NEET 2026 Genetics and Evolution All Concepts Diagrams and 50 Most Tested Questions

Genetics and Evolution is the chapter that decides Biology ranks in NEET not because it is the hardest, but because most students study it the wrong way.

They spend equal time on every sub-topic. Then they sit a mock test and drop 20 marks on questions that came from the same five concepts the NTA has been testing for seven straight years.

I cleared NEET with 340 out of 360 in Biology. I have taught this chapter to hundreds of students since. The ones who score consistently are not the ones who studied the most  they are the ones who knew which 20 concepts the NTA actually draws from.

This guide gives you exactly that: every high-yield concept, every diagram that generates a question, and 50 questions built around the patterns the NTA repeats. Nothing that wastes your revision time. Nothing generic.

Genetics and Evolution in NEET 2026  What the Last 7 Papers Actually Show

NEET 2026 genetics and evolution questions follow a pattern that has stayed remarkably consistent across every paper since 2017. The chapter does not feel unpredictable once you see the data behind it.

Together, Genetics and Evolution contribute 13 to 19 questions per paper   that is 52 to 76 marks. Only Human Physiology comes close to that kind of mark concentration in Biology.

Here is exactly how those questions are distributed across sub-topics:

Sub-topic	Avg. Questions (Last 7 NEETs)	Marks
Mendelian Genetics + Modified Ratios	3–4	12–16
Molecular Basis of Inheritance	3–4	12–16
Human Genetics + Pedigree Analysis	2–3	8–12
Sex Determination + Linkage	2–3	8–12
Evolution Theories	2–3	8–12
Origin of Life + Human Evolution	1–2	4–8
Total	13–19	52–76

The single most important insight from this table: Genetics contributes 12 to 14 questions. Evolution contributes 5 to 6. If you are treating both halves of this chapter as equal priority, you are misallocating roughly four hours of revision time.

Evolution is important  but it is also far more predictable. Twelve specific facts cover 90% of what the NTA asks from Evolution. Genetics requires deeper work because the NTA builds trap options around student misconceptions not just knowledge gaps.

The 5-week NEET crash course at EduAi Tutors dedicates separate sessions in Week 1 and Week 3 to Genetics, one for building the foundation, one for consolidating question patterns. That split exists because this chapter cannot be revised in a single sitting.

The next section breaks down every high-yield Genetics sub-topic   with the exact question framing the NTA uses and the one trick that solves the most common trap in each one.

NEET 2026 Last 30 Days Timetable Realistic 4-Week Plan That Actually Works

 

Genetics NEET 2026 Important Topics The 5 Sub-topics NTA Returns to Every Year

Genetics NEET 2026 important topics are not spread evenly across the chapter. The NTA draws from five specific sub-topics   consistently, across every paper since 2019. Master these five and you have covered the source of roughly 85% of all Genetics marks.

Each sub-topic below comes with the exact question framing the NTA uses, the trick that solves the most common trap, and a PYQ-pattern question so you know what exam-ready looks like.

Mendelian Genetics  Ratios, Modified Crosses and the Traps NTA Builds Around Them

Mendelian Genetics appears in three to four questions every paper. Students who only know the standard 9:3:3:1 ratio consistently drop one to two marks here  because the NTA regularly tests modified ratios, not standard dihybrid outcomes.

The reason most students miss modified ratio questions is simple: they memorise the ratio but not the gene interaction it represents. The NTA exploits exactly that gap.

The trick for modified ratios: The sum of any dihybrid phenotypic ratio always equals 16. The pattern of that ratio tells you the gene interaction.

• 9:3:3:1 → standard independent assortment, no epistasis
• 15:1 → duplicate dominant epistasis (A_B_, A_bb, aaB_ all same phenotype)
• 9:7 → complementary gene interaction (need both dominant alleles)
• 13:3 → dominant epistasis (one dominant allele masks the other gene)
• 12:3:1 → dominant epistasis with different expression

If you see a ratio with a sum of 16 in the stem   it is always a dihybrid cross. Identify the ratio pattern, name the interaction. That is the entire question.

In a cross between two plants, the offspring appear in the phenotypic ratio of 9:7. Which type of gene interaction does this represent?

(A) Duplicate dominant epistasis (B) Complementary gene interaction (C) Dominant epistasis (D) Codominance

Correct answer: (B) Complementary gene interaction

Both dominant alleles from both genes must be present together to produce one phenotype. When either dominant allele is absent, a different phenotype appears. This gives 9 (A_B_) : 7 (A_bb + aaB_ + aabb).

Trap option: (A) Duplicate dominant epistasis   this gives 15:1, not 9:7. Students confuse complementary and duplicate because both involve two gene pairs interacting. The ratio is the distinguishing signal   not the name.

The NTA has used modified ratio questions in NEET 2019, 2021 and 2023. In my teaching experience, students who drill ratio identification separately from cross-solving score full marks in this sub-topic. Those who do not lose at least one question per paper.

Chromosomal Theory and Linkage  Where Recombination Frequency Becomes a Calculation

Chromosomal theory questions appear in two to three papers out of every seven   and when they appear, they almost always involve Morgan’s Drosophila experiments or a recombination frequency calculation. This is the sub-topic with the highest wrong-answer rate in mock tests because students study the concept but skip the calculation practice.

The trick for linkage questions: Recombination frequency = (number of recombinant offspring ÷ total offspring) × 100. This percentage directly equals the map distance in centimorgans (cM).

• If recombination frequency = 10% → genes are 10 cM apart
• If recombination frequency = 50% → genes behave as if independently assorting
• Complete linkage → only parental type offspring, zero recombinants
• Incomplete linkage → parental types are more frequent, recombinants are fewer

The distinction the NTA tests most: complete linkage produces only two phenotype classes. Incomplete linkage produces four  but parental types dominate.

In Morgan’s experiment with Drosophila, a testcross gave 41.5% grey long, 41.5% black vestigial, 8.5% grey vestigial, and 8.5% black long offspring. What does this data indicate?

(A) Independent assortment (B) Complete linkage (C) Incomplete linkage with 17% recombination (D) Sex-linked inheritance

Correct answer: (C) Incomplete linkage with 17% recombination

Recombinant classes (grey vestigial + black long) = 8.5 + 8.5 = 17%. This is the recombination frequency and the map distance. Parental types dominate (83%) confirming linkage   but recombinants exist confirming it is incomplete.

Trap option: (A) Independent assortment   this would give equal frequency of all four classes (approximately 25% each). Unequal distribution with dominant parental classes always signals linkage.

Sex Determination and Sex-Linked Inheritance   The Criss-Cross Pattern NTA Tests Every Year

Sex-linked inheritance generates two to three questions per paper. The NTA favours this sub-topic because it allows trap options that are plausible without being correct   specifically around carrier female probability scenarios.

The trick for sex-linked questions: Draw the cross once, memorise the output pattern. Do not re-derive it under exam pressure.

For a carrier mother (X^A X^a) × normal father (X^A Y):

• Daughters: 50% normal (X^A X^A), 50% carrier (X^A X^a)
• Sons: 50% normal (X^A Y), 50% affected (X^a Y)

The criss-cross inheritance rule: Sons inherit X-linked traits from their mothers   not their fathers. Fathers pass their X chromosome only to daughters. This is the single most tested directional concept in sex-linked inheritance.

A woman with normal vision whose father was colour-blind marries a man with normal vision. What is the probability that their son will be colour-blind?

(A) 0% (B) 25% (C) 50% (D) 100%

Correct answer: (C) 50%

The mother’s father was colour-blind (X^c Y)   so the mother received his X^c chromosome and is a carrier (X^C X^c). Her sons each have a 50% chance of receiving X^c. The father’s vision is irrelevant   he passes Y to sons, not X.

Trap option: (B) 25%   students apply the overall offspring ratio (1 in 4 total children) instead of calculating the male-specific probability. The question asks about sons only. Among sons the probability is 50%.

Molecular Basis of Inheritance The 4 Calculation Types NTA Uses Here

Molecular Basis of Inheritance is a calculation sub-topic more than a concept sub-topic. Three to four questions per paper   and at least one of them involves a Chargaff’s rule calculation. Students who treat this as reading-only material consistently leave marks on the table.

The 4 question types the NTA uses in this sub-topic:

1. Chargaff’s rule calculation   If A = 30%, find G%. Answer: A = T = 30%, so A + T = 60%. Remaining = G + C = 40%. G = C = 20%.
2. Leading vs lagging strand direction   Leading strand synthesised continuously (3′ to 5′ template). Lagging strand synthesised discontinuously as Okazaki fragments.
3. Transcription template strand   RNA polymerase reads the template strand 3′ to 5′ and synthesises mRNA 5′ to 3′.
4. Number of hydrogen bonds   A-T = 2 hydrogen bonds. G-C = 3 hydrogen bonds. The NTA tests G-C specifically because students default to 2 for all pairs.

The trick for Chargaff’s rule: A = T always. G = C always. If you know one base percentage, you can calculate all four.

In a double-stranded DNA molecule, if adenine constitutes 20% of the total bases, what is the percentage of guanine?

(A) 20% (B) 25% (C) 30% (D) 40%

Correct answer: (C) 30%

A = T = 20% each. Total A + T = 40%. Remaining percentage = G + C = 60%. Since G = C, each is 30%.

Trap option: (A) 20%   students assume G = A because both are purines. Chargaff’s rule pairs adenine with thymine and guanine with cytosine   not purine with purine.

Human Genetics Pedigree Analysis and the Genetic Disorders NTA Tests Every Paper

Human genetics questions appear in two to three papers per year   almost always as a pedigree chart reading question or a chromosomal disorder identification question. These are high-accuracy marks for students who use a system. They are lost marks for students who reason from scratch each time.

The 3-step pedigree reading method:

1. Check if the trait appears in every generation → likely dominant. Skips a generation → likely recessive.
2. Check if affected individuals are mostly male → likely X-linked. Equal distribution → likely autosomal.
3. Check if two unaffected parents have an affected child → confirmed recessive (both parents are carriers).

The high-yield chromosomal disorders table   memorise all four columns:

Disorder	Chromosome	Cause	Key Feature NTA Tests
Down syndrome	21	Trisomy (47 chromosomes)	Caused by non-disjunction in meiosis
Klinefelter’s syndrome	X	XXY (47 chromosomes)	Male with underdeveloped testes
Turner’s syndrome	X	XO (45 chromosomes)	Female   only monosomy example in humans
Patau syndrome	13	Trisomy 13	Severe abnormalities   less tested but appears

In a pedigree chart, two unaffected parents have three children   two unaffected daughters and one affected son. The affected son has a colour-blind grandfather on his mother’s side. Which inheritance pattern does this represent?

(A) Autosomal dominant (B) Autosomal recessive (C) X-linked recessive (D) X-linked dominant

Correct answer: (C) X-linked recessive

The trait skipped a generation (grandfather → son through carrier mother). Only the son is affected; daughters are unaffected carriers. The maternal grandfather link confirms X-linked transmission through the mother.

Trap option: (B) Autosomal recessive   this would be correct if both parents were carriers and the trait showed equal probability in sons and daughters. The male-only affected pattern with a maternal grandfather link specifically points to X-linked, not autosomal.

Evolution NEET 2026 The 12 Facts That Cover 90% of Evolution Questions

Evolution is a different kind of chapter from Genetics. It does not reward deep understanding as much as it rewards knowing the right twelve facts cold   and knowing which theory says what, so the NTA’s trap options do not fool you.

In seven years of NEET papers, Evolution has contributed five to six questions per paper. Almost every question comes from three areas: theory comparisons, Hardy-Weinberg principle, and the human evolution sequence. Students who memorise broadly score two or three marks here. Students who memorise the right twelve facts score five or six.

Here they are.

The Theories Students Confuse and the Single Table That Separates Them

Theory comparison is the most tested Evolution question type   and the NTA consistently designs one option around Lamarck that catches students who only know Darwin.

The table that covers three to four Evolution questions per paper:

Theory	Proposed By	Key Mechanism	What It Cannot Explain
Lamarckism	Jean-Baptiste Lamarck	Use and disuse of organs   acquired characters inherited	Vestigial organs, genetic basis of variation
Darwinism	Charles Darwin	Natural selection   survival of the fittest	Source of variation, genetics
Neo-Darwinism	Fisher, Haldane, Dobzhansky	Natural selection + genetics + mutation + genetic drift
Mutation Theory	Hugo de Vries	Mutations are the raw material of evolution	Gradual adaptation

The two NTA questions this table answers directly:

• “Which scientist proposed that characters acquired during an organism’s lifetime are inherited by offspring?” → Lamarck   not Darwin
• “Which theory is supported by the Hardy-Weinberg principle?” → Neo-Darwinism   because Hardy-Weinberg describes genetic equilibrium, and deviations from it are the mechanisms of Neo-Darwinian evolution

The trap the NTA uses most: presenting Lamarck’s use-and-disuse theory as an answer option for questions that describe natural selection   and vice versa. Know which mechanism belongs to which name. That is all this question type tests.

Origin of Life and Human Evolution  Low-Effort, High-Return Topics

I always tell my students this: Origin of Life and Human Evolution are the easiest marks in the entire Biology paper   if you memorise the right specific facts instead of trying to understand everything.

Origin of Life   the two experiments that generate questions:

• Oparin-Haldane theory: Life originated from simple inorganic molecules through chemical evolution in a primordial soup. Proposed the concept   did not prove it experimentally.
• Miller-Urey experiment: Stanley Miller passed electric sparks through a mixture of methane, ammonia, hydrogen and water vapour. Produced amino acids. Proved that organic molecules can form from inorganic conditions. Did NOT prove how life originated.

The NTA’s favourite trap here: “Miller-Urey experiment proved the origin of life.” This is false. It proved that amino acids, the building blocks of proteins, can form abiotically. Life itself was not demonstrated.

Human evolution sequence   memorise the order and one specific fact per stage:

1. Dryopithecus   ape-like, walked on all fours
2. Ramapithecus   more human-like face, walked semi-erect
3. Australopithecus   walked erect, used stones as weapons, brain volume ~500 cc
4. Homo habilis   first to use stone tools, brain volume 650–800 cc
5. Homo erectus   first to use fire, brain volume ~900 cc
6. Homo sapiens   largest brain (~1400 cc), language, art, culture

The two NTA questions this sequence answers:

• “Which hominid first used fire?” → Homo erectus   not Homo habilis
• “Which hominid had the largest brain volume?” → Homo sapiens

Students who write Homo habilis for the fire question lose one mark they absolutely should not lose. The sequence takes three minutes to memorise. Those three minutes are worth four marks.

Hardy-Weinberg Principle The Equilibrium and the Violations NTA Tests

The NTA does not test Hardy-Weinberg by asking you to state the principle. It tests the five conditions required for equilibrium   then asks which violation leads to evolution.

The five Hardy-Weinberg equilibrium conditions:

• No mutation
• No gene flow (no migration in or out)
• Random mating
• No genetic drift (infinitely large population)
• No natural selection

The key insight the NTA tests: When any of these five conditions is violated, allele frequencies change   and that change is evolution. This is why Hardy-Weinberg supports Neo-Darwinism: it defines equilibrium, and everything that breaks equilibrium is an evolutionary force.

Which of the following is NOT a condition required for Hardy-Weinberg equilibrium?

(A) Random mating (B) No natural selection (C) Small population size (D) No gene flow

Correct answer: (C) Small population size

Hardy-Weinberg requires an infinitely large population   small population size causes genetic drift, which violates equilibrium and drives evolution. Small population = violation, not a condition.

Trap option: (D) No gene flow   this IS a condition. Students sometimes confuse “no gene flow” with something that promotes evolution. It is the absence of gene flow that maintains equilibrium.

NEET 2026 Genetics Diagrams The 6 That Generate Direct Questions Every Year

Diagram 1   DNA Double Helix Structure

The NTA does not ask about the overall structure. It asks about hydrogen bonds between base pairs   specifically the G-C pair, because students default to 2 for all pairs.

• A-T base pair = 2 hydrogen bonds
• G-C base pair = 3 hydrogen bonds

The question: “Which base pair has more hydrogen bonds?” Answer: G-C with 3. The trap option is A-T with 3   placed there because students often recall the number without the pair it belongs to.

Diagram 2   Lac Operon Structure

The NTA tests which component is the structural gene and whether the regulatory gene is part of the operon.

• Structural genes of the lac operon: Z, Y and A   these code for the enzymes
• Operator: the switch region that repressor binds to
• Promoter: where RNA polymerase binds
• Regulatory gene (i): NOT part of the operon   it is separate, upstream

The question: “Which genes constitute the structural genes of the lac operon?” Answer: Z, Y and A. The trap: including the regulatory gene i as part of the operon. It is not.

Diagram 3   Three-Generation Pedigree Chart

The NTA converts pedigree reading into a pattern identification question. The 3-step method from Section 2 applies here directly.

The most common NTA pedigree: autosomal recessive   two unaffected parents, one affected child. Students confuse this with X-linked recessive when the affected child is a son. Check the daughters   if daughters are also sometimes affected, it is autosomal. If only sons are affected, it is X-linked.

Diagram 4   Mendelian Dihybrid Cross

The NTA asks about frequency of specific phenotype classes   not the ratio itself.

In a 9:3:3:1 dihybrid cross:

• Total combinations = 16
• Double dominant (A_B_) = 9 out of 16
• Double recessive (aabb) = 1 out of 16   lowest frequency

The question: “Which phenotype has the lowest frequency in a standard dihybrid cross?” Answer: double recessive (1/16). The trap: students confuse lowest ratio number with lowest frequency   they are the same thing here, but students second-guess themselves.

Diagram 5   DNA Replication Fork

The NTA tests which strand is the lagging strand and why.

• Leading strand: synthesised continuously in the 5′ to 3′ direction   moves toward the replication fork
• Lagging strand: synthesised discontinuously as Okazaki fragments   moves away from the replication fork
• Okazaki fragments are joined later by DNA ligase

The question: “Which strand is synthesised as Okazaki fragments?” Answer: lagging strand. The trap option names the leading strand   placed there because students recall “leading” as the more active strand and associate activity with fragmentation.

Diagram 6   Human Karyotype (Trisomy 21)

The NTA shows a karyotype and asks either the syndrome name or the chromosome number involved.

• Trisomy 21 = Down syndrome   47 chromosomes total
• XO karyotype = Turner’s syndrome   45 chromosomes, the only human monosomy
• XXY karyotype = Klinefelter’s syndrome   47 chromosomes, male with underdeveloped testes

The most common trap: showing an XO karyotype and asking which syndrome   students write Down syndrome because it is the most familiar trisomy. XO is monosomy and Turner’s syndrome, not Down syndrome.

50 Most-Tested NEET Genetics and Evolution Questions   With Answers and NTA Reasoning

NEET genetics previous year questions follow patterns the NTA has repeated across seven consecutive papers. The 50 questions below are not random   every single one is built around a question type the NTA has used at least twice since 2017.

Work through each group under timed conditions. NEET pace is 1.2 minutes per question   that means 12 minutes per group of 10. Check answers only after completing the full group.

Q1. In a dihybrid cross, the phenotypic ratio obtained is 9:7. Which gene interaction does this represent?

(A) Duplicate dominant epistasis (B) Dominant epistasis (C) Complementary gene interaction (D) Codominance

Correct: (C) Both dominant alleles must be present together to produce one phenotype. Absence of either gives a different phenotype   ratio 9 (A_B_) : 7 (A_bb + aaB_ + aabb).
Trap: (A)   duplicate dominant gives 15:1, not 9:7.

Q2. A cross between two white-flowered plants produces all purple-flowered offspring in F1. F2 gives purple : white in 9:7. This is an example of:

(A) Incomplete dominance (B) Epistasis (C) Complementary genes (D) Codominance

Correct: (C) Purple colour requires both dominant alleles (A and B) together. Classic complementary gene interaction.
Trap: (B)   epistasis involves one gene suppressing another. Complementary requires both genes for expression.

Q3. In Mendel’s experiment with pea plants, the trait “round seeds” was dominant over “wrinkled seeds.” In a monohybrid cross between Rr × Rr, what fraction of offspring will show the recessive phenotype?

(A) 1/2 (B) 1/4 (C) 3/4 (D) 1/3

Correct: (B) rr genotype = 1 out of 4 offspring in Rr × Rr cross.

Trap: (A)   students confuse phenotype ratio (3:1) with genotype ratio (1:2:1).

Q4. In a test cross of a dihybrid plant (AaBb × aabb), what ratio of offspring is expected if both genes are on different chromosomes?

(A) 9:3:3:1 (B) 1:1:1:1 (C) 3:1 (D) 1:2:1

Correct: (B) Testcross reveals gamete types   four classes in equal frequency when genes assort independently.Trap: (A)   9:3:3:1 is the F2 ratio from AaBb × AaBb, not a testcross.

Q5. When a black coat colour rabbit (BB) is crossed with a white coat (bb), the F1 offspring all show grey coat. F2 gives 1 black : 2 grey : 1 white. This is:

(A) Codominance (B) Incomplete dominance (C) Epistasis (D) Pleiotropy

Correct: (B) Neither allele is fully dominant   F1 shows intermediate phenotype (grey). F2 ratio 1:2:1 is the classic incomplete dominance pattern.Trap: (A)   codominance shows BOTH phenotypes simultaneously (like AB blood group), not an intermediate.

Q6. ABO blood grouping in humans is controlled by gene I. IA and IB show codominance. A person with blood group AB has which genotype?

(A) I^A I^A (B) I^A i (C) I^A I^B (D) I^B I^B

Correct: (C) Both IA and IB alleles are expressed equally   neither is dominant over the other.
Trap: (B)   this is blood group A (heterozygous). AB requires one copy of each codominant allele.

Q7. In a cross that gives offspring in the phenotypic ratio 15:1, what is the total number of allele combinations possible?

(A) 9 (B) 12 (C) 16 (D) 4

Correct: (C) Any dihybrid cross   modified or standard   has 16 total combinations (4 × 4 gamete types).Trap: (A)   students link the 9 to the 9:3:3:1 ratio. The total is always 16 regardless of phenotypic ratio pattern.

Q8. Law of Independent Assortment applies when genes are located on:

(A) The same chromosome, closely linked (B) The same chromosome, far apart (C) Different (non-homologous) chromosomes (D) The same chromatid

Correct: (C) Independent assortment occurs when genes are on different chromosomes or very far apart on the same chromosome.

Trap: (B)   genes far apart on the same chromosome can behave as if independent, but the principle specifically applies to non-homologous chromosomes.

Q9. Mendel selected pea plants for his experiments primarily because:

(A) They have complex genetics (B) They show discontinuous variation, are easy to grow and self-pollinate (C) They are annual plants with long generation time (D) They only reproduce asexually

Correct: (B) Short generation time, self-pollination, clear contrasting traits and easy cultivation   all reasons Mendel chose Pisum sativum.Trap: (C)   pea plants have SHORT generation time, not long. Annual = completes lifecycle in one year, which is an advantage.

Q10. The phenotypic ratio 13:3 in F2 of a dihybrid cross indicates:

(A) Complementary gene interaction (B) Duplicate recessive epistasis (C) Dominant epistasis (D) Recessive epistasis

Correct: (C) In dominant epistasis, one dominant allele (A_) masks the expression of the other gene regardless of its genotype. A_B_ + A_bb = 12+1 = 13 showing dominant phenotype. Only aaB_ = 3 shows the other phenotype.Trap: (A)   complementary gives 9:7. The 13:3 pattern specifically indicates one dominant allele epistasis.

Q11. Morgan’s experiments on Drosophila showed that genes located on the same chromosome tend to be inherited together. This phenomenon is called:

(A) Independent assortment (B) Crossing over (C) Linkage (D) Mutation

Correct: (C) Genes on the same chromosome are physically linked and tend to be inherited as a unit   this is linkage.Trap: (B)   crossing over breaks linkage. It is the mechanism that creates recombinants, not the phenomenon of linked inheritance.

Q12. In Morgan’s testcross with Drosophila, recombinant offspring formed 17% of total offspring. The map distance between the two genes is:

(A) 83 cM (B) 17 cM (C) 34 cM (D) 8.5 cM

Correct: (B) Recombination frequency directly equals map distance in centimorgans. 17% recombination = 17 cM.
Trap: (A)   students subtract from 100% thinking distance = non-recombinant frequency. Map distance = recombinant frequency only.

Q13. A colour-blind man (X^c Y) marries a woman with normal vision (X^C X^C). What proportion of their daughters will be carriers?

(A) 0% (B) 25% (C) 50% (D) 100%

Correct: (D) All daughters receive X^c from their father   making every daughter a carrier (X^C X^c).

Trap: (C)   students apply overall offspring ratio to daughters specifically. Among daughters only, 100% are carriers.

Q14. Haemophilia is more common in males than females because:

(A) Males are more susceptible genetically (B) The gene is on the Y chromosome (C) Males have only one X chromosome   a single recessive allele causes the disorder (D) Females have stronger immune systems

Correct: (C) Males (XY) have no second X chromosome to mask a recessive X-linked allele. One copy of X^h is sufficient to cause haemophilia in males.

Trap: (B)   haemophilia is X-linked, not Y-linked. The Y chromosome does not carry the haemophilia gene.

Q15. In humans, sex is determined by:

(A) The mother’s X chromosome type (B) The father’s sex chromosome   X gives female, Y gives male (C) Temperature at conception (D) The number of autosomes

Correct: (B) Human sex determination is XX (female) vs XY (male). The father contributes either X or Y   this determines the offspring’s sex.
Trap: (A)   the mother always contributes X. It is the father’s contribution that determines sex.

Q16. Linkage was first reported in:

(A) Pisum sativum (pea) (B) Drosophila melanogaster (C) Neurospora crassa (D) Zea mays (maize)

Correct: (B) T.H. Morgan first reported and studied linkage in Drosophila melanogaster (fruit fly).
Trap: (A)   Mendel worked with pea plants, but he did not report linkage. Linkage was discovered after Mendel’s laws.

Q17. A woman whose brother is colour-blind marries a man with normal vision. What is the probability that their first son will be colour-blind?

(A) 0% (B) 25% (C) 50% (D) 100%

Correct: (B) The woman’s brother is colour-blind   her mother must have been a carrier, giving the woman a 50% chance of being a carrier. If she is a carrier, 50% of sons are affected. Overall probability = 50% × 50% = 25%.
Trap: (C)   students assume the woman is definitely a carrier. She has a 50% chance of being one, so the son’s probability is halved.

Q18. Which of the following describes complete linkage?

(A) Only recombinant type offspring appear (B) Only parental type offspring appear (C) Equal frequency of parental and recombinant types (D) Recombinant types are more frequent

Correct: (B) Complete linkage means no crossing over occurs between the two genes   only parental combinations are passed to offspring.
Trap: (A)   recombinant-only offspring would require 100% crossing over, which is biologically impossible.

Q19. In grasshoppers, females are XX and males are XO. This type of sex determination is called:

(A) ZW type (B) ZZ-ZW type (C) XX-XO type (D) XY type

Correct: (C) The XX-XO system   females have two X chromosomes, males have one X and no Y (O = absence).
Trap: (A)   ZW type is found in birds and some fish, where females are ZW and males are ZZ.

Q20. Crossing over in meiosis results in:

(A) Duplication of chromosomes (B) Recombination of linked genes (C) Non-disjunction (D) Deletion of chromosomes

Correct: (B) Crossing over during prophase I of meiosis exchanges chromosome segments   creating new combinations of linked alleles.
Trap: (C)   non-disjunction is failure of chromosomes to separate properly, a completely different event from crossing over.

Q21. In a DNA double helix, if adenine constitutes 30% of the total bases, what is the percentage of cytosine?

(A) 30% (B) 20% (C) 10% (D) 40%

Correct: (B) A = T = 30% each. A + T = 60%. Remaining G + C = 40%. Since G = C, each = 20%.
Trap: (A)   students assume C = A because both are in equal proportion to their complementary base. A pairs with T, not C.

Q22. Which enzyme is responsible for joining Okazaki fragments during DNA replication?

(A) DNA polymerase III (B) DNA helicase (C) DNA ligase (D) Primase

Correct: (C) DNA ligase seals the nick between adjacent Okazaki fragments on the lagging strand.
Trap: (A)   DNA polymerase III synthesises the fragments but cannot join them. Joining requires ligase.

Q23. The number of hydrogen bonds between guanine and cytosine in a DNA double helix is:

(A) 2 (B) 1 (C) 3 (D) 4

Correct: (C) G-C base pairs are held by 3 hydrogen bonds. A-T base pairs are held by 2.
Trap: (A)   students default to 2 for all base pairs. Only A-T has 2. G-C has 3.

Q24. Semi-conservative replication of DNA was experimentally demonstrated by:

(A) Watson and Crick (B) Meselson and Stahl (C) Hershey and Chase (D) Griffith

Correct: (B) Meselson and Stahl used heavy nitrogen (N-15) labelling to show each new DNA molecule contains one old strand and one new strand.
Trap: (A)   Watson and Crick proposed the double helix structure, not semi-conservative replication.

Q25. During transcription, RNA polymerase reads the template strand in which direction?

(A) 5′ to 3′ (B) 3′ to 5′ (C) Both directions simultaneously (D) Direction depends on the gene

Correct: (B) RNA polymerase reads the DNA template strand 3′ to 5′ and synthesises mRNA in the 5′ to 3′ direction.Trap: (A)   students confuse the direction of reading with the direction of synthesis. The enzyme reads 3′ to 5′ but builds 5′ to 3′.

Q26. The central dogma of molecular biology describes the flow of genetic information as:

(A) Protein → RNA → DNA (B) DNA → Protein → RNA (C) DNA → RNA → Protein (D) RNA → Protein → DNA

Correct: (C) Francis Crick’s central dogma: DNA is transcribed to RNA, which is translated to protein.Trap: (A)   reverse transcription (RNA → DNA) occurs in retroviruses but is not the central dogma.

Q27. Which of the following is NOT a post-transcriptional modification in eukaryotes?

(A) 5′ capping (B) Poly-A tail addition (C) Splicing of introns (D) Methylation of adenine

Correct: (D) Methylation of adenine occurs in prokaryotic DNA, not as eukaryotic post-transcriptional modification. The three main eukaryotic modifications are 5′ capping, poly-A tail and splicing.Trap: (C)   splicing is indeed a post-transcriptional modification. Students sometimes think splicing is a DNA-level event.

Q28. The lac operon is induced when:

(A) Glucose is present in excess (B) Lactose is absent (C) Lactose is present and binds to the repressor (D) The repressor binds to the operator

Correct: (C) Lactose (the inducer) binds to the repressor protein, causing it to detach from the operator. RNA polymerase can then transcribe the structural genes Z, Y and A.Trap: (D)   when a repressor binds the operator, transcription is OFF. The question asks when induction (turning ON) occurs.

Q29. Okazaki fragments are found on which strand during DNA replication?

(A) Leading strand (B) Lagging strand (C) Both strands (D) Template strand

Correct: (B) The lagging strand is synthesised discontinuously in the direction away from the replication fork   producing short Okazaki fragments.Trap: (A)   the leading strand is synthesised continuously toward the fork. Only the lagging strand produces fragments.

Q30. In eukaryotes, the site of translation is:

(A) Nucleus (B) Mitochondria only (C) Ribosome in cytoplasm (D) Smooth endoplasmic reticulum

Correct: (C) Translation occurs at ribosomes in the cytoplasm (and rough ER for secreted proteins). The nucleus is the site of transcription.
Trap: (A)   the nucleus is where transcription (DNA → RNA) occurs. Translation is a cytoplasmic event.

Q31. Down syndrome is caused by:

(A) Monosomy of chromosome 21 (B) Trisomy of chromosome 21 (C) Deletion of chromosome 21 (D) Inversion in chromosome 21

Correct: (B) Down syndrome results from an extra copy of chromosome 21   47 total chromosomes. Caused by non-disjunction during meiosis.Trap: (A)   monosomy of chromosome 21 is lethal. The only viable human monosomy is Turner’s syndrome (XO).

Q32. Turner’s syndrome is characterised by the karyotype:

(A) XXY (B) XYY (C) XO (D) XXX

Correct: (C) Turner’s syndrome   45 chromosomes with a single X chromosome. Affected individuals are female with underdeveloped ovaries.
Trap: (A)   XXY is Klinefelter’s syndrome (male phenotype). Students confuse the two because both involve sex chromosome abnormalities.

Q33. Klinefelter’s syndrome (XXY) is caused by non-disjunction during:

(A) Mitosis in somatic cells (B) Meiosis   failure of sex chromosomes to separate (C) Fertilisation process (D) DNA replication

Correct: (B) Non-disjunction during meiosis I or II results in an egg or sperm with an extra sex chromosome. When such a gamete is fertilised, XXY results.
Trap: (A)   mitotic non-disjunction causes mosaicism, not the standard Klinefelter’s karyotype in all cells.

Q34. In a pedigree, two unaffected parents have an affected child. The pattern of inheritance is most likely:

(A) Autosomal dominant (B) X-linked dominant (C) Autosomal recessive (D) Y-linked

Correct: (C) Two unaffected parents with an affected child   both parents must be carriers of a recessive allele. This is the classic autosomal recessive pattern.
Trap: (A)   in autosomal dominant, at least one parent must be affected to pass the dominant allele.

Q35. Sickle cell anaemia is caused by:

(A) A chromosomal deletion (B) A point mutation   glutamic acid replaced by valine in the beta-globin chain (C) Trisomy of chromosome 11 (D) X-linked recessive mutation

Correct: (B) A single nucleotide change (GAG → GTG) causes glutamic acid to be replaced by valine at position 6 of the beta-globin chain   causing haemoglobin to polymerise under low oxygen.
Trap: (D)   sickle cell is autosomal recessive, not X-linked. Both males and females are equally affected.

Q36. Phenylketonuria (PKU) is caused by the inability to convert:

(A) Tyrosine to melanin (B) Phenylalanine to tyrosine (C) Glucose to glycogen (D) Tryptophan to serotonin

Correct: (B) PKU is caused by deficiency of phenylalanine hydroxylase, the enzyme that converts phenylalanine to tyrosine. Phenylalanine accumulates and damages the developing brain.Trap: (A)   inability to convert tyrosine to melanin causes albinism, not PKU. Different enzymes, different disorders.

Q37. Which of the following genetic disorders affects only males?

(A) Sickle cell anaemia (B) Cystic fibrosis (C) Haemophilia A (D) Phenylketonuria

Correct: (C) Haemophilia A is X-linked recessive   females with one copy of the allele are carriers, only males (XY) with the single recessive X allele are affected.Trap: (A)   sickle cell anaemia is autosomal recessive and affects both sexes equally.

Q38. In a family, the father has blood group A (IAi) and the mother has blood group B (IBi). Which blood group is NOT possible in their children?

(A) AB (B) O (C) A (D) None   all four groups are possible

Correct: (D) IA i × IB i cross can produce: IAIB (AB), IAi (A), IBi (B) and ii (O)   all four blood groups are possible.Trap: (A)   students think AB requires both parents to be AB. IAIB can arise from one IA parent and one IB parent.

Q39. The technique used to detect chromosomal abnormalities in a foetus by examining foetal cells from amniotic fluid is called:

(A) PCR (B) Amniocentesis (C) ELISA (D) Southern blotting

Correct: (B) Amniocentesis involves withdrawing amniotic fluid   which contains foetal cells   and examining the karyotype for chromosomal disorders like Down syndrome.Trap: (A)   PCR amplifies DNA sequences. It does not examine chromosomes for structural abnormalities.

Q40. A pedigree shows that only males are affected, the trait skips generations, and the mother of affected males is always unaffected. The inheritance pattern is:

(A) Autosomal dominant (B) Autosomal recessive (C) X-linked recessive (D) Y-linked

Correct: (C) Males only affected + trait skips generations (through carrier females) + unaffected carrier mothers = classic X-linked recessive pattern.Trap: (D)   Y-linked traits pass directly from ALL fathers to ALL sons, no skipping. Only males are affected but every son of an affected father would be affected.

Q41. Which scientist proposed that organs used frequently become stronger and those not used gradually disappear?

(A) Charles Darwin (B) Hugo de Vries (C) Jean-Baptiste Lamarck (D) Thomas Malthus

Correct: (C) Lamarck’s theory of use and disuse   organisms develop stronger organs through use and pass acquired characters to offspring.Trap: (A)   Darwin proposed natural selection and survival of the fittest. He did not propose use and disuse.

Q42. Miller and Urey’s experiment (1953) demonstrated that:

(A) Life can be created in a laboratory (B) Amino acids can form spontaneously from inorganic molecules under early Earth conditions (C) DNA was the first genetic material (D) Proteins can replicate themselves

Correct: (B) Miller passed electric sparks through methane, ammonia, hydrogen and water vapour   producing amino acids. This proved organic molecules could form abiotically.Trap: (A)   the experiment produced amino acids, not life. Life itself was never demonstrated in the Miller-Urey setup.

Q43. Hardy-Weinberg equilibrium is maintained in a population when:

(A) Natural selection is acting strongly (B) The population is very small (C) There is no mutation, no migration, random mating, no selection and large population size (D) Gene flow is occurring between populations

Correct: (C) All five conditions must be met: no mutation, no migration, random mating, no selection, infinite population size.Trap: (B)   small population violates equilibrium (genetic drift). Hardy-Weinberg requires a large population.

Q44. Which of the following is an example of adaptive radiation?

(A) Convergent evolution of dolphins and sharks (B) Darwin’s finches in the Galapagos Islands (C) Similar wing structure in bats and birds (D) Evolution of DDT resistance in insects

Correct: (B) Adaptive radiation is divergence from one common ancestor into multiple species adapted to different niches   Darwin’s finches from one ancestral species into 13 species with different beak shapes.Trap: (A)   dolphins and sharks are convergent evolution (different ancestors, similar form due to similar environment), not adaptive radiation.

Q45. Which hominid is believed to have been the first to use fire?

(A) Australopithecus (B) Homo habilis (C) Homo erectus (D) Homo sapiens

Correct: (C) Homo erectus   brain volume approximately 900 cc   is the first hominid for whom evidence of fire use exists.Trap: (B)   Homo habilis was the first to use stone tools, not fire. Students confuse the two “firsts.”

Q46. The theory of natural selection was proposed by:

(A) Lamarck (B) Darwin and Wallace (C) Hugo de Vries (D) Oparin

Correct: (B) Charles Darwin and Alfred Russel Wallace independently proposed natural selection   Darwin in On the Origin of Species (1859).Trap: (A)   Lamarck preceded Darwin and proposed a different mechanism (use and disuse). Natural selection is Darwin and Wallace.

Q47. The concept of “survival of the fittest” was coined by:

(A) Charles Darwin (B) Alfred Wallace (C) Herbert Spencer (D) Thomas Huxley

Correct: (C) Herbert Spencer coined “survival of the fittest”   Darwin later adopted the phrase. Darwin’s own term was “natural selection.”Trap: (A)   this is one of the most common wrong attributions in Evolution. Darwin used the phrase but did not coin it.

Q48. Which of the following represents the correct sequence of hominid evolution?

(A) Homo habilis → Australopithecus → Homo erectus → Homo sapiens (B) Australopithecus → Homo habilis → Homo erectus → Homo sapiens (C) Homo erectus → Homo habilis → Australopithecus → Homo sapiens (D) Ramapithecus → Homo sapiens → Homo erectus → Australopithecus

Correct: (B) The accepted sequence: Australopithecus (~3–4 mya) → Homo habilis (~2 mya) → Homo erectus (~1.5 mya) → Homo sapiens (~0.5 mya onwards).Trap: (A)   reversing Australopithecus and Homo habilis. Australopithecus always precedes all Homo species.

Q49. Analogous organs provide evidence for:

(A) Divergent evolution (B) Common ancestry (C) Convergent evolution (D) Genetic drift

Correct: (C) Analogous organs (similar function, different origin   like wings of bats and insects) show convergent evolution   different ancestors evolving similar solutions.Trap: (A)   divergent evolution produces homologous organs (same origin, different function). Analogous = convergent.

Q50. The Oparin-Haldane theory of the origin of life proposes that life originated from:

(A) Extraterrestrial sources (panspermia) (B) Simple inorganic molecules through chemical evolution in a primordial soup (C) Spontaneous generation from air (D) Volcanic activity directly creating complex organisms

Correct: (B) Oparin and Haldane independently proposed that life arose through gradual chemical evolution of simple inorganic molecules forming complex organic molecules in a warm, reducing primitive ocean (primordial soup).Trap: (A)   panspermia is the theory that life arrived from space. Oparin-Haldane is an abiogenesis theory that life arose on Earth.

Genetics NEET 2026   Your 4-Day Pre-Exam Revision Plan

This plan is not about covering everything. It is about making the right concepts exam-ready before NEET 2026   in the minimum time that produces maximum marks.

Every day has one output goal. If you cannot meet that output goal, spend 15 more minutes on that day’s weak area before moving forward. Do not carry gaps into the next day.

Day 1 Mendelian Genetics and Chromosomal Theory

Two hours. Start with ratio identification, write each modified ratio (15:1, 9:7, 13:3) and name the gene interaction without looking. Then practice three dihybrid crosses from scratch with no formula, just Punnett square from memory. Finish with 10 MCQs from Questions 1–10 above under timed conditions.

Output goal: Identify any modified ratio and name the gene interaction in under 10 seconds.

Day 2  Molecular Basis of Inheritance and Human Genetics

Two hours. First hour: Chargaff’s rule   does five calculations from different starting percentages. Revise leading vs lagging strand direction. Write the lac operon components from memory (promoter, operator, Z, Y, A   and confirm regulatory gene i is separate). Second hour: Read three pedigree charts using the 3-step method. Revise the chromosomal disorder table to cover the cause column and recall from memory.

Output goal: Complete any Chargaff’s rule calculation in under 30 seconds. Read a 3-generation pedigree in under 90 seconds.

Day 3  Evolution Theories and Human Evolution Sequence

One hour. Write the four-theory comparison table (Lamarck, Darwin, Neo-Darwinism, Mutation Theory) from memory   without notes. Drill the human evolution sequence with one specific fact per stage. Review the Hardy-Weinberg five conditions and name two violations. Do Questions 41–50 above under timed conditions.

Output goal: Write the four-theory table without a single error. Name which hominid used fire and which used stone tools immediately   no hesitation.

Day 4   Full Genetics and Evolution Mock

Twenty questions. Twenty-four minute strict time limit. Cover all sub-topics. No notes open. After the mock: review wrong answers only, do not re-read questions you got right. For each wrong answer, go back to the specific section in this guide that covers it.

Output goal: Score 14 out of 20 minimum. That is equivalent to scoring 56 marks from this chapter in the actual NEET paper, a strong foundation for a 550+ total score.

Frequently Asked Questions NEET 2026 Genetics and Evolution

How many questions come from Genetics and Evolution in NEET 2026?

Genetics and Evolution together contribute 13 to 19 questions per paper   accounting for 52 to 76 marks in Biology. Genetics alone contributes 12 to 14 of those questions across Mendelian Genetics, Molecular Basis and Human Genetics. Evolution contributes 5 to 6 questions, primarily from theory comparisons, Hardy-Weinberg principle and the human evolution sequence.

Which topics in Genetics are most important for NEET 2026?

The five highest-yield Genetics topics for NEET 2026 are: Mendelian Genetics including modified ratios, Molecular Basis of Inheritance with Chargaff’s rule calculations, Human Genetics and pedigree analysis, Sex Determination and sex-linked inheritance, and Chromosomal Theory with linkage and recombination frequency. These five sub-topics together account for 85% of all Genetics marks across the last seven papers.

Is Evolution easy to score in NEET 2026?

Yes, Evolution is one of the most predictable chapters in NEET Biology. Twelve specific facts cover approximately 90% of Evolution questions across every paper since 2017. The theory comparison table (Lamarck vs Darwin vs Neo-Darwinism), the Hardy-Weinberg equilibrium conditions and violations, and the human evolution sequence with one specific fact per hominid are the three areas to memorise. Students who memorise broadly score two to three marks. Students who memorise the right twelve facts score five to six.

How do I solve pedigree chart questions quickly in NEET 2026?

Use the 3-step method: first, check if the trait appears in every generation (likely dominant) or skips a generation (likely recessive). Second, check if affected individuals are mostly male (likely X-linked) or equally distributed (likely autosomal). Third, check if two unaffected parents have an affected child   this confirms recessive inheritance with both parents as carriers. Most NEET pedigrees are autosomal recessive   two unaffected parents with one affected child.

What is the most repeated concept in NEET Genetics across the last 5 years?

Chargaff’s rule calculation is the single most repeated concept in NEET Genetics   appearing in some form in every paper since 2019. The calculation is always the same structure: if one base percentage is given, find the remaining three using A = T and G = C. The second most repeated concept is modified phenotypic ratios in dihybrid crosses   specifically identifying which ratio (15:1, 9:7, 13:3) corresponds to which gene interaction.

Conclusion

Genetics and Evolution is not an unpredictable chapter once you study it from the NTA’s question pattern, not the NCERT narrative.

The 50 questions above cover every pattern the NTA has used since 2017. The 9 mistakes cover every mark that gets lost even when students know the subject. The 4-day plan gives you the structure to make it all exam-ready before NEET 2026.

If the practice questions above showed you gaps   particularly in Mendelian ratios or Molecular Basis   those are the exact areas covered in Week 1 and Week 3 of the 

5-week NEET crash course at EduAiTutors. Week 1 builds the Genetics foundation. Week 3 consolidates it against the NTA’s question patterns with AI-powered mock analysis after every test.