Every botany question saved from the test — all of them from a single chapter — arranged by sub-topic within it. Each carries what was given, what was asked, the concept and rule behind it, the steps in full, and a Punnett square or genetics figure wherever one makes the answer visible.
16
Questions lost
8
Attempted, wrong
8
Left blank
1
Chapter involved
All sixteen come from Principles of Inheritance and Variation, so they are grouped here by sub-topic. On NEET marking they were worth 64 marks, and the eight wrong attempts cost 8 more.
This is the tightest cluster in the whole test paper: one chapter, sixteen questions, and a single recurring failure mode. Six of the eight wrong answers were lost to a word or a count — “always”, a ratio written backwards, boxes counted instead of genotypes — not to a gap in the genetics.
Chapters in this paper — red = wrong · amber = blank · green = correct
Mendelian genetics, its cytological basis and its exceptions · 16 questions · 8 wrong · 8 blank
Mendelian crosses, ratios and counting
7 questions · 4 wrong · 3 blank
Q92Which cross gives 3 : 1Marked wrong
Which one of the following crosses shows the phenotypic ratio of 3:1?
AaBb × AaBb
aabb × AABb
KEYAabb × Aabb
MARKEDAaBb × aabb
Given
Four crosses involving two genes A and B.
Asked
Which cross produces a 3 : 1 phenotypic ratio.
Concept to use
Handle each gene separately, then multiply the two ratios together. A 3 : 1 result means one gene must behave as a monohybrid cross (Aa × Aa → 3 : 1) while the other gene contributes nothing to the variation — which happens only when both parents are identical homozygotes for it, so every offspring looks the same for that trait.
Formula to use
Aa × Aa → 3 : 1 | Aa × aa → 1 : 1 | AA × anything → uniform
Baby steps
AaBb × AaBb: gene A gives 3:1 and gene B gives 3:1, so together (3:1)(3:1) = 9:3:3:1. ✗
aabb × AABb: gene A gives all Aa (uniform), gene B gives Bb × bb = 1:1. Result 1:1. ✗
Aabb × Aabb: gene A is Aa × Aa → 3:1. Gene B is bb × bb → all bb, uniform, so it contributes a factor of 1. Result 3 : 1. ✓
AaBb × aabb: this is a test cross — (1:1)(1:1) = 1:1:1:1. ✗
Answer
Aabb × Aabb
Shortcut
Count the heterozygous pairs shared by both parents. Each such gene multiplies the ratio by (3:1); every other gene multiplies by 1 or splits it 1:1. For a clean 3:1 you need exactly one shared heterozygous gene and everything else fixed — Aabb × Aabb is the only cross that fits.
Where it went wrong
AaBb × aabb is a test cross, and test crosses never give 3:1 — they give 1:1 or 1:1:1:1, because the recessive parent contributes only one kind of gamete and cannot mask anything. If a cross has a fully recessive parent, 3:1 is impossible by construction.
Q98Four true/false statements on Mendelian geneticsMarked wrong
Study the following statements and identify them as True (T) or False (F). A. In a dissimilar pair of factors, one member of the pair always dominates the other completely. B. In the F2 generation of dihybrid cross, the ratio is 1:2 between genotypes rrYy and RRyy. C. Nine types of genotypes and 4 types of phenotypes are obtained in F2 generation of the Mendelian dihybrid cross. D. Hereditary variations of Drosophila can be seen with low power microscopes.
MARKEDA-T, B-F, C-T, D-F
A-F, B-T, C-T, D-F
KEYA-F, B-F, C-T, D-T
A-T, B-T, C-F, D-F
Given
Four statements, each to be marked True or False.
Asked
The correct combination of truth values.
Concept to use
Three of these are settled by a single counter-example or a single count. The word “always” in statement A is the tell — absolute claims in genetics are usually false, because incomplete dominance and codominance exist. Statement B needs you to count boxes in the F2 grid rather than trust the order the genotypes are written in.
A. “Always dominates completely” is false — in snapdragon flower colour and in pea starch grain size the heterozygote is intermediate (incomplete dominance), and in ABO blood groups IA and IB are codominant. A is False.
B. In the 16-box F2 grid, rrYy appears in 2 boxes and RRyy in 1. So the ratio is 2 : 1, not 1 : 2 — the statement has it backwards. B is False.
C. The F2 of a dihybrid gives 3 × 3 = 9 genotypes and 2 × 2 = 4 phenotypes. C is True.
D.Drosophila traits such as eye colour, body colour and wing shape are visible under a hand lens or low-power microscope — this is how Morgan’s lab scored them. D is True.
Combination: A-F, B-F, C-T, D-T.
Answer
A-F, B-F, C-T, D-T
Shortcut
In a four-statement T/F question, find the one statement you are most certain about and eliminate on it first. Here C is certain (9 and 4), which removes one option immediately; then A being false removes two more, leaving only one. You never have to adjudicate D.
Where it went wrong
Two errors were made. A was marked True — but Mendel’s own law of dominance is stated as a general tendency, and the very next NCERT section is incomplete dominance, so “always” cannot stand. D was marked False, probably because it sounds too casual to be a real fact. It is a real fact: Morgan scored millions of flies precisely because the variations were visible at low magnification.
Q110Phenotypes in a trihybrid crossNot attempted
In a trihybrid cross between two heterozygous individuals (AaBbCc × AaBbCc), how many different phenotypic combinations are possible, assuming complete dominance?
16
12
4
KEY8
Given
Cross AaBbCc × AaBbCc.
Complete dominance at all three loci.
Asked
Number of different phenotypic combinations.
Concept to use
With complete dominance, each gene offers exactly two phenotypes — dominant or recessive. The genes act independently, so the total number of phenotype combinations is the product: 2 × 2 × 2. The general rule is 2n for n heterozygous genes.
Gene A: offspring are either A_ (dominant) or aa (recessive) → 2 phenotypes.
Gene B: likewise 2. Gene C: likewise 2.
The three genes assort independently, so multiply: 2 × 2 × 2 = 8.
Check with the formula: n = 3, so 23 = 8 phenotypes. (For comparison, genotypes would be 33 = 27 and the grid would have 64 boxes.)
Answer
8
Shortcut
Keep the three powers together on one line: for n heterozygous genes, 2n gametes, 2n phenotypes, 3n genotypes, 4n Punnett boxes. Read which one the question wants and evaluate a single power — no grid ever needs drawing.
16 is 42, the box count for a dihybrid, and 4 is the dihybrid phenotype count — both are right answers to the wrong question. Check n before you compute.
Q120Back cross with the double recessiveNot attempted
In a certain plant, the yellow fruit colour (Y) is dominant to the green fruit colour (y) and the round shape (R) is dominant to the oval shape (r). The two genes involved are located on different chromosomes. Identify the outcome of this particular cross when a plant with genotype YyRr is back-crossed with the double recessive parent?
9 : 3 : 3 : 1 ratio of phenotypes only
9 : 3 : 3 : 1 ratio of genotypes only
1 : 1 : 1 : 1 ratio of phenotypes only
KEY1 : 1 : 1 : 1 ratio of phenotypes and genotypes.
Given
YyRr crossed with yyrr (double recessive).
Genes on different chromosomes.
Asked
The phenotypic and genotypic outcome.
Concept to use
A cross with the double recessive is a test cross. The recessive parent produces just one kind of gamete (ry), so it cannot mask anything — every offspring’s appearance is decided entirely by which gamete came from the heterozygous parent. That means the four gamete types map one-to-one onto four genotypes and four phenotypes, in equal numbers.
Formula to use
Test cross: offspring ratio = gamete ratio of the heterozygous parent = 1 : 1 : 1 : 1
Baby steps
YyRr produces four gamete types in equal proportion: YR, Yr, yR, yr — 25% each.
yyrr produces only ry.
Combining gives four offspring: YyRr, Yyrr, yyRr, yyrr — one of each, so genotypes are 1 : 1 : 1 : 1.
Phenotypes: yellow round, yellow oval, green round, green oval — each genotype gives its own distinct appearance, so phenotypes are also 1 : 1 : 1 : 1.
Both ratios are 1 : 1 : 1 : 1.
Answer
1 : 1 : 1 : 1 ratio of phenotypes and genotypes.
Shortcut
The defining feature of a test cross is that genotypic and phenotypic ratios are identical, because the recessive parent hides nothing. So an option offering only one of the two is incomplete by definition — the answer must be the one naming both.
9 : 3 : 3 : 1 belongs to the F2 of a dihybrid self-cross, not to a test cross. And it is only ever a phenotypic ratio — the genotypic ratio there is 1:2:1:2:4:2:1:2:1, so the second option is doubly wrong.
Q122Gametes from a fully homozygous organismNot attempted
How many types of gametes can be produced by a diploid organism which is homozygous for all its 4 loci?
16 types
8 types
KEYOnly one type
4 types
Given
Diploid organism, four loci, homozygous at every one.
Asked
Number of gamete types.
Concept to use
The formula 2n counts heterozygous loci, not total loci. A homozygous locus such as AA can only ever contribute A — there is no choice to make, so it multiplies the count by 1, not 2. With all four loci homozygous, n = 0 and the answer is 20 = 1.
Formula to use
Gamete types = 2n, where n = number of heterozygous loci
Baby steps
Take the organism as AABBCCDD (any homozygous combination works the same way).
Locus A: both alleles are A, so the gamete gets A. One possibility.
Same for B, C and D — one possibility each.
Total = 1 × 1 × 1 × 1 = one type of gamete, ABCD.
Formula check: heterozygous loci n = 0, so 20 = 1. ✓
Answer
Only one type
Shortcut
Before applying 2n, count the heterozygous loci only. The distractor 16 is 24, which is what you get by counting all four loci — that is the whole point of the question, and it is placed first among the options for exactly that reason.
This is why pure-breeding lines breed true: one gamete type means every offspring of a self-cross is identical to the parent. It is also why Mendel began with pure lines — they gave him a fixed starting point.
Q127How many genotypes in F2 of RR × rrMarked wrong
How many different genotypes are possible from a cross between the parents RR and rr in F2 generation?
MARKEDFour
One
KEYThree
Two
Given
Parents RR × rr.
F2 generation is asked for, not F1.
Asked
Number of different genotypes in F2.
Concept to use
Two steps, and the question is testing whether you take both. RR × rr gives an F1 that is uniformly Rr. The F2 comes from selfing that F1, and Rr × Rr gives the familiar 1 : 2 : 1. The number of genotypes is three; the number of phenotypes is two.
Formula to use
RR × rr → F1 all Rr → F2 1 RR : 2 Rr : 1 rr
Baby steps
Cross the parents: RR gives only R gametes, rr gives only r gametes, so every F1 is Rr — one genotype.
Self the F1: Rr × Rr. Each parent gives R and r.
The four boxes give RR, Rr, Rr, rr.
Distinct genotypes: RR, Rr, rr → three (in the ratio 1 : 2 : 1).
Note the phenotypes number only two (3 dominant : 1 recessive) — the question asked for genotypes.
Answer
Three
Shortcut
For a monohybrid F2 the counts are fixed and worth memorising as a pair: 3 genotypes (1:2:1) and 2 phenotypes (3:1). Whichever the question asks for, one of those two numbers is the answer.
Where it went wrong
“Four” is the number of boxes in the Punnett square, not the number of distinct genotypes — Rr fills two of them and counts once. Filling in the grid and then reading off the box count instead of the distinct entries is the single most common slip in this topic. Circle the repeats before you count.
Q131Counting offspring from TtYy × TtyyMarked wrong
When a cross is made between plants with genotypes TtYy × Ttyy, 800 plants are obtained. How many of them are tall plants with yellow seeds, tall plants with green seeds, dwarf plants with yellow seeds, and dwarf plants with green seeds, respectively?
MARKED200, 200, 200 and 200
KEY300, 300, 100 and 100
250, 250, 150 and 150
300, 100, 300 and 100
Given
Cross TtYy × Ttyy.
Total offspring = 800.
T = tall (dominant), t = dwarf; Y = yellow (dominant), y = green.
Asked
Numbers of each of the four phenotype classes, in the order given.
Concept to use
Split by gene, because the two genes behave differently in this cross. For height both parents are Tt, so that gene runs as a monohybrid cross giving 3 : 1. For seed colour the parents are Yy × yy, which is a test cross giving 1 : 1. Multiply the two ratios to get the four classes.
Read the two genes separately before writing anything. Same heterozygote on both sides → 3:1; heterozygote against homozygous recessive → 1:1. Multiplying 3:1 by 1:1 gives 3:3:1:1, and the total divides by 8. This takes about fifteen seconds and needs no Punnett square at all.
Where it went wrong
Marking 200 each treats the whole thing as a test cross giving 1:1:1:1 — but only the Y gene is a test cross here. The T gene has Tt on both sides, so it cannot give equal numbers. A quick check: if all four classes were equal, tall and dwarf would be 400 each, which contradicts Tt × Tt producing three times as many tall as dwarf.
Mendel’s laws and their chromosomal basis
6 questions · 2 wrong · 4 blank
Q93Bivalent orientation at metaphase INot attempted
A. The maternal chromosomes of both bivalents always move towards one pole and paternal chromosomes towards another pole. B. The maternal chromosome of one bivalent and the paternal chromosome of another bivalent always moves towards a pole and another combination of paternal of one bivalent and maternal of another bivalent to another pole. C. The movement of maternal and paternal chromosomes of a bivalent towards respective poles is independent of the movement of maternal and paternal chromosomes of another bivalent.
Statement A is correct, and statement C is incorrect.
Statement B is correct.
Statement C is incorrect, and statement B is correct.
KEYStatement C is correct.
Given
Three statements about how bivalents orient and separate during meiosis I.
Asked
Which statement is correct.
Concept to use
This is the physical basis of independent assortment. At metaphase I each bivalent lines up on the plate with its maternal and paternal members facing whichever pole they happen to face — and each bivalent decides this independently of every other. The word to attack in statements A and B is “always”: any claim of a fixed pattern contradicts the randomness that independent assortment depends on.
Statement A claims all maternal chromosomes go to one pole together. That is one possible outcome, but saying “always” makes it false — if it were always so, parental combinations would never be reshuffled. Incorrect.
Statement B claims the opposite fixed pattern — always a mixed combination. Also one possible outcome, also not always. Incorrect.
Statement C says each bivalent behaves independently of the others. That is exactly right, and it is what makes both A’s and B’s patterns possible with equal probability. Correct.
Answer: Statement C is correct.
Answer
Statement C is correct.
Shortcut
Statements A and B describe opposite outcomes and both say “always”. Two mutually exclusive claims cannot both be always true, and there is no reason to prefer one, so both must be false — which leaves C without needing to read it carefully.
This is the cytological explanation of Mendel’s third law. With 23 pairs in humans, 223 ≈ 8.4 million orientations are possible — before crossing over is even counted.
Q97Assertion–reason — gametes are never hybridNot attempted
Assertion (A): Gametes are never hybrids in nature. Reason (R): Two alleles of a gene are separated during gamete formation in Anaphase I. Thus, each gamete receives only one of the pair of alleles, not both.
KEYBoth (A) and (R) are true, and (R) is the correct explanation of (A).
Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(A) is true, but (R) is false.
Both (A) and (R) are false.
Given
Assertion: gametes are never hybrid.
Reason: alleles separate at anaphase I, so each gamete gets one allele.
Asked
Truth of each, and whether R explains A.
Concept to use
“Hybrid” means heterozygous — carrying two different alleles. A gamete is haploid, so it holds only one allele of each gene and physically cannot be heterozygous. The reason states the mechanism that makes this so: homologous chromosomes, and therefore the two alleles, are pulled to opposite poles at anaphase I. This is the law of segregation stated cytologically.
Formula to use
Diploid parent (Aa) → anaphase I separates homologues → gametes are A or a, never Aa
Baby steps
Is A true? Yes. Gametes are haploid; a haploid cell carries one allele per gene, so it cannot be a hybrid. Mendel called this the purity of gametes.
Is R true? Yes. At anaphase I the two homologous chromosomes — one carrying A, the other a — move to opposite poles.
Does R explain A? The separation at anaphase I is precisely why each gamete ends up with one allele. Remove R and A would not follow.
So: both true, and R is the correct explanation.
Answer
Both (A) and (R) are true, and (R) is the correct explanation of (A).
Shortcut
Whenever the reason describes the cell-division event behind a Mendelian law, it is almost always the correct explanation — the whole point of the chromosomal theory is that meiosis is the mechanism for Mendel’s rules. Segregation → anaphase I; independent assortment → metaphase I orientation.
Be precise about which anaphase. Alleles of a gene separate at anaphase I (homologues part). Sister chromatids separate at anaphase II. Questions frequently swap these two.
Q104Genes and chromosomes — which claim is too strongNot attempted
Read the following statements. A. Genes occur in pairs, and chromosomes also occur in pairs in diploid organisms. B. Genes and chromosomes segregate at the time of gamete formation. C. Independent pairs of genes and chromosomes segregate independently of each other during gamete formation. D. One pair of genes or chromosomes always segregates independently of another pair. Which of the statements given above is/are incorrect?
A and B only
C and D only
KEYD only
B and D only
Given
Four statements drawn from the chromosomal theory of inheritance.
Asked
Which are incorrect.
Concept to use
Statements A, B and C are the three parallels Sutton and Boveri drew between genes and chromosomes — all standard. Statement D is C with the word “always” smuggled in, and that single word makes it false: linked genes on the same chromosome do not segregate independently. Notice that C is carefully worded — it says independent pairs segregate independently, which is true by definition.
Formula to use
Independent assortment holds for genes on different chromosomes, or far apart on the same one
Baby steps
A. In a diploid, both alleles and homologous chromosomes come in pairs. Correct.
B. During meiosis both the gene pair and the chromosome pair separate. Correct.
C. Note the qualifier “independent pairs” — those that are independent do segregate independently. Correct.
D. Drops the qualifier and asserts it always happens. Morgan’s linked genes in Drosophila disprove this — genes close together on one chromosome travel together. Incorrect.
Answer: D only.
Answer
D only
Shortcut
Compare C and D directly — they are nearly the same sentence, and the difference is the word “always”. When two options in a list differ by one absolute word, the absolute one is the false one. Spotting that pair answers the question without reading A or B.
Words like always, never, all and only are the highest-yield thing to underline in biology statement questions. Genetics is full of exceptions — linkage, incomplete dominance, pleiotropy, polygeny — so absolute claims are usually the planted error.
Q108Statement of the Law of SegregationNot attempted
What does the Law of Segregation state about allele pairs during gamete formation?
The alleles blend together to form a new trait.
Both alleles are passed to each gamete.
KEYThe alleles segregate from each other, with each gamete receiving only one allele.
Only the dominant allele is passed to the gamete.
Given
Four candidate statements of Mendel’s law of segregation.
Asked
The correct statement of the law.
Concept to use
The law of segregation says the two alleles of a gene separate during gamete formation, so each gamete carries exactly one. Its physical basis is the parting of homologous chromosomes at anaphase I. The three wrong options each contradict a different well-known fact.
Formula to use
Parent Aa → gametes: ½ A and ½ a — one allele each, never both, never blended
Baby steps
“Blend together” — this is the discredited blending theory of inheritance, which Mendel’s work overturned. Recessive traits reappear unchanged in F2, which blending cannot explain. Wrong.
“Both alleles passed to each gamete” — that would make gametes diploid and hybrid, and the zygote would double its chromosome number every generation. Wrong.
“Alleles segregate, one per gamete” — this is the law. Correct.
“Only the dominant allele” — if true, recessive phenotypes could never reappear in F2, yet they do at 25%. Wrong.
Answer
The alleles segregate from each other, with each gamete receiving only one allele.
Shortcut
Test each option against one observation: recessive traits reappear in F2. Blending fails it, dominant-only fails it, and both-alleles fails the haploid requirement. A single anchoring fact eliminates three options.
This links directly to Q97 in the same paper — the assertion “gametes are never hybrids” is the same law stated a different way. Recognising two questions as one idea saves time in the exam.
Q128Assertion–reason on independent assortmentMarked wrong
Assertion (A): According to the law of independent assortment, segregation of alleles of a given trait occurs independently of segregation of alleles for another trait. Reason (R): Independent segregation of traits takes place only when the genes are located on different chromosomes or are far apart on the same chromosome.
MARKEDBoth (A) and (R) are true, but (R) is not the correct explanation of (A).
KEYBoth (A) and (R) are true, and (R) is the correct explanation of (A).
(A) is true, but (R) is false.
(A) is false, but (R) is true.
Given
An assertion stating the law of independent assortment.
A reason stating the physical condition under which it holds.
Asked
Whether both are true, and whether R explains A.
Concept to use
The two-part test: first mark each statement true or false on its own, then ask whether R supplies the mechanism behind A. Here R is not a separate fact sitting alongside A — it names the exact physical condition that makes A happen. Genes assort independently because they sit on different chromosomes, which line up at random on the metaphase plate. That is a causal link, so R explains A.
Formula to use
Correct explanation ⇔ R gives the mechanism or condition that produces A
Baby steps
Is A true? Yes — it is Mendel’s third law, stated correctly.
Is R true? Yes — independent assortment requires genes on different chromosomes, or far enough apart on one chromosome that crossing over separates them freely. Closely linked genes do not assort independently.
Does R explain A? Ask what would happen if R were false: if the genes were closely linked, independent assortment would fail. So R is not an incidental extra fact — it is the condition on which A depends.
Therefore both are true and R is the correct explanation.
Answer
Both (A) and (R) are true, and (R) is the correct explanation of (A).
Shortcut
Apply the removal test: if R were not true, would A still hold? If the answer is no, R is the explanation. If A would survive without R, then R is merely a true but unrelated statement. This one test settles almost every assertion–reason item.
Where it went wrong
The chosen option accepted both as true but denied the link. It reads as if R is a limitation on A rather than the reason for it — but a stated condition is a mechanism here. Watch for this shape: when R begins “this happens only when…”, it is usually explaining, not merely qualifying.
Q130Which statement is incorrect — dihybrid factsMarked wrong
Which of the following statements is incorrect?
In a dihybrid cross, the phenotypic ratio observed is 9:3:3:1.
KEYThe Law of Independent Assortment states that the segregation of one pair of characters is dependent on the segregation of another pair of characters.
MARKEDThere are four types of gametes produced in an F1 RrYy plant: RY, Ry, rY, and ry.
The frequency of each type of gamete (RY, Ry, rY, ry) in an F1 RrYy plant is 25%.
Given
Four statements about dihybrid crosses; exactly one is wrong.
Asked
The incorrect statement.
Concept to use
Three of the four are standard, verifiable facts about a dihybrid. The odd one out contains a single reversed word: the law of independent assortment says segregation of one pair is independent of the other, and this option says dependent. One word flips a true statement into a false one.
Formula to use
RrYy → 22 = 4 gamete types, each at 25% | F2 phenotypes 9 : 3 : 3 : 1
Baby steps
9:3:3:1 — the standard dihybrid F2 phenotypic ratio. Correct statement.
“dependent on” — the law says exactly the opposite. The very name of the law is Independent Assortment. This is the incorrect statement.
Four gametes RY, Ry, rY, ry — two heterozygous loci give 22 = 4 combinations. Correct statement.
25% each — four equally likely gamete types, so 1/4 = 25% each. Correct statement.
Answer: the statement about the law.
Answer
The Law of Independent Assortment states that the segregation of one pair of characters is dependent on the segregation of another pair of characters.
Shortcut
When a question asks for the incorrect statement and one option quotes a named law, check that option first — papers most often plant the error there, and usually by flipping a single word (independent/dependent, always/never, increases/decreases). Read the law statements with your finger on the key word.
Where it went wrong
Statement 3 was marked as incorrect, but RY, Ry, rY, ry is exactly right for RrYy — and statement 4, which says each of those four occurs at 25%, quietly confirms it. Two options agreeing with each other cannot both be the single wrong answer, so both could have been eliminated together.
Deviations from simple dominance
2 questions · 1 wrong · 1 blank
Q101The pea B gene shows two dominance patternsNot attempted
In pea plants, the alleles of B gene shows A. Incomplete dominance w.r.t. starch grain size B. Complete dominance w.r.t. seed shape C. Multiple allelism D. Codominance Which of the above are correct?
KEYA and B only
B and D only
C and D only
A and D only
Given
The B gene of pea, considered for two different characters.
Asked
Which statements about it are correct.
Concept to use
This is one of NCERT’s most quoted examples, and it makes a subtle point: whether dominance looks complete depends on how closely you look at the phenotype. At the level of seed shape, BB and Bb both give round seeds, so dominance appears complete. At the level of starch grain size — a finer, more direct measure of the gene product — Bb is intermediate, so dominance is incomplete.
A. Incomplete dominance for starch grain size. The heterozygote Bb makes grains of intermediate size, because one functional allele produces roughly half the starch-branching enzyme. Correct.
B. Complete dominance for seed shape. Bb seeds are as round as BB seeds — enough starch is made to keep the seed smooth. Only bb is wrinkled. Correct.
C. Multiple allelism. That means three or more alleles in the population, as in ABO blood groups. Only B and b are involved here. Incorrect.
D. Codominance. That requires both alleles to be expressed fully and separately, as in AB blood group. Bb is intermediate, not both-at-once. Incorrect.
Answer: A and B only.
Answer
A and B only
Shortcut
Options C and D are the two “other” inheritance patterns and both need more than two alleles or full dual expression — neither applies to a simple B/b pair. Eliminating them at once leaves only one option containing both A and B.
Keep the three heterozygote outcomes distinct: complete dominance — Bb looks like BB; incomplete dominance — Bb is in between; codominance — Bb shows both parental phenotypes side by side. The last two are constantly confused.
Q133Two genes, two dominance patternsMarked wrong
A pure tall pea plant with small starch grains in the seeds (TTbb) is crossed with a pure dwarf pea plant with large starch grains in the seeds (ttBB) and the plants of F1 generation are self-fertilised. Find out the possible phenotype ratio in the F2 generation.
MARKED1:2:1
9:3:3:1
KEY3:6:3:1:2:1
1:1:1:1
Given
Parents TTbb × ttBB → F1 is TtBb, then selfed.
T controls height — complete dominance.
B controls starch grain size — incomplete dominance.
Asked
F2 phenotypic ratio.
Concept to use
This is the standard dihybrid machinery with one modification: starch grain size in pea shows incomplete dominance, so the heterozygote Bb has intermediate-sized grains and is a phenotype of its own. That turns the usual 3 : 1 for that gene into 1 : 2 : 1, and the combined ratio into 2 × 3 = 6 classes instead of 4.
Full ratio: 3 : 6 : 3 : 1 : 2 : 1, totalling 16 as expected.
Number of phenotype classes = 2 × 3 = 6.
Answer
3:6:3:1:2:1
Shortcut
Count phenotype classes before computing anything: complete dominance contributes 2 classes per gene, incomplete dominance contributes 3. Here 2 × 3 = 6, and only one option has six numbers in it. That alone answers the question.
Where it went wrong
1:2:1 is the ratio for the starch gene alone — correct as far as it goes, but it ignores height entirely. When a cross involves two genes, the answer must account for both; a ratio with only three terms cannot describe two genes unless one of them is uniform, and here neither is. Check the number of terms against the number of segregating genes before choosing.
Linkage and gene mapping
1 question · 1 wrong
Q94Who explained recombination frequency as distanceMarked wrong
The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by?
Gregor J. Mendel
KEYAlfred Sturtevant
Sutton – Boveri
MARKEDT.H. Morgan
Given
A named contribution: using recombination frequency to measure distance between genes.
Asked
Which scientist is credited with it.
Concept to use
Morgan and Sturtevant worked together, and NCERT mentions them in the same paragraph — which is exactly why this is asked. The division of labour matters: Morgan discovered linkage and recombination in Drosophila; his student Sturtevant took the recombination frequencies and converted them into a map, using frequency as a measure of distance. The mapping idea is Sturtevant's.
Formula to use
1 map unit (centimorgan) = 1% recombination frequency — Sturtevant’s scale
Baby steps
Mendel worked with pea plants and independent assortment; he knew nothing of chromosomes or linkage. Out.
Sutton and Boveri proposed the chromosomal theory of inheritance — that genes are carried on chromosomes. Related, but not about distance. Out.
Morgan demonstrated linkage and recombination in Drosophila, and showed that linked genes do not assort independently.
Sturtevant, working in Morgan’s laboratory, used those recombination frequencies to determine the order and distance between genes — producing the first genetic map. He is the answer.
Answer
Alfred Sturtevant
Shortcut
Attach one verb to each name: Mendel assorted, Sutton–Boveri located genes on chromosomes, Morgan linked, Sturtevant mapped. The moment a question says “distance”, “map”, or “map units”, the answer is Sturtevant.
Where it went wrong
Morgan is the more famous name and the work happened in his lab, so he is the natural guess — that is precisely the trap. NCERT credits the mapping explicitly to Sturtevant. When two collaborators appear as options, ask which one did the specific thing named in the question.
What the sixteen have in common
One chapter, and the errors fall into three groups. None of them is a gap in the
genetics itself.
1 · Absolute words went unchallenged — Q98, and the same trap sat unattempted in Q93 and Q104
Q98 statement A said one allele always dominates completely, and it was marked
True — but incomplete dominance is in the same chapter, and Q101 and Q133 in this
very paper both depend on it. Q93 and Q104 are built on exactly the same trick
(“always move towards one pole”, “always segregates independently”)
and both were left blank. Fix: underline always, never, all, only on
sight. In genetics they mark the planted error roughly every time.
2 · Ratios read in the wrong direction or from the wrong object — Q98, Q127, Q131
Q98 asked whether rrYy : RRyy is 1:2, and it is 2:1 — the grid decides, not the
order the names are written in. Q127 answered “Four”, which is the number of
boxes, not distinct genotypes. Q131 applied a test-cross 1:1:1:1 to a cross
where only one of the two genes is a test cross. Fix: before answering, say out
loud what is being counted — boxes, genotypes, or phenotypes — and for a
two-gene cross always split by gene first, then multiply.
3 · The two dominance patterns were not separated — Q133, with Q101 skipped
Q133 needed complete dominance for height and incomplete dominance for starch grain
size, giving 2 × 3 = 6 phenotype classes and the ratio 3:6:3:1:2:1. The answer
given was 1:2:1 — the starch gene alone. Q101, which sets up that exact
distinction for the pea B gene, was left blank. Reading Q101 first would have made
Q133 straightforward. Fix: for each gene in a cross, write “2 classes”
or “3 classes” above it before doing anything else, then multiply.
4 · The blanks were mostly one-line questions
Q108 is the statement of a law. Q110 is 2³. Q122 is 2⁰. Q120 is the definition
of a test cross. Four of the eight blanks are worth 16 marks and about three minutes
in total. This matches what the Chemistry and Physics sets showed — the
unattempted questions are not the difficult ones.
Six rules that cover all sixteen
Powers of n: for n heterozygous loci — 2n
gametes, 2n phenotypes, 3n genotypes, 4n boxes.
Homozygous loci count as 1. (Q110, Q122, Q127)
Split by gene, then multiply. Never draw a 16-box grid for a
question a pair of ratios will answer. (Q92, Q131, Q133)
Test cross: genotypic and phenotypic ratios are identical,
because the recessive parent hides nothing. (Q120)
Heterozygote tells you the dominance type: looks like the dominant
→ complete; in between → incomplete; shows both → codominance.
(Q101, Q133)
Names to verbs: Mendel assorted, Sutton–Boveri located,
Morgan linked, Sturtevant mapped. (Q94)
Meiosis stages: alleles separate at anaphase I; sister
chromatids at anaphase II; random bivalent orientation at metaphase I is
independent assortment. (Q93, Q97, Q128)
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 Punnett squares and genetics figures have been drawn fresh, and every ratio in this file was verified by computing the cross before it was written in.