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D3.2

Inheritance

Theme D · Continuity and change · Organisms · SL and HL · plus additional higher level

Genetics questions look intimidating and are actually among the most reliable marks in the course, because they follow rules. Write the parental genotypes, work out the gametes, fill a Punnett grid, read off the ratios. Almost every mistake comes from skipping one of those steps, or from using symbols carelessly. SL covers single genes, sex linkage, pedigrees and continuous variation; HL adds two genes at once, linkage, and the chi-squared test.

🎯What you need to be able to do

  • Explain inheritance through haploid gametes and a diploid zygote.
  • Describe how genetic crosses are carried out in flowering plants, using the terms P, F1, F2 and Punnett grid.
  • Use the terms genotype, phenotype, allele, homozygous, heterozygous, dominant and recessive; explain phenotypic plasticity.
  • Explain phenylketonuria, multiple alleles and ABO blood groups, codominance and incomplete dominance.
  • Explain sex determination and sex linkage, using haemophilia, and deduce patterns of inheritance from pedigree charts.
  • Explain continuous variation due to polygenic inheritance and environment, and construct and interpret box-and-whisker plots.
  • AHL Explain independent assortment, and predict 9:3:3:1 and 1:1:1:1 ratios with dihybrid Punnett grids.
  • AHL Explain autosomal linkage and recombinants, and use a chi-squared test on dihybrid cross data.

📚The biology

Gametes and zygotes

In every eukaryote with a sexual life cycle, inheritance follows the same pattern. Each parent produces haploid gametes by meiosis, carrying one copy of each gene. At fertilization two gametes fuse to form a diploid zygote, which therefore has two copies of each autosomal gene — one from each parent. (Autosomes are all the chromosomes other than the sex chromosomes.)

Genetic crosses in flowering plants

Pollen contains the male gametes and the female gametes are in ovules inside the ovary, so a cross is made by pollination: pollen from one chosen plant is transferred to the stigma of another. Plants such as peas produce both male and female gametes on the same plant, so they can self-pollinate and therefore self-fertilize. To make a controlled cross, the anthers of the female parent are removed before they release pollen, and the flower is covered to prevent stray pollen arriving.

P generation
the parents of a cross
F1 generation
the offspring of the P generation (first filial)
F2 generation
the offspring of crossing or self-pollinating the F1
Punnett grid
a table with the gametes of one parent along the top and the other down the side, showing all possible offspring genotypes

Genetic crosses are widely used to breed new varieties of crop and ornamental plants.

Genes, alleles and genotype

A gene is a heritable factor that influences a characteristic, found at a particular position (locus) on a chromosome. Alleles are different versions of the same gene, differing slightly in base sequence. The genotype is the combination of alleles an organism has for a gene.

Homozygous
two identical alleles of a gene (e.g. TT or tt)
Heterozygous
two different alleles of a gene (e.g. Tt)

Phenotype

The phenotype is the set of observable traits of an organism, resulting from its genotype and environmental factors. In humans:

Genotype only
ABO blood group; eye colour largely; phenylketonuria.
Environment only
scars; tattoos; a language spoken.
Genotype and environment together
height (genes and nutrition); skin colour (genes and sun exposure); body mass.

Dominant and recessive alleles

A dominant allele affects the phenotype whenever it is present, in one or two copies. A recessive allele only affects the phenotype when no dominant allele is present, that is, in the homozygous recessive. Dominant alleles are written with a capital letter, recessive with the lower-case version of the same letter.

Why does a heterozygote look the same as a homozygous dominant? Usually the dominant allele codes for a functional protein such as an enzyme, and the recessive allele codes for a non-functional version or none at all. One copy of the dominant allele produces enough functional protein to give the full phenotype, so TT and Tt look identical.

Phenotypic plasticity

Phenotypic plasticity is the capacity of an organism to develop traits suited to the environment it experiences, by varying patterns of gene expression. It is not due to a change in genotype, and the changes may be reversible during an individual’s life. Examples: human skin tanning with sun exposure; muscles enlarging with exercise; some plants producing thinner, broader leaves in shade.

Phenylketonuria

Phenylketonuria (PKU) is a recessive genetic condition caused by a mutation in an autosomal gene coding for the enzyme that converts the amino acid phenylalanine to tyrosine. Only people homozygous for the recessive allele are affected: they cannot make the functional enzyme, phenylalanine accumulates in the blood and damages the developing brain. Heterozygous carriers have one working allele and are unaffected. Newborns are screened, and affected children follow a low-phenylalanine diet, which prevents brain damage.

SNPs and multiple alleles

Each base substitution that becomes established in a population is a single-nucleotide polymorphism (D1.3). Over time, many different alleles of a gene can accumulate. Any number of alleles of a gene can exist in the gene pool of a population, but an individual diploid organism inherits only two.

ABO blood groups: multiple alleles

The ABO blood group gene has three alleles: IA, IB and i. IA and IB are codominant with each other, and both are dominant to i.

Group A
IAIA or IAi
Group B
IBIB or IBi
Group AB
IAIB
Group O
ii

Codominance and incomplete dominance

Codominance
Both alleles are fully expressed in the heterozygote, which has a dual phenotype showing both traits. Example: blood group AB (IAIB): red blood cells carry both A and B antigens.
Incomplete dominance
Neither allele is fully dominant; the heterozygote has an intermediate phenotype. Example: the four o’clock flower (Mirabilis jalapa): red × white gives pink offspring. Crossing two pinks gives red : pink : white in a 1 : 2 : 1 ratio.

Sex determination and sex chromosomes

Humans have 22 pairs of autosomes and one pair of sex chromosomes: typical female bodies have XX and typical male bodies XY. All eggs carry an X chromosome; sperm carry either an X or a Y. So the sex chromosome in the sperm determines whether the zygote develops certain male-typical physical characteristics (Y, due to a gene on the Y chromosome that triggers development of testes) or female-typical ones (X). The ratio is expected to be 1 : 1.

The X chromosome is much larger and carries far more genes (over 800 coding for proteins) than the Y chromosome (under 100). Genes carried on the X are sex-linked. Because males have only one X, any recessive allele on it is expressed.

Haemophilia

Haemophilia is a sex-linked recessive disorder in which blood fails to clot properly, because a gene on the X chromosome coding for a clotting factor is faulty. Alleles on the X chromosome are shown as superscripts on an uppercase X: XH (normal) and Xh (haemophilia).

Females
XHXH unaffected
XHXh unaffected carrier
XhXh affected (rare)
Males
XHY unaffected
XhY affected (no second X to mask it)

Haemophilia is far more common in males. An affected male cannot pass it to his sons (they get his Y) but all his daughters will be carriers.

Pedigree charts

A pedigree chart shows the inheritance of a condition in a family: circles for females, squares for males, shaded for affected, horizontal lines joining parents, vertical lines to children. Reasoning from the chart:

  • Recessive: two unaffected parents have an affected child — both parents must be carriers.
  • Dominant: two affected parents have an unaffected child; and every affected person has at least one affected parent.
  • Sex-linked recessive: mostly affected males; affected sons of unaffected (carrier) mothers; never passed from father to son.

This uses both kinds of reasoning. Seeing a pattern in some families and generalizing (“this condition is recessive”) is inductive reasoning. Then applying that conclusion to deduce the genotypes of particular individuals in the chart is deductive reasoning.

Pedigrees also show why many societies prohibit marriage between close relatives. Relatives are more likely to carry the same rare recessive alleles, inherited from a common ancestor, so their children have a higher probability of being homozygous recessive for a harmful condition.

Continuous variation

Discrete (discontinuous) variables
Fall into distinct categories with no intermediates. Usually controlled by one gene with little environmental effect. Example: ABO blood group.
Continuous variables
Show a range of values, often forming a normal distribution. Result from polygenic inheritance (many genes, each with a small additive effect) and/or environmental factors. Example: human skin colour, controlled by several genes affecting melanin production, and by sun exposure.

Continuous data are summarized with measures of central tendency: the mean (sum of values ÷ number of values), the median (middle value when ordered) and the mode (most frequent value).

Box-and-whisker plots

A box-and-whisker plot displays the distribution of a continuous variable, such as student height, showing six features:

Minimum
lowest value that is not an outlier (end of lower whisker)
First quartile (Q1)
lower edge of the box: 25% of values are below it
Median
line inside the box
Third quartile (Q3)
upper edge of the box: 75% of values are below it
Maximum
highest value that is not an outlier (end of upper whisker)
Outliers
plotted as separate points

The interquartile range (IQR) = Q3 − Q1. A data point is an outlier if it is more than 1.5 × IQR above Q3 or below Q1.

Segregation and independent assortment AHL

The movements of chromosomes in meiosis explain the outcomes of crosses:

  • Segregation: the two alleles of a gene, on a pair of homologous chromosomes, separate into different gametes when homologues separate in meiosis I.
  • Independent assortment: for two genes on different chromosomes (unlinked), the orientation of one bivalent at metaphase I is independent of the other. So the allele a gamete receives for one gene does not affect which allele it receives for the other. An individual AaBb produces four types of gamete — AB, Ab, aB, ab — in equal proportions.

Dihybrid crosses AHL

A dihybrid cross follows two genes at once. With two unlinked autosomal genes:

AaBb × AaBb
Each parent makes 4 gametes in equal proportions, giving a 4 × 4 Punnett grid of 16 squares. Phenotypes: both dominant traits 9 : dominant A only 3 : dominant B only 3 : both recessive 1.
AaBb × aabb (test cross)
The heterozygote makes 4 gametes in equal proportions; the homozygous recessive makes only ab. Phenotypes: 1 : 1 : 1 : 1.

The 9:3:3:1 and 1:1:1:1 ratios depend on what has been called Mendel’s second law, independent assortment. This “law” only applies if the genes are on different chromosomes, or so far apart on one chromosome that crossing over separates them half the time (a 50% recombination rate). Like other biological “laws”, it has exceptions.

Gene loci in databases AHL

Online databases such as those of the NCBI and Ensembl give the locus (chromosome and position) of each human gene and the polypeptide it codes for. Exploring them, you can find pairs of genes on different chromosomes, which will assort independently, and pairs close together on the same chromosome, which will tend to be inherited together.

Autosomal gene linkage AHL

Linked genes are genes located on the same chromosome. They do not assort independently, because they tend to be inherited together: they move as one unit when the chromosome moves in meiosis. The closer together they are, the more strongly linked.

When showing linked genes, write the alleles alongside vertical lines representing the two homologous chromosomes, so that it is clear which alleles are on the same chromosome. For example, a heterozygote with A and B on one chromosome and a and b on the other is shown with A above B on the left line and a above b on the right line.

Recombinants AHL

In a cross between an individual heterozygous for both genes and one homozygous recessive for both:

  • If the genes are unlinked, the heterozygote produces four gamete types equally, and offspring appear in a 1:1:1:1 ratio.
  • If the genes are linked, the heterozygote mostly produces gametes with the parental combinations of alleles (AB and ab in the example above). Only when crossing over occurs between the two loci are recombinant gametes (Ab and aB) produced, and these are less common.

Recombinants are gametes, genotypes or phenotypes with a combination of alleles different from either parent. In a test cross with linked genes, the two parental phenotypes are most common and the two recombinant phenotypes are least common.

Chi-squared test on dihybrid data AHL

Observed results never match expected ratios exactly, because of chance. The chi-squared test decides whether a difference between observed and expected results is small enough to be due to chance or is statistically significant.

  1. State a null hypothesis: there is no significant difference between observed and expected results (e.g. the genes are unlinked and assort independently). The alternative hypothesis is that there is a significant difference.
  2. Calculate expected numbers from the predicted ratio and the total.
  3. Calculate \( \chi^{2} = \sum \dfrac{(O - E)^{2}}{E} \).
  4. Degrees of freedom = number of categories − 1 (for four phenotypes, 3).
  5. Compare with the critical value at the p = 0.05 significance level (7.815 for 3 degrees of freedom). If \( \chi^{2} \) is less, accept the null hypothesis; if it is greater, reject it.

The F2 generation is a sample used to represent all possible offspring. p = 0.05 means there is a 5% probability that a difference this large would arise by chance alone if the null hypothesis were true.

✏️Worked example

(a) A woman who is a carrier for haemophilia has children with an unaffected man. Using a Punnett grid, deduce the probability that a child is an affected boy, and the probability that a son is affected.
(b) AHL In pea plants, round seed (R) is dominant to wrinkled (r) and yellow seed (Y) is dominant to green (y). Plants heterozygous for both genes were self-pollinated. The F2 contained 315 round yellow, 108 round green, 101 wrinkled yellow and 32 wrinkled green. Use a chi-squared test to decide whether the results fit a 9:3:3:1 ratio.

(a) Parents: XHXh × XHY. Gametes: mother XH or Xh; father XH or Y.

XH (father) + XH
XHXH — unaffected girl
XH (father) + Xh
XHXh — carrier girl
Y (father) + XH
XHY — unaffected boy
Y (father) + Xh
XhY — affected boy

Probability that a child is an affected boy = 1/4 (25%). Probability that a son is affected = 1 of the 2 boys = 1/2 (50%).

(b) Null hypothesis: the genes assort independently, so there is no significant difference from a 9:3:3:1 ratio. Total = 556.

Round yellow
O = 315, E = 9/16 × 556 = 312.75
(O − E)2/E = 0.016
Round green
O = 108, E = 3/16 × 556 = 104.25
(O − E)2/E = 0.135
Wrinkled yellow
O = 101, E = 104.25
(O − E)2/E = 0.101
Wrinkled green
O = 32, E = 1/16 × 556 = 34.75
(O − E)2/E = 0.218
\[ \chi^{2} = 0.016 + 0.135 + 0.101 + 0.218 = 0.47 \]

Degrees of freedom = 4 − 1 = 3; critical value at p = 0.05 is 7.815. Since 0.47 < 7.815, accept the null hypothesis: the differences are not significant and can be attributed to chance. The results fit a 9:3:3:1 ratio, consistent with the genes being unlinked.

Check it. Expected values must add up to the total: 312.75 + 104.25 + 104.25 + 34.75 = 556. And each observed number is within about 4 of its expected value, so a small chi-squared is the expected result.
Using percentages or ratios instead of actual counts. Chi-squared must be calculated from the numbers of individuals observed and expected, never from percentages; the size of the sample is part of the test. And “accept the null hypothesis” does not prove the genes are unlinked — only that there is no evidence they are not.

📝Practise

Work through these on paper, then reveal the answer. Questions 5 and 6 are AHL.

1. Two parents both have blood group A. Their child has blood group O. Deduce the genotypes of the parents and the probability that their next child has blood group O.
A child with group O is ii, so must have received an i allele from each parent. Both parents are group A, so each must be IAi. Cross IAi × IAi: offspring IAIA : IAi : ii in a 1 : 2 : 1 ratio. Probability of group O (ii) = 1/4 (25%) for each child.
2. Distinguish between codominance and incomplete dominance, with an example of each.
In codominance, both alleles are fully expressed in the heterozygote, which shows a dual phenotype with both traits — e.g. blood group AB (IAIB), with both A and B antigens on red blood cells. In incomplete dominance, the heterozygote has an intermediate phenotype, a blend of the two — e.g. pink four o’clock flowers from red × white parents.
3. In a pedigree, two unaffected parents have an affected daughter. Deduce, with reasons, whether the condition is dominant or recessive, and whether it could be sex-linked.
The condition must be recessive: if it were dominant, an affected child would need at least one affected parent. Both unaffected parents must be carriers (heterozygous). It is not sex-linked recessive: an affected daughter would need to be XaXa, receiving an Xa from her father, but a father with XaY would himself be affected. Since the father is unaffected, the condition is autosomal recessive.
4. The heights of 13 students have Q1 = 160 cm, median = 163 cm and Q3 = 168 cm. The shortest student is 148 cm and the tallest is 190 cm. Determine whether either is an outlier.
IQR = Q3 − Q1 = 168 − 160 = 8 cm. 1.5 × IQR = 12 cm. Upper limit = 168 + 12 = 180 cm; lower limit = 160 − 12 = 148 cm. The tallest student (190 cm) is more than 1.5 × IQR above Q3, so 190 cm is an outlier. The shortest (148 cm) is exactly at the lower limit, not more than 1.5 × IQR below Q1, so it is not an outlier.
5. AHL Explain why a dihybrid cross between two double heterozygotes gives a 9:3:3:1 ratio only if the genes are unlinked.
If the genes are on different chromosomes, the two bivalents orient independently at metaphase I (independent assortment), so a double heterozygote (AaBb) produces four types of gamete — AB, Ab, aB and ab — in equal proportions. Combining these in a 4 × 4 Punnett grid gives phenotypes in a 9:3:3:1 ratio. If the genes are linked (close together on the same chromosome), they tend to be inherited together, so parental combinations of alleles are over-represented in gametes and recombinant gametes are produced only by crossing over, less often. The four gamete types are no longer equal, so the ratio is not 9:3:3:1.
6. AHL In a test cross AaBb × aabb, the offspring were 42 AaBb, 40 aabb, 9 Aabb and 9 aaBb. Explain these results.
If the genes were unlinked, a 1:1:1:1 ratio (25 of each) would be expected. Instead, the parental types AaBb and aabb are far more common and Aabb and aaBb are rare, so the genes are linked, with A and B on one chromosome and a and b on the other in the heterozygous parent. Most gametes from that parent carry the parental combinations AB or ab. Aabb and aaBb are recombinants, produced from Ab and aB gametes formed by crossing over between the two loci in meiosis. Recombinants make up 18 of 100 offspring, a recombination frequency of 18%.

🔗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.

  • NCBI Gene and Ensembl — databases showing the chromosomal locus and protein product of every human gene.
  • Learn.Genetics (University of Utah) — interactive pedigree and heredity activities.
  • HHMI BioInteractive — genetics problem sets with worked chi-squared analyses.