Inheritance: WJEC A level Biology cram notes

Lesson 1 of 19

Genes and alleles

Fast revision of topic 4.3, Inheritance, in WJEC A level Biology Unit 4. It follows the WJEC specification.

A is a length of DNA at a that codes for a polypeptide. are different forms of the same . cells have two copies of every , one on each chromosome of a pair.

The diagram below shows one pair carrying three .

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.

Aa: two different , .
BB: . dd: .

Key terms
Term	Meaning
	The present, for example Bb
	The characteristics expressed, from and environment
	Expressed whenever present
	Only expressed when
	A chromosome that is not a

Lesson 2 of 19

Monohybrid crosses

A follows one . Lay it out as parental , parental , gametes (one each, ringed), a , offspring and , and the ratio.

The cross below follows seed shape in the pea, round R to wrinkled 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.

RR × rr gives all Rr, round.
Rr × Rr gives 1 RR : 2 Rr : 1 rr, so 3 round : 1 wrinkled.

: the two of a separate into different gametes, because separate at anaphase I.

Lesson 3 of 19

The test cross

To find whether a - organism is or , cross it with a individual.

The diagram below shows both possible results for a black labrador.

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.

All offspring show the : the parent was , BB.
About half show the , a 1 : 1 ratio: the parent was , Bb.

Lesson 4 of 19

Codominance

are both expressed in the , which has a third . Use a capital letter with superscripts for the .

The cross 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.

Red × white gives all roan: red and white hairs mixed.
Roan × roan gives 1 red : 2 roan : 1 white.

ABO blood groups: the A and B are , giving group AB.
A removes one class, turning 1 : 2 : 1 into 2 : 1.

Lesson 5 of 19

Independent assortment

line up randomly at metaphase I, so of on different chromosomes end up in gametes in every combination.

The diagram below shows the two ways two pairs can line up.

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.

Arrangement 1 gives AB and ab; arrangement 2 gives Ab and aB.
An AaBb individual makes all four types of in equal numbers.

Number of types = 2ⁿ, where n is the number of on different chromosomes.

Lesson 6 of 19

Dihybrid crosses

A follows two . For unlinked each carries one of each , and AaBb × AaBb needs a four by four .

The cross below follows stem colour and leaf shape in tomato.

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.

AaCc makes AC, Ac, aC and ac gametes.
The ratio is 9 : 3 : 3 : 1.

Ratios to know
Cross	Ratio
AaBb × AaBb	9 : 3 : 3 : 1
AaBb × aabb	1 : 1 : 1 : 1
AaBb × AaBb, completely	3 : 1

Lesson 7 of 19

Linkage

are on the same chromosome, so they do not assort independently.

The diagram below shows the gametes of DdEe with D to 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.

No : only DE and de gametes (parental).
A between the gives a few De and dE gametes ().

Recognise when parental classes are far more common than a 9 : 3 : 3 : 1 or 1 : 1 : 1 : 1 ratio predicts. = ÷ total × 100.

Lesson 8 of 19

The chi-squared test

: there is no difference between observed and expected numbers; any difference is due to chance. Use only for counts in categories.

Critical values at p = 0.05
Degrees of freedom	Critical value
1	3.84
2	5.99
3	7.82
4	9.49

= number of categories − 1.
below the : p > 0.05, accept, the difference is due to chance.
above the : p < 0.05, reject, the difference is .

Lesson 9 of 19

Sex determination

Humans have 22 pairs of and one pair of : XX female, XY male. Eggs all carry X; carry X or Y in equal numbers, so the ratio is 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.

They match only at their tips.
The Y carries , which triggers male development.
Most X have no partner on the Y, so males have one copy of each.

Lesson 10 of 19

Sex linkage: haemophilia and DMD

() and () are caused by on the X. Males have one X, so one is expressed. Write on the X only, for example Xᴴ and Xʰ.

The cross below shows 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.

Daughters: 1 unaffected : 1 .
Sons: 1 unaffected : 1 with .

Fathers never pass to sons.
Every daughter of an affected father is at least a .
Affected females are rare.

Lesson 11 of 19

Pedigrees

The diagram below shows the three patterns to spot.

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.

Unaffected parents, affected child: .
Affected parents, unaffected child: .
Mostly males affected, through mothers: .

Father to son transmission proves the is .
An affected girl with an unaffected father rules out .

Lesson 12 of 19

Mutations and mutagens

A is a random, rare change in the amount, arrangement or structure of DNA. change one ; change a chromosome's structure or the number of chromosomes. Only in -forming cells are inherited.

Mutagens
Mutagen	Effect
X-rays and gamma rays	Break DNA strands
Ultraviolet light	Joins neighbouring bases, so bases are copied wrongly
Tar, benzene	Slot between bases, so DNA is copied wrongly

Lesson 13 of 19

Gene mutations and sickle cell anaemia

The diagram below shows 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.

: at most one amino acid changes, or none if silent.
Addition or deletion: , so every later triplet is read differently.

The diagram below follows from DNA to 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.

A in the beta globin : GAG becomes GTG.
Glutamic acid is replaced by valine at position 6.
The haemoglobin forms fibres at low oxygen, so red cells sickle and block capillaries.

Hbᴬ and Hbˢ are : have sickle cell trait; Hbˢ Hbˢ gives .

Lesson 14 of 19

Chromosome mutations and Down's syndrome

The diagram below shows of chromosome 21.

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.

At anaphase I the pair fails to separate.
A with two chromosome 21s is fertilised by a normal .
The has three copies, 21, and 47 chromosomes: .

The risk rises steeply with the age of the mother, because oocytes stay paused in meiosis for years. is an extra or missing chromosome; is extra whole sets.

Lesson 15 of 19

Carcinogens, oncogenes and cancer

The diagram below shows the two routes 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.

mutates to an : switched on permanently.
(for example TP53) mutated or : switched off.
Either way, mitosis is uncontrolled and a forms.

A causes cancer; many are , such as tar in tobacco smoke, UV light and X-rays.

Lesson 16 of 19

Epigenetics: methylation and histones

is the of expression without changing the DNA base sequence.

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.

Methyl groups are added to cytosine, often in the .
RNA polymerase and transcription factors cannot bind, so the is switched off.

The diagram below shows .

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.

Acetylated hold DNA loosely, so the can be transcribed.
Removing acetyl groups makes DNA coil tightly, so the is switched off.

Lesson 17 of 19

Effects of epigenetic changes

Tags are copied at mitosis, so differentiated cells keep their pattern.
Identical twins become epigenetically different with age.
A mother's diet can change , and so coat colour, in her offspring.
Early trauma can change expression of stress hormone .
can silence .
: one X in each female cell is switched off at random, giving tortoiseshell cats.

Lesson 18 of 19

The specified practical

segregation in maize with a : count the kernel classes on an cob and compare them with the expected ratio.

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 kernel classes on an maize cob and test the fit to 9 : 3 : 3 : 1 with .

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

F2 maize cob
maize cobEach kernel is one offspring.
Marker pen or pin
Marker pen or pinMarks the starting row.
Tally counter
Tally counterCounts each class.
Calculator
CalculatorFor expected numbers and .

Safety

Mouldy or treated cobsRiskIrritation; harmful chemicals.ControlUse dry cobs, do not eat kernels, wash hands.

Variables

Independent variableKernel class

Four classes.

Dependent variableNumber of kernels in each class

Raw counts.

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

Cross that produced the cob

Same × cross.

Rules for classifying

Agreed before counting.

Person counting

One person per cob.

Lighting

Same bright light.

Method, step by step

1)
Mark the first row of the cob.
Why?
Stops kernels being counted twice or missed.
2)
Classify every kernel by colour and shape and count each class.
The four classes of kernel on an F2 maize cob from a dihybrid cross. Original BioWelsh diagram.
Open full size
3)
Calculate expected numbers from the total and 9 : 3 : 3 : 1.
4)
Calculate and compare it with 7.82 (3 , p = 0.05).

Results and analysis

Example data

Kernel counts on one cob

Kernel class	Observed	Expected	(O − E)² ÷ E
Purple, smooth	219	225	0.16
Purple, wrinkled	83	75	0.85
Yellow, smooth	72	75	0.12
Yellow, wrinkled	26	25	0.04

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

= 1.17, with 3 and a of 7.82.

1.17 < 7.82, so p > 0.05: accept the . The difference is due to chance and the are unlinked.

Worked calculations

Expected = 9/16, 3/16, 3/16 and 1/16 of 400

= 0.16 + 0.85 + 0.12 + 0.04 = 1.17

Evaluation

Borderline kernels are hard to classify.

Effect on resultsCounts in two classes change.

ImprovementAgree rules and use one person.

One cob is a small sample.

Effect on resultsChance has a bigger effect.

ImprovementPool several cobs.

Going further

Other ways

cob, 3 : 1, 1 degree of freedom.
Tobacco seedlings, green and albino, 3 : 1.
Computer simulation of crosses.

Model answer: describe the method

Describe how you would test whether kernel colour and shape are inherited as unlinked .

Original practice response · 6 indicative marks

Mark the first row and count every kernel in each of the four classes.
Expected numbers from 9 : 3 : 3 : 1.
: no difference between observed and expected.
= Σ (O − E)² ÷ E.
3 , 7.82 at p = 0.05.
Below 7.82: accept, due to chance, unlinked. Above: reject.

What the answer needs to cover

Count four classes
Expected from 9 : 3 : 3 : 1
calculated
3 df and 7.82
Correct decision

Exam practice

Why is chi-squared suitable?Show answer

Counts in categories compared with expected numbers.

Categorical data
Observed against expected

Lesson 19 of 19

Summary

Ratios: 3 : 1, 1 : 1, 1 : 2 : 1, 9 : 3 : 3 : 1, 1 : 1 : 1 : 1.
makes parental classes too common.
: divide by E, df = categories − 1, compare with the at p = 0.05.
: males affected, female, never father to son.
Sickle cell: . Down's: , 21.
on, suppressor off.
: off, on, sequence unchanged.