Genetics

Monohybrid inheritance involves the inheritance of one gene (one characteristic).

Example: Fur Colour in Mice

• Brown fur (B) is dominant over white fur (b)
• Possible genotypes: BB (brown), Bb (brown), bb (white)

Punnett Square — Cross 1: Heterozygous × Heterozygous (Bb × Bb)

Parents: Both are Bb (brown, heterozygous/carriers)

Offspring genotype ratio: 1 BB : 2 Bb : 1 bb Offspring phenotype ratio: 3 brown : 1 white

Probability of white offspring: $\frac{1}{4}$ or 25%

Punnett Square — Cross 2: Heterozygous × Homozygous Recessive (Bb × bb)

Offspring ratio: 2 Bb : 2 bb → 1 brown : 1 white (50:50)

For exam questions, you should present genetic diagrams clearly:

Example: Two parents heterozygous for tongue rolling (Rr × Rr)

• Parents' phenotypes: Both can roll tongue
• Parents' genotypes: Rr × Rr
• Gametes: Each parent produces gametes with R or r
• Punnett square:

• Offspring genotypes: 1 RR : 2 Rr : 1 rr
• Offspring phenotypes: 3 can roll tongue : 1 cannot
• Probability of non-roller: $\frac{1}{4}$ = 25%

Cystic Fibrosis (Recessive)

• Caused by a recessive allele (f)
• Genotype ff = affected; Ff = carrier (unaffected but carries one copy); FF = unaffected
• Symptoms: Thick, sticky mucus in lungs and digestive system; difficulty breathing; frequent infections
• Both parents must carry at least one recessive allele for a child to be affected

Cross: Two carriers (Ff × Ff)

• Probability of affected child (ff): $\frac{1}{4}$ = 25%
• Probability of carrier child (Ff): $\frac{2}{4}$ = 50%
• Probability of unaffected non-carrier (FF): $\frac{1}{4}$ = 25%

Polydactyly (Dominant)

• Caused by a dominant allele (D)
• Genotype DD or Dd = affected (extra fingers or toes)
• Genotype dd = unaffected
• Only one copy of the dominant allele is needed to show the condition

Cross: Affected heterozygous × Unaffected (Dd × dd)

• Probability of affected child: $\frac{2}{4}$ = 50%

Humans have 23 pairs of chromosomes. The 23rd pair are the sex chromosomes:

• Female: XX
• Male: XY

Punnett Square for Sex

Probability of male: $\frac{2}{4}$ = 50% Probability of female: $\frac{2}{4}$ = 50%

The father determines the sex of the child — he can contribute either an X or a Y chromosome, while the mother always contributes an X.

Some genes are carried on the X chromosome — these are sex-linked traits.

Example: Colour Blindness

• The allele for colour blindness is recessive and located on the X chromosome
• Notation: $X^N$ = normal vision, $X^n$ = colour blind

Males are more likely to be colour blind because they only have one X chromosome — if it carries the recessive allele, they will be affected (there's no second X to mask it).

Question: Two parents are both carriers of cystic fibrosis (Ff). What is the probability that their child will have cystic fibrosis? Show your working using a genetic diagram. (4 marks)

Solution:

Parents: Father Ff × Mother Ff

Gametes: Father: F or f | Mother: F or f

Punnett Square:

Offspring genotypes: 1 FF : 2 Ff : 1 ff

Offspring phenotypes:

• FF = unaffected (not a carrier)
• Ff = unaffected (carrier)
• ff = has cystic fibrosis

Probability of child having cystic fibrosis = $\frac{1}{4}$ = 25%

• Always use capital letters for dominant and lowercase for recessive alleles.
• In Punnett squares, clearly show parents' genotypes, gametes, and the resulting offspring.
• Remember: genetic diagrams show probabilities, not certainties — a 25% chance doesn't mean exactly 1 in 4 children will be affected.
• For sex-linked traits, use the notation $X^A$ and $X^a$ to show alleles on the X chromosome.

• Alleles are different versions of a gene; they can be dominant (expressed when one copy present) or recessive (only expressed when two copies present).
• Homozygous = two same alleles (BB or bb); Heterozygous = two different alleles (Bb).
• Punnett squares predict the probability of offspring genotypes and phenotypes.
• Cystic fibrosis is caused by a recessive allele; polydactyly by a dominant allele.
• Sex is determined by X and Y chromosomes; sex-linked traits are carried on the X chromosome.