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A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection
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A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Dec 30, 2015

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Page 1: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

A2 Biology

unit F215 – exam 1hour 45 mins50% of marks for A2

Population genetics

includes

Hardy Weinberg equations,

natural selection and artificial selection

Page 2: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Call for European Cystic Fibrosis healthcare gap to be closed

Press release 19 March 2010 by Bristol University• A healthcare gap amounting to a ‘death sentence’ for

Cystic Fibrosis (CF) children born in Eastern Europe must be closed say researchers from the EuroCareCF Coordination Action for Cystic Fibrosis.

• “We know that this disease occurs randomly in about 1 in 4,000 children born to healthy parents across the EU,” he said.  “Despite this, the team encountered many fewer people with CF in poorer countries.  CF patients there die far younger than in long standing EU countries.”

• Understanding the inheritance of characteristics and of population genetics can help governments to plan healthcare.

Page 3: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

2 inherited characters in humans – both involve autosomal recessive alleles

Cystic fibrosis – excessive mucus in bronchioles reduces gas exchange + few digestive enzymes from pancreas reach gut

PKU = phenylketonuria – enzyme to convert phenylalanine into tyrosine not present; child has light skin and hair + brain damage (now blood test at birth + special diet)

Page 4: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

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

Page 5: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Hardy was a British

Mathematician;

Weinberg was a

German doctor

Page 6: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

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?

Page 7: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Hardy Weinberg equations

• 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

Page 8: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Hardy Weinberg equations

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

Page 9: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Tables for Hardy-Weinberg calculations

decimal %

allele

p

allele

q

decimal%

genotype

p2

genotype

2pq

genotype

q2

Page 10: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.
Page 11: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Hardy Weinberg - answers

1) p = 0.3 and q = 0.7;

Genotype frequencies = 9% 49% 42%

2)a) q = 0.3 or 30% b) 0.49 or 49%

c) 0.42 or 42%

3) q = 0.008 or 0.8%

Page 12: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Hardy Weinberg - answers

4)a) q = 0.42 or 42% b) p = 0.58 or 58%

c) 33% and 49%

5)a) q = 0.07 so p = 0.93

b) 0.49% c) 0.13 or 13%

6)a) q = 0.01 b) 0.0198 or 1.98%

Page 13: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Hardy Weinberg - answers7) a) q = 0.4 homozygotes = 0.36 or 36%

heterozygotes = 0.48 or 48%

b) Allele for sickle cell anaemia gives resistance to malaria = an environmental selection pressure

People with allele for normal haemoglobin and for sickle cell haemoglobin have selective advantage where malaria is present so more likely to survive and pass on alleles to offspring

c) If malaria eradicated no advantage to have sickle cell allele

Also need large population, no immigration or emigration, random mating, equal fertility and no mutations

Page 14: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Hardy Weinberg - answers8)a) q = 0.6

Homozygotes = 0.16 or 16%

Heterozygotes = 0.48 or 48%

b) No selective advantage to be tongue roller – remains same if all the conditions of Hardy-Weinberg equation are fulfilled

9)a) no – until very recently people with cystic fibrosis have not reproduced

b) Termination of pregnancies may reduce allele frequencies – but gene may still mutate

Page 15: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Genetic drift in small populations p7

Page 16: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Genetic drift = effect of chance factors on allele frequency in gene pool

Page 17: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Genetic drift in frog-hoppers on the Scilly Islands

Page 18: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Genetic drift and frog hoppers

1) St Agnes + St Mary’s – all striped;Other islands – striped and melanic but proportions vary;On all but one island more striped than melanic

2) Small population on island; may only have genes for striped = founder effect;Or chance means some genotypes reproduce more than

others – large effect on phenotypes as small number of individuals

3) Difficult!Allow small populations to breed for several generations in

lab and record resultsIn field could investigate environmental factors which might

affect breeding e.g. temperature, shelter, camouflage

Page 19: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Natural selection – stabilising and directional

Page 20: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Speciation – development of a new species

Needs reproductive isolation of a population so individuals cannot interbreed

Alleles do not pass from one gene pool to the other

Allopatric speciation = geographical isolation

Sympatric speciation = behavioural / seasonal isolation

Page 21: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Reproductive isolation by courtship behaviour = sympatric

Page 22: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Reproductive isolationby time of flowering =

sympatric

Page 23: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Peppered moths – evolution in action http://www.biologycorner.com/worksheets/pepperedmoth.html

Which moths survive? Why are the moths

Why? different colours in different areas?

Page 24: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

The peppered moth Biston betulariaStabilising and directional selection p10

Page 25: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Definition of a species p11

Biological definition

Phylogenetic / evolutionary / cladistic definition

Why have 2 definitions?

Are they equally useful?

Page 26: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Artificial selection and natural selection Similarities and differences p13

Both select the organisms to reproduce

Humans choose the characteristics in artificial selection and it usually produces lower genetic diversity in the gene pool

Selection pressures in natural selection are less likely to reduce genetic diversity of the

species

Page 27: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Artificial selection

Page 28: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Selective breeding of cattle

Page 29: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Artificial selection – dairy cows

Page 30: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Selective breeding = humans select the organisms that will reproduce

(this is not genetic engineering)

• Choose parents which show the desired characteristic (animals or plants)

• e.g. resistance to disease, speed of growth, height, milk yield

• Breed the parents • Select the offspring with the desired

characteristic• Use them for breeding• Continue for many generations

(detailed records needed)

• Result - desired features are more obvious but genetic diversity is decreased by the inbreeding

Page 31: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.
Page 32: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.
Page 33: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Improving food production by selective breeding

Page 34: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Recent selective breeding of wheat

Page 35: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Producing bread wheat Triticum aestivum by artificial selection

Page 36: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Producing bread wheat Triticum aestivum by artificial selection

1) Wild einkorn wheat 2N = 14 chromosomes = AA genome (haploid number = 7)

2) Einkorn wheat – artificial selection of desired characters by early farmers 2N = 14 chromosomes = AA genome

3) Crossed with wild grass 2N = 14 chromosomes = BB genome (happened by chance)Gametes fusing = A and B genomesProduced sterile hybrid 14 chromosomes = AB genome (no homologous pairs so meiosis cannot produce gametes)

Page 37: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Producing bread wheat Triticum aestivum by artificial selection

4) Random mutation chromosome number doubles tetraploid 4N = 28

Emmer wheat has genome AABB

5) Chance cross with goat grass 2N = 14 chromosomes = DD genome

Produced sterile hybrid – gametes fusing are AB and D – meiosis unsuccessful

6) Random mutation chromosome number doubles hexaploid 6N = 42

Modern bread wheat Triticum aestivum has genome AABBDD

Page 38: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Spartina grass on edge of sea

Page 39: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

How are these 2 species reproductively isolated?

Page 40: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Cloning in animals whole organism or stem cells(reproductive or non-reproductive cloning) p6 - 8

Page 41: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Dolly the Sheep – a product of nuclear transfer

Page 42: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Where are the enucleate egg and the surrogate mother?

Page 43: A2 Biology unit F215 – exam 1hour 45 mins 50% of marks for A2 Population genetics includes Hardy Weinberg equations, natural selection and artificial selection.

Cloning animals – advantages or disadvantages?

• Many animals with desired characters produced

• Infertile animals can reproduce• Rare animals conserved

• Animals genetically identical• Very labour intensive and needs sterile

facilities and highly trained staff• Breeding can occur at any time of year

• Genetically modified animals can be used to produce pharmaceutical chemicals