symbols to map out who has the trait and who doesn't, and how they are related.

The NCERT textbook defines it as a method to study the inheritance of a trait in several generations of a family. It is a tool, not a formula. You don't calculate anything; you observe patterns.

Why Do We Need It?

Humans cannot be bred in a laboratory like fruit flies or pea plants. You cannot ask two cousins to have children just to see if a disease is inherited. So pedigree analysis becomes the only ethical, practical way to study how traits pass through human families.

It helps in:

Predicting the likelihood that a child might inherit a genetic disorder
Tracing whether a trait is dominant or recessive
Identifying carriers — people who carry a gene for a disease but do not show it themselves
Offering genetic counselling to families with a history of inherited conditions

The Basic Symbols (What You Need to Know)

In a pedigree chart:

Males are represented by squares
Females are represented by circles
A horizontal line connecting a square and a circle shows a marriage (mating)
A vertical line descending from that couple leads to their children
Shaded (filled) symbols indicate individuals who show the trait
Unshaded (empty) symbols indicate individuals who do not show the trait
A half-shaded symbol often indicates a carrier (someone who has the gene but does not express the trait)

Note

In NCERT, you will see that a pedigree is drawn from the top (oldest generation) to the bottom (youngest). The first generation is labelled I, the second II, and so on. Individuals within a generation are numbered from left to right (e.g., I-1, I-2, II-3).

How to Read a Pedigree: The Core Idea

You are essentially looking for a pattern. Ask yourself:

Does the trait appear in every generation? If yes, it might be dominant — because a dominant trait only needs one copy of the gene to show up, so it rarely skips a generation.
Does the trait skip a generation? If yes, it might be recessive — because a recessive trait needs two copies (one from each parent), and carriers can pass it on without showing it themselves.
Does the trait affect mostly males? If yes, it might be sex-linked (carried on the X chromosome). For example, colour blindness and haemophilia are much more common in males because they have only one X chromosome.

Important

The most important rule from NCERT: A recessive trait can skip a generation, but a dominant trait cannot. If you see a child with a trait whose parents do not have it, the trait is almost certainly recessive. The parents are carriers.

A Simple Example (Without Numbers)

Suppose you see a pedigree where:

A grandfather has a condition (shaded square)
His daughter does not have it (empty circle)
But his grandson (the daughter's son) has it (shaded square)

What does this tell you? The trait skipped a generation (from grandfather to grandson, missing the daughter). That suggests it is recessive. Also, it appears only in males in this case, so it might be X-linked recessive — the daughter is a carrier, and she passed the gene to her son. …

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A woman whose father is colour blind is a carrier of the recessive allele, and when she marries a normal visioned man, her sons have a 50% chance of being colour blind, while her daughters will be either normal or carriers but never colour blind.

Colour blindness is a classic example of X-linked recessive inheritance. The gene responsible for red-green colour blindness is located on the X chromosome, and it is recessive — meaning a male (who has only one X chromosome) will express the trait if he inherits the defective allele, while a female needs two copies (one on each X) to be colour blind. A female with one normal and one defective allele is a carrier, with normal vision.

Now, consider the woman in the question. Her father is colour blind. Since a father passes his only X chromosome to his daughters, this woman must have inherited that defective X from her father. Her mother, not mentioned, could be normal or a carrier — but regardless, the woman herself has at least one defective X. Because she has normal vision, she must have received a normal X from her mother. So the woman is a carrier: genotype X^c X (where X^c denotes the colour blind allele, and X the normal allele).

She marries a normal visioned man. A normal visioned man has the genotype X Y — his single X is normal. So the cross is between a carrier woman (X^c X) and a normal man (X Y).

Let us draw the pedigree chart mentally. The father (colour blind) is at the top, with a filled square (since males are squares, colour blind is shaded). He has a daughter (the woman in question) who is a carrier — shown as a half-filled circle (circle for female, half-filled to indicate carrier status). She marries a normal man (unfilled square). Their children are the next generation.

The possible offspring from this union are:

Sons: A son inherits his Y chromosome from his father and his X chromosome from his mother. The mother can pass either her normal X or her defective X^c. So half the sons (50%) will get X^c and be colour blind (genotype X^c Y), and half will get X and be normal (X Y). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$Q. If a woman with colour blindness marries a man with normal vision, the expected vision in their daughters is (A) Normal vision - homozygous (B) Colour blind (C) Normal vision - carrier (D) Total blindness$
›Reveal solutionSolution
Colour blindness is an X-linked recessive trait. A colour-blind woman (X $^c$ X $^c$ ) and a normal-vision man (X $^C$ Y) will produce daughters who all receive one defective X from the mother and one normal X from the father — so every daughter is a carrier with normal vision.

The key to this problem is understanding X-linked recessive inheritance. Colour blindness (most commonly red-green) is caused by a recessive allele on the X chromosome. A woman has two X chromosomes; a man has one X and one Y.

A woman with colour blindness must have the defective allele on both her X chromosomes — she is homozygous recessive (X $^c$ X $^c$ ). A man with normal vision has the normal dominant allele on his single X chromosome (X $^C$ Y). The Y chromosome carries no colour vision gene.

Now let’s trace the inheritance to the daughters.

Mother’s contribution. The colour-blind mother can only pass on an X chromosome carrying the recessive allele (X $^c$ ). Every egg she produces carries X $^c$ .

Father’s contribution. The normal-vision father passes on either his X chromosome (X $^C$ ) or his Y chromosome. For a daughter, the father must contribute an X chromosome — because the Y chromosome produces a son (XY). So every daughter receives the father’s X $^C$ .

Daughter’s genotype. Each daughter gets X $^c$ from the mother and X $^C$ from the father. Her genotype is X $^C$ X $^c$ — one normal allele and one colour-blind allele. …
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TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
$Q. If mother is carrier for haemophilia and father is normal, chance of their children to be haemophilic is (A) 100% (B) 75% (C) 50% (D) 25%$
›Reveal solutionSolution
Haemophilia is an X-linked recessive disorder. With a carrier mother (X\textsuperscript{H}X\textsuperscript{h}) and a normal father (X\textsuperscript{H}Y), the chance of a child being haemophilic is 25% — only sons can be affected, and each son has a 50% risk.

The key to this problem lies in understanding X-linked recessive inheritance. Haemophilia is caused by a recessive allele on the X chromosome. Because males have only one X chromosome (XY), a single recessive allele on that X will cause the disease. Females, with two X chromosomes, need two recessive alleles to be affected; with one, they are carriers.

The mother is a carrier — her genotype is X\textsuperscript{H}X\textsuperscript{h} (where X\textsuperscript{H} is the normal allele and X\textsuperscript{h} is the haemophilia allele). The father is normal — his genotype is X\textsuperscript{H}Y. The Y chromosome carries no allele for this gene, so the father contributes either an X\textsuperscript{H} or a Y to his child.

Let’s work through the possible offspring:

Mother’s gametes: She produces two types of eggs — half carry X\textsuperscript{H} and half carry X\textsuperscript{h}.

Father’s gametes: He produces two types of sperm — half carry X\textsuperscript{H} and half carry Y.

Punnett square analysis:

If an X\textsuperscript{H} egg meets an X\textsuperscript{H} sperm → daughter with genotype X\textsuperscript{H}X\textsuperscript{H} (normal, not a carrier).

If an X\textsuperscript{H} egg meets a Y sperm → son with genotype X\textsuperscript{H}Y (normal).

If an X\textsuperscript{h} egg meets an X\textsuperscript{H} sperm → daughter with genotype X\textsuperscript{H}X\textsuperscript{h} (carrier, like mother).

If an X\textsuperscript{h} egg meets a Y sperm → son with genotype X\textsuperscript{h}Y (haemophilic). …
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TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$50\%$
›Reveal solutionSolution
Haemophilia is an X-linked recessive disorder. The woman is a carrier ( $X^HX^h$ ), and the man is normal ( $X^HY$ ). Their sons have a $50\%$ chance of inheriting the haemophilic allele from their mother and thus being haemophilic.

Haemophilia is a classic example of an X-linked recessive genetic disorder. Understanding how X-linked traits are inherited is key to solving this problem.

Here's the core idea:

X-linked traits are carried on the X chromosome.

Recessive means that two copies of the affected allele are needed for a female to show the trait, but only one copy is needed for a male.

Males have one X and one Y chromosome ( $XY$ ). They inherit their X chromosome only from their mother and their Y chromosome only from their father. This means whatever allele is on the X chromosome they receive from their mother will determine their phenotype for an X-linked trait.

Females have two X chromosomes ( $XX$ ). They inherit one X from their mother and one X from their father. If a female inherits one normal allele and one recessive allele, she will be a "carrier" and typically not show the trait, but can pass it on to her children.

Let's denote the normal allele for blood clotting as $X^H$ and the haemophilic allele as $X^h$ .

Determine the genotypes of the woman's parents.

The woman's father is haemophilic. Since males have only one X chromosome, his genotype must be $X^hY$ .

The woman's mother is homozygous normal. This means both her X chromosomes carry the normal allele, so her genotype is $X^HX^H$ .

Determine the genotype of the woman.

The woman inherits one X chromosome from her father and one X chromosome from her mother.

From her haemophilic father ( $X^hY$ ), she must inherit $X^h$ .

From her homozygous normal mother ( $X^HX^H$ ), she must inherit $X^H$ .

Therefore, the woman's genotype is $X^HX^h$ . She is a carrier for haemophilia.

Watch out

A common mistake is to assume that because the woman's father is haemophilic, she must also be haemophilic. However, since the trait is recessive and she inherited a normal X from her mother, she is only a carrier.

Determine the genotype of the normal man she married.

The man is described as normal. Since haemophilia is an X-linked recessive trait, a normal man must have the genotype $X^HY$ .

Determine the possible genotypes of their children.

We need to consider the cross between the woman ( $X^HX^h$ ) and the man ( $X^HY$ ).

The woman can produce two types of gametes: $X^H$ and $X^h$ .

The man can produce two types of gametes: $X^H$ and $Y$ . …
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Q.A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.

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