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Mendelian Inheritance Mesut Erzurumluoglu Mendelian Inheritance Mesut Erzurumluoglu

Mendelian Inheritance Mesut Erzurumluoglu - PowerPoint Presentation

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Mendelian Inheritance Mesut Erzurumluoglu - PPT Presentation

epmmeebristolacuk 1 Published review Erzurumluoglu et al Apr 2015 BioMed Research International 2 Who was Gregor Mendel Augustinian monk Czech Republic Founder of modern genetics ID: 935124

disease allele dominant affected allele disease affected dominant recessive genotype inheritance gene alleles generation frequency linked colour phenotype breeding

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Slide1

Mendelian Inheritance

Mesut Erzurumluoglu

epmmee@bristol.ac.uk

1

Slide2

Published review

Erzurumluoglu et al. Apr 2015. BioMed Research International

2

Slide3

Who

was

Gregor

Mendel?Augustinian monk, Czech Republic.Founder of ‘modern’ geneticsStudied segregation of traits in the garden pea (Pisum sativum) beginning in 1854Presented & published his theory of inheritance in 1865.“Versuche über Pflanzen-Hybriden”“Experiments in Plant Hybridization”Mendel was “rediscovered” in 1902Posthumously Death: 1884

3

Slide4

Why did Mendel choose peas?

4

Source:

Mawer S. Gregor Mendel: Planting the Seeds of Genetics

Slide5

Experiment – Flower colour

5

Slide6

6

In his breeding experiments, Mendel did the following, and tracked heritable characteristics for three

generations:

1) Mated two true breeding parents to produce a hybrid.  True-breeding parents are called the P generation. Hybrid offspring are called the F1 generation.2) He then allowed the F1 generation to self-pollinate. The offspring of this group are called the F2 generation.

Mendel’s experiments

Slide7

Purple

Flowers

Parental

Generation(True breeding plants)XF2 Generation(Ratio 3:1)All plants hadpurple flowersWhiteFlowersF1 Generation(Hybrids)

7

Slide8

8

Slide9

9

Locus: A

location in the genome (e.g.

‘flower colour locus’, Chr 1 at position 5005). Plural: lociAllele: Alternate forms of a gene (e.g. pink flower allele, white flower allele, P, p)Genotype: Both alleles at the locus form a genotype (e.g. PP, Pp, pp) Homozygous: Contains identical alleles for a character e.g. AA, TT, CC, GG (or PP, pp)

Heterozygous

Contains

two different alleles for a

character

e.g. AT, GC, CA (or Pp)

Phenotype: 

An

organism’s

traits (e.g. eye

colour

, height, disease)

Terms

Slide10

10

Similarities

between parent and offspring are due to the transmission of discrete elements called

‘genes’.There are multiple versions of the same gene (each alternate version of the gene is called an allele).Each parent has two alleles. The two alleles separate during sex cell formation (i.e. ‘Meiosis’)  Segregation (1st Law)Each organism inherits two (2) alleles; one allele from each parent.

Which

allele an organism inherits from a parent is random (50% probability

)

Independent assortment

(2

nd

Law)

Mendel’s Laws

Slide11

Purple

Flowers

P

XF2 GenerationPurple flowersWhiteFlowersF1 Generation

PP

pp

Pp

p

P

P

p

PP

Pp

pP

pp

Punnett

Square

11

Slide12

12

If a single copy of an allele results in the same phenotype as two copies irrespective of the second allele, the allele is said to be

dominant

over the second allele.E.g. PP and Pp both produce purple flowers. Therefore P is dominant to p.If an allele must occur in both copies of the gene to yield a phenotype, then it is recessive.E.g. only pp produces white flowers. Therefore p is recessive to P.*Dominant and Recessive alleles*This is Mendel’s third law (i.e. Law of Dominance)

Slide13

13

Smooth

Ss

WrinkledssXYou have two pea plants. One is true breeding for wrinkled seeds, the other is a round seed/wrinkled seed hybrid. Smooth seeds are dominant to wrinkled seeds. If we were to cross the two, what is the expected ratio of smooth to wrinkled seeds. Use a Punnett Square to illustrate your answer.

Exercise 1

Slide14

14

S

s

ssSs

Ss

ss

ss

Smooth

Wrinkled

Smooth

Wrinkled

Smooth : wrinkled

1

:

1

Answer

Slide15

15

Let us assume for the moment that eye colour is a simple monogenic trait which follows simple Mendelian inheritance (the reality

is slightly more complicated*)

Let us assume that the eye colour locus consists of two alleles, a brown allele and a blue allele. Brown is dominant to blue, therefore heterozygous individuals (Bb) exhibit brown eyes.If this is the case, why do so many people have blue eyes?~70% of white populationProblem*see http://www.ncbi.nlm.nih.gov/pubmed/20944644

Slide16

16

Using a

Punnett

Square to reveal genotype: Carry out a ‘testcross’                                   By breeding an organism of unknown genotype with an organism with a homozygous recessive trait, we can determine the genotype of the unknown individual.  The ratio of phenotypes in the offspring (F1 generation) is used to determine unknown genotype.                          Test Cross

Slide17

Dominant Phenotype

X

All offspring purple

Genotype Pp?Recessive phenotype

(Genotype PP or Pp?)

(Genotype pp)

p

p

P

P

Pp

Pp

Pp

Genotype PP?

1:1 ratio of colours

p

p

P

p

Pp

Pp

pp

Pp

pp

17

Slide18

18

We are unable to manipulate mating patterns of humans for experimentation.

Traits are studied by gathering information and placing it into a family tree.

The inter-relationships among parents and children across generations are called the family pedigree.Mendelian inheritance in Humans

Slide19

19

Example Pedigree

Deceased male

Unaffected male

Affected female

Affected male

Unaffected female

Slide20

20

For monogenic traits, we can specify the underlying genetic model with

three (sometimes *four)

parameters:Mode of inheritance (e.g. autosomal, X-linked, mitochondrial)Disease allele frequency (e.g. using H-W equation)Penetrance, or probability of being affected given a certain disease locus genotypeSlide in appendixComplex traits will be discussed in next shortcourse!Slide in appendix *for complications see appendix

Disease locus model

Slide21

21

First address for Mendelian geneticists!

Slide22

22

Affected individuals

are usually heterozygous (e.g. Pp)

Disease will occur in all generationsRare homozygotes can be equally or more severely affectedDisease affects both genders equallyRecurrence risk for sibs and children is ½Often have late onset (and reduced penetrance)Affected people often able to reproduce even if disease is fatal.Example: Huntington’s disease

dd

Dd

dd

dd

dd

dd

Dd

Dd

Dd

Dd

Dd

D = Dominant disease allele

d = Recessive normal allele

Mode of Inheritance (

MoI

) 1:

Autosomal dominant

Slide23

23

Affected individuals are homozygous; heterozygotes are u

naffected

Both sexes equally affectedSometimes causes ‘heterozygote advantage’E.g. Sickle cell disease and malariaCan ‘skip’ generations

On average, 25% of children of two carriers are affected

Consanguinity

is often high between parents of patients

Especially for very rare AR disorders

Often onset is in childhood

Examples:

Cystic

fibrosis, AR intellectual disability

Dd

Dd

Dd

dd

DD

D = Dominant normal allele

d = Recessive disease allele

MoI

2:

Autosomal recessive (AR)

Slide24

24

Males = XY, Females = XX

For genes carried on X chromosome inheritance depends

on: (i) Whether the mother or father was the mutation carrier (ii) Sex of the offspring (iii) Dominant or recessiveSex-linked inheritance

Slide25

25

Affected females are

heterozygous (mostly),

affected males hemizygous Frequency of affected females is twice the frequency of affected malesDisease is transmitted by mothers to children of both sex; by fathers, only to daughters50% of children of affected mothers, and 100% of daughters of affected fathers, are affectedE.g. Fragile X syndrome

d / -

D / d

D / d

d / -

d / -

d / d

D / -

d / -

d / -

D / d

D = Disease allele

d = Normal allele

MoI

3:

X-linked Dominant

Slide26

26

Usually only males are affected

Disease is transmitted by carrier mothers to 50% of sons

On average, 50% of daughters of carrier mothers are also carriers100% of daughters of affected males are carriersE.g. Red-green colour blindness (Slide in appendix)

.

.

.

D / d

D / -

d / -

D / d

D / -

D / D

d / -

D / -

D / d

D / -

MoI

4:

X-linked Recessive

D = Disease allele

d = Normal allele

Slide27

27

Why no

MoI

5?Y-linked or mitochondrial?No disorder linked to Y chromosome so farThink about how it would be (slide in appendix)…Mitochondrial DNA (mtDNA) not nuclear thus does not fit into Mendelian inheritance (slide in appendix)Strictly maternally inheritedTo all children – regardless of sexComplex disorders are called ‘complex’ because they do not follow a distinct inheritance pattern (slide in appendix) – due to polygenicity and environmental factors

Slide28

28

Genotype frequencies and allele frequencies are stable across

generations

If population obeys H-W equation, it is said to be in H-W equilibriumH-W equilibrium assumptions in appendixGenotype frequencies are equal to the product of the relevant allele frequencies. If locus alleles are:A = with frequency pa = with frequency q where p + q = 1

Then

the three genotypes:

AA

Aa

aa

have respective frequencies:

p

2

2pq

q

2

(p

2

+ 2pq + q

2

= 1)

Hardy-

Weinburg

equation

Slide29

29

How many people carry Cystic Fibrosis causal allele?

The

prevalence of cystic fibrosis in Caucasians is about 1 in 2500Assuming that all cases of cystic fibrosis are due to the action of one particular recessive mutation, what is the allele frequency of this mutation in the population? HINT: Use the Hardy-Weinberg equationWhat is the frequency of the heterozygotes (i.e. carriers) in the population?

Exercise 2

Slide30

30

C

= Dominant

wild-type allele c = Recessive disease allelep = frequency of wild-type allele q = frequency of cystic fibrosis alleleFreq. of diseased individuals = Freq(genotype = cc) =

1

in 2500

So q

2

= 1/2500

Therefore q =

1/50

and p = 49/50 (1 – 1/50)

Freq.

of heterozygote carriers = 2pq

=

2 x (49/50) x (1/50)

.0392

≈ 4%

Approx. 1

in 25

White Europeans

is a

carrier (heterozygote)!!!

Answer

Slide31

Learning Objectives

Be aware of Mendel and the significance of his early breeding experiments

Know Mendel’s principles of segregation and independent assortment and how they relate to current genetic

knowledgeBe able to discuss concepts of DominanceSex-linked inheritanceBe able to identify features of (monogenic) autosomal or X-linked, and dominant or recessive diseases31

Slide32

Thank You

Any questions?

Please look back at the slides

again once you complete the short-course(s)32

Slide33

Appendices

Test your knowledge at:

http://

biology.kenyon.edu/courses/biol114/TUTORIAL/inherit1/inherit1c.html33

Slide34

34

Penetrance is the conditional probability of having the phenotype P given the genotype G:

Pr

(P|G)We have encountered examples of penetrance parameters before, they are the familiar dominant and recessive examplesi.e. For a Autosomal Dominant Mendelian disease where D denotes the disease allele:Pr(P | G = DD) = 1Pr(P | G = Dd) = 1Pr(P | G = dd) = 0Penetrance may depend on various factors, such as the genotype at another locus, sex, age, imprinting

Penetrance

Slide35

35

Random

mating (i.e. no consanguinity or endogamy)

No migrationNo inbreedingNo selection related to genotypeA large population (preferably infinite)No new mutationsHardy-Weinburg equilibrium assumptions

Strong assumption!

Slide36

36

Protanopia

is a

colour vision deficiency in which the red retinal photoreceptors are absent affecting red-green hue discriminationDeuteranopia is a colour vision deficiency in which the green retinal photoreceptors are absent affecting red-green hue discriminationBoth due to a defective gene cluster on the X chromosomeMuch more common in malesRed-Green Colour blindness

Slide37

37

Y-Linked

MoI

?

Slide38

38

Mitochondrial

MoI

Slide39

39

Females have two copies of genes on their X chromosome (XX) whereas males only have one (XY)

To ensure the same gene dosage between males and females, one of these chromosomes is inactivated in females

Which chromosome is inactivated may differ from cell to cell and is random in most mammalsThe Calico CatComplications: X chromosome inactivation

Slide40

40

One gene affects the expression of another gene

Phenotype depends on the interaction between genotypes at different loci.

In the case of 2 loci, the familiar 9:3:3:1 ratio will not be observedE.g. Black and Yellow LabradorAlleles at the different loci may still assort independentlyComplications: Epistasis

Slide41

41

Complications: Epistasis

Slide42

42

B = Black

b = brown

E/e = gold locusTwo loci:Independent assortment, but ratio is not what we would predict (9:4:3)Complications: Epistasis

More info:

www.nature.com/scitable/topicpage/epistasis-gene-interaction-and-phenotype-effects-460

Slide43

43

Complementation

O

ccurs when organisms with different homozygous recessive mutations produce the wild-type phenotypeE.g. a change in wing structure in fliesNo simple examples in humans (so far)!Co-dominance Incomplete dominance*see www.biologycorner.com/bio2/genetics/notes_incomplete_dominance.html

*Many other complications!

Slide44

44

Complex disorders/traits

(e.g. Obesity, Height)

PolygenicMore than one gene involvedSometimes tens or even hundreds of genes involvedEnvironmental factors can also affect outcomeE.g. smoking and lung cancerThink of how a complex trait may look if we were to analyse a family pedigree…