Linkage Recombination Mapping

8 MCQs1 revision card9-step worked example
Source: NCERT Structural Organisation in AnimalsPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

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 80) 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.


Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Linkage was experimentally demonstrated for the first time by which scientist, using which organism?

MCQ 2Easy RecallPractice

What is the unit of genetic map distance, and what does 1 unit represent?

MCQ 3Easy RecallPractice

During which sub-stage of meiosis does crossing over occur between homologous chromosomes?

MCQ 4Direct ApplicationPractice

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?

MCQ 5Direct ApplicationPractice

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?

MCQ 6Direct ApplicationPractice

If two genes show a recombination frequency of 50% in a test cross, what can be concluded?

MCQ 7Concept TrapPractice

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?

MCQ 8CalculationPractice

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.

Quick recall before you leave

Worked Example

  1. 1

    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

  2. 2

    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.

  3. 3

    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.

  4. 4

    Formula

    RF = (recombinants / total progeny) × 100

  5. 5

    Substitution

    Total recombinants = 75 + 85 = 160 Total progeny = 800 RF = (160 / 800) × 100

  6. 6

    Calculation

    RF = 0.20 × 100 = 20% Note: 800 and 160 are exact counting integers and do not affect significant-figure considerations.

  7. 7

    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%).

  8. 8

    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.

  9. 9

    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.)* ---

Before solving, remember these

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. 80

Formulas

Mendel's monohybrid ratio

F2 ratio in monohybrid cross — products of independent assortment of two alleles per locus.

SymbolQuantitySI Unit
ratioF2 progeny ratio-

Valid when

  • Single gene with complete dominance
  • Pure-bred parents

Mendel's dihybrid ratio

F2 ratio in dihybrid cross with two independently-segregating loci, complete dominance, no linkage.

SymbolQuantitySI Unit
ratioF2 phenotype ratio-

Valid when

  • Two unlinked loci
  • Complete dominance

Hardy-Weinberg equation

In an idealised population (no mutation, drift, selection, gene flow, random mating), allele and genotype frequencies remain constant.

SymbolQuantitySI Unit
pfreq of dominant allele A-
qfreq of recessive allele a-

Valid when

  • Idealised population
  • All five conditions met
  • Diploid, autosomal, biallelic locus

Recombination frequency (genetic mapping)

Proportion of recombinant offspring measures genetic distance between linked loci. Capped at 50% (independent assortment).

SymbolQuantitySI Unit
RFrecombination frequency%

Valid when

  • Linked loci on same chromosome

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Category: Sign Convention

DNA polymerase synthesises only 5'→3'. Leading strand: continuous, same direction as fork. Lagging: discontinuous (Okazaki), opposite to fork.

When it triggers

Question on Okazaki, leading vs lagging, primer direction.

How to avoid

Reading template 3'→5'; synthesising 5'→3'. Lagging strand needs short fragments because it can't run continuously against fork direction.

Category: Negative Marking

Five forces disturb HW: mutation, gene flow, drift, selection, non-random mating. ANY of these violates equilibrium.

When it triggers

Question asks which factor maintains/disturbs HW.

How to avoid

Random mating + no other forces → equilibrium. ANY of mutation/migration/drift/selection/assortative mating → disequilibrium.

Category: Similar Terms

Monohybrid: genotype 1:2:1 (AA:Aa:aa); phenotype 3:1.

When it triggers

Question asks for one ratio while presenting cross details.

How to avoid

Always note dominance: phenotype merges Aa + AA; genotype keeps them separate.

Category: Similar Terms

X-linked recessive (haemophilia, colour-blindness): affects males predominantly; carrier mother → 50% sons affected; affected father → all daughters carriers but not affected.

When it triggers

Pedigree question; carrier vs affected.

How to avoid

Sex chromosomes: XX vs XY. Recessive on X needs both copies (XaXa) in female, only one (XaY) in male.

Category: Similar Terms

Allopatric: geographic isolation. Sympatric: same area, no physical barrier (e.g. polyploidy in plants, host-shift).

When it triggers

Question gives speciation scenario and asks which type.

How to avoid

If geographic barrier mentioned → allopatric. If population overlaps → sympatric.

Past Year Questions

51 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Identify the statement that is NOT correct.

1Constant region of heavy and light chains are located at C-terminus of antibody molecules
2Each antibody has two light and two heavy chains.
3The heavy and light chains are held together by disulfide bonds.
4Antigen binding site is located at C-terminal region of antibody molecules.
NTA Answer: Option 4(final)
NEET 2025

Given below are two statements : Statement I : In the RNA world, RNA is considered the first genetic material evolved to carry out essential life processes. RNA acts as a genetic material and also as a catalyst for some important biochemical reactions in living systems. Being reactive, RNA is unstable. Statement II : DNA evolved from RNA and is a more stable genetic material. Its double helical strands being complementary, resist changes by evolving repairing mechanism. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is incorrect but statement II is correct
2Both statement I and statement II are correct
3Both statement I and statement II are incorrect
4Statement I is correct but statement II is incorrect
NTA Answer: Option 2(final)
NEET 2024

Which one of the following can be explained on the basis of Mendel's Law of Dominance? A. Out of one pair of factors one is dominant and the other is recessive. B. Alleles do not show any expression and both the characters appear as such in F generation. 2 C. Factors occur in pairs in normal diploid plants. D. The discrete unit controlling a particular character is called factor. E. The expression of only one of the parental characters is found in a monohybrid cross. Choose the correct answer from the options given below:

1A, B and C only
2A, C, D and E only
3B, C and D only
4A, B, C, D and E
NTA Answer: Option 2(final)
NEET 2024

Which of the following statement is correct regarding the process of replication in E.coli?

1The DNA dependent DNA polymerase catalyses polymerization in one direction that is 3’ → 5’
2The DNA dependent RNA polymerase catalyses polymerization in one direction, that is 5’ → 3’
3The DNA dependent DNA polymerase catalyses polymerization in 5’ → 3’ as well as 3’ → 5’ direction
4The DNA dependent DNA polymerase catalyses polymerization in 5’ → 3’ direction
NTA Answer: Option 4(final)
NEET 2023

The phenomenon of pleiotropism refers to

1Presence of two alleles, each of the two genes controlling a single trait
2A single gene affecting multiple phenotypic expression
3More than two genes affecting a single character
4Presence of several alleles of a single gene controlling a single crossover
NTA Answer: Option 2(final)
NEET 2023

Expressed Sequence Tags (ESTs) refers to

1All genes that are expressed as proteins.
2All genes whether expressed or unexpressed.
3Certain important expressed genes.
4All genes that are expressed as RNA.
NTA Answer: Option 4(final)
NEET 2023

Which of the following statements are correct about Klinefelter’s Syndrome? A. This disorder was first described by Langdon Down (1866). B. Such an individual has overall masculine development. However, the feminine developement is also expressed. C. The affected individual is short statured. D. Physical, psychomotor and mental development is retarded. E. Such individuals are sterile. Choose the correct answer from the options given below:

1C and D only
2B and E only
3A and E only
4A and B only
NTA Answer: Option 2(final)
NEET 2022

Given below are two statements : Statement I : Mendel studied seven pairs of contrasting traits in pea plants and proposed the Laws of Inheritance. Statement II : Seven characters examined by Mendel in his experiment on pea plants were seed shape and colour, flower colour, pod shape and colour, flower position and stem height. In the light of the above statements, choose the correct answer from the options given below :

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 2(final)
NEET 2022

Statements related to human Insulin are given below. Which statement(s) is/are correct about genetically engineered Insulin? (a) Pro-hormone insulin contain extra stretch of C-peptide (b) A-peptide and B-peptide chains of insulin were produced separately in E.coli, extracted and combined by creating disulphide bond between them. (c) Insulin used for treating Diabetes was extracted from Cattles and Pigs. (d) Pro-hormone Insulin needs to be processed for converting into a mature and functional hormone. (e) Some patients develop allergic reactions to the foreign insulin. Choose the most appropria

1(c), (d) and (e) only
2(a), (b) and (d) only
3(b) only
4(c) and (d) only
NTA Answer: Option 3(final)
NEET 2021

Match List-I with List-II. (2) (a)-Replication; (b)-Transcription; (c)-Transduction; (d)-Protein (3) (a)-Translation; (b)-Replication; (c)-Transcription;(d)-Transduction (4) (a)-Replication; (b)-Transcription; (c)-Translation; (d)-Protein Answer (4) Choose the correct answer from the options given 127. Which of the following are not secondary below. metabolites in plants? (a) (b) (c) (d)

1Rubber, gums (1) (ii) (i) (iv) (iii)
2Morphine, codeine (2) (ii) (iv) (i) (iii)
3Amino acids, glucose (3) (iv) (iii) (ii) (i)
4(iii) (i) (iv) (ii) (4) Vinblastin, curcumin
NTA Answer: Option 2(final)
NEET 2020

Which of the following refer to correct example(s) of organisms which have evolved due to changes 97. ÁŸêŸ ◊ ∑§ÊÒŸ, ∞ ‚ ¡ËflÊ ∑ § ‚„Ë ©ŒÊ„⁄UáÊÊ ∑§Ê ‚ ŒÁ÷¸Ã ∑§⁄UÃÊ „Ò in environment brought about by anthropogenic ¡Ê ◊ÊŸfl ∑§Ë Á∑˝§ÿÊ•Ê mÊ⁄UÊ flÊÃÊfl⁄UáÊ ◊ ’Œ‹Êfl ∑ § ∑§Ê⁄UáÊ action ? Áfl∑§Á‚à „È∞ „Ò? (a) Darwin’s Finches of Galapagos islands. (a) ªÒ‹Ê¬ÒªÊ mˬ ◊ «UÊÁfl¸Ÿ ∑§Ë Á» §ø (b) Herbicide resistant weeds. (b) π⁄U¬ÃflÊ⁄UÊ ◊ ‡ÊÊ∑§ŸÊ‡ÊË ∑§Ê ¬˝ÁÃ⁄UÊ œ (c) Drug resistant eukaryotes. (c) ‚‚Ë◊∑ §ãŒ˝∑§Ê ◊ ŒflÊßÿÊ ∑§Ê ¬˝ÁÃ⁄UÊ œ (d) Man-created breeds of domesticated animal

1∑ §fl‹ (d) (1) only (d)
2∑ §fl‹ (a) (2) only (a)
3(a) ∞fl (c) (3) (a) and (c)
4(b), (c) ∞fl (d) (4) (b), (c) and (d) Hindi+English 23 H3
NTA Answer: Option 4(final)
NEET 2020

•ŸÈ‹ πŸ ∑ § ‚◊ÿ «UË.∞Ÿ.∞. ∑§Ë ∑È §«U‹Ë ∑§Ê πÊ ‹Ÿ ◊ ∑§ÊÒŸ‚Ê

1RNA polymerase ∞ ¡Êß◊ ◊ŒŒ ∑§⁄UÃÊ „Ò?
2DNA ligase (1) •Ê⁄U.∞Ÿ.∞. ¬ÊÚÁ‹◊⁄ U$¡
3DNA helicase
4DNA polymerase (2) «UË.∞Ÿ.∞. ‹Êߪ $¡ (3) «UË.∞Ÿ.∞. „Ò‹Ë∑ §$¡ 117. Snow-blindness in Antarctic region is due to : (4) «UË.∞Ÿ.∞. ¬ÊÚ‹Ë◊⁄ U$¡ (1) Damage to retina caused by infra-red rays (2) Freezing of fluids in the eye by low
NTA Answer: Option 1(final)
NEET 2020

¡ËŸ ‘I’ ¡Ê ABO ⁄UÄà flª¸ ∑§Ê ÁŸÿ òÊáÊ ∑§⁄UÃÊ „Ò ©‚∑ § ‚ Œ÷¸ 121. Identify the wrong statement with reference to the gene ‘I’ that controls ABO blood groups. ◊ ª‹Ã ∑§ÕŸ ∑§Ê ¬„øÊÁŸ∞–

1Allele ‘i’ does not produce any sugar. (1) ‘i’ ∞ ‹Ë‹ ∑§Ê ߸ ÷Ë ‡Ê∑¸§⁄UÊ ©à¬ÛÊ Ÿ„Ë ∑§⁄UÃÊ–
2The gene (I) has three alleles. (2) ¡ËŸ (I) ∑ § ÃËŸ ∞ ‹Ë‹ „Ê Ã „Ò –
3A person will have only two of the three (3) ∞∑§ √ÿÁÄà ◊ ÃËŸ ◊ ‚ ∑ §fl‹ ŒÊ ∞ ‹Ë‹ „Ê ª – alleles.
4¡’ IA ∞fl IB ŒÊ ŸÊ ß∑§_ „Ê Ã „Ò , ÿ ∞∑§ ¬˝∑§Ê⁄U ∑§Ë (4) When IA and IB are present together, they ‡Ê∑¸§⁄UÊ •Á÷√ÿÄà ∑§⁄Uà „Ò – express same type of sugar.
NTA Answer: Option 4(final)
NEET 2020

≈˛UÊ ‚‹ ‡ÊŸ (•ŸÈflÊŒŸ/SÕÊŸÊ Ã⁄UáÊ) ∑§Ë ¬˝Õ◊ •flSÕÊ ∑§ÊÒŸ ‚Ë 123. The first phase of translation is : „Ê ÃË „Ò?

1Recognition of an anti-codon (1) ∞∑§ ∞ ≈UË-∑§Ê «UÊÚŸ ∑§Ë ¬„øÊŸ
2Binding of mRNA to ribosome (2) ⁄UÊß’Ê ‚Ê ◊ ‚ mRNA ∑§Ê ’㜟
3Recognition of DNA molecule (3) «UË.∞Ÿ.∞. •áÊÈ ∑§Ë ¬„øÊŸ
4Aminoacylation of tRNA (4) tRNA ∑§Ê ∞ ◊ËŸÊ ∞‚Ë‹ ‡ÊŸ
NTA Answer: Option 4(final)
NEET 2020

◊ «U‹ Ÿ Sflà òÊ M§¬ ‚ ¬˝¡ŸŸ ∑§⁄UŸ flÊ‹Ë ◊≈U⁄U ∑ § ¬ÊÒœ ∑§Ë 130. How many true breeding pea plant varieties did Mendel select as pairs, which were similar except Á∑§ÃŸË Á∑§S◊Ê ∑§Ê ÿÈÇ◊Ê ∑ § M§¬ ◊ øÈŸÊ ¡Ê Áfl¬⁄UËà Áfl‡Ê ∑§Ê in one character with contrasting traits ? flÊ‹ ∞∑§ ‹ˇÊáÊ ∑ § •‹ÊflÊ ∞∑§ ‚◊ÊŸ ÕË?

18 (1) 8
24 (2) 4
32 (3) 2
414 (4) 14 H3 30 Hindi+English
NTA Answer: Option 4(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Sources

NCERT refs: Class 12 Biology Chapter 4, p.80

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