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Sex-Linked Inheritance Virtual Lab

Cross fruit flies with an X-linked trait, compare reciprocal crosses, test sex-specific ratios with chi-squared and infer a hidden mode of inheritance from your own data.

  • Genetics
  • Secondary to first-year university
  • About 60 minutes
  • Simulation, guided tasks, exam and notes

1Learn: what you will be able to do

Some genes sit on the X chromosome, and that changes how they are passed to sons and daughters. By the end of this laboratory you will predict the offspring of any cross, tell an X-linked trait from an autosomal one, test your data with the chi-squared test and answer exam-style questions on the topic.

Learning outcomes

  • Explain how X and Y chromosomes decide sex and why a son always receives his X from his mother.
  • Draw Punnett squares for X-linked crosses that show the sex and phenotype of every offspring.
  • Predict and explain the results of reciprocal crosses for X-linked and autosomal traits.
  • Describe carrier females and explain why X-linked recessive traits are more common in males.
  • Carry out chi-squared tests on sex-specific ratios, and compare two crosses with a test of homogeneity.
  • Infer from offspring data whether a trait is autosomal or X-linked, and dominant or recessive.
  • Describe X-inactivation in outline, and give introductory human examples such as colour vision deficiency and haemophilia.

Equipment in this virtual laboratory

  • A fruit fly cross simulator with three traits: white eye (X-linked recessive), sepia eye (autosomal recessive) and a mystery trait whose mode of inheritance is hidden.
  • Menus for the genotype of the mother and of the father, and broods of 10 to 1,000 offspring shown sorted into daughters and sons.
  • A one-click reciprocal cross, a data table with CSV download and three linked graphs.
  • Analysis tools: a Punnett-square calculator with full working, a chi-squared test with a distribution graph, a reciprocal-cross and mode-of-inheritance tester, and a best-fit line with uncertainty.
Limits of the simulation. The model follows one gene at a time. Every fertilisation is random, all offspring survive equally well, phenotypes are scored without error, and there is no linkage, mutation or non-disjunction. Real crosses are messier. The human examples are general and introductory and are not advice for any family.

Where this topic appears in courses

Sex-linked inheritance is part of the genetics content of secondary and first-year university biology courses in the UK and Cambridge pathways, WAEC/WASSCE, the IB Diploma and North American high-school and introductory college biology. The depth varies: some courses stay with simple Punnett squares and pedigrees, while others require chi-squared tests, X-inactivation and population frequencies. The three level tags used in the notes (Foundation, Core, Extension) let you match the material to your course. This page supports, and does not replace, the specification you are following. Check it for the exact requirements and the formulae you are given in the exam.

2Prepare: ideas and predictions

Scientists predict before they measure. Commit to an answer for each question, then open it to check your reasoning.

P(class) = P(egg) × P(sperm)    E = N × P    χ² = Σ (O − E)² / E

A son gets his X from his mother and his Y from his father. A daughter gets one X from each parent. A male has a single X, so a recessive X-linked allele shows in him. Everything in this laboratory follows from these statements.

Prediction 1. A white-eyed female fruit fly is crossed with a red-eyed male. What do the daughters and sons look like?
All daughters are red-eyed carriers (XWXw) and all sons are white-eyed (XwY). The sons copy their mother because their only X came from her. Test this in the simulator with the white eye trait and a white-eyed mother.
Prediction 2. What happens in the reciprocal cross (red-eyed female, white-eyed male)?
All offspring are red-eyed: daughters are carriers (XWXw) and sons are XWY. The two reciprocal crosses give completely different results, which is how X-linkage is recognised. Observe task 1 checks this with your own data.
Prediction 3. A carrier female is crossed with a red-eyed male. Will any daughters be white-eyed? What fraction of sons?
No daughter will be white-eyed, because each receives her father's XW. Half of the sons will be white-eyed, because each son gets one of his mother's two X chromosomes with equal chance.
Prediction 4. Two sepia-eye heterozygotes (autosomal) are crossed. Do sons and daughters differ?
No. An autosomal trait is inherited identically in both sexes, so about 25% of sons and 25% of daughters will be sepia-eyed. If the two sexes gave clearly different percentages, you would suspect sex linkage.
Prediction 5. If a hypothesis is exactly right, how often will a chi-squared test at the 5% level still reject it?
About 1 time in 20, because the 5% level is chosen so that a true hypothesis is rejected 5% of the time by chance. Try it: repeat the same cross many times in the simulator and look at how often p falls below 0.05.

3Experiment: the fly room

Choose a trait and the parents, then press Make the cross. The offspring appear one by one, sorted into daughters and sons, and the finished brood is added to the data table.

4Observe: guided investigations

Five short investigations, each checked against the crosses in your own table. Complete them in order, using the simulator above.

5Analyse: predict, test and infer

Use the calculator to predict a brood with full working. Then test your data with chi-squared, compare reciprocal crosses, infer the mode of inheritance and check your counts against theory with a line of best fit.

6Examine: practical examination

Ten multiple-choice questions at three levels and five calculations with numbers generated for you each time. Marking is instant, and you are told which notes to review.

7Notes: the complete theory

Eighteen sections from principles to exam technique, with worked examples and diagrams. Use the search box, or open everything and print.

A. Overview and learning outcomes

Sex-linked inheritance is the pattern shown by genes that sit on the sex chromosomes. In humans and in fruit flies (Drosophila melanogaster) a female normally has two X chromosomes and a male has one X and one Y. Because the sexes carry different numbers of X chromosomes, a gene on the X is passed on differently to sons and to daughters. The same trait can therefore appear in different proportions in the two sexes, and a cross and its reciprocal can give different offspring.

This page uses the white-eye mutation of the fruit fly as its main model. It is a recessive mutation on the X chromosome, and it was one of the first traits shown to be linked to a particular chromosome. You can set up any cross, watch the offspring appear sorted by sex, compare reciprocal crosses and test your counts with the chi-squared test. A hidden mystery trait lets you practise the real task of a geneticist: deciding from data whether a trait is autosomal or X-linked, and dominant or recessive.

The notes work in three layers. Foundation material needs only the ideas of chromosomes, genes and simple fractions. Core material matches typical upper-secondary and pre-university biology courses: genotypes, Punnett squares, pedigrees, probability and the chi-squared test. Extension material goes towards first-year university genetics: allele frequencies for X-linked genes, X-inactivation, non-disjunction and exact statistical tests.

The simulation starts with a white-eyed mother (XwXw) and a red-eyed father (XWY), 200 offspring and the white-eye trait. Before you press the button, predict what you expect to see in daughters and in sons, then compare.

Learning outcomes

Foundation

  • State which sex chromosomes a typical female and a typical male have, and which parent decides the sex of a child.
  • Explain why a son always receives his X chromosome from his mother and his Y chromosome from his father.
  • Describe a carrier and explain why carriers of an X-linked recessive allele are usually female.
  • Give two human examples of X-linked recessive traits and state that they are more often seen in males.

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

  • Write genotypes for X-linked traits using X and Y symbols, and complete Punnett squares that show the sex of each offspring.
  • Predict the phenotype ratios in sons and in daughters for any cross, including reciprocal crosses.
  • Distinguish autosomal from X-linked inheritance, and dominant from recessive, using offspring data and pedigrees.
  • Carry out a chi-squared goodness-of-fit test and a test comparing two crosses, stating the hypothesis, degrees of freedom, p-value and conclusion.
  • Explain X-inactivation in outline and why a female carrier can show a mixture of two cell types.

Extension (university)

  • Derive expected frequencies of affected males, affected females and carriers from an allele frequency q.
  • Explain non-disjunction as evidence that genes lie on chromosomes, and describe pseudoautosomal, Y-linked and ZW inheritance.
  • Choose between chi-squared, Yates-corrected and exact tests, and quote an uncertainty for an observed proportion.
B. Core principles

Chromosomes and sex Foundation

Most human body cells have 46 chromosomes in 23 pairs. Twenty-two pairs are autosomes, which look the same in both sexes. The last pair is the sex chromosomes. A typical female has two X chromosomes (XX). A typical male has one X and one Y (XY). When eggs form, every egg carries one X. When sperm form, half carry an X and half carry a Y. The sperm that fertilises the egg therefore decides the sex of the child: an X sperm makes a daughter and a Y sperm makes a son. About half of children are daughters and half are sons.

Fruit flies work in the same way as far as these notes need: females are XX and males are XY. (Deep down, the sex-determining mechanism of flies is different from that of humans, as the extension notes explain, but the crosses behave identically.)

Hemizygous males Foundation

The X chromosome is large and carries hundreds of genes. The Y chromosome is small and carries few genes, and most X genes have no matching copy on the Y. A female has two copies of each X gene, just like an autosomal gene. A male has only one copy. A male is said to be hemizygous for X-linked genes. Whatever allele is on his single X shows up in his phenotype, whether the allele is dominant or recessive.

Five rules of X-linked inheritance Core

  1. A son receives his X from his mother and his Y from his father. He never receives an X from his father.
  2. A daughter receives one X from each parent. She always receives her father's X.
  3. A father passes his X-linked allele to all of his daughters and to none of his sons.
  4. A mother passes each of her two X chromosomes to half of her sons and half of her daughters, on average.
  5. For a recessive trait, a mother who is affected (XwXw) has sons who are all affected.
Carrier mother crossed with a red-eyed father A two by two Punnett square. Eggs from the mother are X W and X w. Sperm from the father are X W and Y. The four squares are a red-eyed daughter, a red-eyed son, a red-eyed carrier daughter and a white-eyed son. Mother XWXw x father XWY sperm XW sperm Y egg XW egg Xw XWXW red-eyed daughter XWY red-eyed son XWXw carrier daughter XwY white-eyed son Each square has probability 1/4. Shaded square: the only affected child.

Autosomal or X-linked? Core

An autosomal gene lies on a chromosome that both sexes carry in pairs. Both sexes inherit it in the same way, so the two sexes show the same ratios and a cross gives the same result as its reciprocal. An X-linked gene breaks these rules. The sexes can differ, and swapping the phenotypes of mother and father changes the offspring. A reciprocal cross is a second cross in which the phenotypes of the two sexes are exchanged, for example a white-eyed female with a red-eyed male, and then a red-eyed female with a white-eyed male. Doing both crosses is the standard first test for sex linkage.

Dominant and recessive X-linked traits Core

For an X-linked recessive trait, the first generation from a mutant mother and a normal father shows the trait in all sons and in no daughters. Daughters are carriers. The trait seems to cross over from mother to son, which is why this is called criss-cross inheritance. For an X-linked dominant trait the pattern is different: a mutant father and a normal mother give affected daughters and unaffected sons, while a mutant mother (homozygous) gives only affected children.

Carriers Foundation

A carrier is a person or animal who has one copy of a recessive allele and one normal copy, and who usually shows no sign of the condition. For X-linked recessive traits only females can be carriers in this sense. A male either has the allele and shows the trait, or does not have it.

Other ways genes can sit on sex chromosomes Extension

A few genes lie in pseudoautosomal regions at the tips of X and Y, where the two chromosomes pair and exchange material in male meiosis. These genes are inherited like autosomal genes. Genes found only on the Y are Y-linked and pass from father to son only. In birds, butterflies and some other animals the sex chromosomes are Z and W, and the female is ZW while the male is ZZ. All the rules then swap sexes: a Z-linked recessive trait shows more often in females, and the criss-cross pattern runs from father to daughter.

C. Equations and derivations

Gametes and probabilities Core

Each parent makes two kinds of gametes in equal numbers. A carrier mother XWXw makes eggs XW and Xw, each with probability 1/2. A father XWY makes sperm XW and Y, each with probability 1/2. Egg and sperm combine at random, so the four squares of the Punnett square each have probability 1/2 × 1/2 = 1/4.

P(class) = P(egg type) × P(sperm type) = 1/2 × 1/2 = 1/4

For an X-linked gene the sex of the offspring is read directly from the sperm: an X sperm gives a daughter and a Y sperm gives a son. For an autosomal gene, sex and genotype are decided independently, so each of the four squares of the Punnett square splits equally into daughters and sons. Each square-and-sex combination then has probability 1/4 × 1/2 = 1/8.

Expected counts Core

If the brood has N offspring and a class has probability p, the expected number in the class is

E = N × p

The observed number O varies from brood to brood. For a class with probability p the standard deviation of the count is √(Np(1 − p)), so about two-thirds of broods fall within one standard deviation of E.

Chi-squared Core

To compare observed and expected counts across all classes, add up the squared differences, each scaled by the expected count:

χ² = Σ (O − E)² / E

Scaling by E is sensible because a difference of 8 matters more when E is 20 than when E is 500. The number of degrees of freedom is the number of classes used minus 1 (when the expected counts are fixed by the hypothesis alone). If the hypothesis is true, χ² follows the chi-squared distribution, and the p-value is the probability of a χ² at least as large as the one you found.

Comparing two crosses Core

To ask whether two crosses (for example reciprocal crosses) give the same distribution, arrange the counts in a table with one row per cross. Under the hypothesis that the rows come from the same distribution, the expected count in each cell is

E = (row total × column total) / grand total

and χ² is calculated in the same way. Here df = (rows − 1) × (columns − 1). Columns in which both crosses have zero offspring are left out before counting the columns.

Why the mother's X decides her sons Core

Derivation. A son is formed when a Y sperm meets an egg. The Y sperm carries no allele for the gene, so the son's only allele is the one on the egg's X. A homozygous mutant mother has only Xw eggs, so every son is XwY. A carrier mother has XW and Xw eggs in equal numbers, so half of her sons are XwY.

Population frequencies for an X-linked recessive allele Extension

Let q be the frequency of the recessive allele and p = 1 − q the frequency of the normal allele. With random mating and the same allele frequency in both sexes at the start, the genotype frequencies are

Males: XWY = p, XwY = q    Females: XWXW = p², XWXw = 2pq, XwXw = q²

A male has one X, so the proportion of affected males equals q. A female needs two copies, so the proportion of affected females is q², far smaller. The ratio of affected males to affected females is q / q² = 1/q. For q = 0.08 this ratio is 12.5.

D. Worked examples

Every number in these examples has been checked by calculation. Work through each one on paper before reading the steps.

Worked example 1 (Core): carrier mother, red-eyed father

A carrier female fruit fly (XWXw) is crossed with a red-eyed male (XWY). Predict the offspring in a brood of 400.

  1. Eggs: XW and Xw. Sperm: XW and Y. Each has probability 1/2.
  2. Four equally likely squares: XWXW, XWXw, XWY, XwY, each with probability 1/4.
  3. Expected counts: 400 × 1/4 = 100 for each.

100 red-eyed non-carrier daughters, 100 red-eyed carrier daughters, 100 red-eyed sons and 100 white-eyed sons. Phenotypes: 200 red-eyed daughters, 0 white-eyed daughters, 100 red-eyed sons, 100 white-eyed sons. Half of the sons are white-eyed, and no daughter is.

Worked example 2 (Core): reciprocal crosses and the F2

Cross A: white-eyed female (XwXw) with red-eyed male (XWY). Cross B: red-eyed female (XWXW) with white-eyed male (XwY). Predict the F1 of each. Then predict the F2 from crossing F1 flies of cross A together, if the F2 has 800 flies.

  1. Cross A: every egg is Xw. Daughters receive the father's XW, so they are XWXw, red-eyed. Sons receive the father's Y, so they are XwY, white-eyed.
  2. Cross B: every egg is XW. Daughters are XWXw, red-eyed. Sons are XWY, red-eyed. All F1 are red-eyed.
  3. F2 of cross A: XWXw female × XwY male. Eggs: XW, Xw. Sperm: Xw, Y. The four squares are XWXw (red female), XwXw (white female), XWY (red male), XwY (white male).
  4. Each square has probability 1/4, so 800 × 1/4 = 200 flies in each class.

Cross A gives red daughters and white sons; cross B gives all red-eyed flies. Reciprocal crosses differ, so the gene is X-linked. The F2 is 1 : 1 : 1 : 1 for red females : white females : red males : white males, that is 200 of each.

Worked example 3 (Core): chi-squared for sons

From a carrier mother, 100 sons are counted: 58 white-eyed and 42 red-eyed. Test the hypothesis that sons occur in the ratio 1 : 1.

  1. H0: white-eyed : red-eyed sons = 1 : 1. Expected: 50 and 50.
  2. χ² = (58 − 50)²/50 + (42 − 50)²/50 = 64/50 + 64/50 = 1.28 + 1.28 = 2.56.
  3. Classes used = 2, so df = 1. The 5% critical value is 3.84. The p-value is 0.110.

χ² = 2.56 with 1 degree of freedom, p = 0.110. Because 2.56 < 3.84, do not reject H0: the data are consistent with a 1 : 1 ratio. This does not prove the ratio; it says that a split as uneven as 58 : 42 happens by chance about 11 times in 100 even when the true ratio is 1 : 1.

Worked example 4 (Core): all four classes

A brood of 200 from a carrier mother and red-eyed father contains 96 red-eyed daughters, 0 white-eyed daughters, 52 red-eyed sons and 52 white-eyed sons. Test the Mendelian prediction.

  1. Prediction (example 1): red daughters 1/2, white daughters 0, red sons 1/4, white sons 1/4. Expected: 100, 0, 50, 50.
  2. The class with expected 0 contains 0 observed, so it is left out. Three classes are used, so df = 2.
  3. χ² = (96 − 100)²/100 + (52 − 50)²/50 + (52 − 50)²/50 = 0.16 + 0.08 + 0.08 = 0.32.
  4. The 5% critical value for df = 2 is 5.99. The p-value is exp(−0.32/2) = 0.852.

χ² = 0.32, df = 2, p = 0.85. The data fit the prediction very well. Any white-eyed daughter would have been an immediate sign of trouble, because the prediction forbids them.

Worked example 5 (Core): are reciprocal crosses different?

Cross A (white-eyed female × red-eyed male) gives 104 red daughters, 0 white daughters, 0 red sons and 96 white sons. Cross B (red-eyed female × white-eyed male) gives 98 red daughters, 0 white daughters, 102 red sons and 0 white sons. Test whether the crosses give the same distribution.

  1. Arrange a table with columns red daughters, white daughters, red sons, white sons. The white-daughter column is zero for both crosses, so it is removed. Three columns remain.
  2. Column totals are 202, 102 and 96 and the grand total is 400. Each row total is 200, so every expected count is the column total divided by 2: 101, 51, 48.
  3. χ² = (104 − 101)²/101 + (0 − 51)²/51 + (96 − 48)²/48 + (98 − 101)²/101 + (102 − 51)²/51 + (0 − 48)²/48 = 198.18.
  4. df = (2 − 1)(3 − 1) = 2. The 1% critical value is 9.21, so p is far below 0.0001.

χ² = 198.2, df = 2, p < 0.0001. The reciprocal crosses differ hugely. The result depends on which parent carries the mutation, which is strong evidence that the gene is on the X chromosome.

Worked example 6 (Core): a family with haemophilia

Haemophilia A is an X-linked recessive condition in which blood clots slowly. A woman who is a carrier and an unaffected man plan to have children. What is the chance that a child is an affected son? If they have three sons, what is the chance that at least one is affected?

  1. Mother XHXh, father XHY. The four classes each have probability 1/4: unaffected daughter, carrier daughter, unaffected son, affected son.
  2. So the chance that a child is an affected son is 1/4.
  3. Given that a child is a son, he is affected with probability 1/2. For three sons, the probability that none is affected is (1/2)³ = 1/8 = 0.125.
  4. The probability that at least one is affected is 1 − 0.125 = 0.875.

1/4 for a child of unknown sex; 0.875 (87.5%) that at least one of three sons is affected. Each pregnancy is independent, so earlier children do not change the chances for later ones. This is a probability, not a prediction for any single family.

Worked example 7 (Extension): colour vision deficiency in a population

Red-green colour vision deficiency is X-linked recessive. Suppose 8% of males in a population are affected. Estimate the proportion of affected females and of female carriers, assuming random mating.

  1. A male has one X, so the allele frequency is q = 0.08 and p = 0.92.
  2. Affected females: q² = 0.08² = 0.0064, or 0.64%.
  3. Carrier females: 2pq = 2 × 0.92 × 0.08 = 0.1472, or 14.7%.
  4. Ratio of affected males to affected females: 0.08 / 0.0064 = 12.5.

About 0.64% of females are affected and about 14.7% are carriers. Real populations differ: allele frequencies vary between groups and the real figure for affected women is often quoted as a little under 1%. The calculation shows why a trait with 1 affected woman in about 160 can still affect 1 man in 12.5.

Worked example 8 (Extension): a mosaic carrier

In a female embryo, suppose X-inactivation happens in 8 progenitor cells that will form a tissue, each switching off the XW or the Xw with equal probability. How variable is the fraction of cells in which XW stays active?

  1. The number of cells with active XW is binomial with n = 8 and p = 0.5. The mean fraction is 0.5.
  2. The standard deviation of the fraction is √(0.5 × 0.5 / 8) = 0.177.
  3. A range of two standard deviations either side is 0.5 ± 0.354, that is 0.15 to 0.85.
  4. Chance that 2 or fewer of the 8 cells keep XW active: (1 + 8 + 28) / 256 = 37/256 = 0.145.

The fraction of cells with active XW is usually between 0.15 and 0.85, and in about 14% of such tissues only 2 or fewer of 8 founder cells keep XW active. This is a simplified model, but it shows why carriers vary and why a few carriers show mild signs of an X-linked condition.

E. Graphs and data displays

Mutant percentage by sex Foundation

This graph plots, for each recorded cross, the percentage of mutant daughters and the percentage of mutant sons. Read it like a fingerprint. If the two sexes lie at very different heights in the same cross, the trait is probably X-linked. If they lie at about the same height, the trait is probably autosomal. For the white-eye trait with a carrier mother you should see sons near 50% and daughters at 0%.

Running proportion Core

The running-proportion graph follows one brood as the offspring are counted. It plots the share of sons that are mutant after the first son, the second, and so on. The line jumps about at the start and settles near the expected percentage. This is the law of large numbers: the more offspring you count, the closer the observed proportion usually is to the true probability. It shows why a brood of 10 cannot confirm a ratio but a brood of 400 usually can.

Observed against expected Core

Each point shows the expected count of one class (horizontal axis) against the observed count (vertical axis). If theory is right, the points scatter around a straight line through the origin with gradient 1. The analysis tool fits a least-squares line through the origin and reports the gradient with its standard error. A gradient within about two standard errors of 1 is consistent with theory. Because the points with small counts are less reliable, do not read much into a single point.

Chi-squared distribution, 2 degrees of freedom A curve falling steeply from a high value at zero towards zero. The area to the right of chi-squared equal to 5.99 is shaded and marked as the 5 percent tail. critical value 5.99 shaded: 5% of the area 0 4 8 12 chi-squared probability density

The chi-squared curve Core

The Analyse tool draws the chi-squared distribution for the degrees of freedom of your test. The yellow dashed line is the 5% critical value, and the red line is your statistic. If the red line is to the right of the yellow one, the area beyond it is smaller than 5%, so p < 0.05 and the hypothesis is rejected. If it is to the left, the hypothesis is not rejected. Notice how the curve changes shape with the degrees of freedom: with 1 degree of freedom it is very steep at zero, while with 4 or more it develops a hump.

F. Real-world limits

The Mendelian ratios in these notes are ideals. Real crosses, real pedigrees and real families bring complications. This section lists the main ones so that you can judge when a ratio that looks wrong is actually telling you something.

Differences in survival Core

A ratio is a statement about what fertilisation produces. What you count is what survives. Some mutant alleles reduce the chance that an embryo reaches adulthood, or reduce the success of mutant adults in mating and in feeding. If white-eyed flies survive less well, white-eyed flies will be under-represented among the adults you count, and chi-squared may reject a correct 1 : 1 prediction. The simulation assumes equal survival of all classes.

Penetrance and expressivity Core

Not every individual with a disease genotype shows the condition (reduced penetrance), and those who do may show it to different degrees (variable expressivity). Genes elsewhere and the environment both contribute. The simulation assumes complete penetrance and a clear, two-way classification of phenotype.

Linkage and recombination Extension

Two genes on the same X chromosome do not assort independently. The closer they are, the more often their alleles stay together. This is the basis of mapping genes on the X. The simulation follows one gene at a time, so linkage does not arise.

Non-disjunction and unusual karyotypes Extension

Sometimes chromosomes fail to separate in meiosis, giving gametes with two sex chromosomes or none. This explains rare exceptional offspring, for example a white-eyed daughter from a white-eyed mother and a red-eyed father. It also explains human conditions such as 47,XXY and 45,X, which affect the number of X chromosomes. These events are rare and are not included in the simulation.

X-inactivation and carriers Extension

A female carrier is a mosaic of two cell types, because one X is silenced in each cell. If the silencing happens to favour one X, a carrier can show mild signs of an X-linked condition. Textbook labels such as "unaffected carrier" simplify this variability.

New mutations and family history Core

A family can have a first case of an X-linked condition because of a fresh mutation in an egg or sperm. A clear pedigree is therefore not needed for sex linkage, and the absence of a family history does not rule it out. In pedigrees, small families and few children also make patterns hard to see.

Small samples and chance Core

With few offspring, chance variation can mimic or hide a pattern. In one human family, 4 affected sons out of 4 is entirely compatible with a 1/2 chance each time (probability 1/16). Only large samples, such as hundreds of fruit flies, allow a clear statistical conclusion, and even then a correct hypothesis will be rejected about 1 time in 20 at the 5% level.

Genetic information about people Foundation

The human examples here are for learning about patterns of inheritance. Information about a person's genes is private and can affect the whole family. Real families should receive support and clear explanation from qualified genetic counselling services, not from a simulation or a classroom exercise.

G. Planning and carrying out the investigation

A research question and hypotheses Core

A good question names the variables and the relationship. For example: Does the percentage of white-eyed offspring depend on both the sex of the offspring and on which parent carries the white-eye allele? Turn the question into two statements that can be tested. The null hypothesis H0 says that there is no such dependence: both sexes show the same percentage and the reciprocal crosses give the same result. The alternative hypothesis H1 says that the sexes and the reciprocal crosses differ. Then write a prediction from the X-linked model: cross A should give white-eyed sons only and cross B should give no white-eyed offspring at all.

Variables Foundation

KindIn this laboratory
Independent variableThe cross: genotype of the mother and of the father, and which parent carries the mutation.
Dependent variableCounts of daughters and sons, wild type and mutant.
Controlled variablesTrait, number of offspring, the gene and organism used, and the method of scoring.
Random variationWhich egg and which sperm meet in each fertilisation.

How many offspring? Core

The standard error of a proportion p measured from n offspring is √(p(1 − p)/n). For p = 0.5 and n = 100 this is 0.05, so the proportion is uncertain by about 5 percentage points. To halve the uncertainty you need four times as many offspring. Make a broad plan: at least 100 offspring per cross to see the pattern, and 200 to 400 when you plan to carry out a chi-squared test with expected counts of at least 5 in every class.

Order of crosses Core

  1. Start with true-breeding parents: a mutant line and a wild-type line.
  2. Make cross A (mutant mother × wild-type father) and its reciprocal cross B (wild-type mother × mutant father). Record the F1.
  3. Compare the reciprocal crosses. If they differ, suspect the X chromosome. If they agree, suspect an autosome.
  4. Make an F2 by crossing F1 flies from the same cross. The F2 usually separates the models that the F1 left open.
  5. Repeat each cross to check that the result is reproducible.

Recording Foundation

Plan a table before you begin, with one row per cross and columns for the parents, total offspring and the four classes. Record raw counts, not percentages. Add a notes column for anything unusual. The Experiment section does this for you and lets you download the table as a CSV file.

Planning a real fly cross Extension

A real cross takes more care. Females must be virgin, because a female fly stores sperm from earlier matings. Virgins are collected within a few hours of emerging from the pupa, before they are able to mate. Parents are placed in a fresh vial, removed after several days so that they are not confused with their offspring, and the offspring are anaesthetised and counted over a set period. At 25 °C one generation takes about ten days, and longer at lower temperatures.

H. Data analysis and uncertainty

Proportions and their uncertainty Core

Suppose k of n sons are mutant. The observed proportion is p̂ = k/n and its standard error is

SE = √( p̂ (1 − p̂) / n )

For 58 white-eyed sons out of 100, p̂ = 0.58 and SE = 0.049. A rough 95% interval is p̂ ± 2 SE, which gives 0.48 to 0.68. The Wilson interval, used by the Analyse tool, behaves better for small samples and for proportions near 0 or 1; for 58 out of 100 it gives 0.482 to 0.672. The hypothesised value 0.5 lies inside, so the data are consistent with 1/2.

The chi-squared goodness-of-fit test Core

  1. State H0, for example "the sons occur in the ratio 1 white-eyed : 1 red-eyed".
  2. Work out the expected counts from the ratio and the total, E = N × fraction.
  3. Calculate χ² = Σ (O − E)²/E over the classes.
  4. Find df = classes − 1.
  5. Compare χ² with the critical value, or find the p-value. If p is below the chosen level (usually 0.05), reject H0. Otherwise, do not reject it.
  6. State the conclusion in words, in the context of the cross.
df5% critical value1% critical value0.1% critical value
13.846.6310.83
25.999.2113.82
37.8111.3416.27

What a p-value means Core

The p-value is the probability of getting a result at least as far from the expectation as yours, if the hypothesis were true. A p-value of 0.11 does not mean there is an 11% chance that the hypothesis is true. A small p-value means the data are surprising under the hypothesis. A large p-value means they are not. Failing to reject a hypothesis is not the same as proving it: several different hypotheses can all be consistent with the same data, especially in small samples.

Comparing two crosses Core

To test whether two crosses give the same distribution, use the contingency-table test of homogeneity described in section C. To ask whether phenotype depends on sex within one cross, use a 2 × 2 table (sex by phenotype) with df = 1. In an X-linked cross the sexes differ sharply; in an autosomal cross they usually do not. If any expected count is below 5, the approximation is poor: collect more offspring, combine classes, or use an exact test. For a 2 × 2 table with moderate counts you may use Yates's continuity correction, which subtracts 0.5 from each |O − E| before squaring and makes the test slightly more cautious.

Inferring the mode of inheritance Core

Use reciprocal crosses of true-breeding lines. The F1 patterns predicted by each model are:

Mode of inheritanceCross A: mutant mother × wild-type fatherCross B: wild-type mother × mutant father
Autosomal recessiveAll wild typeAll wild type
Autosomal dominantAll mutantAll mutant
X-linked recessiveDaughters wild type, sons mutantAll wild type
X-linked dominantAll mutantDaughters mutant, sons wild type

Each model gives a different pair of patterns, so the two crosses together identify the model. The Analyse tool adds the chi-squared values of the two crosses for each model, adds their degrees of freedom, and reports a p-value for every model. A model is rejected immediately if offspring appear in a class that it forbids. Even when the right model is used, about 1 experiment in 20 will be rejected by chance at the 5% level.

Line of best fit through the origin Extension

When observed counts are plotted against expected counts, theory predicts a straight line through the origin with gradient m = 1. The least-squares gradient and its standard error are

m = Σxy / Σx²     SE(m) = √( Σ(y − mx)² / ((n − 1) Σx²) )

where x is the expected count and y the observed count. If |m − 1| is less than about 2 SE(m), the data are consistent with theory. The line does not test each class separately; chi-squared does that. The two analyses complement each other.

Random and systematic effects Core

Random variation is built into inheritance and does not stop with careful technique. Systematic errors can come from the method: wrongly scored phenotypes, mixed-up parental lines, parents left in the vial too long, or lower survival of one class. A real experiment should include a check on scoring, for example a second person counting a sample of vials.

I. Apparatus and measurement methods

Why fruit flies? Foundation

The fruit fly Drosophila melanogaster is small and cheap to keep, has a short life cycle, produces many offspring and has only four pairs of chromosomes. Its sexes are easy to tell apart, and many mutations with clear visible effects are available. Thomas Hunt Morgan and his students used it from about 1910 to show that genes lie on chromosomes. Morgan reported a white-eyed male among red-eyed flies and found that the trait followed the sex of the flies in a way that matched the behaviour of the X chromosome.

Handling flies Core

Flies are kept in vials with a standard food medium. To count them they are briefly anaesthetised, usually with carbon dioxide or another approved agent, and examined under a hand lens or stereo microscope. A fine brush moves them. They wake in a few minutes. Counts are tallied by sex and eye colour.

Telling the sexes apart Core

FeatureFemaleMale
SizeUsually largerUsually smaller
Rear of abdomenPointed, with several dark stripesRounded, with a dark, fused tip
Front legsNo sex combA small dark sex comb on each front leg

Scoring phenotype Core

Eye colour is scored under steady light, with the same person or the same instructions for every vial. Wild-type eyes are dull red. White-eyed flies have no red pigment, so their eyes look white. Sepia flies have dark brown eyes. Borderline individuals should be recorded in a notes column and not forced into a class. In a hidden-genotype experiment, the person who counts should not know which cross a vial came from.

In the virtual laboratory Foundation

The simulation replaces the vial and the microscope. You choose the parents, the number of offspring and the playback speed. Each offspring is drawn as a circle in the daughter row or the son row, with its eye colour. The program uses a random number generator for each fertilisation, so no two broods are the same. The table records the counts, and the analysis tools use only the counts that you have recorded.

Human genetics: tools of the clinic Extension

Human X-linked conditions are studied with pedigrees, with laboratory tests for the relevant protein or gene, and with chromosome analysis. Colour vision can be screened using patterns of coloured dots on plates (Ishihara-style plates). Clotting factors can be measured in a blood test. Chromosome counts use cells stained and photographed during division. Interpretation of such results belongs with a doctor or genetic counsellor.

J. Safety and limits of the simulation

What the simulation assumes Foundation

  • One gene is followed at a time. Linkage to other genes does not matter.
  • Each parent makes the two kinds of gametes in exactly equal proportions, and egg and sperm combine at random.
  • All genotypes survive equally well, so what is counted is what was produced.
  • Phenotypes are scored without error, and the mutant allele is fully penetrant.
  • There are no new mutations, no non-disjunction, and no maternal effects on the offspring phenotype.
  • Sex is decided by X and Y sperm in equal numbers. The Y chromosome carries no allele of the gene studied.

Limits of the simulation. The model is a clean version of genetics. Real crosses have lower counts, mis-scored flies, differences in survival and occasional rare events. A simulated chi-squared test shows you what the test does when the assumptions hold. A real experiment must also check that they do.

Safety in a real fly laboratory Core

Follow your school or institution's written procedures. In general: use only approved anaesthetics and apply them in a ventilated space; keep food and drink away from cultures; wash hands after handling; dispose of flies as instructed, often by freezing or placing them in a fly-killing trap; and keep vials contained so that flies do not escape. Mould on fly food can trigger allergies, so tell your teacher about any allergy.

Using the human examples Core

This page gives introductory, general information about inheritance. It does not diagnose any person, estimate any family's risk or replace advice from a doctor or a genetic counsellor. A person's genetic information is personal and can have implications for relatives, so it should only be shared with consent.

K. Common mistakes
  1. Passing an X-linked allele from father to son. A father gives his Y to his sons. Check every Punnett square: sons are formed by Y sperm only.
  2. Giving a Y chromosome an allele. The Y carries no copy of most X genes. Write XwY, not XwYw.
  3. Calling a male a carrier. Males with one X-linked recessive allele show the trait. Carriers of X-linked recessive alleles are females.
  4. Quoting a ratio without the sex. "1 : 1 red : white" is incomplete. Say: "all daughters red-eyed; sons 1 red : 1 white".
  5. Trusting a pooled 3 : 1. Two heterozygotes for an autosomal gene give 3 : 1 in both sexes. A carrier mother with a normal father also gives 3 red-eyed : 1 white-eyed when both sexes are pooled, but every white-eyed fly is a son: sons are 1 : 1 and daughters all red. The sex-specific ratios, not the pooled one, show the difference.
  6. Using the wrong group. If the question concerns sons, use the number of sons as the total, not the whole brood.
  7. Using percentages in chi-squared. Use counts.
  8. Wrong degrees of freedom. Use the number of classes minus 1 for a goodness-of-fit test, and (rows − 1) × (columns − 1) for a table. Classes with expected count zero do not count.
  9. Saying "accept the null hypothesis" or "prove". Say "do not reject" and "the data are consistent with".
  10. Confusing sex-linked with sex-limited or sex-influenced. A sex-limited trait, such as milk production, is expressed in only one sex though its genes may be autosomal. A sex-influenced trait is dominant in one sex and recessive in the other, though its gene is autosomal. A sex-linked trait has its gene on a sex chromosome.
  11. Assuming small families follow the ratios. A carrier mother with four sons might have none, some or all of them affected. Ratios describe long-run averages.
  12. Mixing up pedigree symbols. Squares are males and circles are females. Filled symbols are affected. A dot in a circle marks a carrier.
L. Exam technique

Command words Foundation

WordWhat is expected
StateA short fact with no explanation.
ExplainGive reasons, using the inheritance of the X and Y chromosomes.
CalculateShow working: the formula, the numbers and the units or the label.
DeduceReach a conclusion from the data or pedigree given.
PredictGive expected genotypes or ratios from a stated model.
TestState a hypothesis, carry out the statistical test and conclude.

Writing a genetic diagram Core

  1. Define your symbols: for example XW = red-eye allele, Xw = white-eye allele.
  2. Write the parental phenotypes and genotypes. Include Y for males.
  3. Write the gametes, circled or listed. Show X and Y sperm separately.
  4. Draw the Punnett square or use a branching diagram.
  5. Write the offspring genotypes and phenotypes, with the sex.
  6. State the ratio or probability, and answer the question that was asked.

Chi-squared answers Core

Write the hypothesis, the table of O, E and (O − E)²/E, the total χ², the degrees of freedom, the critical value or p-value, and a one-sentence conclusion in the context of the cross. Mark schemes usually give a mark for each step even if an earlier number is wrong, so show all the steps.

Reading a pedigree Core

First look at the affected people. Are they mostly male? Do affected males have unaffected parents (a recessive trait, with a carrier mother)? Does the trait ever pass from father to son? If it does, X-linkage is ruled out. If an affected daughter has an unaffected father, X-linked recessive is ruled out. Test each possibility on the whole pedigree and name the evidence for each decision.

General advice Foundation

  • Answer using the labels given in the question and the language of the topic: allele, genotype, phenotype, hemizygous, carrier.
  • Say "the probability that a child is" versus "the probability that a son is". The wording changes the answer.
  • Check a final answer against common sense: probabilities lie between 0 and 1, and the expected counts in a brood add up to the brood size.
  • Marks in this laboratory are for learning, not an official grade.
M. Curriculum map

The topic names below show where this laboratory fits. Course content and the depth required change from year to year and from one examination body to another. Check your own current specification or syllabus for the exact statements, the equations you are given and the sections that are assessed. This page supports your course and does not replace it.

Course familyTopic names where this material usually appears
UK and Cambridge (GCSE, IGCSE, AS and A Level style courses)Inheritance; sex determination; genetic diagrams and Punnett squares; sex-linked characteristics; pedigrees; the chi-squared test and other statistical tests in biology; meiosis and the behaviour of chromosomes.
WAEC and WASSCE BiologyHeredity and variation; Mendel's laws; sex determination; sex-linked characters; genetic crosses and ratios; genetic disorders; chromosomes and genes.
International Baccalaureate Diploma BiologyGenetics and evolution (or the heredity section of the course); Mendelian inheritance; genetic crosses and gene linkage; sex linkage and sex chromosomes; pedigree charts; chi-squared in the use of statistics; meiosis.
US and North America (high school, AP Biology, introductory college biology and genetics)Heredity; chromosomal inheritance; sex-linked traits; patterns of inheritance beyond simple Mendelian ratios; pedigree analysis; X-inactivation; chi-square analysis of genetic data.

Using the three levels Foundation

Foundation material suits learners who are meeting inheritance for the first time. Core material matches the main body of pre-university courses. Extension material goes to the edge of first-year university work and is not needed for most school examinations. If your course does not include a topic, such as population allele frequencies, use the tags to skip it.

N. Applications

Human X-linked recessive conditions Foundation

Several well-known human conditions are inherited in an X-linked recessive pattern. They are described here in general terms only.

  • Red-green colour vision deficiency. The genes for the light-sensitive pigments that help detect red and green light lie on the X chromosome. People with the deficiency have difficulty telling certain reds, greens and related colours apart. It is common compared with most genetic conditions: roughly 1 in 12 males and far fewer females in some populations, though the figures differ between populations. It usually does not affect health.
  • Haemophilia (A and B). The genes for two blood-clotting proteins lie on the X chromosome. When one of the proteins is missing or reduced, blood clots slowly and bleeding can last longer than usual. Treatment replaces the missing clotting protein, and good medical care and support allow many people to lead full lives.
  • Duchenne muscular dystrophy. A gene for a muscle protein lies on the X chromosome. Mutations in it lead to progressive muscle weakness, mostly in males.

In all three cases, males are affected more often because they have a single X. Women can also be affected, rarely, and carriers may sometimes show mild signs because of X-inactivation.

Pedigree of an X-linked recessive condition Generation I: an affected man and an unaffected woman. Generation II: an unaffected son and a carrier daughter, who has children with an unaffected man. Generation III: an affected son, a carrier daughter, an unaffected son and an unaffected daughter. I II III square: male circle: female filled: affected dot: carrier affected son carrier daughter

Read the pedigree. The affected grandfather (I-1) has no affected son, because he gave his Y to him. His X went to his daughter (II-2), making her a carrier. Her affected son (III-1) received her Xh, and her carrier daughter (III-2) received the same chromosome. The condition skipped a generation in the males and passed through a female carrier: a typical X-linked recessive pattern.

X-linked dominant conditions Core

X-linked dominant traits are rarer. X-linked hypophosphataemia, a form of rickets in which the body loses phosphate, is one example. In a pedigree, an affected father has all daughters affected and no sons affected, while an affected heterozygous mother has about half of her children affected, whatever their sex. The trait appears in every generation, and affected females are more numerous than affected males.

X-inactivation in action Core

In female mammals, one of the two X chromosomes in each cell is mostly switched off early in development, an idea first put forward by Mary Lyon in 1961. The choice is random in each cell, and the choice is copied to all descendants of that cell. The switched-off X becomes a compact Barr body that can be seen in the nucleus. Because the silencing is random, a heterozygous female is a mosaic of cells with one allele active and cells with the other active.

The best-known visible example is the coat of a female tortoiseshell or calico cat. A gene for orange or non-orange fur lies on the X chromosome. In a female with one orange and one non-orange allele, patches of orange and black fur reflect which X was active in the skin cells of each patch. Male cats, with one X, are usually all orange or all black. (Calico cats also have white patches, which depend on a different, autosomal gene.)

Agriculture and breeding Core

In poultry the female is ZW and the male is ZZ. Some feather-colour genes lie on the Z chromosome. A breeder can cross a hen with one type of Z allele and a cock with the other so that chicks of the two sexes hatch with different down colours, and so tell them apart immediately. This is the same criss-cross inheritance as in flies, with the sexes reversed.

Genetic counselling and screening Extension

Genetic counsellors use pedigrees and tests to explain the chance that a family member carries or inherits an X-linked allele. They explain probabilities, which describe a population of similar families and not the outcome for one family. The conditions that are tested, the tests offered and the choices open to families differ between countries, and decisions are personal.

Statistics beyond genetics Core

The chi-squared test used in this laboratory is a general tool. Scientists use it whenever they compare counts in categories with expected counts, for example to see whether the numbers of seeds, patients or species in different groups agree with a prediction. The reasoning is the same as here: state a hypothesis, compute expected counts, measure the distance with χ², and judge it with the p-value.

O. Extension topics

How the sex of a fly is decided Extension

In humans and other mammals, a gene on the Y chromosome (SRY) switches development towards the male pattern. In Drosophila the Y does not decide sex. What matters is the ratio of X chromosomes to sets of autosomes: a ratio of 1 gives a female, and a ratio of 0.5 gives a male. A fly with two X chromosomes and a Y (XXY) is therefore a normal-looking female, while a fly with a single X and no Y (XO) is a male that is sterile. The Y is needed in male flies for fertility. For single-gene crosses the difference does not matter, but it shows why sex determination cannot be assumed to work the same way in every animal.

Non-disjunction and the chromosome theory Extension

Calvin Bridges, a student of Morgan, studied rare exceptional offspring in the white-eye cross, in the 1910s. Rare white-eyed daughters and red-eyed sons appeared among the offspring of white-eyed mothers and red-eyed fathers. Bridges showed that the exceptions arose when the two X chromosomes failed to separate in meiosis in the mother. An egg with two Xw chromosomes fertilised by a Y sperm gives a white-eyed female (XwXwY), and an egg with no X fertilised by an XW sperm gives a red-eyed, sterile male (XWO). This matching of the genetic exception with a visible chromosome event was an important step in showing that genes are carried on chromosomes.

Dosage compensation Extension

Females have two X chromosomes and males have one, so the dose of X genes would differ if nothing compensated. Different animals solve this differently. In mammals, one X is largely silenced in each cell of a female; a non-coding RNA called XIST has a central role in starting the silencing. In fruit flies, the single X of the male is more strongly expressed. In the nematode Caenorhabditis elegans, the two X chromosomes of the hermaphrodite are each partly turned down. The simulation does not model dosage, but the existence of these mechanisms explains why a missing or extra X has such large effects.

Pseudoautosomal regions and Y-linkage Extension

In humans the X and Y pair at the tips and exchange material in the male germ line. Genes in this pseudoautosomal region recombine between X and Y and are inherited like autosomal genes. Genes that lie in the rest of the Y chromosome have no X counterpart and pass from father to son only. Y-linked traits are rare because the Y carries few genes.

Statistics: exact tests and likelihoods Extension

The chi-squared test is an approximation. For small samples, an exact test such as the binomial test (for two classes) or Fisher's exact test (for a 2 × 2 table) is better. Another approach compares the likelihood of the data under competing models, as the mode-of-inheritance tool does in a simple way. When several models are tested, the chance that at least one rejects a true model rises, so scientists adjust their significance level or decide the tests in advance.

Linked genes on the X Extension

Two genes on the same X chromosome tend to be inherited together. In a female, crossing over between the two X chromosomes during meiosis separates them at a frequency that rises with the distance between the genes. In a male, only one X is present, so the gametes of a female can be read directly from the male offspring. Experiments of this kind were used to build the first genetic maps.

P. Key points
  • Females are XX and males are XY. Sperm decide the sex of the child.
  • A son gets his X from his mother and his Y from his father. A daughter gets her father's X and one of her mother's two X chromosomes.
  • Males are hemizygous for X-linked genes, so recessive alleles show in males.
  • A father passes an X-linked allele to all of his daughters and none of his sons.
  • A mutant mother and a normal father give a recessive X-linked trait in all sons and in no daughters (criss-cross inheritance).
  • A carrier mother and a normal father give 1 white-eyed son : 1 red-eyed son and no white-eyed daughters.
  • Reciprocal crosses differ for X-linked traits and agree for autosomal traits.
  • Always state sex in the phenotype ratio for X-linked crosses.
  • χ² = Σ (O − E)² / E. Use counts. df = classes − 1 (or (rows − 1)(columns − 1) for a table).
  • Do not reject if p ≥ 0.05. A test never proves a hypothesis.
  • Two reciprocal crosses of true-breeding lines identify the mode of inheritance.
  • For an X-linked recessive allele with frequency q: affected males q, affected females q², carrier females 2q(1 − q).
  • In female mammals one X is randomly silenced in each cell, so carriers are mosaics.
Q. Glossary
Allele
One of the alternative forms of a gene.
Autosome
Any chromosome that is not a sex chromosome. Both sexes carry two copies of each autosome.
Barr body
The condensed, inactive X chromosome seen in the nucleus of cells of female mammals.
Carrier
A heterozygous individual who has one copy of a recessive allele and usually does not show the trait.
Chi-squared test
A test that measures whether observed counts differ from expected counts by more than chance would explain.
Criss-cross inheritance
The pattern in which a recessive X-linked trait passes from a mother to her sons, and from a father to his daughters' sons.
Degrees of freedom
The number of values in a chi-squared calculation that can vary freely. For a goodness-of-fit test it is the number of classes minus 1.
Dosage compensation
A mechanism that evens out the activity of X-linked genes between the sexes.
Genotype
The alleles that an individual carries for a gene.
Hemizygous
Having only one copy of a gene, as a male has for X-linked genes.
Heterozygous
Having two different alleles of a gene.
Homozygous
Having two identical alleles of a gene.
Mosaic
An individual with two or more genetically or functionally different cell populations, as in a female with random X-inactivation.
Non-disjunction
The failure of chromosomes to separate correctly in meiosis.
Null hypothesis (H0)
The hypothesis of no difference or no effect, which a statistical test tries to reject.
p-value
The probability of a result at least as extreme as the one observed, if the null hypothesis were true.
Phenotype
The observable characteristics of an individual.
Reciprocal cross
A cross in which the phenotypes of the two sexes are swapped compared with an earlier cross.
Sex chromosomes
The pair of chromosomes that differ between the sexes: XX in female and XY in male mammals and flies.
Sex-linked
Located on a sex chromosome, so that inheritance differs between the sexes. Usually means X-linked.
True-breeding
Giving offspring identical to the parents for the trait when crossed with the same type.
Wild type
The common, normal form of a trait in a natural population.
X-inactivation
The random silencing of one X chromosome in each cell of a female mammal.
R. References and further reading

These titles are for further reading. Only the names of the books, articles and publishers are given, with the publisher's home page. Editions change, so check which edition your course uses. This list does not mean that any publisher or examination body endorses this laboratory.

  • Morgan, T. H. (1910). Sex limited inheritance in Drosophila. Science. American Association for the Advancement of Science: science.org
  • Lyon, M. F. (1961). Gene action in the X-chromosome of the mouse (Mus musculus L.). Nature. Nature Portfolio: nature.com
  • Biology 2e. OpenStax: openstax.org (a free textbook; see the chapters on inheritance and chromosomal theory)
  • Introduction to Genetic Analysis, Griffiths and colleagues. Macmillan Learning: macmillanlearning.com
  • Genetics: From Genes to Genomes, Hartwell and colleagues. McGraw Hill: mheducation.com
  • Campbell Biology. Pearson: pearson.com

Examination boards and curriculum bodies publish current specifications and past papers on their own websites. Find the page for your course and use that version.

8Practise: extra problems

Assume equal numbers of each kind of gamete, complete dominance and no new mutations. Attempt each problem before opening the answer.

Problem 1 (Foundation)

A white-eyed female fruit fly (XwXw) is crossed with a red-eyed male (XWY). Give the genotypes and phenotypes of the sons and the daughters.

Show the answer

Every egg carries Xw. Sperm are XW or Y. Daughters (XW sperm) are XWXw: red-eyed carriers. Sons (Y sperm) are XwY: white-eyed. All daughters are red-eyed and all sons are white-eyed.

Problem 2 (Core)

A carrier female (XWXw) is crossed with a white-eyed male (XwY). How many flies of each kind do you expect among 600 offspring?

Show the answer

Eggs: XW, Xw. Sperm: Xw, Y. The four squares are XWXw (red female), XwXw (white female), XWY (red male) and XwY (white male), each with probability 1/4. Expected: 600 × 1/4 = 150 of each kind. Half the daughters are white-eyed in this cross, unlike the cross with a red-eyed father.

Problem 3 (Core)

A carrier female is crossed with a red-eyed male. Among 240 offspring there are 118 red-eyed daughters, 0 white-eyed daughters, 62 red-eyed sons and 60 white-eyed sons. Test the Mendelian prediction with a chi-squared test.

Show the answer

Expected: red daughters 1/2 × 240 = 120, white daughters 0, red sons 60, white sons 60. The class with expected 0 has 0 observed, so it is not used. χ² = (118 − 120)²/120 + (62 − 60)²/60 + (60 − 60)²/60 = 0.033 + 0.067 + 0 = 0.10. Three classes are used, so df = 2 and p = exp(−0.05) = 0.95. Do not reject: the data are consistent with the prediction.

Problem 4 (Core)

A woman who is a carrier of red-green colour vision deficiency (X-linked recessive) has children with a man who has the deficiency. What is the probability that a child is affected? What is the probability that exactly two of four children are affected?

Show the answer

Mother XCXc, father XcY. The four equally likely squares are XCXc (carrier daughter), XcXc (affected daughter), XCY (unaffected son) and XcY (affected son). Two of four squares are affected, so the probability that a child is affected is 1/2. For exactly two of four children: 4C2 × (1/2)⁴ = 6/16 = 0.375. These are probabilities, not predictions for any particular family.

Problem 5 (Core)

Cross A (white-eyed female × red-eyed male) gives 61 red daughters, 59 white sons and no other kinds. Cross B (red-eyed female × white-eyed male) gives 58 red daughters, 62 red sons and no other kinds. Test whether the reciprocal crosses differ, and say what you conclude.

Show the answer

Table of the three non-empty classes (red daughters, red sons, white sons): A is 61, 0, 59 and B is 58, 62, 0. Column totals 119, 62, 59; each row total is 120 and the grand total is 240, so the expected counts are 59.5, 31, 29.5 in both rows. χ² = 2 × (1.5)²/59.5 + 2 × 31 + 2 × 29.5 = 0.08 + 62 + 59 = 121.1. df = (2 − 1)(3 − 1) = 2, so p is far below 0.001. The reciprocal crosses differ, which points to X-linkage.

Problem 6 (Extension)

An X-linked recessive allele has frequency q = 0.05 in a population in which mating is random. Find the expected proportion of affected males, affected females and carrier females, and the ratio of affected males to affected females.

Show the answer

Affected males: q = 0.05, or 5%. Affected females: q² = 0.0025, or 0.25%. Carrier females: 2q(1 − q) = 2 × 0.05 × 0.95 = 0.095, or 9.5%. The ratio of affected males to affected females is q/q² = 1/q = 20. Rarer alleles give larger ratios, because q² falls faster than q.