Error Solutions Workbook
Botany — Genetics

• Mother is a carrier: XᴴXʰ
• Father is unaffected: XᴴY
• The first son is already known to be haemophilic.

The probability that the second son — not the second child — is haemophilic.

A son takes his single X from his mother and his Y from his father. The father’s X never reaches a son, so the father is irrelevant to the answer. The mother carries one normal and one mutant X and passes one of them at random, so each son has a 1 in 2 chance.

Fertilisations are independent events. The first son’s genotype changes nothing about the second — that sentence is in the question to test exactly this, not to be used.

1. Write the cross. XᴴXʰ × XᴴY.
2. List the four equally likely children. XᴴXᴴ, XᴴXʰ, XᴴY, XʰY.
3. Read the question again. It says second son. So restrict to the two sons: XᴴY and XʰY.
4. Count. One of those two is affected → 1/2 = 50%.
5. Ignore the first son. Independent trials; no updating.

• “Son” halves the sample space. If the question says child, the answer is 25%. If it says son, the answer is 50%. The distractors on this paper are exactly those two numbers.
• A carrier mother always gives her sons a straight coin toss. Memorise that one line and every haemophilia/colour-blindness numerical falls out.
• Any sentence about a previous child in a genetics probability question is a decoy unless the question asks you to infer a genotype from it.

• Two named chromosomal abnormalities: aneuploidy and polyploidy.
• Two candidate mechanisms: failure of segregation, and failure of cytokinesis.

Which option pairs each abnormality with the right mechanism.

Aneuploidy is a change of one or a few chromosomes: 2n+1 (trisomy) or 2n−1 (monosomy). It comes from non-disjunction — homologous chromosomes or sister chromatids failing to separate at anaphase, in meiosis or mitosis. The cell still divides; the chromosomes are just handed out unevenly.

Polyploidy is a change of whole sets: 3n, 4n. DNA replicates, the chromosomes separate normally, but the cell never splits — cytokinesis fails, so one cell keeps both sets. Common in plants.

1. Ask what changed. A few chromosomes → aneuploidy. Entire sets → polyploidy.
2. Match the failure to the scale. Uneven hand-out of individual chromosomes needs a segregation failure. Doubling of everything needs a division (cytokinesis) failure.
3. Check option 4 separately. Trisomy is a gain (2n+1), monosomy is a loss (2n−1). That option has them backwards, so it is out on its own.
4. Check option 3. Non-disjunction happens in mitosis too, so “meiosis only” is false.

• Two words, two scales. “Segregation fails → a few chromosomes move wrong. Cytokinesis fails → the whole cell stays one.”
• Options 1 and 2 are the same two clauses with the subjects swapped. When you see that shape, you only have to get one of the two right — then the other is forced.

• Honeybee sex determination is haplodiploid.
• Four claims, one true.

The single correct statement.

In honeybees sex is decided by ploidy, not by a sex chromosome. A fertilised egg (2n) becomes a female — queen or worker. An unfertilised egg (n) develops parthenogenetically into a drone.

The queen is a normal diploid animal, so she makes her eggs by ordinary meiosis. The drone is haploid, so he cannot halve his chromosome number again — he makes sperm by mitosis. A drone therefore has no father, but he certainly has a maternal grandfather: his mother the queen came from a fertilised egg.

1. Fix the ploidy. Drone = haploid male. Queen and worker = diploid females.
2. Test option 4. It states the reverse of that → false.
3. Test option 3. Workers are females, so they come from fertilised eggs → false.
4. Test option 2. No father, yes grandfather (through the mother) → false.
5. Test option 1. The queen is diploid, so gametogenesis is meiosis → true.

• The famous line is “a drone has no father but has a grandfather.” The paper prints the half of that sentence that is false. Recite the whole line before deciding.
• Drone makes sperm by mitosis; queen makes eggs by meiosis. That asymmetry is the most examined fact in this topic.

• Four claims about genetic material and the RNA–DNA relationship.
• The options ask which pair is correct or incorrect — read that word carefully.

Which two statements are true.

The NCERT criteria for a molecule to serve as genetic material are: it should replicate, be stable chemically and structurally, allow slow mutation for evolution, and be able to express itself in the form of Mendelian characters. Statement A is that fourth criterion verbatim.

On origins, the order is RNA first. RNA was the earliest genetic material; it is both a carrier of information and a catalyst, but it is reactive and unstable. DNA evolved from RNA — not the other way round — through two modifications: the 2′-OH is removed (deoxyribose), and uracil is replaced by thymine (5-methyl uracil). Both changes make DNA more stable.

1. A — one of the four stated properties of genetic material. TRUE.
2. B — RNA was the first genetic material, not DNA. FALSE.
3. C — the arrow points the wrong way; DNA evolved from RNA. FALSE.
4. D — thymine in place of uracil is exactly the stabilising change. TRUE.
5. So A and D are correct → option 4. (Option 1 “C and D are incorrect” is half right, which is why it is there.)

• “RNA came first, DNA came stable.” Every statement that reverses that arrow is false.
• Tick each lettered statement T or F in the margin first, then go to the options. Never try to evaluate an option and a statement at the same time.
• Watch for options phrased as incorrect when the ones you ticked are correct — you must flip your list before matching.

• Both sexes must carry two sex chromosomes (so no XO system).
• The male’s pair must be identical (homogametic).
• The female’s pair must be different (heterogametic).

The organism whose female is the heterogametic sex.

In the XY systems (humans, fruit fly) the male is XY — two allosomes of two kinds — and the female is XX. In the XO system (grasshopper) the male has only one allosome, X, and the female has XX. Only the ZW system reverses the roles: male ZZ (two of the same), female ZW (two different). Birds, and also butterflies and moths, use ZW.

1. Eliminate on the count first. Grasshopper males are XO — one allosome, not two. Out immediately.
2. Eliminate on who is heterogametic. Humans and fruit fly: the male is XY. The question wants the male to be the uniform one. Both out.
3. Birds: male ZZ, female ZW. Male uniform, female mixed → fits every condition.

• Z and W mean the female is the odd one out. If a question describes a heterogametic female, the answer is birds (or moths/butterflies) every time.
• Grasshopper is the standard distractor here because XO also “looks unequal” — but the stem insists on two allosomes in both sexes, which XO fails.

• The patient is male — so a Y chromosome is present.
• Signs: tall stature, sterility, gynaecomastia (development of breast tissue).

The chromosomal cause.

Klinefelter’s syndrome is 47, XXY — an extra X alongside the normal XY. The Y keeps the individual male in external appearance; the extra X produces feminine development such as gynaecomastia, together with tall build and sterile (non-functional) testes.

The contrasting disorder, Turner’s syndrome, is 45, X — a missing X, in a female who is short, sterile and lacks secondary sexual development. The paper’s first option describes Turner, not Klinefelter.

1. Use the word ‘male’. A Y is present, so “absence of Y” is impossible → option 2 out.
2. Use gynaecomastia. Feminising signs in a male mean extra X material, not less.
3. Match the triad. Tall + sterile + gynaecomastia = Klinefelter = 47, XXY.
4. Option 4 (XYY) gives tall stature but neither gynaecomastia nor this pattern of sterility → out.

• Extra X feminises; missing X shortens. XXY → tall male with breast development. X0 → short sterile female.
• Gynaecomastia is the single word that decides this question. Circle it.

• Four statements about pleiotropy.
• The answer is a count, not a selection — so every statement must be judged.

How many of A–D are true.

Pleiotropy is one gene affecting many traits. The mechanism is that the gene product sits in a metabolic pathway, and disturbing that one step ripples out into several phenotypes.

The NCERT example is phenylketonuria: a mutation in the single gene coding for phenylalanine hydroxylase. Phenylalanine accumulates, and the one defect produces mental retardation plus reduced hair and skin pigmentation — several traits from one gene. That is pleiotropy in action, and it is why statement C is false.

1. A — the definition of pleiotropy. TRUE.
2. B — the stated mechanism (metabolic pathways). TRUE.
3. C — PKU is a single-gene disorder. FALSE.
4. D — “a single phenotype” contradicts A. FALSE.
5. Count the ticks: 2.

• A and D are direct opposites. In any “how many are true” item, first look for a contradictory pair — exactly one of them is true, which instantly bounds your count.
• PKU = one gene, many effects. A statement that makes PKU multigenic is always false.
• Write T/F beside each letter and then count the T’s. Never estimate the count by feel.

• X — shaded circle → affected female.
• Y — unshaded square → unaffected male.
• The couple produces both affected and unaffected children.
• Myotonic dystrophy is autosomal dominant.

The genotypes of X and Y, in that order.

For a dominant disorder the allele A causes disease. An affected person is AA or Aa; an unaffected person must be aa, because a single A would show.

X is affected but has unaffected children. An AA mother crossed with aa would give every child Aa — all affected. Since unaffected children exist, X must be heterozygous: Aa.

1. Fix the mode. Dominant → unaffected = aa, no exceptions.
2. Y is unaffected → aa. That alone kills two options.
3. X is affected → AA or Aa.
4. Use the children. Aa × aa → 1 affected : 1 unaffected ✓. AA × aa → all affected ✗. So X = Aa.
5. Write them in the order asked: X then Y → Aa and aa.

• In dominant pedigrees, start with the unaffected people. Their genotype is certain (aa); the affected ones need the children to pin down.
• Affected parent + unaffected child = heterozygous parent. That single line answers most autosomal dominant pedigree items.
• Before shading in an answer, re-read which letter comes first in the stem.

1. In any lettered-statement question, write T or F beside each letter on the paper before reading a single option. Q107 and Q126 were both lost for want of four pen strokes.
2. When a question asks for two things “respectively”, write them as two labelled lines (X =, Y =) before looking at the options. Q132 and Q97 were both arrangement errors.
3. Read every clause of the stem as a separate filter. Q113 had three conditions and only the second was used.