Biology Error Notes — AITS

Principles of Inheritance and Variation 24 · Evolution 9 · Human Reproduction 6 · Reproductive Health 4. On NEET marking these forty-three were worth 172 marks, and the twenty-four wrong attempts cost 24 more.

The broadest set in this series — four chapters — and the attempt rate matches physics and chemistry: twenty-four attempted against nineteen blank. The errors are unusually repetitive: five statement-type questions, five wrong, and four more lost to right content in the wrong arrangement.

Q1 Matching genotypes to pea phenotypes Marked wrong

Match List I (Genotypes) with List II (phenotypes) w.r.t. garden pea plant.

List I	List II
(A) RrYy	(I) Wrinkled Green seed
(B) rrYy	(II) Round Green seed
(C) RRyy	(III) Round Yellow seed
(D) rryy	(IV) Wrinkled Yellow seed

A-IV, B-III, C-II, D-I
KEYA-III, B-IV, C-II, D-I
A-II, B-III, C-IV, D-I
MARKEDA-III, B-IV, C-I, D-II

Given

R = round (dominant), r = wrinkled; Y = yellow (dominant), y = green.
Four genotypes to be matched to four phenotypes.

Asked

The correct matching.

Concept to use

Read each genotype one gene at a time. A capital letter anywhere in a pair gives the dominant phenotype; only a double lowercase pair gives the recessive. Doing shape first and colour second turns each genotype into two independent yes/no decisions.

Formula to use

R present → round, rr → wrinkled | Y present → yellow, yy → green

Baby steps

A. RrYy — R present so round; Y present so yellow → round yellow = III.
B. rrYy — rr so wrinkled; Y present so yellow → wrinkled yellow = IV.
C. RRyy — R present so round; yy so green → round green = II.
D. rryy — both recessive → wrinkled green = I.
Matching: A-III, B-IV, C-II, D-I.

Answer

A-III, B-IV, C-II, D-I

Shortcut

Start with D, the double recessive — it can only be wrinkled green, and that single anchor eliminates most options at once. Then A, the double heterozygote, must be the fully dominant phenotype. The two extremes settle the question without touching B or C.

Where it went wrong

The chosen answer gets A and B right but swaps C and D. RRyy has a capital R, so it must be round — only rryy can be wrinkled. Checking each pair for the presence of a capital letter, gene by gene, prevents this. Two options here differ only in the C/D positions, which is the paper signalling where the decision lies.

Q2 Counting double heterozygotes in F2 Not attempted

In a Mendelian dihybrid cross, how many seeds out of 640 in the F₂ generation are expected to be double heterozygous (i.e., genotype AaBb)?

60
80
120
KEY160

Given

Mendelian dihybrid cross, F₂ generation.
Total 640 seeds.

Asked

Number expected to be AaBb.

Concept to use

In the 16-box F₂ grid the double heterozygote AaBb is the commonest single genotype, filling 4 of the 16 boxes. That is because each gene independently gives Aa in half the offspring, and ½ × ½ = ¼.

Formula to use

P(Aa) = 1/2 and P(Bb) = 1/2 → P(AaBb) = 1/2 × 1/2 = 4/16

Baby steps

For gene A alone, the F₂ is 1 AA : 2 Aa : 1 aa, so P(Aa) = 1/2.
The same for gene B: P(Bb) = 1/2.
Independent assortment lets you multiply: P(AaBb) = 1/2 × 1/2 = 1/4.
Number of seeds = 640 × 1/4 = 160.
Grid check: AaBb occupies 4 of the 16 boxes, and 4/16 = 1/4. ✓

Answer

160

Shortcut

Multiply the single-gene probabilities rather than drawing a grid. The genotype ratio 1:2:1 per gene means Aa is 1/2 and AA or aa is 1/4 each — from those three numbers any dihybrid genotype frequency follows in one multiplication.

The distractors correspond to 1/16 × 640 (40), 2/16 (80) and 3/16 (120) — the frequencies of other genotype classes. Naming which genotype you are counting before multiplying is what keeps them apart.

Q12 Counting pink flowers in snapdragon Marked wrong

A heterogenous pink flowered Antirrhinum plant on self-pollination produced 125 white flowered plants. What is the number of plants with pink flowers?

125
KEY250
MARKED375
400

Given

Pink-flowered Antirrhinum (heterozygous Rr) self-pollinated.
125 white-flowered plants produced.

Asked

Number of pink-flowered plants.

Concept to use

Snapdragon flower colour shows incomplete dominance, so the F₂ ratio is 1 red : 2 pink : 1 white, not 3 : 1. White is the double recessive at 1/4 of the total, which lets you scale up to the whole population, and pink is 2/4 — exactly twice the number of whites.

Formula to use

Rr × Rr → 1 RR (red) : 2 Rr (pink) : 1 rr (white)

Baby steps

White (rr) is 1/4 of the offspring, and there are 125 of them.
Total plants = 125 × 4 = 500.
Pink (Rr) is 2/4 = 1/2 of the total.
Pink = 500 × 1/2 = 250.
Equivalently: pink is twice white, so 2 × 125 = 250 in a single step. ✓

Answer

250

Shortcut

In a 1:2:1 ratio the middle class is simply double either end class. Given the whites, the pinks follow by doubling — no need to find the total at all.

Where it went wrong

375 is 3 × 125, which is the answer you would get from a 3 : 1 ratio — treating the cross as ordinary complete dominance. But Antirrhinum is the textbook example of incomplete dominance, and the question tells you so by describing the heterozygote as pink. Whenever the heterozygote has its own name, the ratio is 1:2:1.

Q15 Red crossed with pink in snapdragon Not attempted

When red flowered and pink flowered plants are crossed in Snapdragon, how many pink flowered plants would be obtained in the progeny of 400 plants?

100
300
400
KEY200

Given

Red (RR) × pink (Rr) in snapdragon.
400 progeny plants.

Asked

Number of pink-flowered plants.

Concept to use

With incomplete dominance the genotypes map one-to-one onto phenotypes: RR is red, Rr is pink, rr is white. A cross of RR × Rr is therefore a simple 1 : 1 — half the offspring get R from the pink parent and half get r.

Formula to use

RR × Rr → 1 RR (red) : 1 Rr (pink)

Baby steps

Red parent RR gives only R gametes.
Pink parent Rr gives R and r in equal numbers.
Offspring: half RR (red), half Rr (pink) — a 1 : 1 ratio.
Pink = 400 × 1/2 = 200.
Note that no white plants appear, since neither parent can contribute two r alleles.

Answer

200

Shortcut

A homozygote crossed with a heterozygote always gives 1 : 1, whatever the dominance pattern. The only thing incomplete dominance changes is that the two classes now look different rather than identical — so the phenotype ratio is 1 : 1 too.

Read this alongside Q12 in the same paper. Q12 is Rr selfed, giving 1:2:1; this is RR × Rr, giving 1:1. Identifying which two parents are crossed is what separates them, and both appear in this one paper.

Q22 Outcome of RRTt × rrtt Not attempted

In a plant species where red fruit (R) is dominant over yellow (r), and tallness (T) over dwarfness (t), what would be the outcome of a cross between a plant with genotype RRTt and one with genotype rrtt?

75% will be tall with red fruits.
All offsprings will be tall and with red fruits.
25% will be tall with red fruits.
KEY50% will be tall with red fruits.

Given

RRTt × rrtt.
R dominant over r; T dominant over t.

Asked

The outcome for tall red-fruited plants.

Concept to use

Split by gene and multiply. The R gene is homozygous dominant × homozygous recessive, so every offspring is Rr and therefore red — that gene contributes a factor of 1. The T gene is a test cross, Tt × tt, giving 1 tall : 1 dwarf. So the answer is decided entirely by the second gene.

Formula to use

RR × rr → all Rr (100% red) | Tt × tt → 1 Tt : 1 tt (50% tall)

Baby steps

Fruit colour: RR × rr gives all Rr → 100 % red.
Height: Tt × tt gives half Tt (tall) and half tt (dwarf) → 50 % tall.
Multiply the two independent probabilities: 100 % × 50 % = 50 % tall with red fruit.
The other 50 % are dwarf with red fruit — every plant is red, so colour never varies.

Answer

50% will be tall with red fruits.

Shortcut

Check each gene for whether it varies at all. A cross of homozygotes gives a uniform result and contributes a factor of 1, so it can be ignored — the answer then comes from the single gene that is actually segregating.

“All offspring tall and red” would be right if both genes were homozygous crosses, and 75 % would come from a Tt × Tt cross. Reading the two genotypes gene by gene, rather than as whole words, is what keeps these apart.

Q34 Proportion of heterozygous tall plants Marked wrong

In a monohybrid cross between a homozygous tall plant (TT) and a homozygous dwarf plant (tt), the F₂ generation shows a phenotypic ratio of 3:1. Then, what is the proportion of heterozygous tall plants?

1/4
MARKED1/1
KEY1/2
3/4

Given

TT × tt → F₁ all Tt → F₂ shows 3 : 1.

Asked

The proportion of heterozygous tall plants.

Concept to use

The F₂ genotypic ratio is 1 TT : 2 Tt : 1 tt. The heterozygous talls are the Tt class, which is 2 out of 4 of the whole F₂ generation. The 3 : 1 phenotypic ratio conceals this, because TT and Tt look identical — you have to go back to genotypes.

Formula to use

F₂ genotypes: 1 TT : 2 Tt : 1 tt → Tt = 2/4 = 1/2

Baby steps

F₁ is all Tt; selfing gives F₂ = 1 TT : 2 Tt : 1 tt.
Heterozygous tall = Tt = 2 parts out of 4.
Proportion = 2/4 = 1/2.
(If the question had asked what fraction of the tall plants are heterozygous, the answer would be 2/3 — a different question.)

Answer

1/2

Shortcut

Convert the phenotypic 3 : 1 back to the genotypic 1 : 2 : 1 the moment a question mentions heterozygotes. The middle class is always half the total, and the two homozygous classes a quarter each — three numbers that answer most questions of this type.

Where it went wrong

1/1 would mean every plant is heterozygous tall, which is true of the F₁ but not the F₂. Check which generation is being asked about: F₁ is uniform, F₂ is where the 1:2:1 appears. The question names F₂ explicitly.

Q10 Matching cross types to their outcomes Not attempted

Match List I with List II.

List I	List II
A) Monohybrid Cross	I) 2 genotypes and 2 phenotypes
B) Dihybrid Cross	II) 3 phenotypes and 3 genotypes
C) Incomplete dominance	III) 2 phenotypes and 3 genotypes
D) Monohybrid Test Cross	IV) 4 phenotypes and 9 genotypes

KEYA-III, B-IV, C-II, D-I
A-II, B-IV, C-III, D-I
A-III, B-II, C-IV, D-I
A-I, B-IV, C-II, D-III

Given

Four cross types and four counts of genotypes and phenotypes.

Asked

The correct matching.

Concept to use

Each cross has a fixed pair of counts, and the decisive contrast is between a monohybrid F₂ (3 genotypes but only 2 phenotypes, because the heterozygote looks like the dominant) and incomplete dominance (3 genotypes and 3 phenotypes, because the heterozygote looks different). The number of genotypes is the same; only the phenotype count changes.

Formula to use

Monohybrid F₂: 3 genotypes, 2 phenotypes | Incomplete dominance: 3 and 3 | Dihybrid F₂: 9 and 4 | Test cross: 2 and 2

Baby steps

A. Monohybrid cross — F₂ gives AA, Aa, aa (3 genotypes) but only dominant and recessive appearances (2 phenotypes) → III.
B. Dihybrid cross — 9 genotypes, 4 phenotypes → IV.
C. Incomplete dominance — the heterozygote is visibly intermediate, so 3 genotypes give 3 phenotypes → II.
D. Monohybrid test cross — Aa × aa gives Aa and aa only: 2 genotypes, 2 phenotypes → I.
Matching: A-III, B-IV, C-II, D-I.

Answer

A-III, B-IV, C-II, D-I

Shortcut

Anchor on the dihybrid, which is unmistakable at 9 and 4 — only one entry has those numbers. Then separate the two “3 genotype” entries by asking whether the heterozygote is visible: hidden means 2 phenotypes, visible means 3.

The whole question turns on entries II and III, which share the genotype count and differ only in the phenotype count. That is the pair to check first — the other two matches are free.

Q20 Matching crosses to their ratios Not attempted

Match the types of crosses in List I with their characteristic phenotypic ratios and test cross ratios in List II.

List I	List II
(A) F₂ phenotypic ratio in a monohybrid cross	(I) 1:1:1:1
(B) F₂ phenotypic ratio in a dihybrid cross	(II) 1:1
(C) Test cross ratio in a monohybrid cross	(III) 3:1
(D) Test cross ratio in a dihybrid cross	(IV) 9:3:3:1

A-III, B-I, C-II, D-IV
A-IV, B-III, C-II, D-I
A-II, B-I, C-II, D-IV
KEYA-III, B-IV, C-II, D-I

Given

Four cross types and four ratios.

Asked

The correct matching.

Concept to use

Four ratios that should be held as a single block. The F₂ ratios come from selfing a heterozygote — 3:1 for one gene, 9:3:3:1 for two. The test cross ratios come from crossing with the double recessive, which contributes nothing and therefore simply reveals the gametes — 1:1 for one gene, 1:1:1:1 for two.

Formula to use

Monohybrid F₂ 3:1 · Dihybrid F₂ 9:3:3:1 · Monohybrid test cross 1:1 · Dihybrid test cross 1:1:1:1

Baby steps

A. Monohybrid F₂ → 3:1 = III.
B. Dihybrid F₂ → 9:3:3:1 = IV.
C. Monohybrid test cross — Aa × aa gives half dominant, half recessive → 1:1 = II.
D. Dihybrid test cross — AaBb × aabb reveals the four gamete types equally → 1:1:1:1 = I.
Matching: A-III, B-IV, C-II, D-I.

Answer

A-III, B-IV, C-II, D-I

Shortcut

Split by type before matching: the two F₂ ratios contain a 3 or a 9, the two test-cross ratios are all ones. Sorting into those two families first means you only ever choose between two candidates at a time.

A test cross ratio is always all ones, because the recessive parent contributes a single kind of gamete and hides nothing — so the offspring simply mirror the gametes of the tested parent. Knowing why makes the pair impossible to forget.

Q30 Genotypes absent from back cross and test cross Marked wrong

Which of the following genotypes is not found in progeny of monohybrid back cross and test cross respectively?

KEYtt and TT
TT and tt
Tt and Tt
MARKEDtt and tt

Given

A monohybrid back cross (F₁ Tt × dominant parent TT).
A monohybrid test cross (F₁ Tt × recessive parent tt).

Asked

The genotype absent from each, in that order.

Concept to use

A back cross is with either parent; here with the dominant parent TT, which always contributes a T — so tt can never appear. A test cross is specifically with the recessive parent tt, which always contributes a t — so TT can never appear. Each cross excludes the homozygote of the opposite type.

Formula to use

Back cross Tt × TT → TT, Tt (no tt) | Test cross Tt × tt → Tt, tt (no TT)

Baby steps

Back cross with TT: gametes T from one parent, T or t from the other → offspring TT and Tt. tt is impossible.
Test cross with tt: gametes t from one parent, T or t from the other → offspring Tt and tt. TT is impossible.
The question asks for them “respectively”, so the order is (back cross, test cross).
Answer: tt and TT.

Answer

tt and TT

Shortcut

Each cross has one parent contributing a fixed allele, so the offspring can never be homozygous for the other allele. State it as a rule: cross with TT excludes tt; cross with tt excludes TT. The answer is then the two exclusions in the order asked.

Where it went wrong

“tt and tt” would require both crosses to exclude the same genotype, but a test cross with tt produces tt in half its offspring — so tt cannot be the absent one there. Checking each half of a “respectively” answer separately is what catches this; the two halves are answering two different questions.

Q38 Matching crosses to phenotypic ratios Marked wrong

Match List I (cross) with List II (phenotypic ratio).

List I	List II
A) Monohybrid test cross	I) 9:3:3:1
B) F₂ complete dominance	II) 1:1
C) F₂ incomplete dominance	III) 1:2:1
D) F₂ dihybrid cross	IV) 3:1

KEYA-II, B-IV, C-III, D-I
MARKEDA-II, B-III, C-IV, D-I
A-IV, B-I, C-II, D-III
A-II, B-IV, C-I, D-III

Given

Four cross types and four phenotypic ratios.

Asked

The correct matching.

Concept to use

The whole question turns on separating complete from incomplete dominance in the F₂. Both start from a heterozygous F₁ and both give the genotypic ratio 1:2:1 — but with complete dominance the heterozygote is hidden, collapsing the phenotypes to 3:1, whereas with incomplete dominance it is visible, leaving 1:2:1.

Formula to use

Complete dominance F₂: 3:1 | Incomplete dominance F₂: 1:2:1 | Test cross: 1:1 | Dihybrid F₂: 9:3:3:1

Baby steps

A. Monohybrid test cross → 1:1 = II.
B. F₂ complete dominance — the heterozygote looks like the dominant, so 3:1 → IV.
C. F₂ incomplete dominance — the heterozygote is its own phenotype, so 1:2:1 → III.
D. F₂ dihybrid → 9:3:3:1 = I.
Matching: A-II, B-IV, C-III, D-I.

Answer

A-II, B-IV, C-III, D-I

Shortcut

Fix the two easy ones first — test cross 1:1 and dihybrid 9:3:3:1 — then decide only between 3:1 and 1:2:1 for the two remaining entries. One question settles it: is the heterozygote visible?

Where it went wrong

B and C were swapped, giving 3:1 to incomplete dominance and 1:2:1 to complete dominance. Note that the two options differ only in those positions — the paper is testing exactly this distinction. The rule to hold: 1:2:1 phenotypes means the heterozygote is visible, and that only happens with incomplete dominance.

Q44 Which statement about a dihybrid test cross is false Not attempted

Which statement about a dihybrid test cross is not true?

It tests whether an unknown individual is homozygous or heterozygous
The test individual is crossed with a homozygous recessive individual
If the tested individual is heterozygous the progeny will have 1:1:1:1 phenotypic ratio
KEYIf the test individual is homozygous, the progeny will have a 1:1:1:1 phenotypic ratio

Given

Four statements about the dihybrid test cross; one is false.

Asked

The statement that is not true.

Concept to use

The entire purpose of a test cross is to distinguish a homozygote from a heterozygote — so the two cases must give different results. A heterozygous AaBb produces four gamete types and therefore a 1:1:1:1 ratio; a homozygous AABB produces only one gamete type and therefore uniform offspring. If both gave 1:1:1:1 the test would reveal nothing.

Formula to use

AaBb × aabb → 1:1:1:1 | AABB × aabb → all AaBb, uniform

Baby steps

Statements 1 and 2 state the purpose and the method of a test cross correctly.
Statement 3: a heterozygous AaBb gives four gamete types in equal numbers, so the offspring are 1:1:1:1. True.
Statement 4: a homozygous individual gives only one gamete type, so all offspring are identical — a uniform result, not 1:1:1:1. False.
Answer: the fourth statement.

Answer

If the test individual is homozygous, the progeny will have a 1:1:1:1 phenotypic ratio

Shortcut

Statements 3 and 4 assign the same outcome to opposite conditions — and a test that gives the same result either way would be useless. Spotting that contradiction identifies the false statement without any crosses being worked out.

The uniform result from a homozygote is itself the informative outcome: if a test cross gives offspring that all look alike, the tested individual was homozygous. Both results carry information — that is why the test works.

Q25 Which statements about dominance are incorrect Marked wrong

Read the following statements.
A) Sometimes, F₁ do not resemble either of two parents for flower colour in some plants.
B) Occasionally, a single gene product may produce more than one phenotype.
C) After maturation of pea seeds, those with BB genotype are round, while those with Bb genotype are wrinkled.
D) Dominance is an autonomous feature of a gene.
Identify and choose the set of incorrect statement/s.

MARKEDC only
A, C, and D only
KEYC and D only
B and C only

Given

Four statements about dominance and gene expression.

Asked

Which statements are incorrect.

Concept to use

Statement D is the subtle one. Dominance is not a property of a gene by itself — it depends on the gene product and on the pathway it acts in, and the same gene can appear dominant for one trait and incompletely dominant for another. NCERT makes this point explicitly. Statement C is straightforwardly wrong: B is dominant, so BB and Bb are both round.

Formula to use

Dominance depends on the gene product and the trait examined — it is not an intrinsic property of the allele

Baby steps

A. This describes incomplete dominance, where the F₁ is intermediate — as in snapdragon. Correct statement.
B. This is pleiotropy, where one gene affects several traits — as in the starch gene affecting seed shape and starch grain size. Correct statement.
C. B is dominant, so BB and Bb are both round; only bb is wrinkled. Incorrect.
D. Dominance is not autonomous — it depends on the gene product and on which level of the phenotype is examined. Incorrect.
Incorrect statements: C and D.

Answer

C and D only

Shortcut

Statement D contains the absolute word “autonomous”, and statements A and B in the same list are themselves examples of dominance not behaving autonomously. The list is self-contradicting if D is accepted — which points straight at D as one of the errors.

Where it went wrong

Only C was marked incorrect, so statement D slipped through. The same B gene appears in NCERT as both completely dominant (seed shape) and incompletely dominant (starch grain size) — which is direct evidence that dominance depends on the trait examined. Absolute-sounding words like “autonomous” deserve the same suspicion as “always” and “only”.

Q39 Dominance and the ABO blood group Marked wrong

Statement I: Not all characters show true dominance.
Statement II: ABO blood group shows dominance, incomplete dominance and multiple allelism.

KEYStatement I is correct, but statement II is incorrect.
Statement I is incorrect, but statement II is correct.
Both statement I and statement II are incorrect.
MARKEDBoth statement I and statement II are correct.

Given

Two statements about dominance and the ABO system.

Asked

Which statements are correct.

Concept to use

The ABO system illustrates dominance (I^A and I^B over i), multiple allelism (three alleles in the population) and codominance (I^A and I^B both fully expressed in AB blood). What it does not show is incomplete dominance — in AB blood both antigens appear at full strength, rather than the blending that incomplete dominance would give.

Formula to use

Codominance: both alleles fully expressed | Incomplete dominance: heterozygote is intermediate

Baby steps

Statement I. Incomplete dominance and codominance both exist, so not every character shows true dominance. Correct.
Statement II. ABO shows dominance, multiple allelism and codominance — but not incomplete dominance. The word is wrong. Incorrect.
Answer: I correct, II incorrect.

Answer

Statement I is correct, but statement II is incorrect.

Shortcut

Test the AB genotype specifically. Is AB blood intermediate between A and B, or does it show both antigens fully? It shows both — which is codominance by definition, not incomplete dominance. One example settles the statement.

Where it went wrong

Both statements were accepted, so the substitution of “incomplete dominance” for “codominance” went unnoticed. The two are constantly confused: incomplete = blended (pink from red and white), codominant = both visible (A and B antigens side by side). This distinction was also missed in an earlier botany paper, so it is worth a dedicated line in the notes.

Q11 Assertion–reason on linked genes Not attempted

Assertion (A): Mendel's law of Independent assortment does not hold good for the genes that are located closely on the same chromosome.
Reason (R): Closely located genes assort independently.

Both (A) and (R) are true, but (R) is not the correct explanation of (A).
KEY(A) is true, but (R) is false.
(A) is false, but (R) is true.
Both (A) and (R) are true, and (R) is the correct explanation of (A).

Given

Assertion: independent assortment fails for closely linked genes.
Reason: closely located genes assort independently.

Asked

Judge each statement.

Concept to use

The two statements contradict each other. Genes lying close together on the same chromosome are linked — they tend to travel to the same gamete, so they do not assort independently. The assertion says exactly this and is correct; the reason asserts the opposite and is false.

Formula to use

Closely linked genes → low recombination → assort together, not independently

Baby steps

Is A true? Yes. Morgan's work on Drosophila showed that genes on the same chromosome, close together, are inherited together. Independent assortment fails for them. True.
Is R true? No. It claims closely located genes assort independently, which is the definition of not being linked. False.
Notice A and R directly contradict one another, so they cannot both be true.
Answer: A is true, R is false.

Answer

(A) is true, but (R) is false.

Shortcut

When the assertion and the reason make opposite claims, the answer is always one of the two “one true, one false” options — that halves the list before any biology. Then decide which one is right, and here the assertion states the standard textbook fact.

Assertion–reason questions have been the single most consistent weakness across every paper in this series. The routine to build: write T or F beside each statement before looking at the options, then check whether A and R are even compatible.

Q18 Linkage strength and distance Not attempted

Assertion (A): The strength of linkage will be more when the distance between two genes is greater.
Reason (R): Crossing over will be relatively less frequent for distantly located genes.

Both (A) and (R) are true, and (R) is the correct explanation of (A).
Both (A) and (R) are true, and (R) is not the correct explanation of (A).
(A) is true, but (R) is false.
KEYBoth (A) and (R) are false.

Given

Assertion: linkage strength increases with distance.
Reason: crossing over is less frequent for distant genes.

Asked

Judge each statement.

Concept to use

Both claims are the reverse of the truth. The further apart two genes lie on a chromosome, the more chance a crossover falls between them — so recombination is more frequent and linkage is weaker. This is the entire basis of gene mapping: recombination frequency is used as a measure of distance precisely because the two rise together.

Formula to use

Greater distance → more crossing over → higher recombination frequency → weaker linkage

Baby steps

Statement R. Crossing over is more, not less, frequent for distantly located genes — a longer stretch of chromosome offers more places for a crossover to occur. False.
Statement A. Since crossovers separate the genes more often, linkage is weaker at greater distance, not stronger. False.
Both statements are false.

Answer

Both (A) and (R) are false.

Shortcut

Anchor on the mapping principle: 1 map unit = 1 % recombination. Distance and recombination frequency are directly proportional by definition, so anything claiming they move oppositely is false. That one relation settles both statements at once.

Sturtevant's whole method depends on this. If distant genes recombined less, recombination frequency could not be used to order genes on a chromosome. Recognising that the statements would make gene mapping impossible is a quick way to reject them.

Q23 Identifying a recombinant from a chromosome diagram Marked wrong

Following diagram represents which of the following F₂ member in Morgan's observations? — the figure shows two chromosomes, one carrying w⁺ and m, the other carrying w and m.

MARKEDParental type female
KEYRecombinant female
Parental type male
Recombinant male

Given

A diagram showing two full chromosomes.
Alleles: w⁺ with m on one, w with m on the other.

Asked

Which F₂ class the diagram represents.

Concept to use

Two decisions. Sex comes from the number of chromosomes drawn — two matching X chromosomes means a female. Parental versus recombinant comes from whether the alleles are in the combination the parents supplied: in Morgan's crosses the parental arrangements are w⁺m⁺ together and w m together, so any chromosome carrying w⁺ alongside m is a new combination produced by crossing over.

Formula to use

Parental: w⁺m⁺ or w m | Recombinant: w⁺m or w m⁺

Baby steps

Count the chromosomes: two complete X chromosomes are shown → female.
Read the first chromosome: it carries w⁺ together with m.
Compare with the parental combinations, w⁺m⁺ and w m. The pairing w⁺m matches neither.
A new allele combination means a crossover has occurred → recombinant.
Answer: recombinant female.

Answer

Recombinant female

Shortcut

Check whether the superscripts match along each chromosome. Both wild-type (w⁺m⁺) or both mutant (w m) means parental; a mixture means recombinant. It is a two-second visual test once you know to look for it.

Where it went wrong

The sex was read correctly but the allele combination was not. The w⁺ and m sitting on the same chromosome is precisely the signature of a crossover — that mixture cannot come from either parent intact. Read the alleles along each chromosome individually rather than taking in the picture as a whole.

Q32 Morgan's recombination percentages Marked wrong

In Morgan's experiments on Drosophila linkage, the percentage of white-eyed, brown-body recombinants and the percentage of red-eyed, miniature-winged recombinants in F₂ generation, respectively

MARKED37.2% and 1.3%
KEY1.3% and 37.2%
62.8% and 37.2%
98.7% and 1.3%

Given

Two of Morgan's Drosophila gene pairs.
White eye with yellow body, and white eye with miniature wing.

Asked

The two recombination percentages, in the order stated.

Concept to use

Morgan found that white eye and yellow body are very tightly linked, showing only 1.3 % recombination, while white eye and miniature wing are more loosely linked, showing 37.2 %. The pair of numbers is standard NCERT content; the only difficulty is attaching each to the right gene pair.

Formula to use

Tightly linked → low recombination (1.3 %) | loosely linked → high recombination (37.2 %)

Baby steps

White eye and body colour are extremely close together on the X chromosome, so crossovers between them are rare: 1.3 %.
White eye and miniature wing are further apart, so crossovers between them are common: 37.2 %.
The question asks in that order, so the answer is 1.3 % and 37.2 %.

Answer

1.3% and 37.2%

Shortcut

You only need to remember that the two numbers are 1.3 and 37.2, and that the body colour pair is the tight one. Attaching “tight = small number” is easier than memorising which value goes with which trait in the abstract.

Where it went wrong

The two values were transposed — the right pair of numbers in the wrong order. That is the same failure mode as Q1 and Q28 in this paper: correct content, wrong arrangement. When a question says “respectively”, write the two items in a column and match them one at a time rather than reading the option as a phrase.

Q37 The basis of gene mapping Not attempted

Mapping the position of genes on the chromosome is based on the

frequency of recombination between gene pairs on a non-homologous chromosome pair.
frequency of recombination between gene pairs on only somatic chromosomes.
KEYfrequency of recombination between gene pairs on the same chromosome.
frequency of recombination between gene pairs on only X and Y chromosomes.

Given

Genetic mapping of gene positions.

Asked

What mapping is based on.

Concept to use

Recombination frequency measures distance only for genes on the same chromosome. Genes on different chromosomes assort independently and always show 50 % recombination regardless of position, so they carry no distance information at all. Mapping therefore works within a chromosome, never between them.

Formula to use

1 map unit (centimorgan) = 1 % recombination frequency, for genes on the same chromosome

Baby steps

Genes on different chromosomes assort independently, giving a fixed 50 % recombination — the same value however far apart they are, so no distance can be inferred.
Genes on the same chromosome recombine only when a crossover falls between them, and that becomes more likely the further apart they are.
So recombination frequency is a usable measure of distance only within a chromosome.
Answer: gene pairs on the same chromosome.

Answer

frequency of recombination between gene pairs on the same chromosome.

Shortcut

Two options can be dismissed for being too narrow — mapping is not restricted to somatic chromosomes or to the sex chromosomes. Of the remaining two, “non-homologous” means different chromosomes, which gives no information. That leaves one.

Sturtevant devised this method, and the distinction from Morgan matters: Morgan discovered linkage and recombination, Sturtevant turned the frequencies into a map. The pair was tested in an earlier paper (Q94) and missed there too.

Q40 Reading Morgan's Cross A Marked wrong

The diagram represents cross A of Morgan — a yellow-bodied, white-eyed female (y w / y w) crossed with a wild-type male (y⁺ w⁺). Identify the correct statement related to this cross.

Non-allelic gene pair y and w are tightly linked and hence show high % of recombination.
Non-allelic gene pair y and w are loosely linked and hence show low % of recombination.
MARKEDAllelic gene pair y and w are tightly linked and hence show low % of recombination.
KEYNon-allelic gene pair y and w are tightly linked and hence show low % of recombination.

Given

Morgan's Cross A: yellow-bodied white-eyed female × wild-type male.
Genes y (body colour) and w (eye colour).

Asked

The correct statement about y and w.

Concept to use

Three things must all be right, and each option gets exactly one of them wrong. Allelic versus non-allelic: y and w control different traits, so they are different genes — non-allelic. Tight versus loose: they lie very close together, so tightly linked. High versus low recombination: tight linkage means crossovers are rare, so recombination is low (1.3 %).

Formula to use

Different genes → non-allelic | close together → tightly linked → low recombination

Baby steps

y controls body colour and w controls eye colour — two different traits, so two different genes: non-allelic. (Alleles are alternative forms of the same gene.)
Morgan found only 1.3 % recombination between them, so they must lie very close: tightly linked.
Tight linkage means few crossovers fall between them: low recombination percentage.
The only option combining all three correctly: non-allelic, tightly linked, low recombination.

Answer

Non-allelic gene pair y and w are tightly linked and hence show low % of recombination.

Shortcut

Check the three attributes as a checklist and cross out on the first failure. Two options fail on “tight but high” or “loose but low”, which are internally contradictory — tight linkage means low recombination, so an option pairing them wrongly is self-defeating.

Where it went wrong

The chosen option says allelic, and everything after that is correct. But y and w govern different characters, so they cannot be alleles of one gene. Alleles are alternative versions of the same gene — y and y⁺ are alleles; y and w are not. When an option has three attributes, verify each one separately.

Q27 Six statements about dihybrid crosses Marked wrong

Read the following statements.
i. In the F₂ generation of a dihybrid cross, the round trait appeared in 12 out of 16 plants.
ii. Mendel proposed the law of independent assortment based on monohybrid crosses.
iii. The law of independent assortment states that segregation of one pair of traits is independent of the other.
iv. Only two types of gametes are formed in F₁ plants (RrYy).
v. The probability of a Y allele being present in a gamete from RrYy is 25%.
vi. Punnett square helps derive genotypic ratios in dihybrid crosses.

KEYi, iii, and vi only
MARKEDi, iii, v, and vi only
i, ii, and v only
iii, iv, and vi only

Given

Six statements about dihybrid crosses.

Asked

Which statements are correct.

Concept to use

Three of the six are wrong for specific, checkable reasons. Independent assortment came from dihybrid crosses, not monohybrid. RrYy produces four gamete types (2²), not two. And the probability of Y in a gamete is 50 %, not 25 % — Y and y are equally likely from a Yy pair.

Formula to use

n heterozygous loci → 2ⁿ gamete types; each allele of a heterozygous pair appears in 1/2 of gametes

Baby steps

i. In 9:3:3:1, round appears in 9 + 3 = 12 of 16. Correct.
ii. Independent assortment requires two genes, so it came from dihybrid crosses. Incorrect.
iii. This is a correct statement of the law. Correct.
iv. RrYy has two heterozygous loci, giving 2² = 4 gamete types (RY, Ry, rY, ry). Incorrect.
v. From Yy, half the gametes carry Y — 50 %, not 25 %. Incorrect.
vi. The Punnett square is exactly the tool for deriving genotypic ratios. Correct.
Answer: i, iii and vi only.

Answer

i, iii, and vi only

Shortcut

Statements iv and v are both about the gametes of RrYy and can be checked together in one move: four gamete types, each at 25 %, so each gamete type is 25 % but each allele is 50 %. Settling that one distinction disposes of two statements at once.

Where it went wrong

Statement v was accepted. The 25 % figure is correct for a gamete type such as RY, but the statement asks about the Y allele, which appears in two of the four gamete types (RY and rY) and so occurs in 50 %. The same number attached to the wrong object — read what is being counted before accepting a percentage.

Q28 Ordering four dihybrid quantities Marked wrong

Arrange the following in the increasing order based on the combinations obtained in the F₂ progeny of dihybrid cross in a garden pea.
I. Number of parental phenotype plants
II. Number of recombination plants in phenotype
III. Types of phenotypes
IV. Types of genotypes

MARKEDI, III, IV, II
KEYIII, II, IV, I
II, I, IV, III
IV, II, III, I

Given

F₂ of a dihybrid cross, ratio 9:3:3:1 out of 16.

Asked

The four quantities in increasing order.

Concept to use

Every number needed comes from the 9:3:3:1 grid. Parental phenotypes are the two that resemble the original parents — the 9 and the 1, totalling 10. Recombinant phenotypes are the two new combinations, 3 + 3 = 6. Types of phenotypes is 4 and types of genotypes is 9. Write all four down, then sort.

Formula to use

9:3:3:1 → parental 9 + 1 = 10 · recombinant 3 + 3 = 6 · phenotypes 4 · genotypes 9

Baby steps

I. Parental phenotype plants = 9 + 1 = 10.
II. Recombinant phenotype plants = 3 + 3 = 6.
III. Types of phenotypes = 4.
IV. Types of genotypes = 9.
Sort in increasing order: 4 < 6 < 9 < 10, i.e. III < II < IV < I.

Answer

III, II, IV, I

Shortcut

Write the four numbers in the margin before looking at any option — 10, 6, 4, 9. Sorting four numbers is trivial; the difficulty is entirely in extracting them, so do that first and separately.

Where it went wrong

The chosen order puts I (10) first, which would be the decreasing order for that item. Mixing up increasing and decreasing is a common slip when four items must be ranked. Underline the word “increasing” in the question, and check your first and last entries against the smallest and largest numbers you wrote down.

Q41 What occurs in pairs and segregates Not attempted

Read the following statements.
a) Occur in pairs
b) Segregate at the time of gamete formation such that only one of each pair is transmitted to a gamete
The above statements are applicable to

only genes but not chromosome
only chromosomes but not genes
KEYboth genes and chromosomes
all chromosomes and some genes

Given

Two properties: occurring in pairs, and segregating during gamete formation.

Asked

What these statements apply to.

Concept to use

This is the heart of the chromosomal theory of inheritance. Sutton and Boveri noticed that genes and chromosomes behave in exactly parallel ways — both are present in pairs in a diploid organism, and both separate so that a gamete receives one member of each pair. That parallel is what led to the conclusion that genes are carried on chromosomes.

Formula to use

Sutton–Boveri parallel: genes and chromosomes both occur in pairs and both segregate at gamete formation

Baby steps

In a diploid cell, chromosomes exist as homologous pairs and genes exist as allele pairs. Both occur in pairs.
At meiosis, homologous chromosomes separate to opposite poles, and with them the alleles they carry. Both segregate.
So each statement is true of genes and of chromosomes.
Answer: both genes and chromosomes.

Answer

both genes and chromosomes

Shortcut

Recognise the question as the Sutton–Boveri parallel and the answer is immediate. Whenever a question lists behaviours shared by genes and chromosomes, the point being made is that the two move together — which is the whole basis of the chromosomal theory.

The same parallel was tested in an earlier botany paper (Q104), where the trap was an extra statement adding the word “always”. Here the statements are correctly worded, so both apply.

Q42 Chromosomes versus genes segregating independently Not attempted

Statement I: Independent pairs of chromosomes segregate independently of each other.
Statement II: One pair of genes always segregates independently of another pair.

Both statement I and statement II are correct.
Both statement I and statement II are incorrect.
KEYStatement I is correct, but statement II is incorrect.
Statement I is incorrect, but statement II is correct.

Given

Two statements about independent segregation.

Asked

Which statements are correct.

Concept to use

The word “always” in Statement II is what makes it false. Different chromosome pairs do line up independently at metaphase I, so Statement I holds. But two gene pairs on the same chromosome are linked and travel together — so genes do not always segregate independently. Statement I is about chromosomes and is safe; Statement II overgeneralises to genes.

Formula to use

Chromosomes: always independent | Genes: independent only if on different chromosomes (or far apart on one)

Baby steps

Statement I. Each bivalent orients at random on the metaphase plate, independently of the others. Correct.
Statement II. Linked genes on one chromosome do not assort independently — Morgan's Drosophila work established this. The word “always” makes it false. Incorrect.
Answer: I correct, II incorrect.

Answer

Statement I is correct, but statement II is incorrect.

Shortcut

Scan for absolute words. “Always” in a genetics statement is almost invariably the planted error, because linkage, incomplete dominance and pleiotropy are all exceptions waiting to be applied. One word decides this question.

Note the careful wording of Statement I: it says independent pairs of chromosomes, which is true by definition. Statement II drops that qualifier and applies it to genes, where it fails. Comparing the two statements word by word is what exposes the difference.

Q43 Assertion–reason on dihybrid progeny Marked wrong

Assertion (A): All F₁ progeny of dihybrid cross are dihybrids.
Reason (R): Only 4/16 of the F₂ progeny of dihybrid cross are dihybrids.

Both (A) and (R) are true, and (R) is the correct explanation of (A).
KEYBoth (A) and (R) are true, but (R) is not the correct explanation of (A).
MARKED(A) is true, but (R) is false.
Both (A) and (R) are false.

Given

Assertion about the F₁, reason about the F₂.

Asked

Judge each statement and the link between them.

Concept to use

Both statements are true, but they describe different generations. Crossing two pure lines gives an F₁ that is uniformly AaBb — all dihybrids. Selfing that F₁ gives an F₂ in which AaBb occupies 4 of the 16 boxes. The F₂ figure is a consequence of the F₁ being dihybrid, not a reason for it, so R does not explain A.

Formula to use

F₁: 100 % AaBb | F₂: AaBb = 4/16 = 25 %

Baby steps

Is A true? A dihybrid cross starts with AABB × aabb, so every F₁ receives AB from one parent and ab from the other — all AaBb. True.
Is R true? In the F₂ grid, AaBb fills 4 of 16 boxes. True.
Does R explain A? Apply the removal test: the F₂ proportion follows from the F₁ being dihybrid, so it cannot be the cause. The logic runs forwards, not backwards.
Both true, but R is not the explanation.

Answer

Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Shortcut

Check the time order. A reason must come before, or underlie, the assertion. Here R describes a later generation than A, so it cannot explain it — a later fact never causes an earlier one. Spotting the F₁/F₂ mismatch answers the question without any genetics.

Where it went wrong

R was marked false, but 4/16 is exactly right for AaBb in the F₂ — the same figure calculated in Q2 of this paper. This is the mirror image of the usual error: instead of wrongly accepting a link, a true statement was rejected. Verify each statement's truth first, and only then consider whether one explains the other.

Q46 Hardy-Weinberg: percentage showing the dominant trait Not attempted

In a population in Hardy-Weinberg equilibrium, if the frequency of dominant allele at an autosomal locus is 60%, the percentage of dominant individuals in that population would be

KEY84%
16%
36%
64%

Given

p = frequency of the dominant allele = 0.6, so q = 0.4.
Population in Hardy-Weinberg equilibrium.

Asked

Percentage of individuals showing the dominant phenotype.

Concept to use

“Dominant individuals” means everyone who shows the dominant trait — the homozygous dominants and the heterozygotes, since a single dominant allele is enough. So the answer is p² + 2pq, not p² alone.

Formula to use

p² + 2pq + q² = 1 → dominant phenotype = p² + 2pq = 1 − q²

Baby steps

p = 0.6, so q = 1 − 0.6 = 0.4.
Homozygous dominant: p² = 0.36 → 36 %.
Heterozygous: 2pq = 2(0.6)(0.4) = 0.48 → 48 %.
Both show the dominant trait, so add: 0.36 + 0.48 = 0.84 = 84 %.
Cross-check: 1 − q² = 1 − 0.16 = 0.84. ✓

Answer

84%

Shortcut

Compute the recessives instead. Only q² = 0.16 shows the recessive trait, so everyone else — 1 − 0.16 = 0.84 — shows the dominant one. Squaring one number and subtracting is faster and less error-prone than adding two terms.

Every distractor is a partial answer: 36 % is p² alone (heterozygotes forgotten), 16 % is q² (the recessives), 64 % is 1 − p². Naming which group you are counting before computing keeps them apart.

Q71 Where genetic drift operates Not attempted

Genetic drift operates in

large populations and is stochastic
KEYsmall populations and is stochastic
large populations and is deterministic
small populations and is deterministic

Given

Genetic drift as an evolutionary mechanism.

Asked

The population size in which it operates and its nature.

Concept to use

Genetic drift is random change in allele frequency by chance sampling. In a large population, chance fluctuations average out and have little effect. In a small population, a few chance events can shift allele frequencies dramatically — which is why drift matters most there. And because it is driven by chance, it is stochastic, not deterministic.

Formula to use

Drift ∝ 1/population size | random sampling → stochastic, not predictable in direction

Baby steps

Drift arises from random sampling of alleles between generations.
In a large population, random deviations largely cancel out, so the effect is small.
In a small population, a handful of chance events can change frequencies sharply — the founder effect and bottleneck effect are examples.
Being chance-driven, drift is stochastic: its direction cannot be predicted.
Answer: small populations, stochastic.

Answer

small populations and is stochastic

Shortcut

Two independent decisions, each halving the list. Drift is random, so “deterministic” is wrong in any option — that removes two. Then drift matters in small populations, which settles it. Contrast with natural selection, which is directional and therefore deterministic in a way drift is not.

The founder effect and the bottleneck effect are both instances of drift, and both involve a population becoming small. Recognising them as examples of the same mechanism makes this section considerably shorter to learn.

Q60 Flying squirrel and flying phalanger Marked wrong

Flying squirrel and flying phalanger exemplify

divergent evolution
KEYconvergent evolution
MARKEDadaptive radiation
artificial selection

Given

Flying squirrel (a placental mammal) and flying phalanger (a marsupial).

Asked

What they exemplify.

Concept to use

These two animals are unrelated — one placental, one marsupial — yet they have independently evolved gliding membranes because they occupy similar niches. Different ancestors arriving at similar form is convergent evolution, and the resulting structures are analogous. Divergent evolution is the opposite: one ancestor giving rise to differing forms, producing homologous structures.

Formula to use

Convergent: different ancestors → similar form (analogous) | Divergent: one ancestor → different forms (homologous)

Baby steps

The flying squirrel is a placental mammal; the flying phalanger is an Australian marsupial. Their common ancestor is very distant.
Both have independently evolved a gliding membrane, because both occupy an arboreal gliding niche.
Similar structures from unrelated ancestors = convergent evolution, giving analogous organs.
Answer: convergent evolution.

Answer

convergent evolution

Shortcut

Ask whether the two organisms are closely related. Unrelated but alike → convergent. Related but different → divergent. The placental/marsupial pairing is the clearest signal of convergence in the syllabus, and NCERT lists several such pairs.

Where it went wrong

Adaptive radiation is a related but distinct idea: it means one ancestral stock diversifying into many forms in one geographical area — Darwin's finches, or Australian marsupials as a group. Here we have two separate lineages converging, not one diversifying. Adaptive radiation is a form of divergence; this question is about the opposite direction.

Q83 Evolution resulting from human action Not attempted

Which one of the following are the result of evolution by anthropogenic action?
I. Microbes against which we employ antibiotics.
II. Eukaryotic organisms against which we employ drugs.
III. Evolution of more number of non melanic moths in industrially polluted areas.
IV. Evolution of vegetarian finches.

I, II, III and IV
I, II and III
KEYOnly I & II
Only III & IV

Given

Four proposed examples of evolution caused by human activity.

Asked

Which are genuinely anthropogenic.

Concept to use

Two of these are the standard NCERT examples of evolution driven by human action — antibiotic resistance in microbes and drug resistance in eukaryotic parasites. Statement III has the industrial melanism example backwards: in polluted areas the melanic (dark) moths increased, not the non-melanic ones. And Darwin's finches evolved naturally on the Galápagos, with no human involvement.

Formula to use

Anthropogenic evolution: resistance arising because humans applied a selective agent

Baby steps

I. Antibiotic-resistant bacteria arise because we use antibiotics — a direct human-imposed selection pressure. Correct.
II. Drug-resistant eukaryotes, such as resistant malarial parasites, arise the same way. Correct.
III. In polluted industrial areas the dark, melanic moths became commoner because soot-blackened trees hid them. The statement says non-melanic, which is the reverse. Incorrect.
IV. The finches evolved by natural selection on the Galápagos, without human involvement. Incorrect.
Answer: only I and II.

Answer

Only I & II

Shortcut

Ask of each example: did a human action create the selection pressure? Antibiotics and drugs are things we apply, so yes. Finches are a textbook case of natural selection, so no. That question alone sorts the list without needing the moth details.

The industrial melanism story is worth getting exactly right, because it is often reversed. Before industrialisation the light moths were commoner (lichen-covered trees); after, the dark melanic moths predominated (soot-covered trees). Statement III inverts this.

Q56 Theory of special creation Marked wrong

According to theory of special creation, Earth is about

MARKED4000 million years old
400 million years old
KEY4000 years old
4000 billion years old

Given

The theory of special creation.

Asked

The age of the Earth according to that theory.

Concept to use

This asks what the theory of special creation claims, not what the Earth's actual age is. That theory holds that all life was created as it is, and that the Earth is about 4000 years old. The scientifically established figure is roughly 4600 million years — a completely different number, and the question is testing whether you supply the one that was asked for.

Formula to use

Special creation: Earth ≈ 4000 years old | scientific estimate: ≈ 4600 million years

Baby steps

Identify what is being asked: the claim of a particular theory, not the modern scientific value.
The theory of special creation states the Earth is about 4000 years old.
Contrast with the scientific figure of roughly 4600 million years — a factor of about a million larger.
Answer: 4000 years old.

Answer

4000 years old

Shortcut

When a question names a specific theory, answer from within that theory. NCERT gives the special-creation figure as 4000 years, and the presence of both “4000 years” and “4000 million years” among the options is the paper checking whether you noticed which was requested.

Where it went wrong

4000 million years is close to the true age of the Earth, so it feels like the right answer. But the question asks what special creation says, and that theory is precisely the one modern geology contradicts. Read the opening clause — “according to” signals that the frame of reference is not your own knowledge.

Q62 Ancestors of the first amphibians Not attempted

Which of the following animals evolved into first amphibians that lived on both land and water?

Jawless fish
Ichthyosaurs
KEYLobefins
Pelycosaurs

Given

The evolutionary origin of the first amphibians.

Asked

Which group gave rise to them.

Concept to use

Lobefins were fish with fleshy, bone-supported fins — a structure that could bear weight and so served as a precursor to limbs. NCERT states that lobefins evolved into the first amphibians that lived on both land and water. The other options are all in the wrong part of the tree: ichthyosaurs were marine reptiles, pelycosaurs were early reptiles, and jawless fish are far more primitive.

Formula to use

Lobefins → first amphibians → reptiles → birds and mammals

Baby steps

Lobefins had muscular, bone-supported fins capable of supporting the body out of water.
This is the structural precursor of the tetrapod limb, which is why they, and not ray-finned fish, made the transition.
They gave rise to the first amphibians.
The other three options are reptiles or primitive fish and sit elsewhere on the tree.

Answer

Lobefins

Shortcut

Screen the options by group before thinking about timing. Ichthyosaurs and pelycosaurs are reptiles, which come after amphibians — a descendant cannot be an ancestor. That eliminates two options immediately and leaves a choice between two fish.

Keep the vertebrate sequence as one chain: jawless fish → jawed fish → lobefins → amphibians → reptiles → birds and mammals. Most questions in this section are answered by locating the named group on that chain.

Q76 When cellular life first appeared Marked wrong

When did the first cellular forms of life appear on earth?

KEY2000 mya
500 mya
4500 mya
MARKED3000 mya

Given

The appearance of the first cellular forms of life.

Asked

How long ago.

Concept to use

NCERT gives the figure as about 2000 million years ago for the first cellular forms of life. This is a straight recall item, and the surrounding numbers — the Earth forming about 4500 mya, and the first non-cellular forms much earlier — are what the distractors are drawn from.

Formula to use

Earth formed ≈ 4500 mya · first cellular life ≈ 2000 mya (NCERT)

Baby steps

The Earth itself formed roughly 4500 million years ago — life cannot predate that, which rules out that option.
NCERT states that the first cellular forms of life appeared about 2000 million years ago.
500 mya is far too recent — that is around the era of invertebrate diversification.
Answer: 2000 mya.

Answer

2000 mya

Shortcut

Anchor on the two ends: Earth 4500 mya and cellular life 2000 mya. Everything else in this timeline falls between them, and holding just those two numbers lets you place most options by plausibility.

Where it went wrong

3000 mya is a near-miss on the right order of magnitude — the correct scale but the wrong figure. Dates in this chapter are pure recall with no way to derive them, which makes a written timeline the only reliable preparation. It is worth noting that 4500 mya was also on the list and is the age of the Earth, not of life.

Q81 The disappearance of the dinosaurs Not attempted

Dinosaurs suddenly disappeared from the earth

about 200 mya due to climatic changes
KEYabout 65 mya as most of them evolved into birds
about 100 mya due to a meteorite hit
50 mya as they moved into seas to evolve into fish-like reptiles

Given

The extinction of the dinosaurs.

Asked

When and why they disappeared.

Concept to use

Two facts must both be right: the date, 65 million years ago, and the explanation NCERT offers — that some dinosaurs evolved into birds. The options mix correct dates with wrong reasons and vice versa, so each has to be checked on both counts.

Formula to use

Dinosaurs disappeared ≈ 65 mya; NCERT notes that some evolved into birds

Baby steps

The date is 65 million years ago, at the end of the Cretaceous — so options giving 200, 100 or 50 mya are out on the date alone.
NCERT's account is that dinosaurs suddenly disappeared and that most of them evolved into birds.
Only one option gives both the right date and that reason.
Answer: about 65 mya, as most of them evolved into birds.

Answer

about 65 mya as most of them evolved into birds

Shortcut

Filter on the date first, since it is a single number and is unambiguous. 65 mya appears in only one option here, so the reason need not even be evaluated. Checking the more clear-cut of two attributes first is generally the faster route in questions of this shape.

Birds are widely regarded as living dinosaurs, and Archaeopteryx is the classic transitional fossil linking reptiles to birds. Connecting those two facts gives this answer a foundation rather than leaving it as bare recall.

Q74 Statements on the origin and evolution of man Not attempted

Which of the following statements is correct about the origin and evolution of men?

Agriculture came around 50,000 years back.
The Dryopithecus and Ramapithecus primates existing 15 million years ago, walked like men.
Homo habilis buried their dead.
KEYNeanderthal men lived in Asia between 1,00,000 and 40,000 years back.

Given

Four statements about human evolution.

Asked

The correct statement.

Concept to use

Each wrong option alters one specific detail. Agriculture began around 10,000 years ago, not 50,000. Dryopithecus and Ramapithecus were hairy and walked like gorillas and chimpanzees, not like men. It was Neanderthals, not Homo habilis, who buried their dead. The Neanderthal date range of 100,000 to 40,000 years is as NCERT gives it.

Formula to use

Key dates: Neanderthals 1,00,000–40,000 years ago · Homo sapiens arose in Africa · agriculture ≈ 10,000 years ago

Baby steps

Agriculture at 50,000 years: the accepted figure is about 10,000 years. Wrong.
Dryopithecus and Ramapithecus walking like men: NCERT describes them as hairy and walking like gorillas and chimpanzees. Wrong.
Homo habilis burying their dead: burial of the dead is attributed to Neanderthals. H. habilis probably did not even eat meat. Wrong.
Neanderthals in Asia, 1,00,000–40,000 years: matches the standard account. Correct.

Answer

Neanderthal men lived in Asia between 1,00,000 and 40,000 years back.

Shortcut

This section is almost entirely about attaching the right date or behaviour to the right name. Build a single timeline card — Dryopithecus/Ramapithecus 15 mya, Homo habilis 2 mya, Homo erectus 1.5 mya, Neanderthals 1,00,000–40,000, agriculture 10,000 — and most questions here become lookups.

Burial of the dead is a distinctive Neanderthal trait and appears in questions repeatedly. Pair it with the date range in the same memory item, since both concern the same species.

Q59 Control of LH release in males Marked wrong

In males, release of LH is stimulated by ___X___ and inhibited by ___Y___. Mark the correct set.

	X	Y
1	FSH	GnRH
2	GnRH	FSH
3	GnRH	testosterone
4	testosterone	inhibin

1
2
KEY3
MARKED4

Given

Regulation of LH secretion in the male.

Asked

What stimulates and what inhibits LH.

Concept to use

The male axis runs hypothalamus → pituitary → testis, with feedback closing the loop. GnRH from the hypothalamus stimulates the pituitary to release LH. LH then acts on the Leydig cells to produce testosterone, and testosterone feeds back negatively to suppress LH. Note that the feedback partner is specific: testosterone inhibits LH, while inhibin inhibits FSH.

Formula to use

GnRH → LH → testosterone ⊣ (negative feedback on LH) | inhibin ⊣ FSH

Baby steps

GnRH from the hypothalamus stimulates the anterior pituitary to secrete both LH and FSH. So X = GnRH.
LH acts on Leydig cells, which secrete testosterone.
Rising testosterone exerts negative feedback on the hypothalamus and pituitary, reducing LH. So Y = testosterone.
Set 3 matches: stimulated by GnRH, inhibited by testosterone.

Answer

Shortcut

Pair each pituitary hormone with its own feedback inhibitor: LH with testosterone, FSH with inhibin. Once that pairing is fixed, and GnRH is recognised as the universal stimulator of both, every question of this type resolves in one step.

Where it went wrong

Option 4 pairs testosterone as the stimulator and inhibin as the inhibitor — both wrong, though both hormones are genuinely part of the system. Inhibin is real but belongs to the FSH loop, not the LH loop. Naming the correct partner for each hormone is what this question is testing.

Q85 What the interstitial spaces contain Marked wrong

Statement I: The interstitial spaces between the seminiferous tubules contain Sertoli cells.
Statement II: Other immunologically competent cells are also present in the interstitial spaces.

MARKEDBoth statement I and statement II are correct.
Both statement I and statement II are incorrect.
Statement I is correct but statement II is incorrect.
KEYStatement I is incorrect but statement II is correct.

Given

Two statements about the interstitial spaces of the testis.

Asked

Which statements are correct.

Concept to use

The testis has two compartments and every cell type belongs to exactly one. Inside the seminiferous tubules: germ cells and the Sertoli cells that nourish them. Outside, in the interstitial spaces: Leydig cells, blood vessels and immunologically competent cells. Sertoli cells are anchored inside the tubule, so Statement I places them in the wrong compartment.

Formula to use

Inside tubule: germ cells + Sertoli | Interstitial space: Leydig + blood vessels + immune cells

Baby steps

Statement I. Sertoli cells lie inside the seminiferous tubules, attached to the basement membrane and extending towards the lumen. They are not interstitial. Incorrect.
Statement II. NCERT lists immunologically competent cells among the contents of the interstitial spaces, alongside Leydig cells and blood vessels. Correct.
Answer: I incorrect, II correct.

Answer

Statement I is incorrect but statement II is correct.

Shortcut

One mnemonic settles every question on this topic: Sertoli sits inSide the tubule; Leydig Lies outside. Sorting the named cell into its compartment is the entire question.

Where it went wrong

This is the third time the Sertoli/Leydig distinction has appeared across these papers — it cost Q59 and Q86 in an earlier zoology set, and it has cost this one too. That makes it the single highest-value fact to fix in the reproduction chapters. Statement II being true adds credibility to the pair and makes it tempting to accept both, so judge each statement alone.

Q51 Fate of the ruptured Graafian follicle Marked wrong

Immediately after ovulation, the ruptured Graafian follicle transforms into

corpus callosum
MARKEDcorpus albicans
KEYcorpus luteum
corpus spongiosum

Given

The Graafian follicle immediately after it ruptures at ovulation.

Asked

What it becomes.

Concept to use

The sequence matters. Immediately after ovulation the ruptured follicle becomes the corpus luteum, a yellow body that secretes progesterone. Only later, if fertilisation does not occur, does the corpus luteum degenerate into the corpus albicans, a white scar. The word immediately in the question picks out the first of these two stages.

Formula to use

Graafian follicle → corpus luteum (secretes progesterone) → corpus albicans (if no fertilisation)

Baby steps

At ovulation the Graafian follicle ruptures and releases the secondary oocyte.
The remaining follicular cells reorganise into the corpus luteum, which secretes large amounts of progesterone to maintain the endometrium.
If fertilisation does not occur, the corpus luteum degenerates after about 10–12 days into the corpus albicans.
The question says immediately after ovulation, so the answer is the corpus luteum.
The other two options are distractors from elsewhere: corpus callosum is in the brain, corpus spongiosum in the penis.

Answer

corpus luteum

Shortcut

Use the colours: luteum = yellow, albicans = white. The yellow, active body comes first and the white scar comes later — so “immediately after” always means corpus luteum.

Where it went wrong

Corpus albicans is the correct later stage, so this is a timing error rather than a knowledge gap. The word immediately is doing the work in this question. Two of the four options come from completely different organ systems, which narrows the real choice to these two — and then only the timing separates them.

Q78 Structure of the ovary Marked wrong

Statement I: Each ovary is connected to the pelvic wall and uterus by ligaments.
Statement II: Each ovary is covered by a thick epithelium which encloses the ovarian stroma.

MARKEDBoth statement I and statement II are correct.
Both statement I and statement II are incorrect.
KEYStatement I is correct but statement II is incorrect.
Statement I is incorrect but statement II is correct.

Given

Two statements about ovarian structure.

Asked

Which statements are correct.

Concept to use

Statement II hinges on one adjective. The ovary is covered by a thin epithelium — a single layer of germinal epithelium — not a thick one. Statement I is correct as written: the ovary is anchored to the pelvic wall and to the uterus by ligaments.

Formula to use

Ovary: thin epithelium covering the ovarian stroma, which divides into peripheral cortex and inner medulla

Baby steps

Statement I. The ovary is held in place by ligaments attaching it to the pelvic wall and the uterus. Correct.
Statement II. The covering is a thin epithelium, not thick. Everything else in the statement is right, but that one adjective makes it false. Incorrect.
Answer: I correct, II incorrect.

Answer

Statement I is correct but statement II is incorrect.

Shortcut

In anatomy statements, the adjectives are where errors are planted — thick versus thin, single versus paired, inner versus outer. Read them separately from the nouns, since the structure named is usually right and only the descriptor is altered.

Where it went wrong

Both statements were accepted, so “thick” passed unchallenged. This is the same failure as Q75 in the previous chemistry paper, where “paired prostate” slipped through — a single wrong adjective inside an otherwise accurate sentence. Underlining every descriptive word before judging is the habit that catches these.

Q73 Product of meiosis I in oogenesis Not attempted

In oogenesis, completion of meiosis I results in the formation of

KEYlarge haploid secondary oocyte and a tiny first polar body
large diploid primary oocyte and a tiny first polar body
tiny haploid secondary oocyte and large second polar body
large diploid secondary oocyte and a tiny first polar body

Given

Completion of meiosis I during oogenesis.

Asked

What is formed.

Concept to use

Three things must all be right. Meiosis I is the reduction division, so the products are haploid. The cytoplasm divides unequally, giving one large cell and one tiny one — the egg hoards cytoplasm for the future embryo. And the products of meiosis I are specifically the secondary oocyte and the first polar body.

Formula to use

Primary oocyte (2n) —meiosis I→ secondary oocyte (n, large) + first polar body (n, tiny)

Baby steps

Meiosis I halves the chromosome number, so both products are haploid — that eliminates every “diploid” option.
Cytoplasmic division is unequal: the secondary oocyte keeps nearly all the cytoplasm and is large; the polar body is tiny.
The polar body formed at this stage is the first polar body; the second appears only after meiosis II.
Answer: large haploid secondary oocyte and a tiny first polar body.

Answer

large haploid secondary oocyte and a tiny first polar body

Shortcut

Check the three attributes as a checklist — ploidy, size, and which polar body — and reject on the first failure. “Diploid” after meiosis I is impossible, which kills two options straight away.

The unequal cytoplasmic division is the defining difference between oogenesis and spermatogenesis, and it explains why one oocyte yields a single ovum while one spermatocyte yields four sperms. The same point was tested in the earlier zoology paper (Q80).

Q55 Source of oxytocin in the foetal ejection reflex Marked wrong

Foetal ejection reflex triggers the release of oxytocin from

MARKEDfoetal neurohypophysis
foetal adenohypophysis
KEYmaternal neurohypophysis
maternal adenohypophysis

Given

The foetal ejection reflex during labour.

Asked

The source of the oxytocin released.

Concept to use

Two decisions. Whose pituitary? The oxytocin that drives labour acts on the mother's uterine muscle and comes from the mother's pituitary — the foetal signal merely triggers the reflex. Which lobe? Oxytocin is stored and released by the neurohypophysis (posterior pituitary), not the adenohypophysis (anterior).

Formula to use

Foetal signal → maternal neurohypophysis releases oxytocin → uterine contraction (positive feedback)

Baby steps

The foetal ejection reflex begins with signals from the fully developed foetus and the placenta.
These trigger release of oxytocin from the maternal pituitary — it must act on the mother's uterus, so it is her hormone.
Oxytocin is a posterior pituitary hormone, i.e. from the neurohypophysis.
The oxytocin causes stronger uterine contractions, which in turn stimulate more oxytocin — a positive feedback loop until delivery.
Answer: maternal neurohypophysis.

Answer

maternal neurohypophysis

Shortcut

Two independent choices, each halving the list. Whose body must respond? The mother's — so maternal. Which lobe releases oxytocin? Posterior — so neurohypophysis. Answering them separately is quicker than evaluating four full phrases.

Where it went wrong

“Foetal” was chosen, presumably because the reflex is named after the foetus. But the foetus initiates the signal while the mother supplies the hormone — the name describes the trigger, not the source. Keep the two posterior pituitary hormones together: oxytocin and vasopressin from the neurohypophysis; everything else is anterior.

Q49 Contraceptive that inhibits ovulation for longer Marked wrong

Which of the following contraceptive method inhibits the ovulation and has longer effective period?

Tubectomy
KEYHormonal injection
Copper releasing IUD
MARKEDContraceptive pills

Given

A contraceptive that both inhibits ovulation and has a long effective period.

Asked

Which method fits both conditions.

Concept to use

Two conditions must be satisfied together. Inhibiting ovulation rules out IUDs, which act mainly by preventing implantation and by affecting sperm motility, and rules out tubectomy, which blocks the oviducts without touching ovulation at all. That leaves the two hormonal methods — and between them, injections last far longer than daily pills.

Formula to use

Pills: taken daily | Injections and implants: effective for months — both inhibit ovulation

Baby steps

Tubectomy is a surgical sterilisation blocking the fallopian tubes. Ovulation continues normally. Fails the first condition.
Copper IUD acts by increasing phagocytosis of sperm and suppressing sperm motility, not by stopping ovulation. Fails the first condition.
Contraceptive pills do inhibit ovulation — but must be taken daily, so the effective period is short. Fails the second condition.
Hormonal injections inhibit ovulation and remain effective for months. Satisfies both.

Answer

Hormonal injection

Shortcut

Apply the two filters in order. First keep only the methods that stop ovulation — that leaves pills and injections. Then pick the longer-acting of the two. Applying the conditions sequentially avoids weighing four options against two criteria at once.

Where it went wrong

Contraceptive pills satisfy the first condition but not the second — they are taken daily, which is the shortest effective period of any hormonal method. When a question states two requirements, check both before committing; an option meeting one of them will always look plausible.

Q53 Assertion–reason on condoms Marked wrong

Assertion (A): Usage of condoms give privacy to the user.
Reason (R): They can be self-inserted.

KEYBoth (A) and (R) are correct and (R) is the correct explanation of (A).
MARKEDBoth (A) and (R) are correct but (R) is not the correct explanation of (A).
(A) is correct but (R) is not correct.
(A) is not correct but (R) is correct.

Given

Assertion: condoms give privacy to the user.
Reason: they can be self-inserted.

Asked

Judge each statement and whether R explains A.

Concept to use

NCERT states explicitly that condoms are popular partly because they are self-inserted and thereby give privacy to the user — the two facts appear in the same sentence, joined causally. Being able to use the device without medical assistance is the reason privacy is possible, so R is the explanation rather than a separate observation.

Formula to use

Self-insertion → no clinical visit or assistance required → privacy

Baby steps

Is A true? Yes — NCERT lists privacy among the advantages of condoms. True.
Is R true? Yes — condoms are self-inserted, unlike IUDs which need a doctor or trained nurse. True.
Does R explain A? Apply the removal test: if a condom required a clinician to fit it, the user would lose the privacy. So self-insertion is precisely what delivers the privacy.
Both correct, and R is the correct explanation.

Answer

Both (A) and (R) are correct and (R) is the correct explanation of (A).

Shortcut

The removal test settles almost every assertion–reason item in one line: if R were false, would A still hold? Here, remove self-insertion and privacy disappears — so R is the explanation.

Where it went wrong

Both statements were accepted as true but the link was denied — the identical shape to Q43 in this paper and to Q149 and Q128 in earlier papers. Across this whole series, when both parts of an assertion–reason pair are true, R has been the correct explanation far more often than not. After marking both true, do not stop: run the removal test before choosing.

Q64 Statements about MTP Not attempted

Choose the correct set of statements about MTP.
A. It is an induced abortion performed by a qualified medical practitioner.
B. It is considered relatively safe during the first trimester.
C. It is performed in the second trimester if the mother's life is at risk.
D. It is legally practiced to promote female foeticide in India.
E. It is primarily used as a regular method of contraception.

KEYA, B and C
B, C and D
All except D
Only B

Given

Five statements about Medical Termination of Pregnancy.

Asked

Which set is correct.

Concept to use

MTP is legal in India under specified conditions and is safest in the first trimester; later terminations are riskier and are performed when there is a serious reason, such as danger to the mother's life. Two statements are clearly false: MTP is legally regulated to prevent female foeticide, not to promote it, and it is not a method of contraception — it terminates an established pregnancy.

Formula to use

MTP: legal, performed by a qualified practitioner, safest in the first trimester, not a contraceptive

Baby steps

A. MTP is an induced abortion carried out by a qualified medical practitioner. Correct.
B. It is relatively safe during the first trimester. Correct.
C. Second-trimester MTPs are performed where there is serious cause, such as risk to the mother. Correct.
D. The law regulating MTP exists partly to prevent misuse for female foeticide. Incorrect.
E. MTP is not a contraceptive method — contraception prevents pregnancy, MTP ends one. Incorrect.
Answer: A, B and C.

Answer

A, B and C

Shortcut

Statements D and E both make claims that run against the purpose of the legislation and against the definition of contraception. Rejecting both leaves only one option containing A, B and C without them — and “All except D” fails because E is also wrong.

“All except D” is the trap for spotting only one of the two false statements. When a question offers an “all except one” option, check whether there is a second false item — that is usually why the option is there.

Q54 Which STIs are curable Marked wrong

Which of the following STIs are curable?

KEYChlamydiasis, Trichomoniasis, Gonorrhoea
Hepatitis B, HIV, Trichomoniasis
MARKEDSyphilis, Trichomoniasis, Genital herpes
Genital herpes, Hepatitis B, HIV

Given

A list of sexually transmitted infections.

Asked

Which set is curable.

Concept to use

The dividing line is the causative organism. Bacterial and protozoan infections — chlamydiasis, gonorrhoea, syphilis, trichomoniasis — are curable if detected early, because antibiotics and antiprotozoals work against them. Viral infections — genital herpes, hepatitis B, HIV — are not curable. So the task reduces to finding the option containing no virus.

Formula to use

Bacterial and protozoal STIs → curable | viral STIs (herpes, hepatitis B, HIV) → not curable

Baby steps

Identify the viral STIs first: genital herpes, hepatitis B and HIV. Any option containing one of these is out.
That removes three of the four options at once.
The remaining option — chlamydiasis (bacterial), trichomoniasis (protozoan), gonorrhoea (bacterial) — contains no virus.
All three are curable if detected and treated early.
Answer: Chlamydiasis, Trichomoniasis, Gonorrhoea.

Answer

Chlamydiasis, Trichomoniasis, Gonorrhoea

Shortcut

Scan for the three viral names and strike out any option containing them. That single pass answers the question — you never need to know which of the remaining infections is bacterial and which protozoal.

Where it went wrong

The chosen option includes genital herpes, which is viral and therefore not curable. Syphilis and trichomoniasis in the same option are both curable, so two thirds of the answer was right — but every member of the set must qualify. In “which set” questions, one wrong member condemns the whole option.

The correct option is marked KEY and the option selected in the test is marked MARKED; questions with no marked option were left unattempted. All figures have been drawn fresh for these notes, and every ratio and count here was verified by computing the cross before it was written in.

Four chapters, forty-three questions

Chapters in this paper — red = wrong · amber = blank · green = correct

Principles of Inheritance and Variation

Mendelian crosses and counting

Cross types and their ratios

Dominance and its exceptions

Linkage, recombination and mapping

Laws and their chromosomal basis

Evolution

Hardy-Weinberg and genetic drift

Evidences and patterns of evolution

Origin of life and the fossil record

Human evolution

Human Reproduction

Male reproductive system and hormones

Ovary and the ovarian cycle

Gametogenesis

Parturition

Reproductive Health

Contraceptive methods

MTP

Sexually transmitted infections

What the forty-three have in common

Eight results that cover the four chapters