Mendel's dihybrid 9:3:3:1 ratio depends on one assumption: the two genes sit on different chromosomes. When two genes sit on the same chromosome, they tend to travel together into gametes — that is linkage. The closer two loci are on a chromosome, the more tightly they are linked and the less likely crossing over will separate them.
T.H. Morgan's work on Drosophila (NCERT Class 12 Biology Chapter 4, page 67) established that genes on the same chromosome do not assort independently. Instead, parental (non-recombinant) combinations appear in offspring far more often than recombinant combinations. When crossing over occurs during meiosis I (specifically at the pachytene stage of prophase I), homologous chromatids exchange segments, producing recombinant gametes. The frequency of this recombination is the basis of genetic mapping.
Recombination frequency (RF) measures how often recombinant offspring appear:
RF = (number of recombinants ÷ total progeny) × 100
One percent recombination frequency equals one centiMorgan (1 cM) of map distance. The upper limit of RF is 50% — at that point the two loci behave as if they are on separate chromosomes (effectively unlinked).
The core trap in this topic is confusing linkage with independent assortment. If a test cross yields parental types far exceeding 50%, the genes are linked. If recombinants approach 50%, the genes are either very far apart on the same chromosome or on different chromosomes entirely.
Watch out: RF gives a relative map distance, not a physical distance in base pairs. Two loci 10 cM apart will show roughly 10% recombinant offspring in a test cross — but double crossovers at large distances can underestimate the true map distance.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Linkage was experimentally demonstrated for the first time by which scientist, using which organism?
Answer: A. T.H. Morgan's experiments on Drosophila provided the first definitive evidence for linkage, showing that certain genes on the same chromosome did not assort independently (NCERT Class 12 Biology Chapter 4, page 67).
Why B is wrong: B is wrong because Mendel worked with independently assorting traits on different chromosomes; he did not discover linkage.
Why C is wrong: C is wrong because Hugo de Vries is associated with the mutation theory, not linkage experiments.
Why D is wrong: D is wrong because Bateson and Punnett observed deviations from independent assortment in sweet pea and coined 'coupling and repulsion,' but Morgan's Drosophila work is credited as the first definitive linkage demonstration in NCERT.
What is the unit of genetic map distance, and what does 1 unit represent?
Answer: B. 1 centiMorgan (cM), also called 1 map unit, equals 1% recombination frequency between two loci in a test cross (NCERT Class 12 Biology Chapter 4, page 67).
Why A is wrong: A is wrong because centiMorgans measure genetic distance (recombination-based), not physical distance in base pairs. There is no fixed relationship between 1 cM and 1 kb.
Why C is wrong: C is wrong because 1 Morgan = 100% recombination frequency, which equals 100 cM. The commonly used unit is the centiMorgan, not the Morgan.
Why D is wrong: D is wrong because 1 map unit equals 1% (not 0.1%) recombination frequency.
During which sub-stage of meiosis does crossing over occur between homologous chromosomes?
Answer: D. Crossing over occurs at the pachytene stage of prophase I, when bivalents form tetrads and non-sister chromatids of homologous chromosomes exchange segments (NCERT Class 12 Biology Chapter 4).
Why A is wrong: A is wrong because leptotene is the condensation stage where chromosomes become visible; no synapsis or crossing over has occurred yet.
Why B is wrong: B is wrong because zygotene is when homologous chromosomes begin to pair (synapsis starts), but crossing over has not yet taken place.
Why C is wrong: C is wrong because diplotene is when synapsed chromosomes begin to separate and chiasmata become visible — but crossing over has already occurred during pachytene. Diplotene reveals the result, not the event itself.
In a test cross involving two linked genes in Drosophila, a total of 1000 progeny are obtained. Of these, 860 show parental combinations and 140 show recombinant combinations. What is the recombination frequency?
Answer: B. RF = (recombinants / total) × 100 = (140 / 1000) × 100 = 14%. The two loci are 14 cM apart on the genetic map (NCERT Class 12 Biology Chapter 4, page 67).
Why A is wrong: A is wrong because 8.6 comes from the parental count, 860, divided by 100. RF uses the recombinants: 140/1000 × 100 = 14%.
Why C is wrong: C is wrong because 43% does not correspond to any correct calculation from these numbers. It may arise from confusing parental with recombinant counts or misapplying the formula.
Why D is wrong: D is wrong because 86% is the percentage of parental types, not recombinants. RF uses recombinant offspring in the numerator.
Two genes A and B are located on the same chromosome. In a test cross, the recombination frequency between them is 32%. A third gene C shows 12% RF with A and 20% RF with B. What is the gene order on the chromosome?
Answer: A. A–C = 12 cM, C–B = 20 cM, and A–B = 32 cM. Since 12 + 20 = 32, gene C lies between A and B. The order is A — C — B (NCERT Class 12 Biology Chapter 4, page 67).
Why B is wrong: B is wrong because if C were outside A, the distance C–B would need to equal C–A + A–B (12 + 32 = 44), not 20. The distances are only additive when C is between A and B.
Why C is wrong: C is wrong because placing C beyond B would require A–C = A–B + B–C (32 + 20 = 52), not 12. This order violates the observed distances.
Why D is wrong: D is wrong because three pairwise distances are enough to order three genes: they add up (12 + 20 = 32) only with C in the middle.
If two genes show a recombination frequency of 50% in a test cross, what can be concluded?
Answer: C. 50% RF is the maximum possible recombination frequency and indicates that the two genes assort independently — either because they are on different chromosomes or so far apart on the same chromosome that at least one crossover always occurs between them (NCERT Class 12 Biology Chapter 4).
Why A is wrong: A is wrong because tightly linked genes show RF well below 50% (typically < 10%). An RF of 50% is the opposite of tight linkage.
Why B is wrong: B is wrong because crossing over can still occur at 50% RF; in fact, for distant loci on the same chromosome, crossing over is so frequent that parental and recombinant types are produced equally.
Why D is wrong: D is wrong because 50% RF does not translate to 100 cM. RF caps at 50% and cannot exceed it regardless of actual physical distance. Map distances above 50 cM require summing shorter intervals.
Morgan crossed a yellow-bodied, white-eyed female Drosophila with a brown-bodied, red-eyed male. The F1 intercross produced a higher proportion of parental phenotype combinations than expected from independent assortment. Which statement best explains this observation?
Answer: C. A higher-than-expected proportion of parental combinations indicates the two genes do not assort independently — they are physically linked on the same chromosome, so they tend to be inherited together (NCERT Class 12 Biology Chapter 4, page 67).
Why A is wrong: A is wrong because genes on different chromosomes assort independently and would produce the expected 9:3:3:1 ratio, not an excess of parental types.
Why B is wrong: B is wrong because dominance determines which phenotype appears in heterozygotes but does not explain deviations from independent assortment ratios.
Why D is wrong: D is wrong because epistasis alters the phenotypic ratio (e.g., 9:3:4 or 12:3:1) by masking one gene's expression, not by producing an excess of parental combinations specifically.
In a test cross of a dihybrid organism heterozygous for two linked genes (AaBb × aabb), 500 progeny are obtained: 200 AaBb, 200 aabb, 50 Aabb, 50 aaBb. Calculate the RF, determine the map distance, and identify which offspring classes are recombinant.
Answer: D. Parental types (AaBb and aabb) total 400; recombinant types (Aabb and aaBb) total 100. RF = (100 / 500) × 100 = 20%. Map distance = 20 cM. The recombinants are the less frequent classes: Aabb and aaBb (NCERT Class 12 Biology Chapter 4, page 67).
Why A is wrong: A is wrong because 10% would result from dividing 50 (one recombinant class) by 500 instead of the total recombinants (100). Both recombinant classes must be summed before dividing by total progeny.
Why B is wrong: B is wrong because 80% represents the parental class proportion, not the recombination frequency. RF uses recombinants in the numerator. Also, AaBb and aabb are the parental classes, not recombinants.
Why C is wrong: C is wrong because while the RF calculation is correct at 20%, it misidentifies the recombinants. AaBb and aabb are the majority classes (parental), not recombinants. Recombinants are always the less frequent classes in a linkage test cross.
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Given
In a test cross involving two linked genes P and Q, 800 total progeny are obtained. The phenotype counts are:• PQ (parental): 310• pq (parental): 330• Pq (recombinant): 75• pQ (recombinant): 85
Required
(a) Calculate the recombination frequency.
(b) Determine the map distance between P and Q.
(c) State whether the genes are tightly or loosely linked.
Concept
Genes on the same chromosome show linkage. The degree of linkage is measured by recombination frequency — the proportion of recombinant offspring in a test cross. Low RF means tight linkage; RF approaching 50% means the genes are far apart or effectively unlinked.
Formula
RF = (recombinants / total progeny) × 100
Substitution
Total recombinants = 75 + 85 = 160
Total progeny = 800
RF = (160 / 800) × 100
Calculation
RF = 0.20 × 100 = 20%
Note: 800 and 160 are exact counting integers and do not affect significant-figure considerations.
Final answer
RF = 20%. Map distance between P and Q = 20 cM. Since RF is well below 50%, the genes are linked (though not tightly — tight linkage typically shows RF < 10%).
Common trap
A frequent error is dividing only one recombinant class (75 or 85) by the total instead of summing both recombinant classes first. This would give 9.4% or 10.6% — an underestimate. Both recombinant phenotype classes must be added before dividing by total progeny.
Similar NEET-style question
In a cross between AaBb (cis configuration, both dominant alleles on the same chromosome) and aabb, 1200 offspring are produced: 480 AaBb, 480 aabb, 120 Aabb, 120 aaBb. What is the map distance between the two loci? *(Answer: RF = 240/1200 × 100 = 20 cM.)*
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Linkage: genes on same chromosome inherited together. Recombination frequency = map distance (cM). Morgan's experiments with Drosophila. Sex linkage: colour blindness (X-linked recessive), haemophilia.
-- NCERT Class 12 Biology, Ch. 4, p. 67Proportion of recombinant offspring measures genetic distance between linked loci. Capped at 50% (independent assortment).
| Symbol | Quantity | SI Unit |
|---|---|---|
| RF | recombination frequency | % |
More in Genetics and Evolution: 21 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
3 questions from NEET 2022, 2023. Answers verified against NTA official keys.
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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