A2 Genetics Revision - AIMS Topics include…….. For use with revision Sheets. Remember – last 3 Papers covered Chi 2, the name epistasis and sex linkage.

A2 Genetics Revision - AIMS

Topics include……..

For use with revisionSheets.Remember – last 3 Papers covered Chi 2,

the name epistasis and sex linkage only.There is always an Ecology question!

• Genetics Terminology• Monohybrid Crosses (as done at school)• Codominance• Dihybrid Crosses• Inheritance of Sex-linked Alleles• Autosomal Linkage• Epistasis• Testing the Significance of the Outcomes

of Breeding Programmes by the Chi ² Test• Continuous and discontinuous variation• Population genetics – Hardy Weinburg• Any other topic we have time for!

Terminology

• GENE – a length of ____ that codes for the production of a ___________ molecule.

• GENETIC CODE – • ALLELES – different forms of one

gene, e.g………………• LOCUS – the position of a gene on

a _______________ and on its homologous partner.

• GENOTYPE – the 2 alleles which an organism possesses for a particular gene, e.g. CC, Cc or cc.

• PHENOTYPE – the expression of the genotype e.g. a person either has cystic fibrosis or not.

• DOMINANT – only one of this allele needs to be present for the allele to be ______________ in the p_________.

• RECESSIVE – two of these alleles need to be present for…………..

CODOMINANCE – both alleles are equally dominant and so both expressed in the _____________.

So both C R (red pigment) and C W (white pigment) are expressed if together in a genotype.

The phenotype will therefore be………..

And finally…….

• LINKAGE - 2 or more genes are located (linked) on the same chromosome.

Tall plant

Smooth seed

Linkage reduces the possible combinations of genes during meiosis and therefore reduces thephenotypes resulting from a cross.

So, no ____________ ______________ at ___________ __

Pea Breeding Programmes

Monohybrid crosses

• Monohybrid crosses only take account of one feature, controlled by one gene at one locus on a chromosome.

• Genetic diagrams (Punnett squares) are used to show predicted genotypes of offspring.

• Use the accepted system for showing dominant and recessive alleles.

• Work through the examples on the next slide and then continue on page 4 and 5.

Easy!

Remember…• Monohybrid crosses – 2 alleles

(letters) in the genotypes but only 1 in the gametes.

R R

r

r

Always ask first:

What are the genotypes of the parents?

Monohybrid cross examples – or do one on revision sheet.

• Cystic fibrosis is caused by a faulty protein produced by a recessive allele c.

• The normal allele for the protein is C (dominant).1. Draw diagrams to show the offspring probabilities when: a) a woman with the condition has children with a healthy

homozygous man. All children are Cc so are carriers.b) Heterozygote parents have children. 1:2:1, CC, Cc, cc, so 1 in 4 chance of cf.2. A woman with genotype CC has children: 50% are ‘carriers’ Cc What is the genotype of the father? Cc – now check pages 4 and 5 in booklet.

What are their codominant genotypes?

See question 1, page 6OR REVISION SHEET

Sickle Cell Anaemia – page 126• http://www.youtube.com/watch?v=9UpwV1tdxcs

• http://www.youtube.com/watch?v=bCOJkpL7MVw&feature=related

• Now lets see how sickle cell can be inherited

Symbols for alleles =HA = healthyHS = sickle cell Complete question 4, booklet page 7

Symbols for alleles = HA = healthyHS = sickle cell

• A heterozygous sickle cell trait mother has a child with a husband who has the sickle cell disease. What is the chance of the child inheriting the disease?

• 50% sickle cell disease and 50% sickle cell trait.

CW CW CB CB

Answer Question 2

Grey Swedish Blue Ducks!

Complete question 2 page 127

• Answer to question 2

• What are genotypes of offspring from a grey duck and drake?

• Grey = CWCB, White = CWCW, Black = CBCB

• How could he always get Swedish blue ducks?

• Swedish blue = grey, so breed CWCW with CBCB to always get CWCB (grey) offspring.

• Now check what a test cross is on your revision sheet.

ABO Blood Groups – Multiple Alleles (showing codominance) for one gene.

• http://www.medindia.net/animation/blood-groups.asp

• http://waynesword.palomar.edu/aniblood.htm

• Allele IA is codominant with IB, but Io is recessive to both.

• Write out all the possible genotypes and their phenotypes

Genotype Phenotype – Blood Group

IA IA ANow completerevision sheet

Dihybrid Cross – investigating 2characteristics(phenotypes) at a time.2 genotypes aretherefore writtenside by side.e.g. SSbb x ssBB

See page 8, 9 and 10and revision sheet.

Dihybrid Cross Animation – see booklet page 8

• http://trc.ucdavis.edu/biosci10v/bis10v/media/ch08/dihybrid_v2.html

• http://science.nhmccd.edu/Biol/dihybrid/dihybrid.html

• Websites for practise• http://www.ksu.edu/biology/pob/genetics/intro.htm• Dihybrid Questions 1, 2 and 5.

• Mono and dihybrid problems for brain warm ups• http://www.biology.arizona.edu/mendelian_genetics/

problem_sets/monohybrid_Cross/monohybrid_cross.html

 Now try an exam question.

Linkage of genes can be on any chromosomesincluding the X and Y sex chromosomes.

Why is this boy still bleeding after a fall?

• Haemophilia – sex-linkage.

Albert

Sex-linkage in theRoyal Family.See page 12

Alice LeopoldLeopold

Beatrice

Queen Victoria

Inheritance of Sex-linked Haemophilia – page 11

There is generally no roomfor any other genes besides sex genes on the Ychromosome.

Answer question 1, page 125, and use page 12See above punnett square! See revision sheet.

Some more examples - Red – green colour blindness.

• What numbers can you see?

• How is it inherited?

• Answer question 2, page 125

Mother! He passes his Xc to his daughter and his Y to a son.

In Fruit flies the gene for eye colour is on the X chromosome.

• Answer question 3 on page 125

See board for answer

Starter – question 2, A4 Genetics Problems sheet and 4 c) ii (A3 sheet)

Duchenne Muscular Dystrophy

Answer question 4/5, page 125

Sex – linkage in Cats.

• How are both black and ginger expressed in female cat coats?

Autosomal linkage – 2 genes on the same non-sex

chromosome.See page 16, 17 and page 123 in the book.

Nail Patella Syndromeis caused by a dominant allele linked to the ABO blood group gene by being found on the same chromosome. (page 16)

Alleles which are linked do not produce the expected ratios of offspring in the next

generation • See revision sheet.• E.g. alleles not linked would produce the

expected dihybrid ratio of _:_:_:_ in the F2

generation.• Linked alleles would produce an expected

ratio of…..? (Use ABab x ABab to find out) Notice the linked AB and ab.

• 1 ABAB: 2 ABab: 1 abab• What phenotypic ratio will that be?• 3 dominant : 1 recessive phenotype

• However, if crossing over occurs between the linked alleles, rare recombinants can result, e.g.

70 ABAB (double dominant), 20 abab (double recessive), 4 AbAb and 6 aBaB, (the 2 rare recombinants).

• The closer the genes are linked, the rarer the chance of crossing over between them.

• Try SAQ 19 and 20 on page 17. A worked example is on page 123 in your book.

• Hint – 2 large numbers of 4 groups of offspring and 2 very small numbers hint at crossing over of linked alleles.

• http://www.ksu.edu/biology/pob/genetics/intro.htm

• Linkage Questions 1, 2 and 5.

Epistasis – how one gene may control the expression of another.

• Key definition – 2 marks – page 18• The interaction of different gene loci,• so that one gene locus masks or suppresses

the expression of another gene locus.• The controlling allele is the epistatic allele.• The masked allele is the hypostatic allele.• There are 3 types.

Recessive Epistasis – recessive allele masks another gene.

See page 133 to findout how eye colouris controlled by at least two genesinteracting together.This example is recessive epistasis, as is the recessive albino allele for skin pigment.(see page 132 - mice)

Expected F2 ratio is9 : 3 : 4

Dominant Epistasis in hens p129

2 gene loci interact for feather colour. I is dominant over i and always gives white feathers even if C for colour rather than c for no colour is present too.

Write down all the genotypes which produce a) white and b) coloured hens.State the expected F2 ratio (page 129)13:3

Dominant allele masksanother gene.

Recessive or dominant epistasis?

Salvia - purebreeding pink and white produce purpleF1. Interbreeding produces purple, pinkand white/ 9:3:4 in theF2 generation.

See page 128 to explain and writedown 3 genotypes for white flowers.

Is this an example of dominant orrecessive epistasis?Recessive.

Complementary Epistasis p18…

• Complementary epistasis – Homozygous recessive genotype at either allele locus masks the expression of the dominant allele at the other locus.

• Colourless substance Yellow Orange •

• i.e. just Enzyme 1 = ______ phenotype• Enzyme 1 and 2 = ______ phenotype• No enzyme 1 = ______ phenotype

Enzyme 1 Enzyme 2

Which offspring ratio suggests complementary epistasis?See page 129.9:7

So, epistatic F2 expected ratios are not

9 : 3 : 3 : 1

• Recessive is suggested by 9 : 3 : 4

• Dominant is suggested by 12 : 3 : 1 or 13 : 3

• Complementary is suggested by 9 : 7 (similar)

• These ratios must be learnt to help solve problems.

See page 20

Whiteboard Quiz! • The F2 phenotypic ratio of 9:3:4 suggests?• Recessive epistasis• The hypostatic allele is the ________ allele locus.• Masked• The F2 phenotypic ratio of 45 orange flowers : 35 yellow

suggests?• Complementary epistasis• The epistatic allele is the _______ allele locus.• Controlling• The F2 phenotypic ratio of 13:3 suggests?• Dominant epistasis• Homozygous recessive genotype at either allele locus masking

the expression of the dominant allele at the other locus is called __________ epistasis.

• Complementary• Dominant epistasis is when…….• A dominant allele locus masks a hypostatic allele locus.

Inheritance of Combs in Domestic Chicken – see page 132 and q 5 + 6 (next slide)

Single

RosePea

Walnut

P + R

pp + RP + rr

pp + rr

Which 2 genotypes should the breeder cross to produce 25% walnut, 25% rose, 25% pea

and 25% single combs?

• This is a 1 : 1 : 1 : 1 ratio.• Therefore like the offspring of a test cross of

double recessive x heterozygote parents.• pprr x PpRr = single x walnut• Try it!• A possible explanation for crossing pea x rose

combs and getting walnut combs (F1) is that CR for rose and CP for pea combs are codominant. Explain how the F2 phenotypes negate that.

• How many phenotypes can be present in the codominant offspring?

• Try crossing a walnut with another walnut parent. How many offspring?

Key definitions page 2

1) A population of a species is:-a group of organisms which have similar characteristics e.g. anatomy, physiology, biochemistry and behaviour and which can interbreed to produce fertile offspring

2) The gene pool is:-all the genes in the gametes of a population from which the genotypes or genomes of the next generation are formed

The Hardy Weinberg principle predicts that:-the frequency of an allele in the gene pool

will not change from one generation to the next (and so the frequency of genotypes / genomes also remains constant)

This is only true if

• the population is very large• there is no immigration or emigration from

the population • mating is random• all genotypes are equally fertile • there are no random mutations

Which could be met by a natural population?

See page 2

Two Hardy Weinberg equations

1. Allele frequency • frequency of dominant allele = p• frequency of recessive allele = q• total frequency of alleles for a character in

the population = 1

p + q = 1

So if r = 0.6 then frequency of R = 0.4

See revision sheet.

Hardy Weinberg equation 2 for genotype frequency

p2 = frequency of homozygous dominant genotype2pq = frequency of heterozygous genotypeq2 = frequency of homozygous recessive genotype

total frequency of all genotypes in population = 1

p2 + 2pq + q2 = 1

So if frequency rr is 0.39 and RR is 0.34 frequency of Rr = 0.27

Tables for Hardy-Weinberg calculations(maybe useful)

decimal %

allele

p

allele

q

decimal%

genotype

p2

genotype

2pq

genotype

q2

Now try some questions page 4, 5 and 6!

• Remember: – • allele frequency of R = p• allele frequency of r = q

• Genotype frequency of RR =• Genotype frequency of Rr =• Genotype frequency of rr =• Always start with frequencies.• E.g. 90/110 (total) = 0.818 = bb (told)• 30/110 = 0.272 = BB + 2Bb

Single letters

Double letters

p2

2pq

q2

A Recessive allele 1 Gene pool remains stable due to selection pressures. Less variation.

B Dominant allele 2 Represents double recessive genotype frequency e.g. rr

C Directional selection 3 Represents recessive allele frequency e.g. r

D Stabilising selection 4 Can only be expressed in phenotype if dominant allele is not in genotype.

E Natural selection 5 Frequency of alleles change, heading in one direction.

F Artificial selection 6 Represents heterozygote genotype frequency e.g. Rr

G q2 7 Represents homozygous dominant genotype frequency e.g. RR

H 2pq 8 Always expressed in phenotype if in genotype.

I p2 9 Man is the selection pressure and controls the breeding

J q 10 Environmental and competitive selection pressures control selection

Genetics Challenge!

• Define autosomal linkage• Black and spotted is dominant over brown and

plain in dogs. A black (B) spotted (S) bitch had a heterozygous genotype. She was mated with a brown plain dog.

1. Show the expected offspring if the genes are on separate chromosomes.

2. Now show the expected offspring if the genes are linked on the same chromosomes.

3. Now show the expected offspring if the spots can only be expressed if the dog is black, i.e. if at least one B is in the genotype. Genes are not linked.

4. What is this an example of? (Work out parental genotypes and gametes.)

1. Show the expected offspring if the genes are on separate chromosomes

BbSs x bbss BS Bs bSbs

BS Bs bS bs x bs bs bBsS bBss bbsS bbss Phenotypes? 1 : 1 :

1 : 1

2. Now show the expected offspring if the genes are linked on the same chromosomes.

BSbs x bsbs BS bs BS bs x bs bs bBSs bbss Phenotypes? How might a few rare recombinants occur?

3. Now show the expected offspring if the spots can only be expressed if the dog is black, i.e. if at least one B is in the genotype. Genes are not linked.

BbSs x bbss BS Bs bS bs BS Bs bS bs x bs bs bBsS bBss bbsS bbss Phenotypes? 1 : 1 : 2

4. What is this an example of? Recessive Epistasis – the homozygous presence of a

recessive allele prevents another being expressed.

Remember the F2 ratio for a recessive epistasis is 9 : 3 : 4