Monohybrid inheritance and sex linkageEdexcel International A Level Biology: Revision notes

• A gene is a length of DNA that codes for a polypeptide; an allele is one version of a gene.
• The genotype is the alleles an organism has; the phenotype is the observable characteristic that results.
• A dominant allele is expressed whenever it is present. A recessive allele is expressed only when no dominant allele is present (two copies).
• Codominant alleles are both expressed in the heterozygote.
• A homozygote has two identical alleles of a gene; a heterozygote has two different alleles.

A heterozygote with a recessive allele that causes a condition but does not show it is a carrier.

A monohybrid cross follows one gene with two alleles. Write the parents' genotypes, their gametes, then combine the gametes in a Punnett square.

Worked example (cystic fibrosis, recessive allele f): two carriers, Ff × Ff.

• Gametes: F or f from each parent
• Offspring: FF, Ff, Ff, ff in the ratio 1 : 2 : 1
• Phenotypes: 3 unaffected : 1 with cystic fibrosis, so the probability of an affected child is 1/4 (25%) and of a carrier is 1/2 (50%)

A cross between a heterozygote and a homozygous recessive (Ff × ff) gives 1 : 1.

With codominance both alleles are fully expressed in the heterozygote. In cattle the alleles $C^{R}$ (red) and $C^{W}$ (white) are codominant, so $C^{R}C^{W}$ cattle are roan, with red and white hairs.

• $C^{R}C^{W}$ × $C^{R}C^{W}$ gives 1 red : 2 roan : 1 white
• $C^{R}C^{W}$ × $C^{W}C^{W}$ gives 1 roan : 1 white

In human blood groups, $I^{A}$ and $I^{B}$ are codominant, so a person with $I^{A}I^{B}$ has group AB.

A pedigree diagram shows a family: males are squares, females are circles and shading shows the individuals who have the condition. Use these rules:

• Two unaffected parents with an affected child: the condition is recessive (both parents are carriers).
• Affected parents with an unaffected child: the condition is dominant.
• An affected daughter with an unaffected father rules out X-linked recessive inheritance, so the condition is autosomal.
• Mostly affected males, with the condition passing through unaffected mothers, suggests X-linked recessive inheritance.

Always state the evidence from named individuals.

Genes on the X chromosome show sex linkage. Males are XY and females XX. Red-green colour blindness is caused by a recessive allele on the X chromosome: $X^{B}$ (normal vision) and $X^{b}$ (colour blind).

• Males have only one X, so one $X^{b}$ gives colour blindness ($X^{b}Y$).
• Females need two ($X^{b}X^{b}$); $X^{B}X^{b}$ women are carriers.
• So the condition is more common in males.
• A son inherits his X from his mother and his Y from his father, so a father cannot pass an X-linked allele to his son. He passes his X to all of his daughters.

Worked example: carrier woman $X^{B}X^{b}$ × normal man $X^{B}Y$. Daughters: all normal vision (half are carriers). Sons: 50% colour blind. Overall 25% of children are colour blind, all of them sons.