What you'll learn
Meiosis, inheritance and genetic crosses covers how gametes are formed, how variation is generated, and how characteristics are passed between generations. At CAPE level you must be able to work confidently with monohybrid and dihybrid crosses, handle the departures from simple dominance — codominance, multiple alleles, sex linkage, autosomal linkage and epistasis — and apply the chi-squared test to decide whether observed results differ significantly from those expected. By the end of this topic you should be able to describe meiosis and how it generates variation, construct and interpret genetic diagrams for every cross type, recognise characteristic ratios and what they indicate, and carry out and interpret a chi-squared test.
Key terms and definitions
Meiosis — nuclear division producing four genetically different haploid cells from one diploid cell
Diploid — having two complete sets of chromosomes
Haploid — having one set of chromosomes
Homologous chromosomes — a pair of chromosomes with the same genes at the same loci, one from each parent
Crossing over — the exchange of sections between non-sister chromatids of a homologous pair
Chiasma — the point at which crossing over occurs
Independent assortment — the random orientation of homologous pairs, so that each gamete receives a random combination
Allele — one of the alternative forms of a gene
Homozygous — having two identical alleles for a gene
Heterozygous — having two different alleles for a gene
Genotype — the alleles an organism possesses
Phenotype — the observable characteristics resulting from the genotype
Codominance — where both alleles of a heterozygote are fully expressed in the phenotype
Sex linkage — where a gene is carried on the X chromosome and has no equivalent on the Y
Epistasis — where one gene affects the expression of another
Chi-squared test — a statistical test comparing observed with expected frequencies
Core concepts
Meiosis and the generation of variation
Meiosis consists of two successive divisions and produces four haploid cells from one diploid cell.
In meiosis I, homologous chromosomes pair up and are separated, halving the chromosome number. In meiosis II, the sister chromatids are separated, as in mitosis.
Three processes generate genetic variation, and all three must be known.
Crossing over occurs in prophase I. Homologous chromosomes pair, and non-sister chromatids exchange corresponding sections at points called chiasmata. This produces new combinations of alleles on the same chromosome, which is why linked genes are not always inherited together.
Independent assortment occurs in metaphase I. Each homologous pair lines up at the equator independently of every other pair, so which member of each pair goes to which pole is random. With n pairs of chromosomes, 2 to the power n different combinations are possible, which for humans with 23 pairs gives over eight million.
Random fertilisation then combines any one of the enormous number of possible male gametes with any one of the female gametes, multiplying the variation further.
The significance of variation is that it provides the raw material on which natural selection acts, allowing populations to adapt when the environment changes.
Monohybrid inheritance
A monohybrid cross concerns a single gene. The standard approach is systematic and should be used every time: state the parental phenotypes and genotypes, show the gametes in circles, construct a Punnett square, then give the offspring genotypes, phenotypes and ratio.
Two heterozygotes crossed give a genotype ratio of one homozygous dominant to two heterozygous to one homozygous recessive, and a phenotype ratio of three dominant to one recessive.
A heterozygote crossed with a homozygous recessive gives a one to one phenotype ratio, and this is the test cross used to determine whether an organism showing the dominant phenotype is homozygous or heterozygous. If any offspring show the recessive phenotype, the parent must have been heterozygous.
Codominance and multiple alleles
In codominance both alleles are expressed fully in the heterozygote, which shows a distinct third phenotype rather than an intermediate one. Both alleles are represented by capital letters with different superscripts or subscripts rather than by upper and lower case.
Sickle cell anaemia illustrates codominance and is particularly relevant in the Caribbean. A homozygote for the normal allele has normal haemoglobin; a homozygote for the sickle allele has sickle cell anaemia; and the heterozygote has sickle cell trait, producing both normal and abnormal haemoglobin. Heterozygotes have increased resistance to malaria, which is why the allele persists at appreciable frequency in populations where malaria has been endemic.
Multiple alleles occur where a gene has more than two forms in the population, although any individual still carries only two. Human ABO blood groups are the standard example, with three alleles: the A and B alleles are codominant with each other and both are dominant to the O allele. This gives four phenotypes from six genotypes.
A cross between two heterozygotes for codominant alleles gives a phenotype ratio of one to two to one, rather than three to one, because the heterozygote is distinguishable.
Sex linkage
The X chromosome is much larger than the Y and carries genes for which the Y has no equivalent. A gene on the X chromosome is sex-linked.
Females have two X chromosomes and therefore two alleles, so a female may be homozygous or heterozygous. A heterozygous female showing the dominant phenotype is a carrier.
Males have only one X chromosome, so they have only one allele for any sex-linked gene and cannot be heterozygous. Whatever allele is present is expressed.
This explains why sex-linked recessive conditions such as haemophilia and red-green colour blindness are far more common in males: a male needs only one recessive allele to be affected, whereas a female needs two.
The notation must show the chromosomes: alleles are written as superscripts on an X, and the Y is written without a superscript to indicate that no allele is present.
A characteristic of pedigree analysis is that an affected male passes the allele to all his daughters, who become carriers, but to none of his sons, since sons receive his Y chromosome.
Dihybrid inheritance and linkage
A dihybrid cross concerns two genes. Where the genes are on different chromosomes they assort independently, and a cross between two double heterozygotes gives the characteristic phenotype ratio of nine to three to three to one.
A test cross of a double heterozygote with a double homozygous recessive gives a one to one to one to one ratio if the genes are unlinked.
Autosomal linkage occurs where two genes lie on the same chromosome and are therefore usually inherited together. They do not assort independently, so the expected ratios are not obtained. Instead, the parental combinations appear far more frequently than expected, and the recombinant combinations appear only at low frequency, produced by crossing over.
Recognising linkage from data is a standard question: an excess of parental types and a deficiency of recombinants, rather than the expected nine to three to three to one or one to one to one to one, indicates that the genes are linked. The closer together two genes lie, the less often crossing over separates them and the rarer the recombinants.
Epistasis
Epistasis occurs where one gene affects the expression of another, so the two do not act independently.
In recessive epistasis, a homozygous recessive genotype at one locus prevents the expression of the second gene entirely, typically modifying the nine to three to three to one ratio to nine to three to four.
In dominant epistasis, a dominant allele at one locus masks the second gene, commonly giving twelve to three to one.
A frequent example is coat colour, where one gene determines whether pigment is produced at all and a second determines which pigment. If the first gene produces no pigment, the second gene's alleles cannot be expressed whatever they are.
Recognising an unusual ratio that sums to sixteen is the clue that epistasis is involved.
The chi-squared test
The chi-squared test determines whether the difference between observed and expected results is statistically significant or could reasonably be due to chance.
The null hypothesis states that there is no significant difference between observed and expected values.
The statistic is calculated by taking, for each category, the difference between observed and expected, squaring it, dividing by the expected value, and then summing these quantities across all categories.
The degrees of freedom equal the number of categories minus one.
The calculated value is compared with the critical value from a table at the 0.05 probability level for the appropriate degrees of freedom.
If the calculated value is less than the critical value, the difference is not significant, the null hypothesis is accepted, and the results are consistent with the expected ratio.
If the calculated value is greater than the critical value, the difference is significant, the null hypothesis is rejected, and some factor other than chance is operating — such as linkage, epistasis or selective mortality of one genotype.
Two conditions must be met: the data must be raw counts rather than percentages, and the expected values should generally be at least five.
Worked examples
Example 1: A sex-linked cross (5 marks)
Haemophilia is caused by a recessive allele on the X chromosome. A carrier female marries an unaffected male. Determine the expected proportions of their children.
The mother is heterozygous, carrying one normal and one haemophilia allele on her two X chromosomes. The father is unaffected, so his single X carries the normal allele, and his Y carries no allele for this gene.
The mother produces two types of egg in equal proportion: one carrying the normal allele and one carrying the haemophilia allele. The father produces two types of sperm: one carrying an X with the normal allele and one carrying a Y.
The four equally likely combinations are: a homozygous normal female, a carrier female, an unaffected male, and an affected male.
The expectation is therefore that all daughters are unaffected, with half of them carriers, and that half the sons are affected. Overall one quarter of the children are expected to have haemophilia, and all of these are male.
Example 2: Identifying linkage from data (5 marks)
A test cross of a double heterozygote produces 412 parental-type offspring and 38 recombinant offspring. Explain what this shows.
If the two genes were on different chromosomes they would assort independently, and a test cross of a double heterozygote would give four phenotypes in an approximately one to one to one to one ratio, so about 112 or 113 of each from 450 offspring.
The observed results depart from this dramatically: the parental combinations greatly outnumber the recombinants. This indicates that the two genes are linked, lying on the same chromosome, so they tend to be inherited together and the parental combinations predominate.
The 38 recombinant offspring arise from crossing over between the two loci during prophase I of meiosis, which separates the linked alleles and produces new combinations. The low proportion of recombinants, about 8 per cent, indicates that the two genes lie close together on the chromosome, since crossing over between them is correspondingly rare.
Example 3: Applying the chi-squared test (5 marks)
A cross is expected to give a three to one ratio. From 400 offspring, 289 show the dominant phenotype and 111 the recessive. The critical value at one degree of freedom is 3.84. Determine whether the results differ significantly from expectation.
The null hypothesis is that there is no significant difference between the observed and expected results.
The expected values for a three to one ratio from 400 offspring are 300 dominant and 100 recessive.
For the dominant category, the difference is 289 minus 300, which is minus 11. Squaring gives 121, and dividing by the expected value of 300 gives 0.403.
For the recessive category, the difference is 111 minus 100, which is 11. Squaring gives 121, and dividing by the expected value of 100 gives 1.21.
The chi-squared value is 0.403 plus 1.21, which is 1.613.
There are two categories, so the degrees of freedom are 2 minus 1, which is 1. The calculated value of 1.613 is less than the critical value of 3.84, so the difference is not significant at the 0.05 level. The null hypothesis is accepted and the results are consistent with a three to one ratio, the deviation being attributable to chance.
Common mistakes and how to avoid them
The most frequent error is careless notation. Capital and lower case letters must be clearly distinguished, and for codominance and sex linkage the correct notation — superscripts on a common capital, or superscripts on X with a bare Y — must be used.
Students often omit the sex chromosomes entirely in sex-linked crosses, which makes it impossible to show why males cannot be carriers.
Another common slip is using percentages rather than raw counts in a chi-squared test. The test requires actual numbers.
Many candidates calculate a chi-squared value correctly but then state the conclusion the wrong way round. A calculated value below the critical value means the difference is not significant.
In linkage questions, answers frequently identify the departure from the expected ratio without explaining that the recombinants arise from crossing over.
Finally, candidates sometimes describe codominance as blending. Codominance expresses both alleles fully and distinctly; it is not an intermediate.
Exam technique for "Meiosis, inheritance and genetic crosses"
Set out every cross in full, with a key defining the symbols, parental phenotypes and genotypes, gametes circled, a Punnett square, and the offspring ratio. Marks are allocated to each stage, so the working is worth more than the answer.
Define your symbols before using them. An unlabelled letter cannot be credited.
Learn the diagnostic ratios: three to one for monohybrid dominance, one to two to one for codominance, nine to three to three to one for unlinked dihybrid, one to one to one to one for a dihybrid test cross, and totals of sixteen with unusual groupings for epistasis.
For chi-squared, lay the calculation out as a table with columns for observed, expected, difference, difference squared, and difference squared divided by expected. It is faster and the examiner can follow it.
State the null hypothesis and the conclusion in full, referring to the critical value and the probability level.
Quick revision summary
Meiosis produces four genetically different haploid cells through two divisions, generating variation by crossing over at chiasmata in prophase I, independent assortment of homologous pairs in metaphase I giving 2 to the power n combinations, and random fertilisation. Monohybrid crosses of two heterozygotes give three to one, and a test cross with a homozygous recessive reveals an unknown genotype. Codominance expresses both alleles distinctly, giving one to two to one, as in sickle cell trait, and multiple alleles such as the ABO system give more phenotypes. Sex-linked genes lie on the X with no equivalent on the Y, so males have a single allele and are more often affected by recessive conditions, and affected males pass the allele to all daughters and no sons. Unlinked dihybrid crosses give nine to three to three to one, while autosomal linkage produces an excess of parental types with recombinants arising from crossing over, rarer when the genes lie closer together. Epistasis modifies ratios to nine to three to four or twelve to three to one. The chi-squared test compares observed with expected raw counts, with degrees of freedom one less than the number of categories, and a calculated value below the critical value means the difference is not significant.