1. A2 Genetics Revision - AIMS
Topics include……..
For use with revision
Sheets.
Remember – last 3
Papers covered Chi 2,
the name epistasis
and sex linkage only.
There is always an
Ecology question!

2. 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!

3. 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.

4. 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…………..

5. 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………..

6. And finally…….
LINKAGE - 2 or more genes are located (linked) on the same chromosome.
Linkage reduces the
possible combinations of
genes during meiosis and
therefore reduces the
phenotypes resulting
from a cross.
Tall plant
Smooth seed
So, no ____________ ______________ at ___________ __

7. Pea Breeding Programmes

8. Monohybrid crosses
Easy!
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.

9. Remember…
Monohybrid crosses – 2 alleles (letters) in the genotypes but only 1 in the gametes. R R
Always ask first:
What are the genotypes of the parents? r r

10. 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 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.

11. What are their codominant genotypes?
See question 1, page 6
OR REVISION SHEET
What are their codominant genotypes?

12. Sickle Cell Anaemia – page 126
Now lets see how sickle cell can be inherited
Symbols for alleles =
HA = healthy
HS = sickle cell
Complete question 4, booklet page 7

13. Symbols for alleles = HA = healthy HS = 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.

14. Answer
Question 2
CB CB
CW CW
Grey Swedish Blue Ducks!

15. 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.

16. ABO Blood Groups – Multiple Alleles (showing codominance) for one gene.
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 A
Now complete
revision sheet

17. Dihybrid Cross –
investigating 2
characteristics
(phenotypes) at
a time.
2 genotypes are
therefore written
side by side.
e.g. SSbb x ssBB
See page 8, 9 and 10
and revision sheet.

18. Dihybrid Cross Animation – see booklet page 8
Websites for practise
Dihybrid Questions 1, 2 and 5.
Mono and dihybrid problems for brain warm ups
 Now try an exam question.

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

20. Why is this boy still bleeding after a fall?
Haemophilia – sex-linkage.

21. Sex-linkage in the
Royal Family.
See page 12
Albert

22. Alice
Leopold
Leopold
Queen Victoria
Beatrice

23. Inheritance of Sex-linked Haemophilia – page 11
There is generally no room
for any other genes besides sex genes on the Y
chromosome.
Answer question 1, page 125, and use page 12
See above punnett square! See revision sheet.

24. 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.

25. In Fruit flies the gene for eye colour is on the X chromosome.
Answer question
3 on page 125
See board for answer

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

27. Duchenne Muscular Dystrophy
Answer question 4/5, page 125

28. Sex – linkage in Cats.
How are both black and ginger expressed in female cat coats?

29. Autosomal linkage – 2 genes on the same non-sex chromosome.
See page 16, 17 and
page 123 in the book.

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

31. What phenotypic ratio will that be? 3 dominant : 1 recessive phenotype
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

32. 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.
Linkage Questions 1, 2 and 5.

33. 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.

34. Recessive Epistasis – recessive allele masks another gene.
See page 133 to find
out how eye colour
is controlled by at
least two genes
interacting together.
This example is
recessive epistasis, as
is the recessive albino
allele for skin pigment.
(see page mice)
Expected F2 ratio is
9 : 3 : 4

35. 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 masks
another gene.

36. Recessive or dominant epistasis?
Salvia - pure
breeding pink and
white produce purple
F1. Interbreeding
produces purple, pink
and white/ 9:3:4 in the
F2 generation.
Is this an example
of dominant or
recessive epistasis?
Recessive.
See page 128 to
explain and write
down 3 genotypes
for white flowers.

37. 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 = ______ phenotype
Enzyme 1
Enzyme 2
Which offspring ratio suggests complementary epistasis?
See page 129.
9:7

38. So, epistatic F2 expected ratios are not 9 : 3 : 3 : 1
See page 20
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.

39. 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.

40. Inheritance of Combs in Domestic Chicken – see page 132 and q 5 + 6 (next slide)
Pea
Rose
P + rr
pp + R
Walnut
P + R
Single
pp + rr

41. 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?

42. 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

43. 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

44. Two Hardy Weinberg equations
See revision
sheet.
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

45. Hardy Weinberg equation 2 for genotype frequency
p2 = frequency of homozygous dominant genotype
2pq = frequency of heterozygous genotype
q = 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

46. Tables for Hardy-Weinberg calculations (maybe useful)
decimal %
allele p q
decimal %
genotype p2
2pq q2

47. 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) = = bb (told)
30/110 = = BB + 2Bb
Single letters
Double letters p2
2pq q2

48. 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

49. 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.
Show the expected offspring if the genes are on separate chromosomes.
Now show the expected offspring if the genes are linked on the same chromosomes.
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.
What is this an example of?
(Work out parental genotypes and gametes.)

50. Show the expected offspring if the genes are on separate chromosomes
BbSs x bbss BS Bs bS bs
BS Bs bS bs x bs bs bBsS bBss bbsS bbss
Phenotypes? : : : 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?

51. 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? : :
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