Dihybrid Crosses and Gene Linkage

A dihybrid cross follows two genes at the same time. Each gene has two alleles, so a parent that is heterozygous for both genes is written AaBb. When such an individual is crossed with another AaBb individual, or with a double-recessive aabb individual, the offspring show combinations of two traits. The numbers of offspring in each combination tell us whether the two genes are passed on independently or travel together on the same chromosome.

This laboratory uses a virtual moth. Gene 1 controls wing colour (silver A is dominant to bronze a) and gene 2 controls antenna shape (feathered B is dominant to thread-like b). You choose the cross, the arrangement of alleles in the parent, the recombination frequency (how often a crossover separates the two genes) and the number of offspring. The simulator then breeds the offspring by random sampling, and the analysis tools test whether the results fit the ratio expected for independent genes.

The notes work in three layers. Foundation material needs ratios, fractions and the idea of dominant and recessive alleles. Core material matches typical Cambridge, WASSCE, IB Diploma and AP Biology courses: genetic diagrams, 9:3:3:1 and its modified forms, linkage, crossing over, the chi-squared test and the calculation of recombination frequency. Extension material (marked clearly) goes towards university genetics: mapping functions, LOD scores, interference and the statistics behind the estimates.

The default settings are an F2 cross, coupling arrangement, a recombination frequency of 20% and 400 offspring. Before you change anything, predict what ratio you expect, then test your prediction.

Learning outcomes

Foundation

• Write the genotypes of gametes made by AaBb, and explain why there are four types when the genes are on different chromosomes.
• Complete a 16-box Punnett square and read the 9:3:3:1 phenotype ratio from it.
• State what a test cross is and what ratio it gives for independent genes.
• Describe linked genes, crossing over and recombinant offspring in words.

Core (A-level, IB, AP, WASSCE)

• Predict gamete and offspring frequencies for linked genes from a recombination frequency, for both coupling and repulsion.
• Calculate expected numbers and perform a chi-squared test against 9:3:3:1 or 1:1:1:1, stating the null hypothesis, degrees of freedom, critical value and conclusion.
• Calculate a recombination frequency and map distance from test cross data, with a standard error.
• Explain how epistasis modifies the 9:3:3:1 ratio into 9:3:4, 9:7 or 12:3:1.
• Plan a breeding experiment, choose a sample size and recognise the limits of the evidence.

Extension (HL, university)

• Derive the F2 class frequencies for linked genes and the standard error of the recombination frequency.
• Convert recombination frequency to map distance with the Haldane and Kosambi mapping functions.
• Calculate and interpret a LOD score, and describe interference and double crossovers.

Genes, alleles and segregation

A gene is a region of DNA at a fixed position (a locus) on a chromosome. Different versions of a gene are alleles. An individual has two alleles of each gene, one from each parent. Mendel's law of segregation says that the two alleles of a gene separate during gamete formation, so each gamete carries only one. An Aa individual makes A gametes and a gametes in equal numbers.

Independent assortment

When two genes are on different chromosomes, the way one pair of homologous chromosomes lines up at metaphase I does not affect the way the other pair lines up. The alleles of the two genes therefore enter gametes independently. This is Mendel's law of independent assortment. An AaBb individual makes four gamete types, AB, Ab, aB and ab, each with probability 1/4.

The 16-box Punnett square

For AaBb x AaBb, each parent makes four gamete types, so the Punnett square has 4 x 4 = 16 boxes. With complete dominance, 9 boxes are A_B_ (both dominant traits), 3 are A_bb, 3 are aaB_ and 1 is aabb. The underscore means that the second allele can be either. The phenotype ratio is 9:3:3:1. The simulator draws this square for your current settings and shades each box by phenotype.

The test cross

A test cross crosses an individual of unknown genotype with a homozygous recessive aabb individual. Because aabb gives only ab gametes, each offspring shows the alleles carried by the gamete from the tested parent. For independent genes, an AaBb parent gives AaBb, Aabb, aaBb and aabb in the ratio 1:1:1:1.

Linkage

Genes on the same chromosome are linked: they tend to be inherited together because they travel on one DNA molecule. Linked genes do not assort independently, and the 9:3:3:1 and 1:1:1:1 ratios fail. The parent produces more of the gamete types that match the original chromosomes (parental types) and fewer of the new combinations (recombinant types).

Coupling and repulsion

The arrangement of alleles on the two homologous chromosomes matters. In coupling (cis), the dominant alleles are on one chromosome and the recessive alleles on the other: AB / ab. In repulsion (trans), each chromosome carries one dominant and one recessive allele: Ab / aB. Both are AaBb, but the parental gametes differ. In coupling they are AB and ab. In repulsion they are Ab and aB.

Crossing over and recombination

In prophase I of meiosis, homologous chromosomes pair up and non-sister chromatids exchange segments at points called chiasmata. If a chiasma forms between two genes, two of the four resulting chromatids carry new allele combinations. The more distance between the genes, the more likely a chiasma lies between them. The recombination frequency r is the fraction of gametes (or test cross offspring) that are recombinant. One map unit (one centimorgan, cM) is defined as 1% recombination.

Figure 1. A crossover between the genes turns two parental chromatids into two parental and two recombinant chromatids. Only half of the chromatids are recombinant even when a crossover always forms, which is why r cannot exceed 50%.

Gene interaction and epistasis

The 9:3:3:1 ratio assumes that each gene affects a separate trait. Often two genes act in the same pathway, and then the F2 classes merge. Epistasis is the masking of one gene's effect by a different gene. The simulator offers three examples, all for unlinked genes:

• Recessive epistasis (9:3:4). The genotype aa blocks the pathway, so aaB_ and aabb look alike. A_B_ is purple, A_bb is red and aa__ is white.
• Complementary gene action (9:7). Both genes must have a dominant allele, as when two enzymes are needed in sequence. A_B_ is purple, and A_bb, aaB_ and aabb are all white.
• Dominant epistasis (12:3:1). A masks the effect of B, so A_B_ and A_bb look alike. A___ is white, aaB_ is yellow and aabb is green.

Epistatic ratios are always still sixteenths. They show that the genes assorted independently, but that the phenotypes were merged. This is different from linkage, which changes the sizes of the four classes.

Gamete frequencies (Core)

Let r be the recombination frequency, from 0 to 0.5. The two parental gamete types share 1 - r equally, and the two recombinant types share r equally:

parental: (1 - r)/2 each    recombinant: r/2 each

In coupling, AB and ab are parental. In repulsion, Ab and aB are parental. When r = 0.5, all four types are 0.25, which is independent assortment. When r = 0 the genes are completely linked and only parental gametes occur.

Test cross frequencies (Core)

The aabb parent gives only ab, so the frequency of each offspring class equals the frequency of the gamete that made it. For coupling with r = 0.12 the expected frequencies are AaBb 0.44, aabb 0.44, Aabb 0.06 and aaBb 0.06.

F2 class frequencies (Core)

For a cross AaBb x AaBb, each box of the Punnett square has probability equal to the product of its two gamete frequencies. Summing boxes gives a shortcut. Let q be the frequency of aabb. Because the A gene always segregates 1:1, half of all gametes carry a, so

P(aa) = (0.5)² = 0.25    P(bb) = 0.25

The aa class is made of aaB_ and aabb, so P(aaB_) = 0.25 - q. In the same way P(A_bb) = 0.25 - q. The remainder is A_B_:

P(A_B_) = 1 - 2(0.25 - q) - q = 0.5 + q

The whole F2 depends on one number, q = (frequency of ab gametes)². In coupling, q = ((1 - r)/2)². In repulsion, q = (r/2)². When r = 0.5, q = 1/16 and the classes are 9/16, 3/16, 3/16 and 1/16.

Estimating r from an F2 (Core)

Inverting the formula for q gives the estimate from the observed fraction f of aabb offspring:

coupling: r = 1 - 2√f     repulsion: r = 2√f

Expected counts and chi-squared (Core)

If N offspring are counted and the hypothesis predicts a proportion p_i for class i, the expected count is E_i = N p_i. The chi-squared statistic compares observed counts O_i with E_i:

χ² = Σ (O_i - E_i)² / E_i     degrees of freedom = classes - 1

A large value means the observed counts are far from the expected counts. The p-value is the probability, if the hypothesis were true, of a chi-squared at least this large. If p is below 0.05, we reject the hypothesis at the 5% level. For 3 degrees of freedom the 5% critical value is 7.81, for 2 it is 5.99 and for 1 it is 3.84.

Recombination frequency and its uncertainty (Core)

From a test cross with R recombinants among N offspring:

r = R/N     SE(r) = √[ r(1 - r)/N ]

This is the standard error of a proportion. About 95% of the time the true value lies within 1.96 standard errors of the estimate. For an F2 estimate through the aabb class, the delta method gives SE(r) = √[(1 - f)/N], which is larger, so the F2 is a poorer way to measure r.

How many offspring are needed? (Core)

If the true class proportions are p_i but you test against independent proportions e_i, the expected chi-squared is about

χ² ≈ df + N Σ (p_i - e_i)² / e_i

The second term grows in proportion to N. This shows why a small departure needs a large sample to detect, and the prediction calculator uses this formula.

Mapping functions (Extension)

Double crossovers hide recombination, so r underestimates the true distance for long intervals. If crossovers occur at random (Poisson) with an average of d per meiosis (in Morgans, 1 M = 100 cM), only an odd number of crossovers gives a recombinant, so

r = ½(1 - e^-2d)     d = -½ ln(1 - 2r)    (Haldane)

The Kosambi function allows for interference (a crossover reduces the chance of a nearby one): d = ¼ ln[(1 + 2r)/(1 - 2r)]. Both give d ≈ r for small r.

LOD score (Extension)

The likelihood of seeing R recombinants in N offspring if the true frequency is r is proportional to r^R(1 - r)^N-R. The LOD score compares the best-fitting r with 0.5:

LOD = log₁₀ [ r^R (1 - r)^N-R / 0.5^N ]

The maximum-likelihood estimate of r is R/N, which is why the simple proportion is also the best estimate. A LOD of 3 means the data are 1000 times more likely under linkage than under independent assortment.

All numbers below have been checked by calculation. Use them as models for your own working.

Observed and expected counts

The bar chart in the simulator shows the observed count in each phenotype class with a dashed outline for the expected count. After a run, the graph tab adds a third bar for the expected count if the genes were independent. If the observed bars follow the second set rather than the third, the data support linkage. Sampling variation makes the observed bars wobble around the expected ones, and the wobble is smaller in proportion for larger N.

Expected class frequencies

Figure 2. Linkage in coupling enlarges the double-dominant and double-recessive classes and shrinks the two single-dominant classes.

F2 classes against recombination frequency

The fourth graph tab draws the theory curves for A_B_, A_bb (equal to aaB_) and aabb from r = 0 to r = 50%. At r = 0 in coupling, A_B_ is 75% and aabb is 25%, because the F2 behaves like a monohybrid cross with only two phenotypes. As r rises the curves move smoothly to 56.25%, 18.75% and 6.25%. Your recorded F2 points should scatter around the curves.

Recombination frequency against map distance

Figure 3. For short distances (below about 10 cM) r and d are nearly equal. For longer distances r flattens towards 0.5 and the map distance is larger than r suggests.

Chi-squared against recombination frequency

The third graph tab plots your chi-squared values against the r you set, with the 5% critical value as a dashed line. Points at small r lie far above the line, and points at r = 50% lie below it about 19 times in 20. A point above the line at r = 50% is a false alarm, which is expected in 1 run in 20.

Recombination found against recombination set

This graph tests the method itself. Test cross recombination frequencies should lie on the line y = x. The calibration tool fits a line of best fit through the origin and reports its gradient with an uncertainty. A gradient consistent with 1 shows that the proportion of recombinant offspring is an unbiased measure of r.

The ratios in this laboratory are idealised. Real crosses depart from them for many reasons, and it is part of scientific skill to recognise them.

• Double crossovers. Two crossovers between the genes restore the parental combination, so r underestimates distance for genes far apart and never exceeds 0.5. Mapping functions correct for this.
• Interference. One crossover usually lowers the chance of another nearby. The simulator assumes that no such effect operates inside one interval.
• Variable recombination rates. Rates differ along a chromosome (low near centromeres, higher in some regions) and between sexes and species. Map distance in cM is not a fixed physical length.
• Unequal survival. Some genotypes are less viable. A missing class changes ratios, for example a 2:1 ratio when one homozygote dies. The simulator assumes all offspring survive and are scored.
• Incomplete dominance and codominance. If heterozygotes look different, the F2 has nine phenotypes (1:2:1:2:4:2:1:2:1) rather than four.
• Environment and penetrance. A genotype may not always show its phenotype. Counting errors and mis-scored individuals also blur class boundaries.
• Sampling variation. Even a perfect model gives random departures from the expected ratio. Chi-squared is a way to judge whether a departure is bigger than chance alone would give. A non-significant result is not proof that the model is right.
• Many alleles and many genes. Real traits often involve many genes and the environment, so Mendelian ratios are the exception.

History gives a useful example. Early in the twentieth century, breeders found crosses that did not give 9:3:3:1 even though no gene interaction was suspected. Work by Thomas Hunt Morgan's group on fruit flies explained it by gene linkage and crossing over, and in 1913 Alfred Sturtevant used recombination frequencies to draw the first genetic map. The approach in this laboratory is a simplified version of that logic.

Start with a question and two hypotheses (Core)

A good genetics investigation begins with a question that data can answer, for example: Do the wing-colour gene and the antenna gene of the virtual moth assort independently? From the question, write a null hypothesis and an alternative hypothesis.

• H₀: the genes assort independently, so the F2 ratio is 9:3:3:1 (or the test cross ratio is 1:1:1:1).
• H₁: the genes are linked, so the parental classes are more common than the recombinant classes.

Decide the significance level (usually 5%) before you look at the data. Choosing it afterwards invites you to see what you hoped to see.

Variables

Choosing the sample size (Core)

The standard error of a recombination frequency is √[r(1 - r)/N]. For r = 0.20 it is 4.0 percentage points when N = 100, 2.0 when N = 400 and 1.0 when N = 1600. Quadrupling the sample halves the uncertainty, so precision becomes expensive. For a chi-squared test, make N large enough that every expected count is at least 5. In an F2 the smallest class is 1/16 of the total, so N should be at least 80, and more is better. Decide N before you run the cross and do not keep adding offspring until the result looks significant.

Procedure in the simulator (Core)

1. Set the cross, arrangement, recombination frequency, gene interaction and N. Look at the Punnett square and the dashed expected outlines, and write down your prediction before running.
2. Press Run cross. The simulator breeds the offspring by random sampling and records the counts in the table.
3. Repeat at least three times for each setting, to see how much results vary by chance alone.
4. Vary one factor at a time: first r (for example 5, 15, 25, 35 and 50%), then N, then coupling against repulsion.
5. Use the analysis tools to calculate chi-squared, estimate r with its standard error, and fit the calibration line.
6. Download the CSV file to keep your data, and write a conclusion that refers to your hypotheses.

Suggested investigations (Core and Extension)

• How strong is the evidence? At r = 30%, find the smallest N for which F2 crosses usually reject 9:3:3:1.
• Does the test cross measure r without bias? Run test crosses at five values of r and fit found against set.
• Which design is more precise? Estimate r at 20% from five test crosses and five F2 crosses with the same N, and compare the spread.
• Do coupling and repulsion differ? Compare the class counts in test crosses with identical r.
• Can epistasis be confused with linkage? Compare a 9:3:4 cross at r = 50% with a linked cross.

The chi-squared test step by step (Core)

1. State H₀ as a ratio, for example 9:3:3:1.
2. Convert the ratio to expected counts: total x proportion for each class.
3. Check that every expected count is at least 5. If not, combine classes or collect more data.
4. Calculate (O - E)²/E for each class and add them up.
5. Find the degrees of freedom: number of classes minus 1.
6. Compare with the critical value at 5% (7.81 for 3 df, 5.99 for 2 df, 3.84 for 1 df), or find p.
7. Write a conclusion in context: reject or do not reject H₀, and what that says about the genes.

Always work with counts, never percentages, because the statistic depends on the number of observations. Use the expected counts from the hypothesis, not the observed counts, in the denominator.

Errors in testing (Core)

A Type I error is rejecting a true H₀. With a 5% level it happens in 1 in 20 tests of unlinked genes. You can see this: run twenty F2 crosses at r = 50% and about one will be "significant". A Type II error is failing to reject a false H₀. It becomes more likely when linkage is loose (r near 0.5) or N is small. The prediction calculator estimates the sample size where the linkage term of chi-squared reaches the critical value.

Estimating r and its uncertainty (Core)

For a test cross, r = R/N with SE = √[r(1 - r)/N]. Report the result as r ± SE in map units, or as an interval r ± 1.96 SE (about 95% confidence). Replicate crosses with the same settings show the spread directly: the standard deviation of replicate estimates should be close to the SE. If it is much larger, something other than chance is operating.

A line of best fit for calibration (Core)

The calibration tool plots r found against r set for your test crosses and fits a straight line through the origin by least squares. The gradient m should be 1. Its standard error comes from the scatter of the points about the line: SE(m) = √[Σ(y - mx)² / ((n - 1) Σx²)]. Judge the result by how many standard errors m is from 1. Within 2 is good agreement. Scatter comes from sampling alone, so it falls when N is increased. The line goes through the origin because zero recombination should be found when none is set.

Testing for linkage with a LOD score (Extension)

The LOD score is more useful than a chi-squared test when you want the strength of evidence for linkage. A LOD of 3 or more is the traditional threshold for declaring linkage in genetic mapping, and a LOD of -2 or less excludes linkage. The threshold is stricter than p < 0.05 because many pairs of genes are tested in a genome and many are unlinked.

Small expected counts and one degree of freedom (Extension)

With 1 degree of freedom (a 9:7 cross or a 1:1 test) some textbooks apply Yates' continuity correction, subtracting 0.5 from each |O - E| before squaring. The correction is conservative. When counts are very small, an exact binomial or Fisher test is preferred.

Presenting results (Core)

Give each calculated value to a sensible precision (usually 3 significant figures), with units or "map units". Put all counts in a table, state expected values, and draw graphs with labelled axes. A conclusion should refer to the hypothesis, quote the statistic and p-value or critical value, and mention the main limitation.

In real laboratories

Linkage was discovered, and is still taught, with organisms that breed quickly and give many offspring. Examples include fruit flies, maize (where each kernel on an ear is an offspring and colour or shape can be scored by eye), fast-cycling Brassica plants with seedling markers, and fungi such as Neurospora where the products of a single meiosis stay together in an ascus. The method is similar in each case:

1. Start with pure-breeding parents that differ in two traits.
2. Cross them to get the F1, all heterozygous. The parents decide whether the F1 is in coupling or repulsion.
3. Either cross the F1 with a double-recessive tester (test cross) or cross F1 siblings (F2).
4. Raise the offspring and score each for both traits, ideally without knowing which cross produced them (blind scoring).
5. Tally the counts by class and analyse them.

Measurement issues

• Scoring accuracy. Traits must be clear. Ambiguous phenotypes (for example faint colour differences) need a written rule agreed before counting.
• Counting. Count by one person, and recount a sample. Mark counted individuals so none is counted twice.
• Sample size. More offspring means smaller uncertainty, but only random errors fall as N rises. A systematic error, such as always mis-scoring a faint phenotype, does not.
• Survival. If some genotypes die before scoring, class sizes change. Record the number of dead or lost individuals.

The virtual equipment

Unlike a real experiment, the virtual moth gives exact counts without scoring errors. The only randomness is the sampling of gametes. The simulator draws each offspring from the gamete frequencies using a pseudo-random number generator.

Safety

The virtual laboratory involves no hazards. If you carry out a real genetics practical, follow your school's or institution's rules and your teacher's instructions. Typical points are listed below, but they do not replace a risk assessment for your own setting.

• Anaesthetising insects must be done with an approved method, and handled humanely. Dispose of organisms as instructed.
• Plant pollen, fungal spores and some growth media can cause allergic reactions. Know about allergies in the group before you start.
• Wash hands after handling organisms or media, and keep food and drink out of the laboratory.
• Organisms must not be released into the environment.

Limits of the simulation

• The traits show complete dominance, and every offspring survives and is scored correctly.
• The recombination frequency is constant. There is no interference, no difference between sexes and no variation along the chromosome.
• The two genes are tested on their own, with no further genes or environment effects. The epistasis options are three textbook patterns, and all assume unlinked genes.
• The recombination frequency is capped at 50%, as in real genetics.
• Sample sizes run from 10 to 2000. Small samples can break the chi-squared test, because the expected counts fall below 5. The tools warn you.
• Offspring are drawn with a pseudo-random generator. Each run uses a new random seed, so two runs with the same settings give different counts.
• The estimation of r from an F2 uses a simple formula that requires you to know whether the F1 is in coupling or repulsion. If the wrong arrangement is used, the estimate is wrong.

1. Using percentages in chi-squared. The statistic needs counts. Percentages treat every sample as if it had 100 offspring.
2. Using the observed value in the denominator. The formula divides by the expected count E, not O.
3. Wrong degrees of freedom. It is classes minus 1, not the total number of offspring or the number of genes. A 9:3:3:1 test has 3.
4. Saying a result "proves" a ratio. A non-significant chi-squared means the data are consistent with the ratio. It never proves it.
5. Confusing "reject" and "accept". Large chi-squared (above the critical value) means reject H₀. Small chi-squared means do not reject.
6. Mixing up coupling and repulsion. Which classes are recombinant depends on the arrangement of alleles in the heterozygous parent. In coupling the recombinants are Ab and aB. In repulsion they are AB and ab.
7. Calling the larger classes "recombinant". Recombinants are the smaller classes whenever r < 50%.
8. Dividing by the wrong total. Recombination frequency is recombinants divided by all offspring, not by the number of parentals.
9. Treating 9:3:4 or 9:7 as evidence of linkage. Epistatic ratios are sixteenths of independently assorting genes. Linkage changes the sizes of the classes without merging them.
10. Expecting r above 50%. A value above 0.5 is impossible. If you calculate it, check which classes you counted.
11. Using the F2 formula for a test cross. The estimate r = 1 - 2√f applies only to an F2 in coupling.
12. Forgetting to write genotypes and gametes in a genetic diagram. Marks are given for parental phenotypes, parental genotypes, gametes, offspring genotypes and phenotypes, and the ratio.
13. Using too few offspring. Small samples give low expected counts, wide uncertainty and unreliable chi-squared values.
14. Picking the result you like. Decide the settings and sample size first, run once, and report every run, including the awkward ones.

Command words

• Explain: give reasons, using the mechanism (independent assortment, crossing over in prophase I).
• Calculate: show the formula, the substitution and the answer with units or a clear label.
• Predict: give an expected ratio or number, with the working that leads to it.
• Suggest: apply your knowledge to an unfamiliar situation, for example a new ratio.
• Evaluate: give strengths and weaknesses of the data and conclusions, including sample size and statistics.

Genetic diagrams

• Define the symbols first: A = silver, a = bronze, B = feathered, b = thread (or whatever the question uses). Use the same letter in capital and lower case for dominant and recessive.
• Write the parental phenotypes, the parental genotypes, the gametes, the offspring genotypes (a Punnett square for 16 combinations), the offspring phenotypes and the ratio.
• For linked genes, write the alleles on each chromosome, as AB / ab, and show the parental and recombinant gametes with their proportions.
• Circle or label gametes, and state the ratio in words, for example "9 silver feathered : 3 silver thread : 3 bronze feathered : 1 bronze thread".

Chi-squared answers

• Start with a hypothesis in words and numbers.
• Draw a small table with O, E, O - E and (O - E)²/E.
• State the chi-squared value, degrees of freedom, critical value (or p) and then the decision.
• Finish with a sentence about the genes. For example: "chi-squared = 32 is greater than 7.81, so reject H₀; the genes are linked".

Recognising the cross from the numbers

Check the arithmetic of ratios: all classes should sum to the total. When a calculation gives an impossible value (r above 0.5, a negative count, a probability above 1) stop and look for the mistake. Marks in this laboratory are for learning, not an official grade.

Dihybrid inheritance, linkage and epistasis appear in most upper-secondary and introductory university biology courses. The table lists topic names only, to help you find the right part of your own course. Specifications change, and exam boards and teachers may include or leave out items such as linkage, epistasis or the chi-squared test. Check the current specification for your own course for the exact content, notation, equations given in the exam and the level of mathematics expected. This page is independent and is not endorsed by any exam board.

Using the levels

• Foundation sections suit lower-secondary and first-course students: gametes, the 16-box square and 9:3:3:1.
• Core sections cover what most upper-secondary courses require: test crosses, linkage, recombination frequency, epistasis ratios and chi-squared.
• Extension sections suit higher-level and university students: mapping functions, LOD scores, interference and the underlying statistics.

Genetic mapping

Recombination frequencies between many pairs of genes can be combined into a genetic map that orders genes along a chromosome. This is how the first maps were drawn from fruit fly crosses, and it remains the basis of linkage mapping in many species. Combined with DNA sequencing, genetic maps help to place a trait on the physical genome.

Plant and animal breeding

Breeders want to combine useful traits, such as disease resistance and yield, in one variety. If the genes are linked in repulsion, the useful alleles sit on different chromosomes, and a rare crossover is needed to bring them together. Knowing the map distance tells a breeder how large a population must be to find the recombinant. Marker-assisted selection uses a DNA marker that is tightly linked to a gene of interest, so that seedlings can be screened without waiting for the trait to appear.

Human genetics

In families, geneticists look for DNA markers that are inherited together with a disease, using LOD scores to measure the evidence for linkage. This approach helped to locate many single-gene conditions to chromosome regions. For common conditions influenced by many genes, other methods that compare large groups of people are used.

Understanding biochemical pathways

Epistatic ratios such as 9:3:4 and 9:7 give clues about how gene products work together in a pathway. Coat colour in mammals and flower colour in plants have been studied this way. A modified ratio is a hint that two genes act in the same pathway, which can then be tested biochemically.

Statistics in science

The chi-squared test is widely used outside genetics to compare counts with a model: surveys, ecology, medicine and quality control. The logic of this laboratory, which is to state a hypothesis, calculate what chance alone would give and then decide whether the data are surprising, applies in every field.

Three-point crosses and gene order (Extension)

With three linked genes, a test cross of a triple heterozygote gives eight classes. The two commonest classes are parental. The two rarest are the double crossovers, because two crossovers in neighbouring intervals are the least likely event. Compare the double-crossover class with the parental class: the gene whose allele has switched is in the middle. Recombination frequencies for each interval then give the map.

Coincidence and interference (Extension)

If crossovers in two adjacent intervals were independent, the expected frequency of doubles would be r₁ x r₂. The coefficient of coincidence is the observed number of doubles divided by this expected number, and interference is 1 minus that value. For intervals of 10 and 20 map units in 1000 offspring, 20 doubles are expected. If 8 are seen, the coincidence is 0.4 and the interference is 0.6. Positive interference is typical: one crossover discourages a neighbour.

Likelihood and the standard error of r (Extension)

The log-likelihood of R recombinants in N offspring is l(r) = R ln r + (N - R) ln(1 - r). Setting dl/dr = 0 gives r = R/N, which is why the simple proportion is the maximum-likelihood estimate. The second derivative gives the Fisher information N / [r(1 - r)], and the standard error is the square root of its inverse: √[r(1 - r)/N]. The same calculation for an F2 works through the aabb class, with the additional factor from the transformation r = 1 - 2√f, and gives the larger standard error √[(1 - f)/N].

Types of gene interaction (Extension)

• Recessive epistasis (9:3:4): the genotype aa masks the effect of gene B.
• Dominant epistasis (12:3:1): the allele A masks the effect of gene B.
• Complementary genes (9:7): both genes need a dominant allele.
• Duplicate genes (15:1): a dominant allele of either gene gives the phenotype, so only aabb differs.
• Duplicate recessive genes and other modifications give further ratios. Each is a regrouping of the nine genotype classes.

Note that a test cross for these patterns gives simple ratios: for example 1:1:2 for recessive epistasis (purple:red:white) and 1:3 for complementary genes.

Penetrance and expressivity (Extension)

Penetrance is the proportion of individuals with a genotype who show the phenotype. Expressivity is how strongly they show it. Incomplete penetrance blurs the classes of a cross and lowers the apparent size of some classes.

Power and multiple testing (Extension)

The power of a chi-squared test is its chance of rejecting a false H₀. It depends on how far the truth is from H₀ (measured by the sum of (p - e)²/e), on N, and on the significance level. When many pairs of genes are tested together, as in a genome scan, some will pass a 5% test by chance, so stricter thresholds are needed. That is why LOD scores of 3 or more are used for mapping.

Genetic and physical maps (Extension)

The distance in cM depends on crossover rates, which vary along the chromosome and between sexes. A physical map measures distance in base pairs. The two maps usually have the same gene order but different spacing. Regions where crossing over is rare appear short on the genetic map and long on the physical map.

• A dihybrid cross follows two genes at once. An AaBb parent that has independent genes makes four gamete types in equal numbers.
• The 16-box Punnett square gives the F2 phenotype ratio 9:3:3:1 for independent genes with complete dominance.
• A test cross with aabb reveals the gametes of the tested parent. For independent genes the ratio is 1:1:1:1.
• Genes on the same chromosome are linked. Crossing over in prophase I produces recombinant gametes with frequency r.
• Recombination frequency = recombinants / total offspring. It ranges from 0 to 0.5. One map unit (1 cM) is 1% recombination.
• Parental classes are the large ones. Which genotypes are parental depends on coupling (AB / ab) or repulsion (Ab / aB).
• F2 class frequencies for linked genes: aabb = q, A_bb = aaB_ = 0.25 - q, A_B_ = 0.5 + q, with q = (frequency of ab)².
• Epistasis merges classes: 9:3:4, 9:7 and 12:3:1 are still sixteenths and show independent assortment.
• Chi-squared = Σ(O - E)²/E, with degrees of freedom = classes - 1. Use counts, not percentages, and expected counts of at least 5.
• A significant result rejects the ratio. A non-significant result is consistent with the ratio but does not prove it.
• SE of a recombination frequency = √[r(1 - r)/N]. Quadrupling the sample halves it.
• Extension: map distances from Haldane or Kosambi functions exceed r for long intervals, and a LOD of 3 or more is the usual threshold for linkage.

Allele

One version of a gene.

Autosome

A chromosome that is not a sex chromosome.

Chi-squared test

A statistical test that compares observed counts with the counts expected from a hypothesis.

Chiasma

The point where non-sister chromatids cross over during prophase I of meiosis.

Coefficient of coincidence

Observed double crossovers divided by the number expected if the intervals were independent.

Complementary genes

Two genes whose dominant alleles are both needed for one phenotype, giving 9:7.

Coupling

Dominant alleles of two genes on the same chromosome, as in AB / ab.

Crossing over

The exchange of segments between non-sister chromatids of homologous chromosomes.

Degrees of freedom

The number of independent comparisons in a test. For a goodness of fit test it is classes minus 1.

Dihybrid cross

A cross that follows the inheritance of two genes.

Epistasis

The masking of the effect of one gene by a different gene.

F1 and F2

The first and second filial generations. F2 is produced by crossing F1 individuals with each other.

Gamete

A reproductive cell (egg or sperm) carrying one allele of each gene.

Genotype

The alleles carried by an individual.

Heterozygous

Having two different alleles of a gene.

Independent assortment

The random distribution of alleles of different genes into gametes, when the genes are on different chromosomes or far apart.

Interference

Reduction in the chance of a second crossover near a first one.

Linkage

The tendency of genes close together on a chromosome to be inherited together.

LOD score

The base-10 logarithm of the odds in favour of linkage against independent assortment.

Map unit (centimorgan)

A genetic distance equal to 1% recombination.

Null hypothesis

The hypothesis of no effect or no difference, which a test tries to reject.

Parental type

An offspring or gamete with the same allele combination as one of the parental chromosomes.

p-value

The probability, if the null hypothesis were true, of a result at least as extreme as the one observed.

Phenotype

The observable trait of an individual.

Recombinant

An offspring or gamete with a new combination of alleles produced by crossing over.

Recombination frequency

The fraction of gametes or test cross offspring that are recombinant.

Repulsion

A dominant allele of one gene and a recessive allele of the other on each chromosome, as in Ab / aB.

Standard error

An estimate of how much a calculated value would vary from sample to sample.

Test cross

A cross of an individual with a homozygous recessive individual to reveal its genotype.

The following titles are named for further reading. Find current editions through your library or the publisher's home page. No page numbers or quotations are given here, so check the original source for details.

• Gregor Mendel, Versuche über Pflanzen-Hybriden (Experiments on plant hybridisation), 1866, in the proceedings of the natural history society of Brünn.
• Karl Pearson, paper on the chi-squared criterion for deviations from expectation, Philosophical Magazine, 1900.
• J. B. S. Haldane, "The combination of linkage values, and the calculation of distances between the loci of linked factors", Journal of Genetics, 1919.
• D. D. Kosambi, "The estimation of map distances from recombination values", Annals of Eugenics, 1944.
• Newton E. Morton, "Sequential tests for the detection of linkage", American Journal of Human Genetics, 1955.
• Textbook, Introduction to Genetic Analysis, Griffiths and colleagues, Macmillan Learning (home page: https://www.macmillanlearning.com).
• Textbook, Genetics: From Genes to Genomes, Hartwell and colleagues, McGraw Hill (home page: https://www.mheducation.com).
• Textbook, Campbell Biology, Pearson (home page: https://www.pearson.com).

Your own exam board's specification and past papers are the best guide to what is required in your course.