Biology · 17 Inheritance

17.4 Monohybrid inheritance

Before you start

A gene is a length of DNA that codes for a protein, and an allele is an alternative form of a gene (17.1).
Body cells are diploid — two sets of chromosomes — so they carry two alleles of every gene, one on each chromosome of a pair.
Meiosis halves the chromosome number, so a gamete carries only one allele of each gene (17.3).
Fertilisation is the fusion of two gamete nuclei, which brings the two alleles back together.
In humans, females have two X chromosomes and males have one X and one Y (17.1).
Writing a ratio in its lowest terms, e.g. $2 : 2$ simplifies to $1 : 1$ .

Syllabus objectives covered (18)

Describe inheritance as the transmission of genetic information from generation to generation
Describe genotype as the genetic make-up of an organism and in terms of the alleles present
Describe phenotype as the observable features of an organism
Describe homozygous as having two identical alleles of a particular gene
State that two identical homozygous individuals that breed together will be pure-breeding
Describe heterozygous as having two different alleles of a particular gene
State that a heterozygous individual will not be pure-breeding
Describe a dominant allele as an allele that is expressed if it is present in the genotype
Describe a recessive allele as an allele that is only expressed when there is no dominant allele of the gene present in the genotype
Interpret pedigree diagrams for the inheritance of a given characteristic
Use genetic diagrams to predict the results of monohybrid crosses and calculate phenotypic ratios, limited to 1 : 1 and 3 : 1 ratios
Use Punnett squares in crosses which result in more than one genotype to work out and show the possible different genotypes
Explain how to use a test cross to identify an unknown genotype
Describe codominance as a situation in which both alleles in heterozygous organisms contribute to the phenotype
Explain the inheritance of ABO blood groups: phenotypes are A, B, AB and O blood groups and alleles are IA, IB and Io
Describe a sex-linked characteristic as a feature in which the gene responsible is located on a sex chromosome and that this makes the characteristic more common in one sex than in the other
Describe red-green colour blindness as an example of sex linkage
Use genetic diagrams to predict the results of monohybrid crosses involving codominance or sex linkage and calculate phenotypic ratios

Part 1 of 9

Inheritance, and the two words examiners test hardest

Inheritance is the transmission of genetic information from generation to generation. That is the definition, and the two phrases that score are transmission (passing on) and from generation to generation (parent to offspring). "Getting things from your parents" is the same idea in words that earn nothing.
What is actually transmitted is alleles. Remember the pair of definitions from 17.1: a gene is a length of DNA that codes for a protein, and an allele is an alternative form of a gene. One gene controls seed colour in peas; that gene has an allele for yellow seeds and an allele for green seeds.
Your body cells are diploid, so you carry the gene twice — once on each chromosome of a pair. That means two alleles of every gene, one inherited from each parent. Everything in this subtopic follows from that one fact.
Genotype is the genetic make-up of an organism, described in terms of the alleles present. A genotype is written as letters: $T T$ , $T t$ or $t t$ .
Phenotype is the observable features of an organism. A phenotype is written as words: tall, short, black fur, blood group A.
The trap is quiet and expensive. If a question says "state the genotype", answering "tall" scores nothing, even though you knew the biology. If it says "state the phenotype", answering $T t$ scores nothing either. Read the word, then decide whether your answer is letters or words.
One more discipline that costs marks all through this topic: say allele when you mean one version of a gene. "He inherited the gene for colour blindness" is wrong — everyone has the gene; what he inherited was a particular allele of it.

Part 2 of 9

Homozygous, heterozygous, and what pure-breeding means

Since you hold two alleles of each gene, there are only two possibilities: the two are the same, or they are different. That is the whole of this section.
Homozygous means having two identical alleles of a particular gene. $T T$ is homozygous, and so is $t t$ . If you need to be precise, say homozygous dominant ( $T T$ ) or homozygous recessive ( $t t$ ).
Heterozygous means having two different alleles of a particular gene — $T t$ .
Now pure-breeding. Two identical homozygous individuals that breed together will be pure-breeding: every offspring shows the same feature as the parents, generation after generation. Watch why, in the gametes.
$T T \times T T ⟶ gametes T and T ⟶ all offspring T T$
There is only one kind of gamete available from each parent, so there is only one possible offspring genotype. The same is true of $t t \times t t$ , which breeds true for the recessive feature.
A heterozygous individual will not be pure-breeding. It makes two kinds of gamete, so it can produce offspring unlike itself.
$T t \times T t ⟶ gametes T and t ⟶ T T, T t, T t, t t$
Notice the short offspring, $t t$ , appearing from two tall parents. That single line is the reason a heterozygous plant cannot be sold as pure-breeding, and it is also the clue you will use later to read pedigree diagrams.

Part 3 of 9

Dominant and recessive alleles

A heterozygote holds two different instructions at once. Which one shows in the phenotype? That is what dominant and recessive mean.
A dominant allele is an allele that is expressed if it is present in the genotype. Present once is enough: $T t$ is tall, exactly as $T T$ is tall.
A recessive allele is an allele that is only expressed when there is no dominant allele of the gene present in the genotype. So the recessive feature appears only in $t t$ .
That gives you a deduction you will use constantly: an organism showing the recessive phenotype has a genotype you can write down immediately — both alleles must be recessive. An organism showing the dominant phenotype could be either $T T$ or $T t$ , and the phenotype alone cannot tell you which.
Notation rules, because sloppy symbols lose marks that the biology had already earned. Use the same letter for both alleles, capital for the dominant one and small for the recessive one: $T$ and $t$ . Never use two unrelated letters such as $T$ and $s$ — they look like two different genes.
Choose a letter whose capital and small forms look different when you are writing fast. $B$ and $b$ are safe. $C$ and $c$ , $S$ and $s$ , $O$ and $o$ are not — a hurried $S$ is unreadable, and an examiner cannot give you credit for a genotype they cannot tell apart.
Last trap in this section: dominant does not mean common, and recessive does not mean rare. Dominance is about which allele is expressed when both are present, nothing else.

Part 4 of 9

The genetic diagram, line by line

A genetic diagram is not decoration — it is the answer, and each line of it is a separate marking point. Learn the layout as a fixed sequence of lines and you will never lose a mark for leaving one out.
Six lines, six marks. A line left out is a mark left behind.
Take a cross between two tall pea plants, each heterozygous. Tall ( $T$ ) is dominant to short ( $t$ ). Line one is the parental phenotypes; line two is the parental genotypes.
$\begin{matrix} parental phenotypes & tall & \times & tall \\ parental genotypes & T t & \times & T t \end{matrix}$
Line three is the gametes, and this is where most marks in the topic are thrown away. Meiosis halves the chromosome number, so a gamete carries one allele of the gene — never two. From $T t$ the gametes are $T$ and $t$ .
$gametes (T) (t) \times (T) (t)$
Circle each gamete, as shown by the brackets above. It takes a second and it makes it physically obvious that each one is a single allele — the habit stops you writing $T t$ in a gamete space.
Line four is the offspring genotypes: combine every gamete from one parent with every gamete from the other. Two kinds of gamete each gives four combinations.
$offspring genotypes T T T t T t t t$
Line five converts genotypes to phenotypes using the dominance rule. $T T$ is tall, $T t$ is tall because one $T$ is enough, $T t$ is tall, and $t t$ is short.
$offspring phenotypes tall tall tall short$
Line six is the phenotypic ratio, counted from line five and written in its lowest terms, with the phenotypes named.
$phenotypic ratio = 3 tall : 1 short$
A ratio is what the question asks for, so give a ratio. "75%", " $\frac{3}{4}$ " or "3 out of 4" is a different answer to a different question, and it is not automatically credited when a ratio was demanded.
The ratio is what you expect, not what is guaranteed. Fertilisation is random: four seeds from this cross could easily all be tall. The larger the number of offspring, the closer the results get to $3 : 1$ — say so if a question asks why real results do not match exactly.

Part 5 of 9

Punnett squares, and the two ratios you must recognise

A Punnett square is the same cross in a grid. It is worth using whenever more than one genotype is possible, because the grid makes it hard to lose a combination — and the syllabus asks you to be able to use one.
One allele per label on the outside — never a genotype such as $T t$ above a column.
Build it in this order. Write the gametes of one parent along the top, one per column. Write the gametes of the other parent down the side, one per row. Only single alleles go on the outside.
$\begin{matrix} \times & T & t \\ T & T T & T t \\ t & T t & t t \end{matrix}$
Fill each box by combining its row allele with its column allele. Write the dominant allele first inside the box — $T t$ , not $t T$ — so that identical genotypes look identical when you count them.
The boxes are the possible offspring genotypes: $1 T T : 2 T t : 1 t t$ . That is the genotypic ratio. Convert to phenotypes before you answer, unless genotypes were what was asked for.
$1 T T : 2 T t : 1 t t ⟶ 3 tall : 1 short$
Now the other cross you must know. A heterozygous tall plant crossed with a short plant: one parent gives $T$ or $t$ , the other can only give $t$ .
$\begin{matrix} \times & T & t \\ t & T t & t t \\ t & T t & t t \end{matrix}$
Counting gives $2 T t : 2 t t$ , which is $2$ tall $: 2$ short. Simplify — a ratio is always given in its lowest terms.
$2 tall : 2 short = 1 tall : 1 short$
Those are the only two ratios the Core objective asks for, and each comes from a recognisable pairing. Heterozygous $\times$ heterozygous gives $3 : 1$ . Heterozygous $\times$ homozygous recessive gives $1 : 1$ . Spotting which one you are looking at is often the fastest route into a question.
The commonest Punnett square error is labelling the outside of the grid with genotypes — writing $T t$ above a column. The outside is for gametes, and gametes carry one allele each.

Part 6 of 9

Reading a pedigree diagram

A pedigree diagram is a family tree that shows which members of a family have a particular characteristic. You are not asked to draw one; you are asked to interpret one.
The conventions are fixed. A square is male and a circle is female. A shaded symbol has the characteristic and an unshaded one does not. A horizontal line joins two parents, and a vertical line drops from that couple to their children, who are drawn side by side on the generation below.
Ask one question first, before anything else: is the characteristic caused by a dominant or a recessive allele? Everything else depends on the answer, and there is a single piece of evidence that settles it.
The evidence is two unaffected parents with an affected child. If the allele were dominant it would be expressed whenever it is present, so an affected child would have to have at least one affected parent. It does not — so the allele must be recessive, and both parents must be heterozygous, carrying it without showing it.
State that reasoning in full when you answer, because the reasoning is where the marks are. "Recessive" on its own is one word; "individuals 1 and 2 do not have the condition but their daughter does, so the allele must be recessive and both parents must be heterozygous" is the whole answer.
Then work out genotypes, using the same two deductions every time. Anyone showing the recessive characteristic is homozygous recessive, $n n$ . Anyone unaffected who has an affected child must be heterozygous, $N n$ , because they passed on a recessive allele.
Where the diagram genuinely does not tell you, say so: an unaffected child of two heterozygous parents is $N N$ or $N n$ , and writing both is the correct answer, not a hedge.
One extra pattern to notice: if nearly everyone affected in the family is male, think sex linkage — the gene is on the X chromosome. That idea is developed in the last section of this lesson.

Part 7 of 9

The test cross: finding a genotype you cannot see

Here is the practical problem. A black rabbit stands in front of you, black ( $B$ ) being dominant to brown ( $b$ ). Its genotype is $B B$ or $B b$ and its appearance cannot tell you which, because one $B$ looks exactly like two.
A test cross answers the question by breeding: cross the unknown individual with one that is homozygous recessive — that is, one showing the recessive phenotype, here a brown rabbit, $b b$ .
The choice of partner is the clever part, so make sure you can justify it. A $b b$ parent can only donate $b$ . That means the offspring's phenotype is decided entirely by which allele came from the unknown parent — the test cross makes the hidden allele visible.
Case one: the unknown is homozygous, $B B$ . Every gamete from it carries $B$ , so every offspring is $B b$ and every offspring is black.
$B B \times b b ⟶ all offspring B b ⟶ all black$
Case two: the unknown is heterozygous, $B b$ . Half its gametes carry $b$ , so about half the offspring are $b b$ and brown.
$B b \times b b ⟶ B b, B b, b b, b b ⟶ 1 black : 1 brown$
So the rule you report is: if any offspring shows the recessive phenotype, the unknown parent must be heterozygous, because that offspring had to receive a recessive allele from it. If all the offspring show the dominant phenotype, the unknown parent is probably homozygous dominant.
Notice the difference in certainty, and write it. One brown rabbit proves heterozygosity outright. All-black offspring is only evidence for $B B$ , and the more offspring there are, the stronger it is — a small litter could be all black by chance even from a heterozygote.

Part 8 of 9

Codominance and the ABO blood groups

Dominance is not the only possible relationship between two alleles. Codominance is the situation in which both alleles in a heterozygous organism contribute to the phenotype — neither is masked.
Notation has to change, because capital-and-small letters would imply one allele beats the other. Use one capital letter for the gene with a superscript for each allele: for coat colour in cattle, $C^{R}$ for red and $C^{W}$ for white.
A heterozygous animal, $C^{R} C^{W}$ , is roan: its coat contains both red hairs and white hairs. Both alleles have contributed. It is not a blend into pink — you can see each allele's product side by side, which is exactly what codominance means.
The example the syllabus names is the ABO blood groups. Three alleles exist in the population — $I^{A}$ , $I^{B}$ and $I^{o}$ — although any one person still carries only two of them, because they have only two copies of the gene.
Careful with that last allele: it is the letter o written small and raised, $I^{o}$ , standing for the group O allele. It is not a zero, and it is not a capital O.
The relationships between the three are simple to state and worth stating in exactly this order: $I^{A}$ and $I^{B}$ are codominant with each other, and both are dominant to $I^{o}$ , which is recessive.
Now every genotype follows without memorising a table. Group A needs at least one $I^{A}$ and no $I^{B}$ ; group AB needs both codominant alleles; group O is the recessive phenotype, so it must be homozygous recessive.
$\begin{matrix} group A & I^{A} I^{A} or I^{A} I^{o} \\ group B & I^{B} I^{B} or I^{B} I^{o} \\ group AB & I^{A} I^{B} \\ group O & I^{o} I^{o} \end{matrix}$
Two of those you can read backwards with certainty. Group AB and group O each have exactly one possible genotype, so a question mentioning either has handed you a parent's genotype for free.
Crosses work exactly as before — one allele per gamete, grid, count. Take a group A father who is $I^{A} I^{o}$ and a group B mother who is $I^{B} I^{o}$ .
$\begin{matrix} \times & I^{A} & I^{o} \\ I^{B} & I^{A} I^{B} & I^{B} I^{o} \\ I^{o} & I^{A} I^{o} & I^{o} I^{o} \end{matrix}$
Read the boxes into phenotypes: $I^{A} I^{B}$ is AB, $I^{B} I^{o}$ is B, $I^{A} I^{o}$ is A, $I^{o} I^{o}$ is O. Four different blood groups from one cross, in equal proportions.
$1 group AB : 1 group B : 1 group A : 1 group O$
That result is worth remembering as a sanity check: two heterozygous parents of groups A and B can have children of every blood group, including groups neither parent has.

Part 9 of 9

Sex linkage and red-green colour blindness

Recall how sex is inherited: a female is $X X$ and a male is $X Y$ . The X and Y chromosomes are the sex chromosomes, and genes sit on them just as they sit on any other chromosome.
A sex-linked characteristic is a feature in which the gene responsible is located on a sex chromosome, and this makes the characteristic more common in one sex than in the other. Both halves of that sentence are marking points — the location of the gene, and the consequence.
Red-green colour blindness is the example to know. The gene involved is on the X chromosome, and the allele that causes colour blindness is recessive.
Your notation must show the chromosome, or the whole answer collapses. Write $X^{B}$ for an X chromosome carrying the normal-vision allele, $X^{b}$ for one carrying the colour-blindness allele, and a bare $Y$ for the Y chromosome — because the Y does not carry an allele of this gene at all.
Which is why the notation has to keep the chromosome visible: $X^{b} Y$ , never $b b$ .
Writing the genotypes as $B B$ , $B b$ or $b b$ is the classic sex-linkage error. It hides the very thing the question is about, and it makes it impossible to show that a male has only one copy.
Now the explanation of why it is more common in males, which is the most frequently asked part of this section. A male is $X^{B} Y$ or $X^{b} Y$ : he has only one X, and no second allele on the Y to mask it, so a single recessive allele is expressed and he is colour blind.
A female has two X chromosomes, so she needs the recessive allele on both — $X^{b} X^{b}$ — to be colour blind. If she is $X^{B} X^{b}$ she has normal vision but is a carrier, able to pass the allele to her children. Needing two copies rather than one is why colour blindness is rarer in females.
Set out a cross the same way as any other, but with the sex chromosomes written in full. Here is a carrier mother with a father who has normal vision.
$\begin{matrix} parental phenotypes & normal vision (carrier) & \times & normal vision \\ parental genotypes & X^{B} X^{b} & \times & X^{B} Y \end{matrix}$
The mother's gametes are $X^{B}$ and $X^{b}$ ; the father's are $X^{B}$ and $Y$ . Each gamete is one sex chromosome — the same one-allele-per-gamete rule you have been using all along.
$\begin{matrix} \times & X^{B} & X^{b} \\ X^{B} & X^{B} X^{B} & X^{B} X^{b} \\ Y & X^{B} Y & X^{b} Y \end{matrix}$
Read the four boxes carefully, because sex-linked answers must name the sex. $X^{B} X^{B}$ is a girl with normal vision; $X^{B} X^{b}$ is a girl with normal vision who is a carrier; $X^{B} Y$ is a boy with normal vision; $X^{b} Y$ is a colour-blind boy.
$3 normal vision : 1 colour blind$
Say the sex-specific version too, because questions usually want it: none of the daughters is colour blind, though half of them are carriers, and half of the sons are colour blind.
One consequence worth carrying into pedigree questions: a father gives his X to every daughter and his Y to every son. A colour-blind man therefore cannot pass the allele to any of his sons, but every one of his daughters receives it.

Worked examples

Example 1. In rabbits, black fur ( $B$ ) is dominant to brown fur ( $b$ ). A heterozygous black rabbit is crossed with a brown rabbit. Use a genetic diagram to predict the phenotypes of the offspring and calculate the phenotypic ratio.

Show the working

Turn each phenotype into a genotype before you draw anything. Heterozygous black is $B b$ . Brown is the recessive phenotype, so it must be homozygous recessive, $b b$ — there is no other option.
$\begin{matrix} parental phenotypes & black & \times & brown \\ parental genotypes & B b & \times & b b \end{matrix}$
Write the gametes, one allele in each. $B b$ gives $B$ or $b$ ; $b b$ can only give $b$ , so both of its gametes are the same.
$gametes (B) (b) \times (b) (b)$
Put them into a Punnett square: the black parent's gametes across the top, the brown parent's down the side.
$\begin{matrix} \times & B & b \\ b & B b & b b \\ b & B b & b b \end{matrix}$
List the offspring genotypes from the boxes, then convert each to a phenotype. $B b$ is black, because a dominant allele is expressed whenever it is present; $b b$ is brown.
$B b, B b, b b, b b ⟶ black, black, brown, brown$
Count and simplify. Two black to two brown is $2 : 2$ , which in its lowest terms is $1 : 1$ . Name the phenotypes in the ratio so it cannot be misread.

Answer: Offspring genotypes $B b$ and $b b$ ; phenotypes black and brown; phenotypic ratio $1$ black $: 1$ brown.

Example 2. A pedigree diagram shows an inherited condition in one family. In generation I, a man (individual 1) and a woman (individual 2), neither of whom has the condition, have three children: a son (individual 3) who does not have the condition, a daughter (individual 4) who has the condition, and a son (individual 5) who does not. Interpret the pedigree: determine whether the allele causing the condition is dominant or recessive, and give the genotypes of individuals 1, 2 and 4. Use $N$ and $n$ for the alleles.

Show the working

Look for the one situation that settles dominance: two unaffected parents with an affected child. Individuals 1 and 2 do not have the condition, but their daughter, individual 4, does.
Reason it through rather than asserting it. If the allele were dominant it would be expressed whenever it was present, so an affected daughter would have to have an affected parent. Neither parent is affected, so the allele must be recessive.
Now use the definition of a recessive allele to fix individual 4's genotype. A recessive allele is expressed only when no dominant allele is present, so an individual with the condition must be $n n$ .
Work backwards to the parents. Individual 4 received one allele from each parent, and both of those alleles are $n$ . So each parent carries at least one $n$ .
But neither parent shows the condition, so neither can be $n n$ . Each must therefore also carry $N$ — both parents are heterozygous, $N n$ . They are carriers.
$N n \times N n ⟶ N N, N n, N n, n n$
Check the unaffected sons against your conclusion, which is a good habit. Individuals 3 and 5 are unaffected, so each is $N N$ or $N n$ — the pedigree cannot distinguish them, and the correct answer says both possibilities.

Answer: The allele is recessive, because individuals 1 and 2 do not have the condition but their daughter does. Individual 4 is $n n$ ; individuals 1 and 2 are both heterozygous, $N n$ .

Example 3. In tomato plants, red fruit ( $R$ ) is dominant to yellow fruit ( $r$ ). A grower has a plant with red fruit and needs to know whether it is homozygous or heterozygous. Explain how a test cross would identify its genotype, and state the result expected in each case.

Show the working

State the problem precisely first: a red-fruited plant is $R R$ or $R r$ , and the phenotype cannot distinguish them because one $R$ is expressed just as fully as two.
Give the method. Cross the unknown red plant with a plant showing the recessive phenotype — a yellow-fruited plant, which must be homozygous recessive, $r r$ .
Justify the choice of partner, because this is the mark most people miss. The $r r$ plant can only supply $r$ gametes, so whatever the offspring show depends only on the allele supplied by the unknown parent.
Predict case one. If the unknown plant is $R R$ , every gamete carries $R$ , so all offspring are $R r$ and all have red fruit.
$R R \times r r ⟶ all R r ⟶ all red$
Predict case two. If the unknown plant is $R r$ , half its gametes carry $r$ , so about half the offspring are $r r$ and have yellow fruit.
$R r \times r r ⟶ R r, R r, r r, r r ⟶ 1 red : 1 yellow$
Finish by saying how to read the outcome, including how sure it makes you. Any yellow offspring proves the parent is heterozygous; all-red offspring indicates it is homozygous, and the more offspring grown, the more reliable that conclusion.

Answer: Cross the unknown red plant with a yellow-fruited plant, $r r$ , which can only donate a recessive allele. If any offspring have yellow fruit the unknown plant is heterozygous, $R r$ , and about $1$ red $: 1$ yellow is expected. If all the offspring have red fruit the unknown plant is homozygous, $R R$ .

Example 4. A man of blood group AB and a woman of blood group O have children. Use a genetic diagram to predict the blood groups of their children and calculate the phenotypic ratio. Explain why none of the children has blood group AB.

Show the working

Write the parental genotypes. Both are groups with only one possible genotype: group AB must be $I^{A} I^{B}$ , and group O is the recessive phenotype so it must be $I^{o} I^{o}$ .
$\begin{matrix} parental phenotypes & group AB & \times & group O \\ parental genotypes & I^{A} I^{B} & \times & I^{o} I^{o} \end{matrix}$
Write the gametes, one allele each. The father produces $I^{A}$ or $I^{B}$ ; the mother can only produce $I^{o}$ .
$gametes (I^{A}) (I^{B}) \times (I^{o}) (I^{o})$
Grid them.
$\begin{matrix} \times & I^{A} & I^{B} \\ I^{o} & I^{A} I^{o} & I^{B} I^{o} \\ I^{o} & I^{A} I^{o} & I^{B} I^{o} \end{matrix}$
Convert genotypes to phenotypes using the two relationships. $I^{A}$ is dominant to $I^{o}$ , so $I^{A} I^{o}$ is group A; likewise $I^{B} I^{o}$ is group B.
Count: two group A to two group B, which simplifies to $1 : 1$ .
$1 group A : 1 group B$
Now answer the last part from the diagram rather than from memory. Group AB requires both $I^{A}$ and $I^{B}$ in the same child, but the mother can only donate $I^{o}$ , so no child can inherit both.

Answer: Children can be group A ( $I^{A} I^{o}$ ) or group B ( $I^{B} I^{o}$ ), in the ratio $1$ group A $: 1$ group B. None can be group AB because that genotype needs $I^{A}$ and $I^{B}$ together, and the group O mother can only pass on $I^{o}$ .

Example 5. Red-green colour blindness is caused by a recessive allele carried on the X chromosome. A woman with normal vision whose father was colour blind has children with a man who is colour blind. Use a genetic diagram to predict the phenotypes of their children.

Show the working

Work out the mother's genotype from the information given, since it is not stated directly. Her father was colour blind, $X^{b} Y$ , and he passes his single X to every daughter — so she must have received $X^{b}$ . She has normal vision, so her other X carries $X^{B}$ : she is a carrier, $X^{B} X^{b}$ .
The father is colour blind. He has one X, which must carry the recessive allele, plus a Y: $X^{b} Y$ .
$\begin{matrix} parental phenotypes & normal vision (carrier) & \times & colour blind \\ parental genotypes & X^{B} X^{b} & \times & X^{b} Y \end{matrix}$
List the gametes as whole sex chromosomes. The mother produces $X^{B}$ or $X^{b}$ ; the father produces $X^{b}$ or $Y$ .
$gametes (X^{B}) (X^{b}) \times (X^{b}) (Y)$
Fill the square, keeping the X chromosomes and the Y visible in every box.
$\begin{matrix} \times & X^{B} & X^{b} \\ X^{b} & X^{B} X^{b} & X^{b} X^{b} \\ Y & X^{B} Y & X^{b} Y \end{matrix}$
Interpret every box, naming the sex as well as the vision. $X^{B} X^{b}$ is a girl with normal vision who is a carrier; $X^{b} X^{b}$ is a colour-blind girl; $X^{B} Y$ is a boy with normal vision; $X^{b} Y$ is a colour-blind boy.
Count the phenotypes for the ratio, then add the sex-specific statement, since that is what makes a sex-linkage answer complete.
$1 normal vision : 1 colour blind$

Answer: Offspring are $X^{B} X^{b}$ (carrier girl, normal vision), $X^{b} X^{b}$ (colour-blind girl), $X^{B} Y$ (boy, normal vision) and $X^{b} Y$ (colour-blind boy) — a ratio of $1$ normal vision $: 1$ colour blind. Half the daughters and half the sons are expected to be colour blind.

Where people go wrong

Mistake: Writing gametes as genotypes — putting $T t$ in a gamete circle, or labelling the top of a Punnett square with $B b$ .

Instead: A gamete carries one allele of the gene. From $T t$ the gametes are $T$ and $t$ , written separately.

Meiosis halves the chromosome number, so a gamete receives one chromosome of each pair and therefore one allele of each gene. There is a separate mark for the gametes line in most genetic-cross questions, and two-letter gametes lose it — and they usually wreck the offspring line as well.

Mistake: Answering "75%", " $\frac{3}{4}$ " or "3 out of 4" when the question asked for the phenotypic ratio.

Instead: Give a ratio, in its lowest terms, with the phenotypes named: $3$ tall $: 1$ short.

The command word tells you the form of the answer that is being marked. A probability is a different quantity, and an examiner marking a ratio point cannot award it for something that is not a ratio. Simplify too — $2 : 2$ is not a finished answer, $1 : 1$ is.

Mistake: Treating a $3 : 1$ ratio as a promise — expecting exactly three tall plants in four seeds, or concluding that a $2 : 2$ result disproves the cross.

Instead: A phenotypic ratio is the expected proportion over a large number of offspring, because fertilisation is random.

Which gamete meets which is chance. Small numbers of offspring scatter around the expected ratio; large numbers approach it. Questions asking why real results differ from the predicted ratio are asking for exactly this sentence.

Mistake: Saying "gene" where the answer needs "allele" — "she inherited the gene for colour blindness", "the dominant gene".

Instead: A gene is a length of DNA coding for a protein; an allele is an alternative form of that gene. Dominant and recessive describe alleles.

Everyone has the gene — that is why it is the same characteristic being discussed at all. What differs between individuals is which alleles of it they carry, so an answer written in terms of genes has not actually said anything about the inheritance.

Mistake: Answering with a phenotype when the genotype was asked for, or the reverse — writing "black" for a genotype, or $B b$ for a phenotype.

Instead: Genotype is written as alleles ( $B b$ ); phenotype is written as the observable feature (black fur).

These are the two words most likely to appear in the question stem, and they are easy to read past under time pressure. Underline which one is being asked for before you write, and check your answer is in the matching form.

Mistake: Writing sex-linked genotypes without the chromosomes — $B b$ , $b b$ — or giving a male two alleles of a sex-linked gene.

Instead: Write $X^{B} X^{b}$ , $X^{b} X^{b}$ , $X^{B} Y$ , $X^{b} Y$ . The Y chromosome carries no allele of the gene, so a male has only one.

The whole explanation of sex linkage depends on the male having a single X. Notation that hides the X and Y makes it impossible to show why the characteristic is more common in males, and the sex of each offspring can no longer be read from the diagram.

Check yourself

Describe what is meant by inheritance.
Answer
The transmission of genetic information from generation to generation.
A pea plant is tall and has the alleles $T t$ . State its genotype and its phenotype, and state which of the two is its genetic make-up.
Answer
Genotype $T t$ ; phenotype tall. The genotype is the genetic make-up, described in terms of the alleles present; the phenotype is the observable feature.
Describe the difference between a homozygous and a heterozygous individual.
Answer
Homozygous means having two identical alleles of a particular gene (e.g. $B B$ or $b b$ ); heterozygous means having two different alleles of that gene (e.g. $B b$ ).
Explain why two homozygous tall pea plants breed true, but two heterozygous tall pea plants do not.
Answer
Two identical homozygous individuals ( $T T \times T T$ ) each produce only one kind of gamete, $T$ , so all the offspring are $T T$ and tall — they are pure-breeding. A heterozygous plant produces $T$ and $t$ gametes, so $T t \times T t$ can give $t t$ offspring, which are short; a heterozygous individual is not pure-breeding.
Describe what is meant by a dominant allele and by a recessive allele, and state which genotype must be present for the recessive feature to appear.
Answer
A dominant allele is expressed if it is present in the genotype. A recessive allele is only expressed when there is no dominant allele of the gene present. The recessive feature appears only in the homozygous recessive genotype, e.g. $t t$ .
Two black mice, both heterozygous ( $B b$ ), are crossed. Use a Punnett square to give the offspring genotypes and calculate the phenotypic ratio, given that black is dominant to brown.
Answer
Gametes $B$ and $b$ from each parent. Offspring genotypes $B B$ , $B b$ , $B b$ , $b b$ — that is $1 B B : 2 B b : 1 b b$ . Phenotypic ratio $3$ black $: 1$ brown.
In a pedigree diagram, two unshaded parents have a shaded child. State what this tells you about the allele responsible, and give the genotypes of the two parents using $A$ and $a$ .
Answer
The allele is recessive — if it were dominant it would be expressed in at least one parent. Both parents must be heterozygous, $A a$ , since each passed a recessive allele to the affected child but neither shows the characteristic.
Describe what codominance means, and use the ABO blood groups to give an example.
Answer
Codominance is when both alleles in a heterozygous organism contribute to the phenotype. $I^{A}$ and $I^{B}$ are codominant, so a person with the genotype $I^{A} I^{B}$ has blood group AB — both alleles are expressed. Both are dominant to $I^{o}$ , and group O is $I^{o} I^{o}$ .
Explain why red-green colour blindness is more common in males than in females.
Answer
The gene is located on the X chromosome, so it is sex-linked. A male is $X Y$ and has only one X, with no allele of this gene on the Y, so a single recessive allele ( $X^{b} Y$ ) makes him colour blind. A female has two X chromosomes and needs the recessive allele on both ( $X^{b} X^{b}$ ); with $X^{B} X^{b}$ she has normal vision and is a carrier.
State what a test cross is, and state the result that would prove an individual showing the dominant phenotype is heterozygous.
Answer
A test cross is a cross between an individual of unknown genotype and one that is homozygous recessive. If any offspring show the recessive phenotype, the unknown individual must be heterozygous, because that offspring received a recessive allele from it.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. In a species of garden plant, flower colour is controlled by a single gene. Purple flowers ( $P$ ) are dominant to white flowers ( $p$ ). A grower crosses a heterozygous purple-flowered plant with a white-flowered plant.

(a) State the genotype of the white-flowered parent plant, and explain how you know it. 2 marks

Mark scheme

B1 $p p$ / homozygous recessive
B1 white is the recessive phenotype, so it is only expressed when no dominant allele is present

(b) Construct a genetic diagram to show this cross. Include the parental genotypes, the gametes and the genotypes of the offspring, and calculate the phenotypic ratio. 4 marks

Mark scheme

B1 parental genotypes $P p$ and $p p$
B1 gametes shown as single alleles: $P$ and $p$ from one parent, $p$ and $p$ from the other
B1 offspring genotypes $P p$ , $P p$ , $p p$ , $p p$ (in a Punnett square or as a list)
B1 phenotypic ratio 1 purple : 1 white, in lowest terms

(c) The grower wants to sell pure-breeding purple-flowered plants. Explain why the heterozygous parent plant is not pure-breeding. 2 marks

Mark scheme

B1 it has two different alleles / it is $P p$ , so it produces two kinds of gamete ( $P$ and $p$ )
B1 so some offspring can inherit two recessive alleles and have white flowers / not all offspring have the same phenotype as the parent

Question 2. A pedigree diagram shows an inherited condition in a family. In generation I, a man and a woman, neither of whom has the condition, are labelled individuals 1 and 2. They have two children, shown in generation II: a daughter (individual 3) who has the condition, and a son (individual 4) who does not. Use the symbols $A$ and $a$ for the alleles.

(a) Explain how the pedigree diagram shows that the allele causing the condition is recessive. 2 marks

Mark scheme

B1 individuals 1 and 2 do not have the condition but their daughter (individual 3) does
B1 a dominant allele would be expressed in at least one parent if it were present, so the allele must be recessive; both parents must carry it without showing it

(b) State the genotype of individual 3 and the genotype of individual 1. 2 marks

Mark scheme

B1 individual 3 is $a a$ / homozygous recessive
B1 individual 1 is $A a$ / heterozygous

(c) Use a Punnett square to determine the phenotypic ratio expected among the children of individuals 1 and 2. 3 marks

Mark scheme

B1 gametes $A$ and $a$ shown for both parents on the outside of the square
B1 offspring genotypes $A A$ , $A a$ , $A a$ , $a a$
B1 3 without the condition : 1 with the condition

(d) Individuals 1 and 2 have four children in total and only one has the condition. Suggest why the numbers in a real family often do not match the ratio predicted by a Punnett square. 2 marks

Mark scheme

B1 fertilisation is random — which gamete fuses with which is a matter of chance
B1 the ratio is only expected over a large number of offspring; the numbers in one small family are too few

Question 3. The ABO blood group system in humans is controlled by a gene with three alleles. A man of blood group A, whose mother was blood group O, has children with a woman of blood group AB.

(a) Describe what is meant by codominance, using the ABO blood groups as your example. 2 marks

Mark scheme

B1 both alleles in a heterozygous organism contribute to the phenotype / both are expressed, neither is masked
B1 a person with the genotype $I^{A} I^{B}$ has blood group AB; accept the statement that $I^{A}$ and $I^{B}$ are codominant and both are dominant to $I^{o}$

(b) Determine the genotype of the man, and explain your reasoning. 2 marks

Mark scheme

B1 $I^{A} I^{o}$
B1 his mother was group O ( $I^{o} I^{o}$ ) so she could only pass on $I^{o}$ ; he is group A so his other allele is $I^{A}$

(c) Predict the blood groups of the children of this couple, and calculate the phenotypic ratio. 4 marks

Mark scheme

B1 parental genotypes $I^{A} I^{o}$ and $I^{A} I^{B}$
B1 gametes as single alleles: $I^{A}$ and $I^{o}$ ; $I^{A}$ and $I^{B}$
B1 offspring genotypes $I^{A} I^{A}$ , $I^{A} I^{B}$ , $I^{A} I^{o}$ , $I^{B} I^{o}$
B1 phenotypes 2 group A : 1 group AB : 1 group B

Question 4. Red-green colour blindness in humans is caused by a recessive allele of a gene carried on the X chromosome. A woman with normal vision who is a carrier has children with a man who has normal vision.

(a) Explain why red-green colour blindness is more common in males than in females. 3 marks

Mark scheme

B1 the gene is located on the X chromosome (it is sex-linked)
B1 a male is $X Y$ — he has only one X chromosome, and the Y carries no allele of this gene
B1 so one recessive allele is enough for a male to be colour blind, whereas a female must inherit the recessive allele on both X chromosomes

(b) Construct a genetic diagram for this couple, showing the parental genotypes, the gametes and the genotypes of the possible children. 3 marks

Mark scheme

B1 parental genotypes $X^{B} X^{b}$ and $X^{B} Y$ , using X and Y in the notation
B1 gametes $X^{B}$ and $X^{b}$ from the mother; $X^{B}$ and $Y$ from the father, each shown singly
B1 offspring genotypes $X^{B} X^{B}$ , $X^{B} X^{b}$ , $X^{B} Y$ , $X^{b} Y$

(c) State the expected proportion of the sons of this couple who are colour blind, and state the phenotype of all of their daughters. 2 marks

Mark scheme

B1 half (1 in 2) of the sons are colour blind
B1 all the daughters have normal vision; accept the addition that half of them are carriers

How this is examined

Constructing a complete monohybrid cross from two described parents: a four- or five-mark Construct or Use — "construct a genetic diagram to show this cross", or a part-drawn Punnett square to complete, ending in "calculate the phenotypic ratio" The marks are awarded line by line and examiners credit them separately: (1) both parental genotypes correct; (2) gametes shown as single alleles, one per circle; (3) all four offspring genotypes; (4) offspring phenotypes matched to those genotypes; (5) the phenotypic ratio in lowest terms with the phenotypes named, e.g. $3$ tall $: 1$ short. Because error carried forward applies, a wrong parental genotype still lets lines 3–5 score if you work correctly from it — so never leave the diagram unfinished.
A sex-linked cross, and why the characteristic is more common in males: a three- to five-mark question on red-green colour blindness or another X-linked condition — usually one Explain part on the sex difference, then a genetic diagram whose answer must be given by sex Notation first: genotypes written as $X^{B} X^{b}$ , $X^{b} X^{b}$ , $X^{B} Y$ , $X^{b} Y$ — a scheme will not credit $B b$ here. For the explanation: "the gene is on the X chromosome", "a male has only one X and the Y carries no allele of this gene", "so one recessive allele is enough for him to be colour blind, whereas a female needs two". For the cross: parental genotypes, gametes as single sex chromosomes, all four offspring, then the outcome stated by sex — "all daughters have normal vision, half are carriers; half the sons are colour blind".
ABO blood-group crosses, and what codominance lets you deduce: a four-mark cross using blood groups, often followed by "explain why one of their children could not be group O" or a disputed-parentage scenario Correct genotypes using the superscript notation $I^{A}$ , $I^{B}$ , $I^{o}$ — with group AB as $I^{A} I^{B}$ and group O as $I^{o} I^{o}$ , the two that are fixed. Then the relationships stated: $I^{A}$ and $I^{B}$ are codominant, both dominant to $I^{o}$ . Then the offspring genotypes converted to named blood groups, and the ratio. For the reasoning parts, argue from which allele a parent can donate: a group O parent can only give $I^{o}$ .
Interpreting a pedigree diagram to decide dominant or recessive, then assigning genotypes: a three- or four-mark Explain attached to a family tree, typically "explain how the diagram shows the allele is recessive" and "give the genotype of individual 4" The reasoning, not just the word: "individuals 1 and 2 do not show the characteristic but their child does, so the allele must be recessive — a dominant allele would be expressed in a parent." Then genotypes with justification: anyone showing the recessive characteristic is homozygous recessive; an unaffected parent of an affected child must be heterozygous. Where the diagram cannot decide, write both possibilities ( $N N$ or $N n$ ) rather than guessing.
Using a test cross to identify an unknown genotype: a three- or four-mark Explain — a breeder or grower with an individual showing the dominant phenotype who needs to know whether it is homozygous Four separable points: cross it with an individual that is homozygous recessive (showing the recessive phenotype); state why — that parent can only donate a recessive allele; if any offspring show the recessive phenotype the unknown parent is heterozygous, with a $1 : 1$ ratio expected; if all offspring show the dominant phenotype it is homozygous dominant. Saying "cross it with another one" without naming the homozygous recessive partner scores nothing.
The definitions: genotype, phenotype, homozygous, heterozygous, dominant, recessive, pure-breeding: one- and two-mark State or Describe parts opening an inheritance question, or a table of terms to match to meanings The syllabus wording, which is tight: genotype is "the genetic make-up, in terms of the alleles present"; phenotype is "the observable features"; homozygous is "two identical alleles of a particular gene"; heterozygous is "two different alleles"; a dominant allele "is expressed if it is present in the genotype"; a recessive allele "is only expressed when there is no dominant allele of the gene present". The word allele must appear — answers written in terms of genes are not credited.

Words to know

Inheritance: The transmission of genetic information from generation to generation.
Genotype: The genetic make-up of an organism, described in terms of the alleles present, e.g. $T t$ .
Phenotype: The observable features of an organism, e.g. tall.
Homozygous: Having two identical alleles of a particular gene, e.g. $T T$ or $t t$ .
Heterozygous: Having two different alleles of a particular gene, e.g. $T t$ .
Pure-breeding: Producing offspring that always show the same feature; two identical homozygous individuals bred together are pure-breeding, and a heterozygous individual is not.
Dominant allele: An allele that is expressed if it is present in the genotype.
Recessive allele: An allele that is only expressed when there is no dominant allele of the gene present in the genotype.
Punnett square: A grid with the gametes of one parent along the top and those of the other down the side, used to show all the possible offspring genotypes.
Test cross: A cross between an individual showing the dominant phenotype, whose genotype is unknown, and a homozygous recessive individual, used to identify that genotype.
Codominance: A situation in which both alleles in a heterozygous organism contribute to the phenotype.
Sex-linked characteristic: A feature whose gene is located on a sex chromosome, which makes the characteristic more common in one sex than in the other.
Carrier: An individual who is heterozygous for a recessive allele, so does not show the characteristic but can pass the allele on, e.g. $X^{B} X^{b}$ .

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