Edexcel International GCSE Biology 4BI1: Inheritance – Study Guide

This study guide teaches sub-topic 3(b), Inheritance (specification points 3.14–3.39), of the Pearson Edexcel International GCSE Biology (4BI1) specification, Issue 3. The qualification is untiered. Points with a B reference (3.16B, 3.17B, 3.18B, 3.21B, 3.35B, 3.36B and 3.37B) are bold in the specification and are examined on Paper 2 only; everything else can appear on either paper. It applies to the June and November series examined under Issue 3.

Use it with the Inheritance revision notes and the Inheritance practice questions. For the whole of Topic 3, including reproduction, see the topic study guide, topic revision notes and topic practice questions. The full course is on the Edexcel IGCSE Biology hub, and you can tick off each point on the printable checklist. To find your weak spots first, try one of the free diagnostics.

What this sub-topic covers

Spec point	What you must be able to do	Paper
3.14–3.15	Define genome and gene; genes are on chromosomes in the nucleus	1 and 2
3.16B–3.17B	Describe DNA as a double helix with paired bases; RNA is single stranded with uracil	Paper 2 only
3.18B	Describe transcription and translation (mRNA, ribosomes, tRNA, codons, anticodons)	Paper 2 only
3.19–3.20	Alleles; dominant, recessive, homozygous, heterozygous, phenotype, genotype	1 and 2
3.21B	Codominance	Paper 2 only
3.22	Most features are polygenic	1 and 2
3.23–3.25	Genetic diagrams, family pedigrees, probabilities from monohybrid crosses	1 and 2
3.26–3.27	Sex determination by XX and XY, shown in a genetic diagram	1 and 2
3.28–3.32	Mitosis and meiosis; random fertilisation; 46 and 23 in humans	1 and 2
3.33–3.34	Genetic and environmental variation; mutation	1 and 2
3.35B–3.37B	How mutations affect phenotype; mutagens	Paper 2 only
3.38–3.39	Darwin’s natural selection; antibiotic resistance	1 and 2

Genes, chromosomes and the genome (3.14–3.15)

The genome is the entire DNA of an organism. A gene is a section of a DNA molecule that codes for a specific protein. Each chromosome is one long DNA molecule, and genes sit at positions along it. Chromosomes are found in the nucleus.

DNA and RNA (3.16B–3.17B) – Paper 2 only

A DNA molecule is two strands coiled to form a double helix. The strands are linked by a series of paired bases:

• adenine (A) pairs with thymine (T)
• cytosine (C) pairs with guanine (G)

An RNA molecule is single stranded and contains uracil (U) instead of thymine (T).

Worked example. One strand of DNA reads ATTGCAGC. Write the other strand.

Given strand:          A T T G C A G C
Complementary strand:  T A A C G T C G

Protein synthesis (3.18B) – Paper 2 only

DNA stays in the nucleus, but proteins are made on ribosomes in the cytoplasm, so a copy of the gene is carried out.

Transcription (in the nucleus)

1. The two DNA strands of the gene separate.
2. Free RNA nucleotides pair with the bases on one DNA strand (the template), using the base-pairing rule with U in place of T.
3. They join to form a single strand of messenger RNA (mRNA), which leaves the nucleus through a pore in the nuclear membrane.

Translation (at the ribosome)

1. The mRNA attaches to a ribosome.
2. The ribosome reads the mRNA three bases at a time. Each triplet of bases on the mRNA is a codon, which codes for one amino acid.
3. Transfer RNA (tRNA) molecules each carry a specific amino acid and have an anticodon of three bases, complementary to one codon.
4. The anticodon pairs with its codon, bringing the correct amino acid into place.
5. Neighbouring amino acids join, and the ribosome moves along to the next codon. The chain of amino acids folds to make the protein.

Worked example. The template strand of part of a gene is TAC GGA TTC. Give the mRNA codons and the tRNA anticodons.

DNA template:     T A C   G G A   T T C
mRNA codons:      A U G   C C U   A A G    (T on DNA → A; A on DNA → U)
tRNA anticodons:  U A C   G G A   U U C    (complementary to each codon)

Three codons, so this section codes for three amino acids. Notice that the anticodons have the same sequence as the DNA template, but with U in place of T.

Alleles and the key terms (3.19–3.21B)

Genes exist in alternative forms called alleles. Different alleles give rise to differences in inherited characteristics. Body cells carry two alleles of each gene.

Term	Meaning
Genotype	The alleles an organism has, e.g. Bb
Phenotype	The characteristic that is expressed, e.g. black fur
Dominant	An allele expressed in the phenotype even when only one copy is present
Recessive	An allele expressed only when two copies are present (no dominant allele)
Homozygous	Two identical alleles, e.g. BB or bb
Heterozygous	Two different alleles, e.g. Bb
Codominance (Paper 2 only)	Both alleles in a heterozygote are fully expressed in the phenotype

Codominance – Paper 2 only

In a codominant heterozygote, neither allele masks the other. Both show. A standard example is coat colour in shorthorn cattle: a red parent (CR CR) crossed with a white parent (CW CW) gives roan calves (CR CW), which have both red hairs and white hairs.

Worked example. Two roan cattle are crossed.

Parents' phenotypes:  roan        ×   roan
Parents' genotypes:   CR CW       ×   CR CW
Gametes:              CR, CW          CR, CW

              CR        CW
   CR       CR CR     CR CW
   CW       CR CW     CW CW

Offspring:  1 red (CR CR) : 2 roan (CR CW) : 1 white (CW CW)

The phenotype ratio is 1 : 2 : 1, matching the genotype ratio, because each genotype has its own phenotype.

Polygenic inheritance (3.22)

Most phenotypic features are polygenic: they result from several genes acting together, not a single gene. Human height and skin colour are examples. Because many alleles add their effects, these features show a range of values rather than a few clear-cut types.

Monohybrid crosses and probabilities (3.23, 3.25)

A monohybrid cross follows one gene. Set out every genetic diagram in the same order:

1. Parents’ phenotypes
2. Parents’ genotypes
3. Gametes (each gamete carries one allele)
4. Punnett square or lines joining gametes
5. Offspring genotypes, then phenotypes, then the ratio or probability

Worked example. In guinea pigs, the allele for black fur (B) is dominant to the allele for white fur (b). A heterozygous black guinea pig is crossed with a white one. Find the probability that an offspring is white.

Parents' phenotypes:  black      ×   white
Parents' genotypes:   Bb         ×   bb
Gametes:              B, b           b, b

              b       b
   B         Bb      Bb
   b         bb      bb

Offspring:  2 Bb (black) : 2 bb (white)

Probability of white = 2 out of 4 = 1/2 (0.5, or 50%). Ratio 1 black : 1 white.

Give a probability as a fraction, decimal, percentage or “1 in 4”; a ratio such as 3 : 1 is not a probability. Each fertilisation is independent, so the probability is the same for every offspring, however many have already been born.

Family pedigrees (3.24)

A pedigree shows a characteristic through generations. Squares are males, circles are females, a shaded symbol is affected, and a horizontal line joins parents.

Worked example. In a family, two unaffected parents (I-1 and I-2) have three children: II-1, an affected daughter; II-2, an unaffected son; II-3, an unaffected daughter.

1. Is the allele dominant or recessive? Two unaffected parents have an affected child. If the allele were dominant, at least one parent would be affected. So it is recessive.
2. Parents’ genotypes. Let A be the dominant (unaffected) allele and a the recessive allele. II-1 is aa, so she received an a from each parent. Both parents are unaffected, so both are Aa.
3. Chance their next child is affected. Aa × Aa gives 1 AA : 2 Aa : 1 aa, so the probability is 1/4.

Key clue: two unaffected parents with an affected child means recessive; two affected parents with an unaffected child means dominant.

Sex determination (3.26–3.27)

Sex is controlled by one pair of chromosomes: XX in a female, XY in a male. All eggs carry an X. Half of the sperm carry an X and half carry a Y, so the sperm decides the sex at fertilisation.

Parents:     female XX   ×   male XY
Gametes:     X, X            X, Y

              X        Y
   X         XX       XY
   X         XX       XY

Offspring: 2 XX (female) : 2 XY (male)

Probability of a boy = 1/2 at every fertilisation.

Mitosis, meiosis and fertilisation (3.28–3.32)

Mitosis divides a diploid cell into two cells with identical sets of chromosomes. It happens in growth, repair, cloning and asexual reproduction.

Meiosis divides a cell into four cells, each with half the number of chromosomes. These are haploid gametes, and they are genetically different from each other.

Random fertilisation: any of many genetically different sperm can fuse with any genetically different egg, producing genetic variation among offspring of the same parents.

In human cells the diploid number is 46 and the haploid number is 23.

Worked example. A plant species has a diploid number of 20. How many chromosomes are in (a) a root-tip cell made by mitosis, (b) a gamete, (c) a zygote?

• (a) Mitosis keeps the number the same: 20.
• (b) Meiosis halves it: 20 ÷ 2 = 10.
• (c) Two gametes fuse: 10 + 10 = 20, restoring the diploid number.

Variation and mutation (3.33–3.37B)

Variation within a species can be:

• genetic – e.g. blood group
• environmental – e.g. a scar, or a language spoken
• a combination of both – e.g. height, where alleles set a potential range and diet affects where in that range a person ends up

A mutation is a rare, random change in genetic material that can be inherited.

Paper 2 only (3.35B–3.37B). A change in the DNA base sequence can change a codon, so a different amino acid may be placed in the protein. The protein may fold into a different shape and work differently, which can change the phenotype. Most mutations have no effect on the phenotype, some have a small effect, and rarely do they have a significant effect. The rate of mutation is increased by ionising radiation (for example gamma rays, X-rays and ultraviolet rays) and by some chemical mutagens (for example chemicals in tobacco).

Natural selection and antibiotic resistance (3.38–3.39)

Darwin’s theory of evolution by natural selection, in steps:

1. Individuals in a population show variation, caused partly by different alleles.
2. More offspring are produced than can survive, so there is competition for resources.
3. Individuals with features better suited to the environment are more likely to survive and reproduce.
4. They pass on their alleles to their offspring.
5. Over many generations, the advantageous allele becomes more common in the population.

Antibiotic resistance follows the same steps. A random mutation may make a bacterium resistant. When an antibiotic is used, non-resistant bacteria die, but resistant ones survive and divide, so the proportion of resistant bacteria rises. Infections caused by these bacteria are difficult to control because the antibiotic no longer kills them.

Common errors

• Writing gametes with two alleles (Bb) instead of one (B or b).
• Using different letters for the two alleles (T and s) instead of one letter in two cases.
• Giving a ratio when the question asks for a probability.
• Saying the mother decides the sex of a child. Eggs all carry X; the sperm decides.
• Saying the antibiotic causes the mutation. The mutation was already there; the antibiotic selects it.
• Saying all mutations are harmful. Most have no effect.
• Mixing up the products: mitosis gives two identical diploid cells; meiosis gives four different haploid cells.

Official syllabus

Pearson Edexcel International GCSE in Biology (4BI1) specification, Issue 3 (September 2024), Pearson Education Limited; first assessment June 2019. Topic 3, Reproduction and inheritance, sub-topic (b) Inheritance, points 3.14–3.39.