Q.What is pedigree analysis? Suggest how such an analysis, can be useful.

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.

1. 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$^c$.

2. 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$^C$.

3. 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. …

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:

1. Mother’s gametes: She produces two types of eggs — half carry X\textsuperscript{H} and half carry X\textsuperscript{h}.
2. Father’s gametes: He produces two types of sperm — half carry X\textsuperscript{H} and half carry Y.
3. 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). …

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 $X^h$.

1. 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 $X^HX^H$.

2. 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.

3. 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 $X^HY$.

4. 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 $Y$. …