Inheritance: WJEC A level Biology revision notes

Lesson 1 of 19

What are genes and alleles?

These notes cover topic 4.3, Inheritance, in WJEC A level Biology Unit 4: , Inheritance and Options. Each lesson follows the WJEC specification, and the exam tips come from BioWelsh analysis of WJEC past papers.

A is a length of DNA that codes for one polypeptide, and it sits at a position on a chromosome called its . Many exist in slightly different versions. Each version is an . of the same differ a little in their base sequence, so they code for slightly different polypeptides, and that is why individuals differ.

Body cells are : they carry two sets of chromosomes, one set from each parent. Chromosomes therefore come in , and every on them is present twice, once on each chromosome of the pair. The two copies can be the same or different .

The diagram below shows one pair and three of the it carries.

Two tall rounded bars side by side, blue on the left labelled paternal and pink on the right labelled maternal, each with a thick black band part way down. Both bars carry three coloured bands at the same heights. The top gold band reads A on the blue bar and a on the pink bar. The middle green band reads B on both. The bottom purple band reads d on both. Labels explain that each band is a locus, that A and a make the individual heterozygous, that BB is homozygous dominant and dd homozygous recessive, and that the black band is the centromere. — A pair of homologous chromosomes carrying three genes. Original BioWelsh diagram.

Both chromosomes carry the same in the same order, so each appears twice.
At the top the differ, A and a. This individual is for that .
At the middle both are B, so the individual is for that .
At the bottom both are d, so the individual is for that .

The genetic terms you need to define
Term	Meaning
	A length of DNA at a particular that codes for a polypeptide
	The position of a on a chromosome
	One of the different forms of the same
	The an organism has, for example Bb
	The characteristics that are expressed, the result of the and the environment
	Having two identical of a
	Having two different of a
	An that is expressed in the whenever it is present
	An that is only expressed in the when no is present, that is, when
	Any chromosome that is not a

The photograph below shows a group of labrador retrievers. Coat colour in these dogs is controlled partly by one with two : B for black, which is , and b for chocolate, which is .

Nine dogs sitting in a row on dry grass in front of bare trees and a wire fence. The dog at each end has a glossy black coat. The seven dogs between them all have brown coats, in shades from dark chocolate to a lighter reddish brown. — Labrador retrievers with black and chocolate coats. Photo: Blaine Hansel, CC BY 2.0. Resized.

The two black dogs at each end could be BB or Bb. Their does not tell you which.
The chocolate dogs must be bb, because the only appears when no B is present.
This is why a always gives away the , while a does not.

Lesson 2 of 19

How do you draw a genetic diagram for a monohybrid cross?

A follows the inheritance of one . The rules were first worked out by Gregor Mendel using the garden pea, Pisum sativum. He chose characteristics controlled by a single with two clear-cut forms, such as round or wrinkled seeds, and he started with plants, which always produce offspring like themselves because they are .

Choose one letter for the : a capital for the and the same letter in lower case for the . Pick a letter whose capital and lower case look different.
Write the parental , then the parental .
Write the gametes, one each, and ring them. Meiosis puts one of each pair into each .
Show every possible fusion of gametes in a .
Write the offspring and , then the ratio.

The genetic diagram below follows seed shape in the garden pea, where the for round seeds, R, is to the for wrinkled seeds, r.

Genetic diagram in two halves separated by a dashed line. On the left, rows labelled parental phenotypes, parental genotypes, gametes, F1 genotype and F1 phenotype show round seed RR crossed with wrinkled seed rr, circled gametes R and r, arrows joining to Rr, and the words all round. On the right, Rr crossed with Rr gives circled gametes R, r, R and r, then a two by two Punnett square with RR round, Rr round, Rr round and rr wrinkled. The three round squares are pale green and the wrinkled square is pale yellow. The ratio line reads 3 round to 1 wrinkled. — A monohybrid cross: seed shape in the garden pea. Original BioWelsh diagram.

The parents are , RR and rr, so each makes only one type of .
Every plant is Rr and has round seeds, because R is .
Each plant makes two types of , R and r, in equal numbers.
Random fusion gives RR, Rr, Rr and rr in the , so the ratio is 3 round to 1 wrinkled.

Mendel explained his results with what is now called the : the two of a separate from each other when gametes form, so each receives only one . We now know this happens because separate at anaphase I of meiosis.

A ratio is a probability. A 3 : 1 ratio that each offspring has a 3 in 4 chance of being round and a 1 in 4 chance of being wrinkled. Real crosses rarely give exactly 3 : 1, because which gametes fuse is a matter of chance, and the smaller the number of offspring, the further the result is likely to be from the ratio.

Lesson 3 of 19

How does a test cross reveal an unknown genotype?

An organism with the could be or , and you cannot tell by looking. A , also called a , finds out. The organism is crossed with one that is for the , because a parent can only pass on , so the offspring show whatever the other parent passed on.

The diagram below shows the two possible results when a black labrador is crossed with a chocolate one.

Two genetic diagrams side by side. The left one is headed if the black dog is BB: black BB crossed with chocolate bb gives gametes B and b and a single Punnett cell Bb black, with the result all offspring black. The right one is headed if the black dog is Bb: black Bb crossed with chocolate bb gives gametes B, b and b and a Punnett square with Bb black and bb chocolate, with the result 1 black to 1 chocolate. A note underneath says the chocolate parent only passes on b. — A test cross to find out whether a black labrador is homozygous or heterozygous. Original BioWelsh diagram.

If the black dog is BB, every carries B, so every puppy is Bb and black.
If the black dog is Bb, half its gametes carry b, so about half the puppies are bb and chocolate.
A single chocolate puppy is enough to show that the black parent was .

The test only works with enough offspring. If a parent has three puppies, there is a 1 in 8 chance, (1/2)³, that all three are black by chance, which would wrongly suggest it is .

Lesson 4 of 19

What happens when neither allele is dominant?

Not every has a and a . When two are , both are expressed in the , so the has a third that shows features of both .

Because neither is , a lower-case letter would be misleading. Instead the gets one capital letter and each a superscript, for example C with superscripts R and W for the red and white coat in shorthorn cattle.

The genetic diagram below follows coat colour in shorthorn cattle.

Genetic diagram in two halves. On the left, red cattle with genotype C superscript R C superscript R are crossed with white cattle C superscript W C superscript W, giving circled gametes of each type and F1 calves that are all C superscript R C superscript W and roan. A note explains that both alleles are expressed, giving red and white hairs mixed. On the right, two roan cattle are crossed in a two by two Punnett square giving one red square, two roan squares and one white square, and the ratio 1 red to 2 roan to 1 white. — Codominance: coat colour in shorthorn cattle. Original BioWelsh diagram.

A red bull crossed with a white cow gives calves that are all .
In the both are expressed: red hairs and white hairs grow mixed together, giving a roan coat.
Crossing two roan cattle gives red, roan and white calves in the ratio 1 : 2 : 1.
With , every has its own , so the ratio is the same as the ratio.

Human ABO blood groups: the has three . The for group A and group B are , so a person with both has group AB, and both are to the for group O.
Sickle cell haemoglobin: a person with one normal and one sickle cell makes both kinds of haemoglobin.
Some are when . The offspring die, often as embryos or seedlings, so one class is missing and a 1 : 2 : 1 ratio becomes 2 : 1.

Lesson 5 of 19

How does meiosis explain independent assortment?

When two are on different chromosomes, they are inherited independently of each other. The reason is the way line up at metaphase I of meiosis. Each pair lines up on the equator independently of every other pair, so which side the maternal and paternal chromosomes face is a matter of chance.

The diagram below follows two , A and B, on two different pairs of chromosomes in a cell that is AaBb.

Two large oval cells side by side, each with a dashed vertical line down the middle. Each cell has a long pair of chromosomes above a short pair, one pink and one blue in each pair, lined up either side of the dashed line. The long pair carries A on the pink and a on the blue. In the left cell the short pair has B on the pink and b on the blue; in the right cell it is the other way round. Arrows lead from each cell to two small round gametes, labelled AB and ab under the left cell and Ab and aB under the right cell. A note says pink came from the mother and blue from the father and that the four gamete types form in equal numbers. — Independent assortment of two pairs of chromosomes at metaphase I. Original BioWelsh diagram.

In arrangement 1 the chromosomes carrying A and B face the same pole, so the gametes are AB and ab.
In arrangement 2 the short pair is the other way round, so the gametes are Ab and aB.
Both arrangements are equally likely, so across many cells an AaBb individual makes AB, Ab, aB and ab gametes in equal numbers.

This is Mendel's second law, the law of : each of one can combine with either of another . It only holds for on different chromosomes, which are called unlinked .

Lesson 6 of 19

How do you work out a dihybrid cross?

A follows two at once. If the are unlinked, each carries one of each , and all four combinations are equally likely. Write each with two letters, one for each , and use a four by four when both parents are for both .

The diagram below follows stem colour and leaf shape in tomato. Purple stem, A, is to green stem, a, and cut leaf, C, is to potato leaf, c.

Text at the top gives the parents purple stem cut leaf AACC and green stem potato leaf aacc, the F1 AaCc, and the four circled gametes AC, Ac, aC and ac. Below is a four by four Punnett square with the gametes along the top and down the side, and a genotype and phenotype written in each of the 16 squares. Nine squares are pale purple for purple stem and cut leaf, three are peach for purple stem and potato leaf, three are pale green for green stem and cut leaf, and one, aacc, is pale yellow for green stem and potato leaf. A key on the right matches the colours and the ratio 9 to 3 to 3 to 1 is written underneath. — A dihybrid cross: stem colour and leaf shape in tomato. Original BioWelsh diagram.

The parents are AACC and aacc, so every plant is AaCc with a purple stem and cut leaves.
Each plant makes four types of : AC, Ac, aC and ac.
The 16 squares hold every combination. Nine have at least one A and one C, three have A but cc, three have aa but C, and one is aacc.
The ratio is therefore 9 : 3 : 3 : 1.

You can reach the same answer without the square by multiplying probabilities. For each alone the ratio is 3 : 1, so the chance of green stem and potato leaf together is 1/4 × 1/4 = 1/16, and the chance of purple stem and cut leaf is 3/4 × 3/4 = 9/16. Multiplying works because the are inherited independently.

Ratios to recognise in a dihybrid cross
Cross	Offspring phenotype ratio
AaBb × AaBb (unlinked)	9 : 3 : 3 : 1
AaBb × aabb, a (unlinked)	1 : 1 : 1 : 1
AaBb × AaBb with the completely , A with B	3 : 1
AaBb × aabb with the completely , A with B	1 : 1

Sometimes two affect the same characteristic. If a flower is only purple when it has at least one of each of two , the of AaBb × AaBb is 9 purple : 7 white, because every other class is white. Ratios made of sixteenths are the clue that two unlinked are involved.

Lesson 7 of 19

What happens when two genes are linked?

on the same chromosome are . They cannot assort independently, because they are carried on the same piece of DNA and move to the same pole at meiosis. the are on the same ; , covered later, a is on a .

The diagram below shows a , DdEe, whose chromosomes carry D with E and d with e.

Two halves separated by a dashed line. Each half shows a pink and a blue chromosome side by side, the pink carrying D above E and the blue carrying d above e. On the left no chiasma forms, and arrows lead to two round gametes containing one chromosome each, labelled DE and de. On the right the two chromosomes cross between the two genes at a point labelled chiasma, and arrows lead to four gametes: DE and de outlined in purple, and De and dE, each with a chromosome half pink and half blue, outlined in red. Text explains that parental gametes are most common and recombinant ones are rarer. — Gametes from linked genes, with and without crossing over. Original BioWelsh diagram.

If no forms between the two , only two types of are made, DE and de, the same combinations as the parental chromosomes.
If a forms between the , swaps the ends of two chromatids, so some gametes are De and dE. These are gametes.
between two particular only happens in some cells, so parental gametes are always more common than ones.

shows up in the results of a cross. Instead of 9 : 3 : 3 : 1 or 1 : 1 : 1 : 1, the two parental classes are much more common than expected and the two classes much rarer. If the are so close that between them never happens, a gives a 3 : 1 ratio, as if only one were involved.

Lesson 8 of 19

How do you use the chi-squared test?

Genetic ratios are predictions, and real counts never match them exactly. The () test tells you whether the difference between the observed numbers, O, and the expected numbers, E, is small enough to be put down to chance, or so large that something other than chance must be involved.

The test starts with a : there is no difference between the observed and expected numbers, and any difference is due to chance. In genetics this amounts to saying the results fit the expected Mendelian ratio.

The data are counts in categories, such as numbers of each . Never use percentages, or measurements.
There is an expected number for each category, usually from a ratio.
The sample is large, and no expected number is very small.

The are the number of categories minus one. Two give 1 degree of freedom, and the four of a give 3.

Critical values of χ²
Degrees of freedom	p = 0.10	p = 0.05	p = 0.01
1	2.71	3.84	6.63
2	4.61	5.99	9.21
3	6.25	7.82	11.34
4	7.78	9.49	13.28

Biologists use the 5 % , p = 0.05.
If is less than the , the probability that the difference is due to chance is more than 5 %. Accept the : the difference is not .
If is greater than the , the probability is less than 5 %. Reject the : the difference is and is not due to chance alone. Look for another explanation, such as or a .

Lesson 9 of 19

How is sex determined in humans?

Human body cells contain 46 chromosomes in 23 pairs. When they are photographed and arranged in pairs in order of size, the result is a . Twenty-two of the pairs are , which are the same in both sexes. The 23rd pair is the : two X chromosomes in a female, and one X and one Y in a male.

The photograph below shows the chromosomes of a human male, stained to show bands and arranged in pairs.

Grey chromosomes on a white background arranged in four rows of pairs, largest at the top and smallest at the bottom. Each chromosome is a slightly curved rod crossed by alternating dark and pale bands, and the two members of each pair have the same length and banding pattern. The last pair at the bottom right is unequal, one medium-sized and one much smaller. — The chromosomes of a human male, stained to show bands and arranged in pairs. Photo: National Human Genome Research Institute, Public domain.

Each pair of contains two chromosomes of the same length with the same banding pattern.
The last pair does not match: the X chromosome is much larger than the Y.
Staining bands let geneticists spot a missing or extra chromosome, or a piece that has moved.

A female makes only one type of egg, carrying an X, so she is called the sex. A male makes two types of in equal numbers, half carrying X and half carrying Y, so he is the . The sex of a child therefore depends on the : XX × XY gives 1 female : 1 male.

The diagram below compares the X and Y chromosomes.

A tall pink chromosome labelled X beside a much shorter blue chromosome labelled Y. Both have gold rounded tips at the top and bottom, labelled as homologous regions where they can pair. Each has a black band for the centromere. The X carries a purple band above its centromere labelled DMD gene for dystrophin and a red band near its lower end labelled F8 gene for clotting factor VIII. A label on the long pink section explains that most of the X has no partner region on the Y, so a male has only one copy of these genes. The Y carries a green band just below its top tip labelled SRY gene, which switches on male development, and its blue middle section is labelled male-specific region. — The human X and Y chromosomes compared. Original BioWelsh diagram.

The X and Y only match at their tips. These small regions let them pair at meiosis.
The Y carries the , which switches on the development of testes and so male characteristics.
Most of the X has no partner region on the Y. A male therefore has only one copy of every in this region, including the for and for clotting .

Lesson 10 of 19

How are sex-linked conditions such as haemophilia and DMD inherited?

A is one carried on a , and in practice this a on the part of the X with no partner on the Y. A female has two copies of such a ; a male has only one. A male therefore expresses whichever is on his single X, even if it is , because there is no second to mask it.

of are written as superscripts on the X, for example Xᴴ for the normal clotting and Xʰ for the . The Y is written on its own, with no , because it does not carry the .

The five genotypes for an X-linked recessive condition such as haemophilia
Genotype	Phenotype
XᴴXᴴ	Unaffected female
XᴴXʰ	female: unaffected, but carries the
XʰXʰ	Affected female (rare)
XᴴY	Unaffected male
XʰY	Affected male

is caused by a of the for a blood clotting protein, , so blood clots very slowly. , , is caused by a of the for , a protein that holds membranes stable; without it, are damaged and gradually wasted, starting in early childhood. Both are on the X chromosome.

The genetic diagram below shows the children of a mother and an unaffected father.

Genetic diagram. A carrier female, X superscript H X superscript h, is crossed with an unaffected male, X superscript H Y. Circled gametes are X superscript H and X superscript h from the mother and X superscript H and Y from the father. A two by two Punnett square has the father's gametes along the top and the mother's down the side, giving an unaffected female, an unaffected male, a pink carrier female and a darker pink male with haemophilia. A line at the bottom reads daughters 1 unaffected to 1 carrier, sons 1 unaffected to 1 with haemophilia. — The inheritance of haemophilia from a carrier mother. Original BioWelsh diagram.

The mother makes two types of egg, one carrying the normal and one carrying the .
Every daughter receives her father's X, which carries the normal , so no daughter has . Half are .
Every son receives his father's Y and one of his mother's X chromosomes, so each son has a 1 in 2 chance of .

A father passes his X to every daughter and his Y to every son, so he never passes an to a son.
An affected son always inherits the from his mother.
Every daughter of an affected father is at least a .
Affected females are rare: a girl needs an affected father and a or affected mother. Boys with rarely live long enough to have children, so girls with are extremely rare.

Lesson 11 of 19

How do you analyse a pedigree diagram?

A is a family tree that shows who has an inherited condition. Squares are males and circles are females, shaded symbols are affected, a horizontal line joins parents, and their children hang from a vertical line below. Generations are numbered with Roman numerals and individuals with ordinary numbers.

The diagram below shows the three patterns you need to recognise.

Three small family trees side by side, each with two parents joined by a horizontal line and three children below. Squares are males and circles are females. The first, autosomal recessive, has two unshaded parents and a shaded daughter. The second, autosomal dominant, has two shaded parents with a shaded son, an unshaded daughter and a shaded daughter. The third, X-linked recessive, has an unshaded father and a half-shaded carrier mother, with a shaded son, an unshaded daughter and an unshaded son. A sentence under each explains the rule. — Three patterns of inheritance to look for in a pedigree. Original BioWelsh diagram.

: two unaffected parents have an affected child, so both parents must be .
: two affected parents have an unaffected child, so both parents must be , and the condition must be .
: mostly males are affected, and affected sons have unaffected mothers who are .

Is it ? Look for unaffected parents with an affected child.
Is it ? Look for affected parents with an unaffected child.
Could it be ? An affected daughter must have an affected father. If an affected girl has an unaffected father, the is .
Could it be ? Every daughter of an affected father would be affected. An unaffected daughter of an affected father rules it out.
Is it ? Passing from father to son proves it, because a son gets his father's Y, not his X.

The below shows three generations of a family with .

Family tree with generations I, II and III and individuals numbered 1 to 13. In generation I an unshaded male 1 and unshaded female 2 have four children: shaded male 4, unshaded female 5, unshaded male 6 and unshaded female 7. Male 4 has a partner, unshaded female 3, and their children are unshaded female 9 and unshaded male 10. Female 7 has a partner, unshaded male 8, and their children are shaded male 11, unshaded female 12 and unshaded male 13. A key shows unshaded and shaded squares and circles for unaffected and affected males and females. — A family with haemophilia over three generations. Original BioWelsh diagram.

Only males are affected, and none of the affected men has an affected father, which fits an .
Individual 2 is unaffected but has an affected son, 4, so she must be a .
Individual 9 is the daughter of an affected man, so she must have received his X and be a .
Individual 7 has an affected son, 11, so she is a too, even though her father, 1, is unaffected.

Lesson 12 of 19

What is a mutation and what increases the rate?

A is a change in the amount, arrangement or structure of the DNA of an organism. happen spontaneously and at random, and they are rare. They are the original source of all new , so they are the raw material of genetic .

A , or , changes the base sequence of one .
A changes the structure of a chromosome or the number of chromosomes, and so affects many .

Only in cells that make gametes can be passed to offspring. A in a body cell, a , affects only that cell and the cells that come from it by mitosis. Many have no effect, some are harmful, and a very few give an advantage.

A is anything that increases the rate of . Most happen when DNA is copied before cell division, so do most damage in cells that divide often.

Mutagens and how they damage DNA
Mutagen	Examples	Effect on DNA
	X-rays, gamma rays	Breaks DNA strands, which may rejoin wrongly
Ultraviolet light	Sunlight, sunbeds	Joins neighbouring bases on one strand, so DNA polymerase may insert the wrong base
Chemicals	Tar in tobacco smoke, benzene, mustard gas	Slot between base pairs or change bases, so they are copied wrongly

Lesson 13 of 19

How do gene mutations change proteins, as in sickle cell anaemia?

The base sequence of a is read in triplets, and each triplet codes for one amino acid. A changes one or a few bases, which can change the amino acid sequence of the polypeptide, then its three-dimensional shape, and then how well it works.

The diagram below shows the same short before and after four kinds of .

Five rows, each showing six DNA triplets in boxes with an amino acid in a rounded label below each. The original row reads ATG AAA CCG TGG GAT TCA, coding Met Lys Pro Trp Asp Ser. In the substitution row CCG becomes CAG, with the A in red, and Pro becomes Gln in pink. In the silent substitution row CCG becomes CCA and the amino acid stays Pro. In the deletion row a C is lost, the triplets read ATG AAA CGT GGG ATT CAG, and the last four amino acids, Arg Gly Ile Gln, are all pink. In the addition row a red T is inserted, the triplets read ATG AAA TCC GTG GGA TTC, and Ser Val Gly Phe are pink. Notes on the right explain each effect. — How point mutations change the amino acid sequence. Original BioWelsh diagram.

A replaces one base with another. At most one triplet and one amino acid change.
A can be silent. Because the genetic code is , CCG and CCA both code for proline, so nothing changes.
A deletion removes a base and an addition inserts one. Every triplet after the is read differently, called a , so the rest of the polypeptide is usually changed completely.

How much a matters depends on the new amino acid and where it is. A similar amino acid away from the may make little difference. An amino acid with different properties, or a change at the of an , can change the shape of the protein and stop it working.

is caused by a single in the for the beta chain of haemoglobin. The diagram below follows the change from DNA to red blood cell.

Two columns of boxes, headed normal allele and sickle cell allele. Row by row they read: DNA coding strand triplet G A G against G T G; template strand C T C against C A C; mRNA codon G A G against G U G, with the changed letter in red each time; amino acid glutamic acid against valine; charged and hydrophilic against non-polar and hydrophobic; stays dissolved against molecules stick into long fibres at low oxygen. At the bottom the left column ends in a round red disc labelled biconcave disc, flexible, and the right in a curved crescent labelled sickle-shaped, rigid, blocks capillaries. — Sickle cell anaemia: one base change, from DNA to red blood cell. Original BioWelsh diagram.

In the sixth triplet, A is replaced by T on the coding strand, so the mRNA codon becomes GUG instead of GAG.
The sixth amino acid of the beta chain becomes valine instead of glutamic acid.
Valine is non-polar. At low oxygen concentrations the haemoglobin molecules stick together into long fibres.
The red blood cells become rigid and sickle-shaped. They block capillaries, carry less oxygen and are destroyed quickly, causing anaemia.

The normal is written Hbᴬ and the sickle cell Hbˢ. They are : a makes both normal and sickle haemoglobin and has sickle cell trait, usually with mild symptoms or none, while a person for the sickle has .

The micrograph below shows a stained blood smear from a person with .

Light micrograph of many pink red blood cells on a pale background. Many cells are long, thin and curved with pointed ends. Others are round or oval, some with a paler centre, and a few are irregular. At the top right is a larger cell with a pale pink edge and a dark purple nucleus divided into lobes. — Stained blood smear from a person with sickle cell anaemia. Photo: Ed Uthman, CC BY 2.0. Resized.

Many red blood cells are long and curved, with pointed ends: these are sickle cells.
Others are round, but many of these are also irregular in shape.
The large cell at the top right with a dark purple, lobed is a white blood cell.

Lesson 14 of 19

How do chromosome mutations happen, as in Down's syndrome?

change the structure of chromosomes or the number of them. Structural changes happen when chromosomes break and rejoin wrongly, for example at : part of a chromosome can be lost, turned round, or attached to another chromosome.

Changes in number usually come from : at anaphase I a pair fails to separate, or at anaphase II a pair of sister chromatids fails to separate. One then gets an extra copy of that chromosome and another gets none. A cell with one chromosome too many or too few is . Much more rarely a whole extra set is gained, called , which is far more common in plants than in animals.

The diagram below shows of chromosome 21 at meiosis I.

Flow diagram from left to right. A large cell holds one pink and one blue chromosome 21, each of two chromatids. Arrows labelled meiosis I, the pair fails to separate, lead to two cells: one holding both chromosomes and one with no chromosome 21. Arrows labelled meiosis II lead to four gametes: two containing a single pink and a single blue chromatid, labelled n plus 1, and two empty ones labelled n minus 1. An arrow labelled fertilisation joins an n plus 1 gamete and a normal sperm with one green chromosome to a large red-edged zygote containing three chromosomes, labelled trisomy 21, 47 chromosomes. Notes explain the colours and that non-disjunction at anaphase II can also cause this. — Non-disjunction of chromosome 21 in meiosis I, leading to trisomy 21. Original BioWelsh diagram.

The pair of chromosome 21s fails to separate at anaphase I, so one cell gets both and the other gets neither.
After meiosis II, two gametes have two copies of chromosome 21 and two have none.
If a with two copies is fertilised by a normal , the has three copies of chromosome 21 and 47 chromosomes in all. This is 21.

21 causes . Every cell of the child has an extra chromosome 21, and the extra copies of its alter development, with effects that include learning difficulties, characteristic facial features and an increased risk of heart defects. In a small minority of cases the extra material is a piece of chromosome 21 attached to another chromosome, a translocation, and then the total number stays at 46.

The diagram below shows the chromosomes of a person with .

Black and white diagram of human chromosomes drawn as banded bars and numbered 1 to 22, in three rows, with a dotted line through the centromeres. Every numbered position has two chromosomes except 21, which has three. At the end of the bottom row are the sex chromosomes, shown as an X and a smaller Y, or as two X chromosomes. — Karyotype diagram of a person with Down's syndrome. Photo: National Human Genome Research Institute, Public domain.

Every pair has two chromosomes except chromosome 21, which has three.
The can be XY or XX: 21 happens in both sexes.

Most extra chromosome 21s come from the egg. The graph below shows how the chance of a baby with changes with the age of the mother.

Line graph. The horizontal axis is the age of the mother at delivery from 20 to 46 years and the vertical axis is babies with Down's syndrome per 1000 births from 0 to 36. Six crosses joined by straight lines sit at about 0.7 at 20, 0.7 at 25, 1.1 at 30, 2.9 at 35, 10 at 40 and 33 at 45, so the line is almost flat to 30 and then rises steeply. — The chance of a baby with Down's syndrome at different ages of the mother, using rounded published screening figures. Original BioWelsh graph.

The chance is low and almost level until about 30, at roughly one baby in a thousand.
It then rises steeply, to about 1 in 100 at 40 and about 1 in 30 at 45.
Oocytes begin meiosis before a woman is born and stay paused in prophase I until , so an older oocyte has been waiting longer and becomes more likely. (For the timing of these divisions, see the human reproduction lesson on oogenesis.)

Lesson 15 of 19

How do carcinogens, oncogenes and tumour suppressor genes cause cancer?

The cell cycle is normally controlled by two groups of working against each other. code for proteins that stimulate cell division when the body needs new cells. code for proteins that stop the cell cycle, give time for damaged DNA to be repaired, and trigger cell death if it cannot be. Cancer begins when this breaks down and a cell divides again and again by mitosis, forming a .

The diagram below shows how a change to either kind of can lead to cancer.

Flow diagram. A green box, proto-oncogene, codes for a protein that stimulates cell division when needed, has an arrow labelled mutation to a red box, oncogene, switched on permanently. Below, a blue box, tumour suppressor gene, codes for a protein that stops the cell cycle, for example p53, has an arrow labelled mutation or methylation to a red box, inactivated, switched off. Both red boxes point to a box labelled uncontrolled mitosis, which points to a box labelled tumour. Two sentences below name carcinogens and say cancer usually needs several mutations in one cell. — How changes to proto-oncogenes and tumour suppressor genes can lead to cancer. Original BioWelsh diagram.

A can switch a on permanently. It is then an , and it stimulates division all the time.
A , or heavy of its , can switch a off, so damaged cells are no longer stopped from dividing.
Either change leads to uncontrolled mitosis, and the cells form a .

The protein made by the tumour suppressor gene TP53
Normal p53 protein	Mutated p53 protein
Stops the cell cycle before DNA is copied if the DNA is damaged	The cell copies its damaged DNA and divides
Gives time for the damage to be repaired	Damage is passed on to the daughter cells
Triggers , programmed cell death, if repair fails	Damaged cells survive and keep dividing

A is anything that causes cancer. Many are , because they cause the in and . Tobacco smoke contains many : chemicals in tar enter the of cells lining the airways and slot between base pairs, so the DNA is copied wrongly. Ultraviolet light, X-rays and gamma rays are too.

A in a that is inherited is present in every cell of the body from the start, so fewer further are needed in any one cell. This is why some families have a much higher risk of certain cancers.

Lesson 16 of 19

What is epigenetics, and how do methylation and histones control genes?

Every cell in your body has the same DNA, yet a skin cell and a cell use very different . Part of the explanation is : the of expression by factors other than changes to the DNA base sequence. Chemical tags on the DNA, or on the proteins it is wound round, decide whether a can be transcribed.

The first tag is on the DNA itself. The diagram below shows .

On the left, two six-sided ring structures labelled cytosine and 5-methylcytosine. Each has an NH2 group at the top, a double-bonded O at the lower left, nitrogen atoms in the ring and a bond to sugar at the bottom. The second has an extra CH3 group, in red, on the upper right carbon. On the right, two double DNA strands each have a gold box marking the promoter at the left end and the word gene further along. In the upper strand, headed promoter not methylated, an oval labelled RNA polymerase sits on the promoter with an arrow labelled mRNA made. In the lower strand, headed promoter heavily methylated, four red lollipop shapes stand on the promoter, labelled CH3 groups, and the text says binding is blocked and the gene is switched off. — DNA methylation of cytosine and its effect on a gene's promoter. Original BioWelsh diagram.

A methyl group, CH₃, is added to cytosine, giving 5-methylcytosine. The base sequence is unchanged: it still pairs with guanine.
Methyl groups often build up in the , the region at the start of a where RNA polymerase and transcription factors bind.
A heavily blocks their binding, so the is not transcribed and is switched off.

The second kind of tag is on the , the proteins that DNA is wound round to form . The diagram below shows how changes packing.

Two halves separated by a dashed line. On the left, headed tightly packed, gene switched off, five blue circles sit close together in a row, each with a purple strand of DNA looped round it and joined to the next. On the right, headed loosely packed, gene switched on, three blue circles are spread far apart along the purple strand, each with three small green circles labelled Ac attached, and a gold oval labelled RNA pol sits on the DNA between two of them. Sentences under each half explain the effect of acetyl groups. — Histone acetylation changes how tightly DNA is packed. Original BioWelsh diagram.

Without acetyl groups, hold the DNA tightly and the are packed close together.
Adding acetyl groups to the loosens their hold, so the DNA is less tightly coiled.
RNA polymerase can then reach the , and it is transcribed. Removing the acetyl groups switches the off again.

The two main epigenetic modifications
Modification	Where	Usual effect on the gene
of DNA	Methyl groups on cytosine, especially in promoters	switched off
of	Acetyl groups on proteins	switched on
Removal of acetyl groups	proteins	DNA coils tightly; switched off

Lesson 17 of 19

What are the effects of epigenetic changes?

tags are copied when a cell divides by mitosis, so a pattern set up in one cell is kept by all the cells that come from it. The environment can change the tags, and some can even be passed through gametes to the next generation.

Cell : as an embryo develops, that a cell type does not need are switched off, so a skin cell makes melanin but not the proteins of a cell.
Identical twins: they have the same base sequence, but their tags become more different as they get older, which helps explain why one twin may develop a disease that the other does not.
Diet: in mice, a mother's diet during pregnancy changes the of a coat colour in her offspring, and so their coat colour. Diets rich in methyl donors such as folic acid increase .
Stress and trauma: early trauma can change the expression of for stress hormones such as , with lasting effects on behaviour.
: some are switched off by in the gametes of one parent, so only the copy from the other parent is expressed.
Cancer: of the of a switches it off, which can lead to uncontrolled cell division.

is a whole-chromosome example. In each cell of a female mammal embryo, one of the two X chromosomes is packed tightly around its and switched off. Which X is inactivated is random, but once chosen, the same X stays inactive in every cell that descends from that cell.

The photograph below shows a tortoiseshell cat, whose patchy coat is the result of .

A cat lying on its back on a fluffy white blanket, looking at the camera with green eyes. Its fur is mostly black with irregular patches of orange, including a narrow orange stripe running down one side of its face, orange patches on its chest and belly, and an orange front paw. — A tortoiseshell cat, whose patchy coat is the result of X inactivation. Photo: Theedecemberblues, CC BY 4.0. Resized.

In cats, a for orange or black fur is on the X chromosome. A tortoiseshell cat has one X carrying the orange and one carrying the black .
In the orange patches the X with the black is inactive, and in the black patches the X with the orange is inactive.
A male has only one X, so he cannot have both . Tortoiseshell cats are therefore almost always female.

Lesson 18 of 19

The specified practical work for this topic

WJEC lists one piece of specified practical work for this topic: an experiment to illustrate segregation, including the use of the to assess the significance of the results. It is set out in full below using maize cobs, with other ways of carrying it out in the teaching notes.

Specified practical

Illustrating gene segregation in maize, with a chi-squared test

Unit 4, section 3, specified practical 1 (BioWelsh entry number)

Aim To count the kernels of each on a maize cob from an generation, and to use a to find out whether the results fit the ratio expected for two unlinked .

Five rounded panels joined by arrows. The first shows a maize cob with purple and yellow kernels, captioned take a cob from an F2 generation. The second shows a roll of tape and a marker pen, captioned mark the first row so no kernel is counted twice. The third shows a hand lens, captioned sort each kernel by colour and shape. The fourth shows a hand tally counter, captioned count every kernel in each class. The fifth shows a calculator, captioned compare with 9 to 3 to 3 to 1 using chi-squared. — Gene segregation in maize and the chi-squared test, at a glance. Original BioWelsh diagram.

Equipment

Maize cob from an F2 generation
Maize cob from an generation1 per pairEach kernel is one offspring, so one cob gives hundreds of results.
Marker pen or pin
Marker pen or pinMarks the first row so no kernel is counted twice or missed.
Hand lens
Hand lens×10Helps to classify small or pale kernels.
Tally counter
Tally counter1 per classKeeps a running count of each class.
Calculator
CalculatorWorks out expected numbers and .
Pencil and ruler
Pencil and rulerDraws the results table before counting starts.

Safety

PinsRiskPricked fingers.ControlPush the pin in firmly between kernels and remove it before passing the cob on.
Stored cobsRiskMould spores can grow on damp cobs and irritate the airways.ControlUse dry cobs, discard any mouldy ones, and wash hands after handling.
Treated seedRiskKernels may have been treated with chemicals.ControlDo not eat the kernels, and wash hands after the practical.

Variables

Independent variableKernel class

Purple smooth, purple wrinkled, yellow smooth or yellow wrinkled.

These are the categories the counts are sorted into.

Dependent variableNumber of kernels in each class

Counted with a tally counter for the whole cob.

needs raw counts, never percentages.

Example: 219 purple smooth out of 400

Controlled variablesCross that produced the cob, Rules for classifying a kernel, Person counting, Lighting

Cross that produced the cob

Use cobs from the same × cross only.

Cobs from a different cross would have a different expected ratio.

Rules for classifying a kernel

Agree what counts as wrinkled and as yellow before starting.

Borderline kernels classified inconsistently would bias the counts.

Person counting

The same person classifies every kernel on a cob.

Different people draw the line between classes in different places.

Lighting

Count under the same bright, white light.

Pale purple kernels can look yellow in poor light.

Method, step by step

1)
Record the cross that produced the cob: plants for kernel colour and kernel shape, crossed with each other.
Why?
The parents' tell you which ratio to expect. For PpSs × PpSs it is 9 : 3 : 3 : 1.
2)
Draw a results table with a column for each kernel class before you start counting.
Why?
Deciding the classes first stops you changing your mind about a kernel halfway through.
3)
Put a pin, or a pen mark, beside the first kernel of one row.
Why?
The cob is round, so without a marked start it is easy to count a row twice or miss one.
4)
Classify each kernel in the row as purple or yellow, and as smooth or wrinkled, using a hand lens where needed.
Why?
Each kernel is a separate offspring with its own , so every kernel counts as one result.
The four classes of kernel on an F2 maize cob from a dihybrid cross. Original BioWelsh diagram.
Open full size
5)
Count each class with a tally counter, then move to the next row until you reach the marked row again.
Why?
Counting every kernel on the cob gives the largest possible sample.
6)
Leave out any gaps or damaged kernels and note how many there are.
Why?
A missing or damaged kernel has no that can be classified reliably.
7)
Add the four totals, then calculate the expected number for each class from the total and the ratio 9 : 3 : 3 : 1.
Why?
Expected numbers must add up to the same total as the observed numbers.
8)
Calculate = Σ (O − E)² ÷ E.
Why?
The test measures whether the difference between O and E is small enough to be due to chance.
9)
Use 3 and the at p = 0.05 to accept or reject the .
Why?
Four classes minus one gives 3 , so the is 7.82.
10)
Pool your results with other groups who counted cobs from the same cross, and test the pooled totals.
Why?
A larger sample makes the test more reliable and chance deviations smaller in proportion.

Results and analysis

Example data

Numbers of each kernel class counted on one cob, with the calculation

Kernel class	Observed (O)	Expected (E)	O − E	(O − E)²	(O − E)² ÷ E
Purple, smooth	219	225	−6	36	0.16
Purple, wrinkled	83	75	8	64	0.85
Yellow, smooth	72	75	−3	9	0.12
Yellow, wrinkled	26	25	1	1	0.04
Total	400	400			= 1.17

Bar chart with pairs of bars for four kernel classes. The vertical axis is number of kernels from 0 to 240. Lilac observed bars and gold expected bars read: purple smooth 219 and 225, purple wrinkled 83 and 75, yellow smooth 72 and 75, yellow wrinkled 26 and 25. A key identifies observed and expected for 9 to 3 to 3 to 1. — Observed and expected numbers of each kernel class on one F2 cob. Original BioWelsh graph.

What the results show

The is that there is no difference between the observed and expected numbers, so any difference is due to chance and the kernels fit a 9 : 3 : 3 : 1 ratio.

The calculated value of is 1.17. There are four classes, so there are 3 and the at p = 0.05 is 7.82.

Because 1.17 is less than 7.82, the probability that the difference is due to chance is greater than 0.05. The is accepted: the difference is not .

The results therefore fit the ratio expected when two unlinked segregate independently. If the had been , the parental classes would have been much larger than expected and would have been far above 7.82.

Worked calculations

Expected purple smooth = 9/16 × 400 = 225

Expected purple wrinkled = 3/16 × 400 = 75

Expected yellow wrinkled = 1/16 × 400 = 25

= 0.16 + 0.85 + 0.12 + 0.04 = 1.17

= 4 − 1 = 3; at p = 0.05 = 7.82

Evaluation

Some kernels are hard to classify, especially pale purple or slightly wrinkled ones.

Effect on resultsKernels may be put in the wrong class, changing O for two classes at once.

ImprovementAgree clear rules first, use a hand lens, and have one person classify every kernel.

Kernels can be counted twice or missed on a round cob.

Effect on resultsThe totals are inaccurate.

ImprovementMark the first row and count row by row; check the total against a separate count of all kernels.

One cob gives only a few hundred results.

Effect on resultsChance has a larger effect on a small sample, so a real difference is harder to detect.

ImprovementPool the results of several cobs from the same cross before testing.

Some kernels are missing or damaged.

Effect on resultsIf one is more likely to fail, that class is under-counted.

ImprovementRecord the gaps, and treat a large number of gaps in one area as a sign of a problem rather than ignoring it.

Going further

Why chi-squared?

The results are counts of individuals in separate categories, and there is an expected number for each category. That is exactly what the is designed for.

A t-test compares the of two sets of measurements, such as lengths, so it does not suit counts of .

Other ways to carry out this practical

A cob with only purple and yellow kernels, tested against 3 : 1 with 1 degree of freedom.
Seeds from a self-pollinated tobacco plant for chlorophyll production, germinated on damp paper in the light: green and albino seedlings in a 3 : 1 ratio.
Crosses of fruit flies with different eye colours or wing shapes, scored under a hand lens after anaesthetising them.
Computer simulations of segregation, which let you repeat a cross thousands of times and watch chance deviations shrink.

What a rejected null hypothesis might mean

The are , so parental combinations are too common.
One is or less likely to survive, so a class is missing or small.
The parents' were not what was assumed.

Model answer: describe the method

Describe how you would use a maize cob from an generation to find out whether kernel colour and kernel shape are inherited as two unlinked .

Original practice response · 6 indicative marks

Mark the first row of the cob with a pin so no kernel is counted twice.
Classify every kernel as purple or yellow and as smooth or wrinkled, and count each of the four classes.
Calculate the expected numbers from the total using the ratio 9 : 3 : 3 : 1.
State the : there is no difference between the observed and expected numbers.
Calculate = Σ (O − E)² ÷ E and compare it with the for 3 at p = 0.05, which is 7.82.
If is less than 7.82, accept the : the are unlinked and the difference is due to chance. If it is greater, reject it, which could the are .

What the answer needs to cover

Count all kernels in each of the four classes
Expected numbers from 9 : 3 : 3 : 1
of no difference between observed and expected
calculated with (O − E)² ÷ E summed
3 , 7.82 at p = 0.05
Correct accept or reject decision to chance

Exam practice

Give one reason why the chi-squared test is a suitable statistical test for these results.Show answer

The data are counts in categories, and the test compares the observed numbers with the numbers expected from a ratio.

Categorical or data
Compares observed with expected numbers

A monohybrid cob gave 312 purple and 88 yellow kernels. Calculate χ² for a 3 : 1 ratio and state your conclusion.Show answer

Expected: 300 purple and 100 yellow. = 12² ÷ 300 + 12² ÷ 100 = 0.48 + 1.44 = 1.92. With 1 degree of freedom the is 3.84. 1.92 is less than 3.84, so p is greater than 0.05 and the is accepted: the difference is due to chance.

Expected 300 and 100
= 1.92
3.84 at 1 degree of freedom
Accept, difference due to chance

Explain why the number of degrees of freedom is 3 for a dihybrid cross.Show answer

are the number of categories minus one. There are four classes, so 4 − 1 = 3.

n − 1
Four classes, so 3

Suggest why pooling the counts from several cobs improves the investigation.Show answer

A larger sample reduces the effect of chance on the proportions in each class, so the test is more reliable at detecting a real difference from the expected ratio.

Larger sample
Chance has less effect, so more reliable

Lesson 19 of 19

Putting the topic together

Inheritance runs from the to the . are passed on through meiosis in predictable ratios, create new , and tags decide which are actually used.

How the ideas in this topic connect

Zoom 100%

Zoom in to read each branch. Swipe or scroll sideways to explore.

Inheritance
- and
  - , , , ,
- Predicting crosses
  - 3 : 1, 1 : 1, 1 : 2 : 1
  - 9 : 3 : 3 : 1 from
  - : parental classes too common
  - : is the difference due to chance?
- - XX female, XY male
  - : and
- - :
  - :
  - , , and cancer
- - and
  - , , diet and disease

An is a form of a ; a is not an .
Each carries one of each .
A uses a partner.
In , divide by E; accept or reject the , never prove it.
The Y chromosome carries no of an .
is 21, caused by .
are switched on; are switched off.
changes alter expression, not the base sequence.