Principles of Inheritance and Variation ISC Class 12 Biology

What Principles of Inheritance and Variation is about

Weightage: Principles of Inheritance and Variation is part of the unit Genetics and Evolution, which carries 15 of the 70 marks in the ISC Biology Paper 1 (Theory), shared with 2 other chapters. Questions on it can appear in any section of the paper.Source: CISCE ISC Biology syllabus, examination year 2027, distribution of marks

Key concepts

Basic terms and Mendel's experiments

Gene: unit of inheritance, a segment of DNA. Alleles: alternative forms of a gene at the same locus on homologous chromosomes (a pair, one from each parent, with the same genes in the same order). Autosomes: chromosomes other than sex chromosomes (22 pairs in humans); sex chromosomes (X, Y) decide sex. Dominant allele expresses itself in the heterozygote (T); recessive only when homozygous (t). Genotype: genetic make-up (TT, Tt, tt); phenotype: observable character (tall, dwarf). Homozygous: identical alleles (TT, tt); heterozygous: different alleles (Tt). Monohybrid cross: one character; dihybrid: two. Back cross: F₁ crossed with either parent; test cross: F₁ (or any dominant phenotype) crossed with the homozygous recessive parent to find its genotype. Mendel's experiments: garden pea (Pisum sativum) with seven pairs of contrasting traits (stem height, flower colour, flower position, pod shape, pod colour, seed shape, seed colour). Reasons for his success: pea was easy to grow, self-pollinating but easily cross-pollinated, short life cycle, many seeds; he chose clear-cut contrasting traits; studied one or two characters at a time; used true-breeding lines; kept careful records and analysed large numbers statistically. Biological importance: explains heredity and variation; basis for plant and animal breeding (hybrid vigour), genetic counselling and predicting disorders.

Worked example

A tall pea plant of unknown genotype is crossed with a dwarf plant. Half the offspring are tall and half dwarf. What is the genotype of the tall parent, and what is this cross called?

Try it first, then show the solutionSolution

1. Dwarf parent is tt and gives only t gametes.
2. A 1:1 ratio means the tall parent gave T and t gametes equally: Tt.
3. Crossing with the homozygous recessive is a test cross.

Answer: Tt; test cross

Laws of dominance and segregation (monohybrid cross)

Law of dominance: characters are controlled by paired factors; when two unlike factors are present in a hybrid, only one (dominant) is expressed. Law of segregation (purity of gametes): the two alleles of a pair do not blend; they separate during gamete formation so that each gamete receives only one allele. Monohybrid cross: TT (tall) × tt (dwarf) → F₁ all Tt (tall). F₁ selfed: Tt × Tt → gametes T and t from each; Punnett square gives TT : Tt : tt = 1 : 2 : 1 (genotype) and tall : dwarf = 3 : 1 (phenotype). Test cross ratio: Tt × tt → 1 Tt : 1 tt (1:1). The law of segregation is universal (it holds even when dominance is incomplete).

Worked example

In 800 F₂ plants of a monohybrid cross for seed colour (yellow Y dominant over green y), how many are expected to be yellow, green, and homozygous yellow?

Try it first, then show the solutionSolution

1. Phenotypes 3:1: yellow = 800 × 3/4 = 600; green = 200.
2. Genotypes 1:2:1: YY = 800 × 1/4 = 200.

Answer: 600 yellow, 200 green, 200 homozygous yellow

Law of independent assortment (dihybrid cross)

When two pairs of traits are combined in a hybrid, the segregation of one pair is independent of the other. Dihybrid cross: round yellow (RRYY) × wrinkled green (rryy) → F₁ RrYy (round yellow). F₁ produces four kinds of gametes in equal numbers: RY, Ry, rY, ry. The 4 × 4 Punnett square gives the F₂ phenotypic ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green; of 16 combinations there are 9 genotypes. Each character separately still shows 3:1 (12:4). Dihybrid test cross: RrYy × rryy → 1:1:1:1. The law applies to genes on different chromosomes (or far apart on one); linked genes do not assort independently.

Worked example

Of 320 F₂ plants from RrYy × RrYy, how many are expected in each phenotypic class? How many are expected to breed true for round yellow seeds?

Try it first, then show the solutionSolution

1. 9:3:3:1 of 320: 180 round yellow, 60 round green, 60 wrinkled yellow, 20 wrinkled green.
2. True-breeding round yellow = RRYY = 1/16 of 320 = 20.

Answer: 180:60:60:20; 20 plants

Incomplete dominance, co-dominance and multiple alleles

Incomplete dominance: neither allele is completely dominant; the heterozygote has an intermediate phenotype. In snapdragon (Antirrhinum), red (RR) × white (rr) → F₁ pink (Rr); F₂: 1 red : 2 pink : 1 white (phenotype ratio = genotype ratio). The same happens in Mirabilis. Co-dominance: both alleles express fully in the heterozygote. ABO blood groups are controlled by gene I with three alleles (multiple alleles): Iᴬ and Iᴮ produce A and B sugars on the RBC surface and are co-dominant with each other; i produces no sugar and is recessive. Six genotypes, four phenotypes: Iᴬ Iᴬ, Iᴬ i → A; Iᴮ Iᴮ, Iᴮ i → B; Iᴬ Iᴮ → AB (co-dominance); ii → O. Multiple alleles: more than two alleles of a gene exist in a population, though an individual carries only two.

Worked example

A man of blood group A (heterozygous) marries a woman of blood group B (heterozygous). What are the possible blood groups of their children and in what proportions?

Try it first, then show the solutionSolution

1. Iᴬi × Iᴮi: gametes Iᴬ, i and Iᴮ, i.
2. Offspring: IᴬIᴮ (AB), Iᴬi (A), Iᴮi (B), ii (O), one each.

Answer: A, B, AB and O, each with probability 1/4

Pleiotropy and polygenic inheritance

Pleiotropy: a single gene affects several phenotypic traits. Phenylketonuria (PKU): a mutation in the gene for the enzyme phenylalanine hydroxylase; phenylalanine is not converted to tyrosine, so it accumulates and is converted to phenylpyruvic acid and other derivatives; effects include mental retardation, reduced hair and skin pigmentation, and excretion of phenylpyruvate in urine. Pea seeds: the gene for starch synthesis has alleles B (efficient starch synthesis: large starch grains, round seeds) and b (less starch, small grains, wrinkled seeds). For seed shape B is dominant, but for starch grain size the alleles show incomplete dominance (Bb grains are intermediate): one gene, more than one effect, and the dominance relation depends on the phenotype considered. Polygenic inheritance (elementary): a trait controlled by three or more genes, each contributing additively, giving continuous variation; environment also affects it. Human skin colour: suppose three genes A, B, C, each dominant allele adding melanin. AABBCC is darkest, aabbcc lightest; AaBbCc is intermediate. AaBbCc × AaBbCc gives seven phenotypic classes in the ratio 1 : 6 : 15 : 20 : 15 : 6 : 1 (out of 64). Height and intelligence are other polygenic traits.

Worked example

If skin colour is controlled by three genes, what fraction of children of two AaBbCc parents will be as dark as AABBCC, and what fraction will have exactly three dominant alleles?

Try it first, then show the solutionSolution

1. Each gene: probability of AA = 1/4; for all three: (1/4)³ = 1/64.
2. Number of dominant alleles follows the distribution 1:6:15:20:15:6:1 over 64; three dominant alleles is the middle class, 20/64 = 5/16.

Answer: 1/64; 20/64 (5/16)

Chromosomal theory, linkage and crossing over

Chromosomal theory of inheritance (Sutton and Boveri, 1902): genes are located on chromosomes; chromosomes occur in pairs, as do alleles; homologous chromosomes segregate at meiosis, explaining segregation; non-homologous pairs assort independently, explaining independent assortment; gametes get one of each pair; fertilisation restores pairs. T.H. Morgan confirmed it with Drosophila (short life cycle, many offspring, easily visible variations, only four chromosome pairs, easy to grow). Linkage: genes on the same chromosome tend to be inherited together; they do not assort independently, so parental combinations exceed recombinants. Crossing over: exchange of segments between non-sister chromatids of homologous chromosomes at pachytene of prophase I (chiasmata seen in diplotene); it produces recombinants. Recombination frequency = (recombinants / total offspring) × 100; the farther apart two genes, the higher the frequency (Sturtevant used it to make chromosome maps: 1% = 1 map unit). Significance: linkage keeps favourable combinations together; crossing over creates new combinations (variation), helps gene mapping and plant breeding.

Worked example

A test cross of a dihybrid fly gives 210 + 210 parental-type and 40 + 40 recombinant offspring. Find the recombination frequency and map distance between the two genes.

Try it first, then show the solutionSolution

1. Total = 420 + 80 = 500.
2. RF = 80/500 × 100 = 16%: the genes are 16 map units apart (linked, as parentals far exceed 25% each).

Answer: 16%; 16 map units

Sex determination and sex-linked inheritance

Henking (1891) observed the X body in insect sperm; it was later named the X chromosome. Humans (XX-XY, male heterogamety): females XX produce only X eggs; males XY produce X and Y sperms in equal numbers, so the sperm decides the sex (probability 1/2 each). Grasshopper (XX-XO): females XX; males have one X (XO), so sperms with and without X form. Birds (ZW-ZZ, female heterogamety): females ZW produce two types of eggs; males ZZ. Honey bees (haplodiploidy): fertilised eggs (2n = 32) develop into females (queen or workers); unfertilised eggs (n = 16) develop into males (drones) by parthenogenesis; drones produce sperm by mitosis, have no father and cannot have sons. Sex-linked (X-linked) inheritance: genes on the X chromosome. Males are hemizygous, so a single recessive allele shows. Criss-cross inheritance: a father passes his X-linked allele to all daughters (carriers) and they pass it to half their sons. Drosophila: Morgan's eye colour (red dominant, white recessive, X-linked) and body colour (brown/grey dominant, yellow recessive, X-linked): white-eyed male × red-eyed female → all F₁ red; F₂: all females red, half the males white. Haemophilia (bleeder's disease): recessive X-linked; blood fails to clot (lack of clotting factor); carrier mothers (XᴴXʰ) pass it to half their sons. Colour blindness: recessive X-linked; red-green cone defect; about 8% of males and 0.4% of females.

Worked example

A colour-blind man marries a woman who is homozygous for normal vision. Their daughter marries a man with normal vision. What fraction of the daughter's sons will be colour blind? Why is this called criss-cross inheritance?

Try it first, then show the solutionSolution

1. XᶜY × XX: every daughter receives Xᶜ from her father and X from her mother, so all daughters are carriers (XXᶜ); all sons are normal (XY).
2. Carrier daughter XXᶜ × normal man XY: her sons receive X or Xᶜ from her, so 1/2 of her sons are colour blind.
3. The allele passes from an affected grandfather, through a carrier daughter, to a grandson: father to daughter to grandson, crossing sexes each generation.

Answer: Half of her sons; the trait passes grandfather → daughter (carrier) → grandson

Mutation, pedigree analysis and human disorders

Mutation: a sudden heritable change in DNA. Spontaneous (natural errors in replication) or induced by mutagens (UV, X-rays, chemicals). Gene (point) mutations: change in one base pair: transition (purine replaced by purine or pyrimidine by pyrimidine, e.g. A→G), transversion (purine replaced by pyrimidine or vice versa, e.g. A→T), frame-shift (insertion or deletion of one or two bases shifts the reading frame, altering every codon after it). Chromosomal aberrations: deletion, duplication, inversion, translocation; aneuploidy (gain or loss of a chromosome, by non-disjunction) and polyploidy. Pedigree analysis: standard symbols (square male, circle female, shaded affected, half-shaded or dotted carrier, horizontal line for mating, vertical line to offspring, generations in Roman numerals). Mendelian disorders: sickle-cell anaemia (autosomal recessive; GAG → GTG in the β-globin gene: a transversion; glutamic acid replaced by valine at position 6 of the β-chain; HbS polymerises at low O₂, RBCs become sickle-shaped; HbᴬHbᴬ normal, HbᴬHbˢ carrier, HbˢHbˢ affected); thalassaemia (autosomal recessive; reduced synthesis of α or β globin chains; abnormal haemoglobin, anaemia; a quantitative problem, unlike sickle-cell); phenylketonuria (autosomal recessive, see pleiotropy); haemophilia and colour blindness (X-linked recessive). Chromosomal disorders: Down's syndrome (trisomy 21, 47 chromosomes: short stature, small round head, furrowed tongue, partly open mouth, palm crease, mental retardation); Klinefelter's syndrome (47, XXY: sterile male with feminine traits such as gynaecomastia, tall); Turner's syndrome (45, XO: sterile female with rudimentary ovaries, short, lacking secondary sexual characters).

Worked example

Sickle-cell anaemia is called a qualitative and thalassaemia a quantitative disorder of haemoglobin. Explain.

Try it first, then show the solutionSolution

1. Sickle-cell anaemia: a point mutation (GAG → GTG) changes the kind of globin made: glutamic acid at position 6 of the β-chain is replaced by valine, so HbS has an altered structure (qualitative).
2. Thalassaemia: mutations or deletions reduce the amount of α- or β-globin chains synthesised, so too little normal haemoglobin is formed (quantitative).

Answer: Sickle-cell: wrong (altered) globin; thalassaemia: too little globin

Principles of Inheritance and Variation in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Sex Determination and Sex-linked Inheritance
• Pedigree Analysis and Mendelian Disorders
• Incomplete Dominance, Co-dominance and Multiple Alleles

Principles of Inheritance and Variation: key facts

Monohybrid F₂ ratios

$\text{phenotype }3:1;\text{genotype }1:2:1$

Test cross 1:1.

Dihybrid F₂ ratio

$9:3:3:1\text{(of 16)}$

9 genotypes; test cross 1:1:1:1.

Incomplete dominance F₂

$1\text{red}:2\text{pink}:1\text{white}$

Phenotype = genotype ratio.

Gametes from a genotype

$\text{kinds of gametes}=2^n$

n = number of heterozygous gene pairs.

Polygenic (3 genes) F₂

$1:6:15:20:15:6:1\text{(of 64)}$

Classes by number of dominant alleles.

Recombination frequency

$RF=\frac{\text{recombinants}}{\text{total progeny}}\times 100$

1% = 1 map unit.

Blood group genotypes

$A:I^A{I}^A,I^{A}i;B:I^B{I}^B,I^{B}i;AB:I^A{I}^B;O:ii$

Six genotypes, four phenotypes.

Principles of Inheritance and Variation: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Principles of Inheritance and Variation

The question patterns to expect, and the method that works for each.

Tricks to remember Principles of Inheritance and Variation

Short memory aids. Say each one aloud once, then use it on a question the same day.

• 9:3:3:1 and 16

  Nine-three-three-one adds to sixteen; each trait alone gives 12:4 = 3:1.

  Use it when: Checking dihybrid answers.

• Incomplete = in-between, Co = coexist

  Pink snapdragon is in-between; AB blood has both.

  Use it when: Naming the deviation.

• Henking found the X

  H for Henking, the X 'hidden' body in 1891.

  Use it when: Discovery of the X chromosome.

• Criss-cross

  Father → daughter → grandson for X-linked traits.

  Use it when: Sex-linked pedigrees.

• Down 21, Klinefelter XXY, Turner XO

  Down has 47 with three 21s; K is the extra X in a male; Turner turns one X off.

  Use it when: Chromosomal disorders.

Principles of Inheritance and Variation: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Principles of Inheritance and Variation

The slips examiners see most often, and what to write instead.

• ✗Mistake: Wrong probability for X-linked traits

  CISCE examiners' comments, 2024 and 2025: for a daughter of a haemophilic man and carrier woman, candidates wrote 25% instead of 50%; in the bleeder's-disease case they wrote autosomal dominant or 'criss-cross' instead of sex-linked, and gave vague answers (less chance, low chance) instead of zero for the sister being affected.

  Wrong working

  Daughter haemophilic: 25%.

  Right working

  Among daughters: XʰXʰ = 1/2 = 50% (25% applies to all children taken together).

  ✓Do this: Note whether the question asks about all children or one sex.

  Usually costs: 1 mark each.

• ✗Mistake: Treating autosomal disorders as sex-linked

  CISCE examiners' comments, 2024: candidates treated thalassaemia as sex-linked and gave an XY genotype as for haemophilia.

  Wrong working

  XᵗY

  Right working

  tt (autosomal recessive).

  ✓Do this: Learn which disorders are X-linked: only haemophilia and colour blindness in this list.

  Usually costs: 1 mark.

• ✗Mistake: Imprecise biochemistry and mutation type in sickle-cell anaemia

  CISCE examiners' comments, 2024: candidates gave the valine-glutamic acid substitution without referring to the β-chain (or the wrong way round), described sickle shape instead of the biochemical cause, did not state the mother's genotype, and wrote 'point mutation' without naming substitution or transversion.

  Wrong working

  Valine is replaced by glutamic acid.

  Right working

  Glutamic acid at position 6 of the β-globin chain is replaced by valine (GAG → GTG, transversion).

  ✓Do this: Learn the exact sentence: Glu → Val, sixth position, β-chain.

  Usually costs: 1 mark each.

• ✗Mistake: Unlabelled ratios and wrong deviation name

  CISCE examiners' comments, 2024: for red × yellow giving orange, candidates said co-dominance or 'masking of colours' instead of incomplete dominance, and quoted 1:2:1 without assigning phenotypes, or wrote red:yellow:orange 1:1:2.

  Wrong working

  Co-dominance; 1:2:1.

  Right working

  Incomplete dominance; 1 red : 2 orange : 1 yellow.

  ✓Do this: Always attach a phenotype to every number in a ratio.

  Usually costs: 1 mark.

• ✗Mistake: Non-standard pedigree symbols and misspelt names

  CISCE examiners' comments, 2024 and 2025: candidates could not draw a proper pedigree with standard symbols for male, female, carrier and affected; and misspelt Henking (Hankin, Hunting) or wrote Morgan for the discoverer of the X chromosome.

  Wrong working

  Stick figures and no key; 'Hankin'.

  Right working

  Squares, circles, shading, dotted carriers, generation numbers and a key; Henking.

  ✓Do this: Practise drawing two pedigrees with a key.

  Usually costs: 1-3 marks.

Practice questions on Principles of Inheritance and Variation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   The F₂ genotypic ratio of a monohybrid cross is:
   (a) 3:1
   (b) 1:2:1
   (c) 9:3:3:1
   (d) 1:1

   Hint

   Genotypes.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   Which is an example of pleiotropy?
   (a) ABO blood groups
   (b) phenylketonuria
   (c) skin colour
   (d) snapdragon flower colour

   Hint

   One gene, many effects.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerHarder board level

   How many kinds of gametes are produced by a plant of genotype AaBBCcDd?

   Hint

   2ⁿ for heterozygous pairs.

   Answer

   Three heterozygous pairs (Aa, Cc, Dd): 2³ = 8.

4. Q4 · 1 mark · Very short answerBoard level

   Name the chromosomal disorder with karyotype 45, XO.

   Hint

   Female.

   Answer

   Turner's syndrome

5. Q5 · 2 marks · Short answerHarder board level

   Distinguish between transition and transversion with examples.

   Hint

   Purine/pyrimidine.

   Answer

   Transition: a purine replaced by another purine or a pyrimidine by a pyrimidine (A→G, C→T). Transversion: a purine replaced by a pyrimidine or vice versa (A→T, G→C).

6. Q6 · 2 marks · Short answerBoard level

   A woman of blood group O marries a man of blood group AB. Give the possible blood groups of their children.

   Hint

   ii × IᴬIᴮ.

   Answer

   Iᴬi (A) and Iᴮi (B), each 50%; no O or AB children.

7. Q7 · 2 marks · Short answerHOTS / competency

   In a cross between AaBb and aabb, what fraction of offspring is aabb if the genes assort independently? What would you conclude if 45% were aabb?

   Hint

   Test cross 1:1:1:1.

   Answer

   Independent assortment: 1/4 (25%). 45% aabb means parental types dominate: the genes are linked (RF = 10%, since recombinants Ab and aB total 10%).

8. Q8 · 2 marks · Short answerBoard level

   Explain the chromosomal theory of inheritance in four points.

   Hint

   Sutton and Boveri.

   Answer

   Genes are on chromosomes; chromosomes and genes occur in pairs; homologous chromosomes (and alleles) segregate at meiosis; non-homologous chromosomes assort independently; fertilisation restores pairs.

9. Q9 · 3 marks · Short answerHarder board level

   Explain the inheritance of skin colour in humans as polygenic inheritance.

   Hint

   Three genes, additive.

   Answer

   Controlled by at least three genes (A, B, C), each dominant allele adding a unit of melanin (additive effect). AABBCC is darkest, aabbcc lightest, AaBbCc intermediate. Crosses between intermediates give seven classes (1:6:15:20:15:6:1), giving continuous variation; environment (sunlight) also affects it.

10. Q10 · 3 marks · Short answerHOTS / competency

    How is sex determined in honey bees? Why can a drone have a grandfather but no father?

    Hint

    Haplodiploidy.

    Answer

    Fertilised eggs (2n = 32) become females (queen or workers); unfertilised eggs (n = 16) develop parthenogenetically into males (drones). A drone develops from an unfertilised egg, so it has no father; its mother (the queen) came from a fertilised egg, so it has a maternal grandfather.

11. Q11 · 5 marks · Long answerHarder board level

    (a) State Mendel's law of segregation and explain why it is called the law of purity of gametes. [2]
    (b) Give three reasons for Mendel's success. [2]
    (c) Why did Mendel not observe linkage? [1]

    Hint

    Seven traits.

    Answer

    (a) The two alleles of a gene separate during gamete formation, each gamete receiving one; gametes are always 'pure' for one allele as the alleles do not blend. (b) Chose pea with clear contrasting traits and true-breeding lines; studied one or two traits at a time; used large samples and statistics and kept careful records. (c) The traits he studied were on different chromosomes or so far apart that they assorted independently.

12. Q12 · 2 marks · Short answerHarder board level

    A colour-blind woman marries a man with normal vision. Predict the phenotypes of their sons and daughters.

    Hint

    XᶜXᶜ × XY.

    Answer

    Mother XᶜXᶜ gives Xᶜ to all children. Sons XᶜY: all colour blind. Daughters XXᶜ: all normal-sighted carriers.

13. Q13 · 3 marks · Short answerHOTS / competency

    Two normal parents have a child with phenylketonuria. (a) What is the pattern of inheritance? (b) What is the chance their next child is normal but a carrier? (c) Why does the disorder show several symptoms?

    Hint

    Autosomal recessive; pleiotropy.

    Answer

    (a) Autosomal recessive (both parents are carriers, Pp). (b) Pp × Pp: carriers 2/4 = 1/2. (c) Pleiotropy: lack of phenylalanine hydroxylase causes accumulation of phenylalanine and its derivatives, affecting the brain (mental retardation), pigmentation and urine composition.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.What are alleles?Show answer

   Alternative forms of a gene at the same locus on homologous chromosomes.

2. 2.Dihybrid F₂ phenotypic ratio?Show answer

   9:3:3:1.

3. 3.Example of incomplete dominance?Show answer

   Flower colour in snapdragon (Antirrhinum).

4. 4.Who proposed the chromosomal theory of inheritance?Show answer

   Sutton and Boveri.

5. 5.When does crossing over occur?Show answer

   At pachytene of prophase I of meiosis.

6. 6.Sex chromosomes of a female bird?Show answer

   ZW.

7. 7.Karyotype of Klinefelter's syndrome?Show answer

   47, XXY.

8. 8.What is a frame-shift mutation?Show answer

   Insertion or deletion of one or two bases, shifting the reading frame of all later codons.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you define genotype, phenotype, homozygous, heterozygous, test cross?
• ☐Can you derive 3:1, 1:2:1, 9:3:3:1 and 1:1:1:1?
• ☐Can you solve ABO blood group problems?
• ☐Can you explain PKU and pea-seed starch as pleiotropy?
• ☐Can you calculate recombination frequency?
• ☐Can you predict haemophilia and colour blindness in sons and daughters?
• ☐Can you draw a pedigree with standard symbols and a key?
• ☐Can you give the karyotype and features of Down's, Klinefelter's and Turner's syndromes?

Principles of Inheritance and Variation: frequently asked questions