Home › Learning Hub › IGCSE Biology › 17b Monohybrid inheritance
Topic 17 · 17.4

Monohybrid inheritance

Core and Extended · Papers 1–6

🎯What you need to be able to do

  • Describe inheritance, genotype, phenotype, homozygous, heterozygous, pure-breeding, dominant and recessive.
  • Interpret pedigree diagrams; use genetic diagrams and Punnett squares to predict 1 : 1 and 3 : 1 ratios.
  • Explain a test cross; describe codominance and the ABO blood groups EXTENDED.
  • Describe sex linkage, with red-green colour blindness, and use genetic diagrams for codominance and sex linkage EXTENDED.

📚The biology

Vocabulary

inheritance: the transmission of genetic information from generation to generation
genotype: the genetic make-up of an organism, in terms of the alleles present (e.g. Pp)
phenotype: the observable features of an organism (e.g. purple flowers)
homozygous: two identical alleles of a gene (PP or pp); two identical homozygous individuals breed true (pure-breeding)
heterozygous: two different alleles (Pp); not pure-breeding
dominant allele: expressed if it is present in the genotype (capital letter)
recessive allele: only expressed when there is no dominant allele of the gene present (lower-case letter)

Genetic diagrams

A genetic diagram for two heterozygous purple-flowered plants, Pp crossed with Pp. Each makes gametes P and p. A Punnett square gives PP, Pp, Pp and pp: genotypes 1 PP to 2 Pp to 1 pp, phenotypes 3 purple to 1 white.
Heterozygous × heterozygous gives a 3 : 1 ratio of phenotypes.

Set out a genetic diagram in the same steps every time: parental phenotypes, parental genotypes, gametes (circled), offspring genotypes (Punnett square), offspring phenotypes, ratio. A 1 : 1 ratio comes from heterozygous × homozygous recessive (Pp × pp).

Ratios are predictions of probability. With small numbers of offspring the actual numbers often differ from the ratio, because fertilisation is random.

Pedigree diagrams

A family pedigree. Squares are males, circles females; filled shapes are affected. Unaffected parents 1 and 2 have three children: an unaffected daughter 3, an affected son 4 and an unaffected daughter 5. Daughter 5 and her unaffected husband 6 have an affected daughter 7 and an unaffected son 8.
Two unaffected parents with an affected child means the condition is recessive and both parents are heterozygous (carriers).

The test cross EXTENDED

Two Punnett squares. If the purple plant is PP, crossing with pp gives all purple Pp offspring. If it is Pp, crossing with pp gives half Pp purple and half pp white: 1 purple to 1 white.
EXTENDED Cross the unknown with a homozygous recessive individual and look at the offspring.

EXTENDED An organism showing the dominant phenotype could be homozygous or heterozygous. Crossing it with a homozygous recessive individual reveals which: all offspring with the dominant phenotype means it is homozygous; any recessive offspring (about half) mean it is heterozygous.

Codominance and blood groups EXTENDED

EXTENDED In codominance both alleles in a heterozygous organism contribute to the phenotype. The ABO blood groups are controlled by three alleles: IA and IB are codominant, and both are dominant to Io. So IAIA and IAIo are group A; IBIB and IBIo are group B; IAIB is group AB; IoIo is group O.

A Punnett square for a father with genotype IA Io, group A, and a mother with IB Io, group B. The offspring are IA IB group AB, IB Io group B, IA Io group A and Io Io group O: 1 A to 1 B to 1 AB to 1 O.
EXTENDED Parents of groups A and B can have children of all four groups.

Sex linkage EXTENDED

EXTENDED A sex-linked characteristic is one where the gene responsible is on a sex chromosome, which makes it more common in one sex than the other. Red-green colour blindness is caused by a recessive allele on the X chromosome. Males have only one X, so a single recessive allele (XbY) causes colour blindness; females need two (XbXb), so it is much more common in males.

A Punnett square for a carrier mother, XB Xb, and a father with normal vision, XB Y. The offspring are XB XB, a normal daughter; XB Xb, a carrier daughter; XB Y, a normal son; and Xb Y, a colour-blind son.
EXTENDED A son gets his X from his mother, so a carrier mother passes colour blindness to half her sons.

✏️Worked example

Use the pedigree above. The allele for the condition is n; the normal allele is N. (a) State the genotypes of person 4 and person 1. [2] (b) Persons 5 and 6 are both heterozygous. Draw a genetic diagram to find the probability that their next child is affected. [4] (c) EXTENDED Suggest how a test cross could show whether a pea plant with purple flowers is pure-breeding. [2]

(a) Person 4 is affected: nn. Person 1 is unaffected but has an affected son: Nn.

(b) Nn × Nn; gametes N, n and N, n; offspring NN, Nn, Nn, nn; one in four is nn: probability 0.25 (25%, 1 in 4).

(c) Cross it with a white-flowered plant (pp). If all the offspring are purple, it is PP (pure-breeding); if some are white, it is Pp.

Check it. Person 7 is affected (nn), so she received an n from each parent — which confirms that 5 and 6 each carry n.
Gametes carry one allele. Writing “Nn” as a gamete, or forgetting to show gametes at all, loses marks in the genetic diagram.

📝Practise

In the style of the multiple-choice and theory papers. EXTENDED marks Supplement content.

1. (Multiple choice.) An organism with genotype Tt is described as: A: homozygous dominant. B: homozygous recessive. C: heterozygous. D: pure-breeding.
C.
2. (Theory.) In pea plants, round seeds (R) are dominant to wrinkled (r). Give the phenotype of a plant with genotype Rr, and of one with rr. [2]
Rr: round. rr: wrinkled.
3. (Theory.) A heterozygous round-seeded plant is crossed with a wrinkled-seeded plant. Draw a genetic diagram and give the expected ratio. [4]
Rr × rr; gametes R, r and r; offspring Rr, Rr, rr, rr; ratio 1 round : 1 wrinkled.
4. (Theory.) From a cross of two heterozygous plants, 120 seeds are produced. Predict how many are wrinkled. [2]
Ratio 3 round : 1 wrinkled, so one quarter: \( 120 \div 4 = 30 \).
5. (Theory.) Explain why two homozygous recessive parents cannot have a child with the dominant phenotype. [2]
Neither parent has a dominant allele, so every gamete carries the recessive allele; every child is homozygous recessive.
6. (Theory.) EXTENDED A man of blood group AB and a woman of group O have children. State the possible blood groups of the children. [3]
IAIB × IoIo: children IAIo or IBIo, so group A or group B (1 : 1); never AB or O.
7. (Theory.) EXTENDED A colour-blind man (XbY) and a woman who is not a carrier (XBXB) have children. What proportion of their daughters are carriers? [2]
All daughters get Xb from the father and XB from the mother: all are carriers (XBXb). All sons are XBY, normal.
8. (Theory.) EXTENDED Explain why red-green colour blindness is more common in males. [2]
The allele is recessive and on the X chromosome. Males have one X, so one recessive allele causes it; females need two, and usually have a dominant allele on the other X.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Learn Genetics (University of Utah) — Punnett squares and pedigrees, interactive
  • Colour Blind Awareness — how colour blindness is inherited