Chapter 3 — Chromosomal Basis of Inheritance
Key Concepts Quick Recap
| Term | Definition |
|---|
| Chromosomal theory | Genes are located on chromosomes (Sutton and Boveri, 1902) |
| Linkage | Genes on the same chromosome tend to be inherited together |
| Crossing over | Exchange of segments between non-sister chromatids of homologous chromosomes during meiosis I |
| Recombination | New gene combinations produced by crossing over |
| Recombination frequency | (Recombinants/Total offspring) × 100; equals map distance in centimorgans (cM) |
| Sex chromosomes | Chromosomes that determine sex (X and Y) |
| Autosomes | Non-sex chromosomes |
| Sex-linked inheritance | Inheritance of genes located on sex chromosomes |
| Barr body | Inactivated X chromosome visible as dense chromatin mass in female somatic cells |
I. Choose the Best Answer (MCQs)
1. The chromosomal theory of inheritance was proposed by: (a) Morgan and Bridges (b) Sutton and Boveri (c) Watson and Crick (d) Beadle and Tatum
Answer(b) Sutton and Boveri Walter Sutton (grasshopper studies) and Theodor Boveri (sea urchin studies) independently proposed that Mendel's factors (genes) are carried on chromosomes (1902). This was the chromosomal theory of inheritance.
2. Morgan conducted his classical linkage experiments on: (a) Pisum sativum (b) Zea mays (c) Drosophila melanogaster (d) Neurospora crassa
Answer(c) Drosophila melanogaster T.H. Morgan used Drosophila (fruit fly) for genetics experiments because of: short generation time (~2 weeks), large number of offspring, easy to culture, only 4 pairs of chromosomes, many visible mutations.
3. Genes that are located on the same chromosome are called: (a) Multiple alleles (b) Linked genes (c) Pleiotropic genes (d) Supplementary genes
Answer(b) Linked genes Linked genes on the same chromosome tend to be inherited together and do not show independent assortment. The strength of linkage depends on their physical distance — closely linked genes rarely separated by crossing over.
4. The unit of map distance between genes is: (a) Angstrom (b) Nanometre (c) Centimorgan (d) Dalton
Answer(c) Centimorgan (cM) 1 centimorgan = 1% recombination frequency. Named after T.H. Morgan. A recombination frequency of 10% means 10 cM distance between genes.
5. In humans, sex is determined by: (a) Temperature (b) XX-XO type (c) XX-XY type (d) ZZ-ZW type
Answer(c) XX-XY type Human females: 44 + XX (46 chromosomes). Human males: 44 + XY (46 chromosomes). The Y chromosome carries the SRY gene that determines male sex. Sperm determines the sex of the offspring (X-bearing sperm → female; Y-bearing sperm → male).
6. Colour blindness in humans is an example of: (a) Autosomal dominant (b) Autosomal recessive (c) Sex-linked recessive (d) Sex-linked dominant
Answer(c) Sex-linked recessive The gene for colour vision is on the X chromosome. Males (XY) need only one recessive allele (X^c Y) to be colour blind. Females need two recessive alleles (X^c X^c) to be colour blind — hence colour blindness is more common in males.
7. The Barr body is: (a) Second X chromosome in males (b) Inactivated X chromosome in female somatic cells (c) Y chromosome (d) Autosome pair
Answer(b) Inactivated X chromosome in female somatic cells In females, one X chromosome is randomly inactivated in each somatic cell (Lyon hypothesis, 1961). The inactivated X condenses into a dark-staining body = Barr body. Number of Barr bodies = number of X chromosomes − 1.
II. Short Answer Questions
Q1. What is the chromosomal theory of inheritance? Who proposed it?
Answer
The chromosomal theory of inheritance states that: 1. Genes are located on chromosomes. 2. Chromosomes behave in a manner parallel to Mendel's factors during cell division. 3. The behaviour of chromosomes during meiosis explains the laws of segregation and independent assortment.
Proposed by: Walter Sutton (1902) studying grasshopper chromosomes, and Theodor Boveri (1902) studying sea urchin eggs independently.
Evidence: Chromosomes occur in pairs (diploid); they segregate during meiosis; different chromosome pairs assort independently — exactly paralleling Mendel's laws.
Q2. What is crossing over? State its significance.
Answer
Crossing over is the exchange of chromosomal segments between non-sister chromatids of homologous chromosomes during pachytene stage of meiosis I.
Process:
- Homologous chromosomes pair (synapsis) at zygotene.
- Bivalents form at pachytene.
- Chiasmata (crossing points) form — breakage and reunion of non-sister chromatids.
- Segments exchanged → new allele combinations on recombinant chromosomes.
Significance: 1. Genetic recombination: Produces new combinations of traits not present in either parent. 2. Increases variation: Raw material for evolution and natural selection. 3. Gene mapping: Recombination frequency used to construct genetic maps (linkage maps). 4. Breaks down unfavourable linkages: Allows favourable alleles to combine independently.
Q3. Explain sex determination in humans.
Answer
Humans have 46 chromosomes (23 pairs):
- 22 pairs = autosomes (same in both sexes)
- 1 pair = sex chromosomes
Female: 44 + XX → gametes all carry X Male: 44 + XY → gametes carry X or Y (equal ratio)
At fertilisation:
- Egg (X) + X sperm → XX → Female
- Egg (X) + Y sperm → XY → Male
The sex of the offspring is determined by the type of sperm that fertilises the egg. The SRY gene (Sex-determining Region of Y) on the Y chromosome triggers testis development and maleness.
Q4. Explain sex-linked inheritance with the example of haemophilia.
Answer
Haemophilia A (bleeder's disease) is caused by a recessive allele (h) on the X chromosome that fails to produce clotting factor VIII.
Genotypes:
- X^H X^H — normal female
- X^H X^h — carrier female (phenotypically normal)
- X^h X^h — haemophilic female (rare)
- X^H Y — normal male
- X^h Y — haemophilic male
Cross: Carrier female × Normal male X^H X^h × X^H Y
| X^H | Y |
|---|
| X^H | X^H X^H (normal female) | X^H Y (normal male) |
| X^h | X^H X^h (carrier female) | X^h Y (haemophilic male) |
Result50% of sons are haemophilic; daughters appear normal but 50% are carriers.
Historical note: Queen Victoria of England was a carrier; haemophilia spread through European royal families.
III. Worked Examples
Example 1 — Calculating recombination frequency
In a dihybrid test cross of Drosophila, out of 2000 offspring:
- Parental types: 900 + 900 = 1800
- Recombinants: 60 + 140 = 200
Recombination frequency = (200/2000) × 100 = 10% = 10 cM
The two genes are 10 centimorgans apart on the same chromosome.
Example 2 — Colour blindness cross
Normal female (carrier) × Colour-blind male X^C X^c × X^c Y
| X^c | Y |
|---|
| X^C | X^C X^c (carrier female) | X^C Y (normal male) |
| X^c | X^c X^c (colour-blind female) | X^c Y (colour-blind male) |
Result50% daughters carrier, 50% daughters colour-blind; 50% sons normal, 50% sons colour-blind.
Common Mistakes
- Confusing recombination frequency (%) with linkage — higher frequency = genes farther apart = less linked.
- Thinking crossing over happens between sister chromatids — it occurs between NON-sister chromatids of homologous chromosomes.
- Forgetting Barr body formula: Barr bodies = X chromosomes − 1 (so XXY male has 1 Barr body).
- Writing sex as determined by the egg — sex is determined by the sperm (X or Y).
Exam Checklist
- Chromosomal theory: Sutton and Boveri (1902).
- Crossing over: pachytene, non-sister chromatids, chiasmata.
- Recombination frequency formula: (recombinants/total) × 100%.
- Sex determination: XX (female), XY (male); sperm determines sex.
- Haemophilia cross with Punnett square.
- Barr body = inactivated X chromosome.
