Q.Define and design a test-cross.

Step 1 — Recall the defect in sickle cell anaemia

Sickle cell anaemia results from a point mutation that changes one amino acid (glutamic acid to valine) in the beta-globin chain — this alters the structure/quality of the globin molecule produced, not the amount made. This is a qualitative defect.

Step 2 — Recall the defect in Thalassemia

Thalassemia is caused by mutations (often deletions or defects in regulatory/synthesis machinery) that reduce or completely stop the synthesis rate of one of the globin chains (alpha or beta) — the globin chains produced are structurally normal, but fewer of them are made. This is a quantitative defect.

Step 3 — Evaluate the options …

Step 1 — Read the pedigree.

• Generation I: an unaffected male × an unaffected female.
• Generation II: five children — an affected son, an affected daughter, and three unaffected children.
• Generation III: an unaffected Gen-II daughter married to an unaffected male produces an affected son plus two unaffected daughters.

Step 2 — Dominant or recessive?

The trait skips: both Gen-I parents are unaffected, yet affected children appear. A dominant allele must be expressed in at least one parent (an affected child of a dominant trait must have an affected parent). Since the phenotype disappears in a generation and reappears, it is carried silently by heterozygotes ⇒ the allele is recessive. Both parents are carriers ($Aa \times Aa \rightarrow \tfrac14\ aa$ affected).

This kills options (B) "Sex-linked dominant" and (D) "Autosomal dominant".

Step 3 — Autosomal or sex-linked (X-linked)?

Test the X-linked recessive hypothesis on the Gen-I family. An affected daughter would need genotype $X^a X^a$, i.e. one $X^a$ from her mother (a carrier, fine) and one $X^a$ from her father. But a father's only X carries the allele he expresses — so an $X^a$-bearing father would himself be affected. The Gen-I father is drawn unaffected. Contradiction. …

1. Statement (A) — check the cross

Haemophilia is a sex-linked (X-linked) recessive disorder. Let $X^H$ = normal allele, $X^h$ = haemophilia allele.

A carrier mother is heterozygous $X^H X^h$ (she is normal, because the dominant $X^H$ masks $X^h$). Cross her with a normal father $X^H Y$:

A son receives his single X from his mother and a Y from his father. With no second X to carry a masking dominant allele, he is hemizygous and expresses whatever is on that X. Hence half the sons of a carrier mother are haemophilic — statement (A) is CORRECT. ✓

2. Why the other three are wrong

(B) "Thalassemia is a qualitative problem." ✗ — inverted. The standard contrast is:

So thalassemia is quantitative, not qualitative. …

Step 1 — The two vocabularies you need

Autosome vs allosome: autosomes are chromosome pairs 1–22; allosomes (sex chromosomes) are the X and Y.

Trisomy vs monosomy: these are aneuploidies, caused by the failure of chromatids to segregate during cell division (non-disjunction):

• Trisomy = one EXTRA chromosome ($2n+1$).
• Monosomy = one chromosome MISSING ($2n-1$).

Separately, a Mendelian disorder is one caused by an alteration/mutation in a SINGLE GENE (not a change in chromosome number).

Step 2 — Work through Column I

i. Autosomal trisomy → (s) Down's Syndrome.

An extra copy of chromosome 21 (trisomy 21) — chromosome 21 is an autosome. Karyotype $47$, i.e. $2n+1$. Features: short stature, small round head, furrowed tongue, partially open mouth, palm crease, retarded mental and physical development.

ii. Allosomal trisomy → (r) Klinefelter's Syndrome.

Karyotype 47, XXY — an extra X on the SEX chromosomes. The individual is masculine in development but has gynaecomastia (feminine development, e.g. breast development) and is sterile.

iii. Allosomal monosomy → (p) Turner's Syndrome.

Karyotype 45, XO — one of the X chromosomes is MISSING, again on the SEX chromosomes. Such females are sterile, have rudimentary ovaries and lack other secondary sexual characters.

iv. Cystic fibrosis → (q) Mendelian disorder. …