Inheritance

A base substitution changes one base in a triplet (codon), so only one codon is altered by the mutation.

The genetic code is degenerate, which means that most amino acids are coded for by more than one codon. As a result, the new (mutated) codon may still code for the same amino acid as the original codon.

If the same amino acid is coded for, the amino acid sequence of the polypeptide is unchanged, so the protein's structure and function are unaffected. This is called a silent (synonymous) mutation.

A recessive allele is one that is only expressed in the phenotype when two copies are present — that is, when the individual is homozygous recessive. It is masked by a dominant allele when only one copy is present.

An unaffected carrier is heterozygous (Ff): they have one dominant allele F, which produces enough working CFTR protein, so they do not have cystic fibrosis, but they carry the recessive f allele.

An affected person is homozygous recessive (ff): they have two faulty alleles and so produce no working CFTR channel, and therefore have cystic fibrosis.

The CFTR gene normally codes for a chloride ion channel in the cell-surface membrane of epithelial cells. A mutation produces a faulty or absent channel, so chloride ions cannot be secreted out of the cells onto the surface mucus. Because sodium ions and water normally follow the chloride ions by osmosis, less water moves out and the mucus becomes thick and sticky.

In the gas exchange system, the sticky mucus lines the airways (bronchi and bronchioles). It is not easily moved by the cilia of the mucociliary escalator, so it accumulates and partially blocks the airways, reducing airflow and the surface area available for gas exchange. The trapped mucus also harbours bacteria, causing repeated chest infections that progressively damage the lung tissue.

In the digestive system, thick mucus blocks the pancreatic duct, so digestive enzymes (such as amylase, lipase and proteases) made by the pancreas cannot reach the small intestine. Food, especially fat and protein, is therefore poorly digested and absorbed, leading to weight loss and malnutrition. This is why patients take enzyme supplements with their meals.

In this way a single change in the CFTR base sequence produces a faulty protein, a cellular ion-transport defect and symptoms across more than one organ system.

The breeder should carry out a test cross: cross the black mouse of unknown genotype with a homozygous recessive (brown, bb) mouse. A brown mouse can only be bb, so the alleles it contributes are always recessive, and the phenotypes of the offspring will therefore reveal the alleles supplied by the unknown black parent.

If the black mouse is homozygous (BB): the cross is BB × bb. All offspring receive a B from the black parent and a b from the brown parent, giving the genotype Bb, so all offspring are black. No brown offspring appear.

If the black mouse is heterozygous (Bb): the cross is Bb × bb. The black parent contributes B or b, so half the offspring are Bb (black) and half are bb (brown), giving an approximate 1 black : 1 brown ratio.

Therefore, if any brown offspring are produced, the black parent must be heterozygous (Bb); if all offspring are black, the black parent is most likely homozygous (BB).

Let X^N represent the normal allele (dominant) carried on the X chromosome and X^n the colour-blindness allele (recessive); the Y chromosome carries no allele for this gene.

The man has normal vision and one X chromosome, so his genotype is X^N Y. The woman is a carrier, so her genotype is X^N X^n.

Gametes: the man produces (X^N) and (Y); the woman produces (X^N) and (X^n).

Combining these in a Punnett square gives four equally likely offspring: X^N X^N, X^N X^n, X^N Y and X^n Y.

Daughters (X^N X^N and X^N X^n): both daughters inherit X^N from their father, so each has at least one normal allele. None of the daughters is colour blind (0%); one of the two is a carrier (X^N X^n).

Sons (X^N Y and X^n Y): one son is X^N Y (normal vision) and one is X^n Y (colour blind). So half (50%) of the sons are expected to be colour blind.

Why the condition is more common in males: a male has only one X chromosome, so a single recessive allele (X^n Y) is enough for him to be colour blind — he cannot be a carrier. A female has two X chromosomes and must inherit two recessive alleles (X^n X^n) to be affected; a single recessive allele is masked by the dominant allele, making her an unaffected carrier. Because only one recessive allele is needed in males but two in females, the condition appears far more often in males.

Both procedures obtain fetal cells so that the fetal DNA or chromosomes (karyotype) can be examined for genetic conditions such as cystic fibrosis, sickle-cell anaemia or Down syndrome. In chorionic villus sampling, a sample of cells is taken from the placenta (chorionic villi) at about 10–13 weeks of pregnancy. In amniocentesis, a sample of amniotic fluid containing fetal cells is withdrawn later, at about 15–18 weeks.

An advantage of CVS over amniocentesis is that it can be carried out earlier in the pregnancy and gives results sooner, so a couple has more time to make decisions. A disadvantage is that CVS carries a slightly higher risk of miscarriage than amniocentesis; both procedures are invasive and put a healthy fetus at some risk.

Ethical issues arise on both sides. In favour, screening gives parents informed reproductive choice, allows them to prepare for an affected child or to consider their options, and can reduce suffering. Against, the result may create pressure to terminate a pregnancy, which raises questions about the value placed on the life of a person with a disability; the procedures risk the miscarriage of a healthy fetus; there are concerns about genetic discrimination by employers or insurers, about confidentiality of genetic information, and some people object on religious or moral grounds.

On balance, the choice of whether to undergo screening, and which procedure to use, should be made by a fully informed couple after genetic counselling, weighing the value of the information gained against the risk of the procedure and their own ethical views.

Adding one base is an insertion.

Because one base has been added, the reading frame is shifted (a frameshift), so every codon after the point of insertion is changed/misread and a completely different, usually non-functional, polypeptide is produced.

Symbols: let T = allele for tall stems (dominant) and t = allele for dwarf stems (recessive).

Parental phenotypes / genotypes: both parents are tall and heterozygous, so the cross is:

Tall (Tt) × Tall (Tt)

Gametes (each in a circle, one allele per gamete from meiosis):

Parent 1: (T) (t) × Parent 2: (T) (t)

Punnett square — combine the gametes:

Offspring genotypes: 1 TT : 2 Tt : 1 tt.

Offspring phenotypes: TT and Tt are both tall (T is dominant), tt is dwarf. So 3 tall : 1 dwarf.

Phenotype ratio = 3 tall : 1 dwarf. This means each offspring has a 3-in-4 (75%) chance of being tall and a 1-in-4 (25%) chance of being dwarf.

A substitution replaces one base with a different base, so only one codon (triplet) is changed; every other codon in the gene is read normally.

Because the genetic code is degenerate (most amino acids are coded for by more than one codon), the new codon may still code for the same amino acid — a silent (synonymous) mutation with no effect on the polypeptide. Even when the amino acid does change (a missense mutation), usually only one amino acid in the primary structure is altered, so the protein often still folds and functions, at least partially.

A single-base deletion, by contrast, removes one base, so the reading frame is shifted (a frameshift). The bases are read in non-overlapping triplets from a fixed starting point, so after the deletion every codon downstream is changed/misread.

This means most or all of the amino acids after the mutation are different. The primary structure is drastically altered, so the polypeptide folds into a completely different tertiary structure and almost always loses its function. A frameshift may also introduce a premature stop codon, truncating the protein.

Therefore the deletion affects the whole sequence downstream and is very likely to destroy function, whereas the substitution affects at most one amino acid and may have no effect at all because of the degenerate code.

Both parents are unaffected carriers, so each is heterozygous, Ff (F = normal dominant allele, f = cystic fibrosis recessive allele). The cross is Ff × Ff.

Gametes: each parent produces (F) and (f).

Combining these in a Punnett square gives offspring 1 FF : 2 Ff : 1 ff:

Only the ff genotype is affected by cystic fibrosis. ff is one of the four equally likely combinations, so the probability that any one child is affected is 1/4 (25%). This justifies the counsellor's figure: at each pregnancy there is an independent 1-in-4 chance of an affected (ff) child and a 3-in-4 chance of an unaffected child (1 FF + 2 Ff).

The actual proportion may differ from exactly 1 in 4 because the ratio is a statistical prediction, not a guarantee. Each fertilisation is an independent event, and the predicted ratio is only approached over a large number of offspring (the law of large numbers). With a small family — say four children — it is perfectly possible to have no affected children, or more than one, purely by chance. The 1-in-4 figure is the probability per pregnancy, not a fixed quota the family must fill.