(a) describe, with the aid of diagrams and photographs, the - PowerPoint Presentation

Slide1

(a) describe, with the aid of diagrams and photographs, the behaviour of chromosomes during meiosis, and the associated behaviour of the nuclear envelope, cell membrane and

centrioles

. (Names of the main stages are expected, but

not

the subdivisions of prophase);

(b) explain the terms

allele, locus, phenotype, genotype, dominant,

codominant

and recessive

;

(c) explain the terms

linkage

and

crossing-over

;

(d) explain how meiosis and fertilisation can lead to variation through the independent assortment of alleles;

(e) use genetic diagrams to solve problems involving sex linkage and codominance;

(f) describe the interactions between loci (

epistasis

). (Production of genetic diagrams is

not

required);

(g) predict phenotypic ratios in problems involving

epistasis

;

(h) use the chi-squared test to test the significance of the difference between observed and expected results. (The formula for the chi-squared test will be provided);

(

i

) describe the differences between continuous and discontinuous variation;

(j) explain the basis of continuous and discontinuous variation by reference to the number of genes which influence the variation;

(k) explain that both genotype and environment contribute to phenotypic variation. (

No

calculations of heritability will be expected);

(l) explain why variation is essential in selection;

(m) use the Hardy–Weinberg principle to calculate allele frequencies in

populationsSlide2

Meiosis

Meiosis results in daughter cells having half of the number of original chromosomes. The

daughter cells are haploid

and can be used for sexual reproduction.

Sexual reproductive cells are known as gametes. They fuse in fertilisation to produce the first cell of a new organism called a zygote.

Meiosis in humans occurs in the testes and ovaries. Diploid (2n) cells contain 2 sets of chromosomes – 1 paternal and 1 maternal set. During meiosis, haploid gametes are formed (n) containing half the normal number of chromosomes. Fertilisation restores the diploid condition.

Meiosis involves 2 divisions called meiosis I and meiosis II. Meiosis I is a reduction division: 2 daughter nuclei result with half the number of chromosomes of the parent nucleus. During meiosis II, the chromosomes behave as in mitosis and each of the haploid daughter cells divide to form 4 haploid daughter cells. Slide3

Meiosis I: Interphase I

DNA replicates

Cell increases in size, making more membrane

Organelles replicated

Nuclear envelope

Nucleolus

Meiosis I: Prophase I

Chromosomes

condense and supercoil

Chromosomes come together in their

homologous pairs

forming a

bivalent

Non-sister chromatids overlap and attach at points called

chiasmata

They may swap sections of chromatids with each other: this is called

crossing over

Nucleolus disappears and nuclear envelope disintegrates.

Homologous pair: One maternal and one paternal chromosome, each with the same genes at the same loci. Slide4

Meiosis I: Metaphase I

Bivalents line up on equator

Spindle fibres attach to centromeres

Chiasmata still present Random Assortment: bivalents line up randomly at the equator, with each member of the homologous pair facing opposite poles, allowing chromosomes to independently segregate when pulled apart in anaphase I.

Meiosis I:Anaphase I

Homologous chromosomes in each bivalent are

pulled to opposite poles by spindle fibres contractingCentromeres do not divide

Chiasmata separate

and lengths of chromatid that have been crossed over remain with the chromatid to which they have become newly attached. Slide5

Meiosis I:Telophase I

New nuclear envelopes form – one around each set of chromosomes at the poles

Plasma membrane constricts; cell divides by cytokinesis

2 cells are producedThere is 1 chromosome from each pair in each daughter cellChromosome number has been halvedEach chromosome still consists of 2 chromatidsSlide6

Meiosis II: Interphase II

Very short – sometimes cell will go straight from telophase I to prophase II.

Chromosomes uncoil

No DNA replication

Meiosis II: Prophase II

Nuclear envelope breaks downNucleolus disappears

Chromosomes condense Centrioles move to poles at right angles to meiosis ISpindle formsMeiosis II: Metaphase II

Chromosomes line up at equator

Spindle fibres attach at centromeres

Random assortment: random orientation of chromosomes on equator Slide7

Meiosis II: Anaphase II

Sister chromatids pulled to opposite poles by spindle fibres contracting

Centromeres divide

Chromatids randomly segregate

Meiosis II: Telophase II

Spindle breaks down

Nuclear envelopes reform around haploid daughter nucleiPlasma membrane constricts and cell divides by cytokinesis

4 haploid daughter cells with unduplicated chromosomes that are genetically different

are produced. Slide8

Sexual reproduction increases genetic variation, as genetic material from cells of 2 unrelated organisms combines. Genetic variation increases the chances of evolution as natural selection favours organisms that are best adapted to the ever changing environment.

How does meiosis increase genetic variation?

Crossing Over

Crossing over occurs during

prophase 1Non-sister chromatids wrap around each other and attach at points called

chiasmata Chromosomes break at these points. Broken ends of the chromatids re-join to the ends of the non-sister chromatid in the same bivalent, resulting in similar sections of chromatid being swapped over. Often these sections will contain the same genes but different alleles.

Crossing over produces new combinations of alleles on the chromatids

The

chiasmata remain in place during metaphase

, holding the maternal and paternal chromosomes together on the equator, so that when they segregate at anaphase I, one member of each pair goes to each pole. Slide9

2. Re-assortment of Chromosomes

Reassortment is the consequence of the random distribution of maternal and paternal chromosomes on the spindle equator in metaphase I

and the subsequent segregation into 2 daughter nuclei at anaphase I.

Each gamete acquires a different mixture of maternal and paternal chromosomes.

3. Re-assortment of ChromatidsReassortment is the consequence of the random distribution of chromatids on the spindle equator in metaphase IIDue to the process of crossing over,

sister chromatids are no longer genetically identicalHow the chromatids align in metaphase II determines how they segregate at anaphase II. Slide10

4. Fertilisation

In humans, 1 ovum is released from an ovary at a time. There are 300 million genetically different spermatozoa, and any one can fertilise the egg.

The genetic information of the 2 unrelated individuals then fuses to form the zygote.

5. Mutation

DNA mutation can occur during DNA replication during interphaseChromosome mutations may occur

Mutation increases variationIf the mutation occurs in the egg or sperm cell that are used in fertilisation, the mutated gene will be present in every cell of the offspring. Slide11

Monohybrid Crosses: Dominant-Recessive

Monohybrid crosses deal with the inheritance of 1 gene.

In this first example of a monohybrid cross, the allele ‘A’ is for normal colouring and the allele ‘a’ is for albino colouring. The albino allele ‘a’ is recessive, and so in order to be expressed in the phenotype, an offspring must inherit 2 copies of it. If the parents are both heterozygous dominant, Aa, the offspring’s phenotype ratio will be 3 normal : 1 albino.

In this second example, the allele Y is dominant for yellow and the allele y is recessive for green. Two yellow peas with the homozygous genotype Yy are crossed. The F

1

phenotypic ratio is 3 yellow : 1 green.

Heterozygous dominant x heterozygous dominant = 3:1 F1 phenotype ratioSlide12

In this example, we have 2 different parental phenotypes: one is homozygous for red, RR, and the other is homozygous for white, WW. The alleles R and W are co- dominant, and so when these 2 plants produce offspring, 100% of will have a genotype RW, giving a phenotype of pink – a mix of red and white.

Monohybrid Crosses: Co-Dominance

Co-dominance occurs when 2 alleles are dominant. Heterozygous offspring will have a mixed phenotype different to both parents. This type of inheritance is most common in plants.

If we then take 2 of the F1 generation, 2 pink flowers with genotype RW, and cross them, we get a different phenotype ratio:

25% red (RR), 25% white (WW) and 50% pink (RW) Slide13

Monohybrid Crosses: Multiple Alleles

There are some situations in which there are more than 2 alleles for a trait. An example is blood group in humans. There are 3 alleles: I

A

, I

B and IO .

Genotype

Phenotype (blood group)

I

A

I

A

A

I

A

I

O

A

I

A

I

B

AB

I

B

I

B

B

I

B

I

O

B

I

O

I

O

O

It is possible for 1 couple to have 4 children with different blood groups. If the mother is heterozygous dominant for A (AO) and the father is heterozygous dominant for B (BO) then there is equal probability that the offspring will be A, AB, B or O:

A and B are co-dominant; O is recessive. Slide14

h

H

H

H

H

H

hh

Monohybrid Crosses: Sex-Linked

A characteristic is sex-linked is the gene that codes for it is found on the X or Y chromosome (as opposed to the autosomes – chromosomes that have nothing to do with determining sex). Most sex-linked genes are found on the X chromosome. Sex linked characteristics include haemophilia, red-green colour blindness and Duchenne Muscular Dystrophy.

In this example, we are looking at haemophilia. The gene for haemophilia is found on the X chromosome: X

H

is the allele for normal clotting. This allele is dominant to the X

h

allele, which denotes haemophilia. If a normal mother (who is not a carrier) and a father, who has haemophilia, reproduce, and the offspring is a girl, there is a 100% chance that she will be a carrier for the gene. Any boys produced will be normal.

HSlide15

Dihybrid Cross

Dihybrid crosses are used when there is an inheritance of 2 genes, each of which has 2 alleles.

In this example, the 2 genes being inherited are:

Colour of seed – this is either yellow or green

Seed Skin Type: round or wrinkled

The alleles involved:For colour, Y is yellow and y is green: Y is dominant to y.For skin type, R is round and r is wrinkled: R is dominant to r.

If we cross a round yellow seed that is homozygous dominant for both traits (RRYY) and a wrinkled green seed that is homozygous recessive for both traits (rryy), all offspring have the genotype RrYy and a phenotype of yellow and round. If we then cross two of the F1 generation, we see that there are 4 different phenotypes displayed in the offspring. They are displayed in the ratio 9

(round, yellow):

3

(round green):

3

(wrinkled yellow):

1

(wrinkled green)

For a heterozygous x heterozygous

in a dihybrid cross, the ratio is always

9:3:3:1 Slide16

Test Cross

A genetic cross is used to determine the genotype of an organism displaying the dominant phenotype.

In this example, there are 2 alleles for plant colour, P for purple (dominant) and p for white (recessive).

This plant displays the dominant phenotype, purple. There are 2 genotypes which would result in this phenotype: PP and Pp. In order to determine the genotype, we cross the plant with plant with a known genotype: a white plant that is definitely homozygous recessive, pp. We can then tell from the phenotypes of the offspring what the genotype of the parent is.

If all of the offspring are purple, we know the parent plant must be heterozygous dominant, PP.

If half of the offspring are white and half are purple, we know the parent plant must be heterozygous dominant, Pp. Slide17

Epistasis

Epistasis is when the expression of a gene depends upon the presence of a particular allele of another gene. It is the interaction of different gene loci so that one gene locus suppresses the expression of another gene locus.

There are 2 types of epistasis:

Recessive Epistasis: when the presence of 2 recessive alleles of one gene prevent the expression of another gene. The alleles at the first locus are epistatic to the alleles at the second locus. Dominant Epistasis: when the presence of a dominant allele of one gene prevents the expression of a second gene.

Complementary Action: Genes can work in a complementary fashion. Bateson and Punnet crossed two white flowered sweet peas, ccRR x CCrr. All of the F1 plants were found to have purple flowers. The then allowed the F1 generation to interbreed and found that the F2 generation had purple flowers and white flowers in a ration of

9:7. This suggests that at least one dominant allele for both gene loci (C-R-)must be present for the flowers to be purple. The homozygous recessive condition at either locus masks the expression of the dominant allele at the other locus.

Colourless Compound

Colourless Intermediate Compound

Final Purple Pigment

Gene R/r

Gene C/c

For example:

a lower has a genotype CCrr. The 1

st

gene will be expressed – the flower remains white. The second gene is not expressed, because there is no dominant allele present. The flower’s final colour is white. Slide18

In ferrets, there are

2 genes controlling fur colour: gene A and gene B. Gene A codes for enzyme A, which turns the white fur present to a brown colour. Gene B codes for an enzyme that will turn brown fur black. Both genes have 2 alleles: for gene A, the dominant A and the recessive a; for gene B, the dominant B and the recessive b.

If the ferret is homozygous recessive for gene A (aa), gene B cannot be expressed. The ferret will therefore have white fur. The homozygous

aa is epistatic to both alleles of gene B. If the ferret is heterozygous for gene A, Aa, or homozygous dominant, AA, gene A will be expressed and enzyme A will cause the white fur to turn to brown. The final colour of the ferret will then depend on the genotype for gene B: if the ferret is homozygous

or heterozygous dominant for gene B (BB or Bb), gene B will be expressed and the ferret will have black fur. If the ferret is homozygous recessive for gene B, it will not be expressed and it will remain brown.

Look for the F

1 phenotype ratio of 9:3:4

How recessive epistasis works:Slide19

How does dominant Epistasis work?

The dominant allele of one gene will prevent the expression of a second gene.

Colourless

Green

Orange

Gene E/e

Gene D/d

Butterflies can either have colourless, orange or green wings. The colour of the wings depends upon 2 genes, ‘D’ and ‘E’. The recessive alleles of these genes, d and e, do not function, and so in order to affect the phenotype, the alleles present must be dominant.

A particular butterfly has the allele ‘D’ for gene D. The butterfly will therefore have green wings. Regardless of what allele is present for gene E, the wings will remain green, because the presence of the dominant allele at this gene locus masks the expression of the allele at the second locus.

A different butterfly has the allele ‘d’ for gene D. This butterfly can therefore display 2 phenotypes, depending on which allele is present on gene E:

If the dominant E allele is present, the butterfly will be orange: if the recessive e allele is present, it will remain white.

Look for the F

1

phenotype ratio of 12:3:1 or 13:3 Slide20

Gene Linkage

Linkage refers to 2 or more genes located on the same chromosome. They are usually inherited together

– meaning that they segregate together during anaphase 1 of meiosis.

However, because we have two of every autosome (chromosomes 1-22), one from the maternal and one from the paternal side, there is a possibility that during prophase I, chiasmata will form and crossing over will occur. This makes a

recombinant chromosome that contains parts of both the maternal and paternal chromosomes. The numbers of recombinants formed is related to the distance between the 2 genes on the chromosome

: if the loci are far apart (ie. there are more base pairs between the genes) they are more likely to be separated and so this will give a higher frequency of recombination. If the genes are very close to each other on the chromosome, there is less chance that the genes will be separated during meiosis I and so there will be a lower frequency of recombination.

Linkage reduces the number of phenotypes resulting from a cross.Slide21

Chi Squared

The Chi-squared test is a statistical test to find out if the difference between observed data and expected data is small enough to be due to chance.

Phenotype

Purple jagged

Purple Smooth

Green jagged

Green smooth

Observed No (O)

86

26

24

8

Expected Ratio

9

3

3

1

Expected number (E)

81

27

27

9

O-E

5

-1

-3

-1

(O-E)

2

25

1

9

1

(O-E)

2

E

0.31

0.04

0.33

0.11

1. Fill in a table of values as shown on the right.

2. To find the chi squared value, we add up the 4 values at the bottom of the table: this comes out as

0.79

3. We then need to decide what the degree of freedom is. To calculate this, we subtract 1 from the number of phenotypes present. In this case, 4-1 =

3

.

4. We then look up in a table the critical value for this degree of freedom. We look at the

0.05 level.

If the Chi squared value is

less than the critical value

, we can say that the difference is due to chance and is not statistically significant and we can

accept the hypothesis

.

If the Chi squared value is

greater than the critical value

at the 0.05 level, there is less than a 5% probability that the results are due to change (meaning that there is a more than 95% probability that the results are due to a wrong hypothesis). The

hypothesis should be rejected.

Slide22

Continuous and Discontinuous Variation

Continuous

Variation

eg. HeightDiscontinuous Variation

eg. Eye ColourCharacteristics that are controlled by many

genes. Characteristics that are controlled by one or two single major genes

There are several alleles at each locus that have small effects on the phenotype. A number of different genes have a combined affect on the phenotype: these are polygenes, and the characteristic they control is known a polygenic

. Each gene provides an

additive component

to the phenotype.

Few different alleles

at a single gene locus

have large effects on the phenotype. Different gene loci have different effects on the phenotype – codominance and dominant/recessive inheritance patterns are examples.

There is a

wide range of phenotypic variation

across a population and there are

no discrete categories

.

Phenotypes fall into

discrete, clear cut categories. There are no intermediate categories. For example, you have blood group O, A, B or AB.Characteristics are

quantitative in nature - for example height and weight that can be measured

Characteristics are qualitative

in nature – for example

eye colour that is described with words.

Environmental factors exert a

large effect

Environmental factors

exert a

small effectSlide23

Genotype + Environment = Phenotype

In humans,

intelligence is partly determined by genes and partly by environment.

Children inherit many genes with alleles from each parent , giving a genetic potential. However, the potential will only be realised with the help of a stimulating learning environment at home and school. It is also aided by good nutrition for growth and development of the brain.

The expression of polygenic traits (like height –continuous variation) is influenced more by the environment than monogenic traits (like eye colour).

Genetic variation

is the basis on which natural selection acts. In a population, there will be a range of different alleles present for many genes – this is the gene pool. Individuals in a species vary, so it is likely that some will be more likely to survive than others. These are the individuals that have the best chance of breeding and passing on their alleles to offspring. Over time, the alleles that confer an advantage become more common, whilst disadvantageous alleles become less common or disappear.

Variation is essential in selection, whether it is the environment or humans that are doing the selection...Slide24

Hardy-Weinberg Principle

The Hardy-Weinberg principle involves the frequencies of alleles, genotypes and phenotypes in a given population. The equation can be used to calculate the frequency of genotypes within a population and predict probabilities.

p + q = 1

p

2 + 2pq + q2

= 1 p = frequency of dominant allele q = frequency of recessive allele

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

Example:

A gene has 2 alleles: A (dominant) and a (recessive). A = long wings and a = short winds in flies. In a population of 500 flies, 480 have long wings and 20 have short wings. Calculate the frequencies of the dominant and recessive allele.

1. Frequency of homozygous recessive (q

2

– short winged) = 20/500 =

0.04

2. Frequency of recessive allele, q, = square root (0.04) =

0.2

3. Frequency of dominant allele, p = 1- q = 1 – 0.2 =

0.8