Topic 3.4 Inheritance 13.4 Theoretical Genetics. Definitions 2 This image shows a pair of homologous...

Topic 3.4 Inheritance

13.4 Theoretical Genetics

Definitions

2

This image shows a pair of homologous chromosomes. Name and annotate the labeled features.


Definitions

3

This image shows a pair of homologous chromosomes. Name and annotate the labeled features.

CentromereJoins chromatids in cell division

Gene lociSpecific positions of genes on a chromosome

AllelesDifferent versions of a gene

Dominant alleles = capital letterRecessive alleles = lower-case letter

Homozygous dominantHaving two copies of the same dominant allele

Homozygous recessiveHaving two copies of the same recessive allele. Recessive alleles are only expressed when homozygous.

Heterozygous Having two different alleles.The dominant allele is expressed.

CodominantPairs of alleles which are both expressed when present.

CarrierHeterozygous carrier of a

recessive disease-causing allele

GenotypeThe combination of alleles of a gene carried by an organism

PhenotypeThe expression of alleles of a gene carried by an organism


Explain this

4

Mendel crossed some yellow peas with some yellow peas. Most offspring were yellow but some were green!

Mendel from:http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm


http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm


Segregation

5

“alleles of each gene separate into different gametes when the individual produces gametes”

The yellow parent peas must be heterozygous. The yellow phenotype is expressed.

Through meiosis and fertilization, some offspring peas are homozygous recessive – they express a green color.

Mendel did not know about DNA, chromosomes or meiosis.

Through his experiments he did work out that ‘heritable factors’

(genes) were passed on and that these could have different

versions (alleles).





Segregation

6


“alleles of each gene separate into different gametes when the individual produces gametes”

F0

F1

Genotype: Y y Y y

Gametes: Y or y Y or y

Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes.

gametesPunnett Grid:

Genotypes:

Phenotypes:

Phenotype ratio:




Monohybrid Cross

7


F0

F1

Genotype: Y y Y y



Fertilization results in diploid zygotes.

A punnett square can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F1).


Genotypes:

Phenotypes:

Phenotype ratio:

Crossing a single trait.



8


F0

F1

Genotype: Y y Y y





gametes Y yY YY Yy

y Yy yy

Punnett Grid:

Genotypes:

Phenotypes:

Phenotype ratio:

Monohybrid Cross Crossing a single trait.



9


F0

F1

Genotype: Y y Y y





Ratios are written in the simplest mathematical form.

gametes Y yY YY Yy

y Yy yy

Punnett Grid:

YY Yy Yy yyGenotypes:

Phenotypes:

Phenotype ratio: 3 : 1

Monohybrid Cross Crossing a single trait.



10

F0

F1

Genotype:


Genotypes:

Phenotypes:

Phenotype ratio:

Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross?

Homozygous recessiveHomozygous recessive

Phenotype:Key to alleles:Y = yellowy = green


11

F0

F1

Genotype: y y y y

gametes y yy yy yy

y yy yy

Punnett Grid:

yy yy yy yyGenotypes:

Phenotypes:

Phenotype ratio: All green


Homozygous recessiveHomozygous recessive



12

F0

F1

Genotype:


Genotypes:

Phenotypes:

Phenotype ratio:


Phenotype:

HeterozygousHomozygous recessive

Key to alleles:Y = yellowy = green


13

F0

F1

Genotype: y y Y y

gametes Y yy Yy yy

y Yy yy

Punnett Grid:

Yy Yy yy yyGenotypes:

Phenotypes:

Phenotype ratio: 1 : 1


HeterozygousHomozygous recessive



14

F0

F1

Genotype:


Genotypes:

Phenotypes:

Phenotype ratio:


Phenotype:

HeterozygousHomozygous dominant

Key to alleles:Y = yellowy = green


15

F0

F1

Genotype: Y Y Y y

gametes Y yY YY Yy

Y YY Yy

Punnett Grid:

YY YY Yy YyGenotypes:

Phenotypes:

Phenotype ratio: All yellow


HeterozygousHomozygous dominant



16

F0

F1

Genotype: R ? r r

Phenotypes:

Test Cross Used to determine the genotype of an unknown individual.The unknown is crossed with a known homozygous recessive.

Homozygous recessiveunknown

Phenotype:Key to alleles:R = Red flowerr = white

Unknown parent = RR Unknown parent = Rr

Possible outcomes:

gametes gametes


17

F0

F1

Genotype: R ? r r

Phenotypes: All red

Test Cross Used to determine the genotype of an unknown individual.The unknown is crossed with a known homozygous recessive.

Homozygous recessiveunknown

Phenotype:Key to alleles:R = Red flowerr = white

Some white, some redUnknown parent = RR Unknown parent = Rr

Possible outcomes:

gametes r rR Rr Rr

R Rr Rr

gametes r rR Rr Rr

r rr rr


18

Pedigree Charts Key to alleles:T= Has enzymet = no enzymePedigree charts can be used to trace family histories and deduce genotypes

and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. Key: female male

affected

Not Affected

deceased

Looks like

Deduce the genotypesof these individuals: A & B C DGenotype

Reason


19

Pedigree Charts Key to alleles:T= Has enzymet = no enzymePedigree charts can be used to trace family histories and deduce genotypes

and risk in the case of inherited gene-related disorders. Here is a pedigree chart for this family history. Key: female male

affected

Not Affected

deceased

Looks like

Deduce the genotypesof these individuals: A & B C DGenotype Both Tt tt Tt

Reason Trait is recessive, as bothare normal, yet have produced an affected child (C)

Recessive traits only expressed when homozygous.

To have produced affected child H, D must have inherited a recessive allele from either A or B


20

Pedigree Charts Key to alleles:T= Has enzymet = no enzymeIndividuals D and $ are planning to have another child.

Calculate the chances of the child having PKU. Key: female male

affected

Not Affected

deceased

Looks like

$

Genotypes: D =

$ =

Gametes

Phenotype ratio

Therefore


21

Pedigree Charts Key to alleles:T= Has enzymet = no enzymeIndividuals D and $ are planning to have another child.

Calculate the chances of the child having PKU. Key: female male

affected

Not Affected

deceased

Looks like

$

Genotypes: D = Tt (carrier)

$ = tt (affected)

Gametes T tt Tt tt

t Tt tt

Phenotype ratio1 : 1 Normal : PKU

Therefore 50% chance of a child with PKU


22

Codominance Some genes have more than two alleles. Where alleles are codominant, they are both expressed.

Human ABO blood typing is an example of multiple alleles and codominance.The gene is for cell-surface antigens (immunoglobulin receptors). These are either absent (type O) or present. If they are present, they are either type A, B or both.

Where the genotype is heterozygous for IA and IB, both are expressed. This is codominance.

Key to alleles:i = no antigens presentIA = type A antigens presentIB = type B antigens present


23

More about blood typing A Nobel breakthrough in medicine.

Images and more information from:http://learn.genetics.utah.edu/content/begin/traits/blood/

Antibodies (immunoglobulins) are specific to antigens. The immune system recognizes 'foreign' antigens and produces antibodies in response - so if you are given the wrong blood type your body might react fatally as the antibodies cause the blood to clot.

Blood type O is known as the universal donor, as it has not antigens against which the recipient immune system can react. Type AB is the universal recipient, as it has no antibodies which will react to AB antigens.

Blood typing game from Nobel.org:http://www.nobelprize.org/educational/medicine/bloodtypinggame/gamev2/index.html


http://learn.genetics.utah.edu/content/begin/traits/blood/

http://learn.genetics.utah.edu/content/begin/traits/blood/

http://nobelprize.org/educational/medicine/landsteiner/readmore.html

http://www.nobelprize.org/educational/medicine/bloodtypinggame/gamev2/index.html

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Sickle Cell Another example of codominance.

Remember the notation used: superscripts represent codominant alleles. In codominance, heterozygous individuals have a mixed phenotype.

The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia.

Complete the table for these individuals:

Genotype

Description Homozygous HbA Heterozygous Homozygous HbS

Phenotype

Malaria protection?


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Remember the notation used: superscripts represent codominant alleles. In codominance, heterozygous individuals have a mixed phenotype.

The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia.

Complete the table for these individuals:

Genotype HbA HbA HbA HbS HbS HbS

Description Homozygous HbA Heterozygous Homozygous HbS

Phenotype normal carrier Sickle cell disease

Malaria protection? No Yes Yes


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Predict the phenotype ratio in this cross:

Key to alleles:HbA = Normal HbHbS = Sickle cell

Therefore 50% chance of a child with sickle cell disease.

F1


Genotypes:

Phenotypes:

F0Genotype:

Phenotype: carrier affected

Phenotype ratio: :


27





F1

gametes HbS HbS

HbA HbAHbS HbAHbS

HbS HbSHbS HbSHbS

Punnett Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA Hbs HbS Hbs

Phenotype: carrier affected

Phenotype ratio:

Carrier & Sickle cell

1 : 1

HbAHbS & HbSHbS


28




F1


Genotypes:

Phenotypes:

F0Genotype:

Phenotype: carrier carrier

Phenotype ratio:


29





F1

gametes HbA HbS

HbA HbAHbA HbAHbS

HbS HbAHbS HbSHbS

Punnett Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA HbS HbA HbS

Phenotype: carrier carrier

Phenotype ratio:

Unaffected & Carrier & Sickle cell

1: 2 : 1

HbAHb & 2 HbAHbS & HbSHbS


30




F1

gametes

HbA

HbS

Punnett Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA HbS

Phenotype: carrier unknown

Phenotype ratio:


31




F1

gametes HbA HbA HbA HbS

HbA

HbS

Punnett Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA HbS HbA HbA or HbA HbS


Phenotype ratio:


32




Therefore 12.5% chance of a child with sickle cell disease.

F1

gametes HbA HbA HbA HbS

HbA HbAHbA HbAHbA HbAHbA HbAHbS

HbS HbAHbS HbAHbS HbAHbS HbSHbS

Punnett Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA HbS HbA HbA or HbA HbS


Phenotype ratio:

3 Unaffected & 4 Carrier & 1 Sickle cell

3 : 4 : 1

3 HbAHbA & 4 HbAHbS & 1 HbSHbS


33

Sex Determination It’s all about X and Y…

Karyotype of a human male, showing X and Y chromosomes:http://en.wikipedia.org/wiki/Karyotype

Humans have 23 pairs of chromosomes in diploid somatic cells (n=2).

22 pairs of these are autosomes, which are homologous pairs.

One pair is the sex chromosomes. XX gives the female gender, XY gives male.

The X chromosome is much larger than the Y. X carries many genes in the non-homologous

region which are not present on Y.

The presence and expression of the SRY gene on Y leads to male development.

SRY

Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome


http://en.wikipedia.org/wiki/Karyotype


http://en.wikipedia.org/wiki/Y_chromosome

34

Sex Determination It’s all about X and Y…

Segregation of the sex chromosomes in meiosis.

Chromosome pairs segregate in meiosis.

Females (XX) produce only eggs containing the X chromosome.

Males (XY) produce sperm which can contain either X or Y chromosomes.

gametes X YX XX XY

X XX XY

Therefore there is an even chance* of the offspring being male or female.

SRY gene determines maleness.

Find out more about its role and just why do men have nipples?

http://www.hhmi.org/biointeractive/gender/lectures.html







35

Sex Determination Non-disjunction can lead to gender disorders.

XYY Syndrome: Fertile males, with increased risk of learning difficulties. Some weak connections made to violent tendencies.

XO: Turner SyndromeMonosomy of X, leads to short stature, female children.

XXX Syndrome:Fertile females. Some X-carrying gametes can be produced.

XXY: Klinefelter Syndrome:Males with enhanced female characteristics

Image from NCBI:http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179

Interactive from HHMI Biointeractive:http://www.hhmi.org/biointeractive/gender/click.html


http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179

http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179

http://www.hhmi.org/biointeractive/gender/click.html

http://www.hhmi.org/biointeractive/gender/click.html

36

Colour Blind cartoon from:http://www.almeidacartoons.com/Med_toons1.html


http://www.almeidacartoons.com/Med_toons1.html

http://www.almeidacartoons.com/Med_toons1.html

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Sex Linkage X and Y chromosomes are non-homologous.

Non-homologous region

Non-homologous region

The sex chromosomes are non-homologous. There are many genes on the X-chromosomewhich are not present on the Y-chromosome.

Sex-linked traits are those which are carried on the X-chromosome in the non-homologous region. They are more common in males.

Examples of sex-linked genetic disorders: - hemophilia- color blindness

X and Y SEM fromhttp://www.angleseybonesetters.co.uk/bones_DNA.html



http://www.angleseybonesetters.co.uk/bones_DNA.html

http://www.angleseybonesetters.co.uk/bones_DNA.html


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What number do you see?




39


What number do you see?

5 = normal vision2 = red/green color blindness




40


How is color-blindness inherited?

The red-green gene is carried at locus Xq28. This locus is in the non-homologous region, so there is no corresponding gene (or allele) on the Y chromosome.

Normal vision is dominant over color-blindness.

Xq28Key to alleles:N = normal visionn = red/green color blindness

XN XN

Xn Xn

XN Xn

XN Y

Xn Y

no allele carried, none written

Normal female Normal male

Affected female Affected male

Carrier female

Human females can be homozygous or heterozygous with respect to sex-linked genes. Heterozygous females are carriers. Chromosome images from Wikipedia:




41


What chance of a color-blind child in the cross between a normal male and a carrier mother?

Key to alleles:N = normal visionn = red/green color blindness

XN Xn XN YNormal maleCarrier female X

F1

Punnett Grid:

F0 Genotype:

Phenotype:




42





XN

Xn

XN YXN XN

XN Xn

XN Y

Xn YF1

Punnett Grid:

F0 Genotype:

Phenotype:




43





XN

Xn

XN YXN XN

XN Xn

XN Y

Xn YF1

Punnett Grid:

F0 Genotype:

Phenotype:

There is a 1 in 4 (25%) chance of an affected child.

Carrier female


Affected male

What ratios would we expect in a cross between: a. a color-blind male and a homozygous normal female?b. a normal male and a color-blind female?




44

Red-Green Color Blindness How does it work?

Xq28

The OPN1MW and OPN1LW genes are found at locus Xq28.

They are responsible for producing photoreceptive pigments in the cone cells in the eye. If one of these genes is a mutant, the pigments are not produced properly and the eye cannot distinguish between green (medium) wavelengths and red (long) wavelengths in the visible spectrum.

Because the Xq28 gene is in a non-homologous region when compared to the Y chromosome, red-green color blindness is known as a sex-linked disorder. The male has no allele on the Y chromosome to combat a recessive faulty allele on the X chromosome.




45

Hemophilia Another sex-linked disorder.

Blood clotting is an example of a metabolic pathway – a series of enzyme-controlled biochemical reactions.

It requires globular proteins called clotting factors. A recessive X-linked mutation in hemophiliacs results in one of these factors not being produced. Therefore, the clotting response to injury does not work and the patient can bleed to death.

XH XH

Xh Xh

XH Xh

XH Y

Xh Y

no allele carried, none written


Affected female Affected male

Carrier female

Human females can be homozygous or heterozygous with respect to sex-linked genes. Heterozygous females are carriers.

Key to alleles:XH = healthy clotting factorsXh = no clotting factor




3.4 Theoretical Genetics 46

Hemophilia results from a lack of clotting factors. These are globular proteins, which act as enzymes in the clotting pathway.

Read/ research/ review:

How can gene transfer be used to treat hemophiliacs?

What is the relevance of “the genetic code is universal” in this process?



47

Hemophilia results from a lack of clotting factors. These are globular proteins, which act as enzymes in the clotting pathway.



Hemophilia This pedigree chart of the English Royal Family gives us a picture of the inheritance of this X-linked disorder.

Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp


http://www.sciencecases.org/hemo/hemo.asp

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Hemophilia Pedigree chart practice


Key: female male

affected

Not Affected

deceased

Key to alleles:H = healthy clotting factorsh = no clotting factor

State the genotypes of the following family members:1. Leopold

2. Alice

3. Bob was killed in a tragic croquet accident before his phenotype was determined.

4. Britney



50



Key: female male

affected

Not Affected

deceased



2. Alice


4. Britney

Xh Y



51



Key: female male

affected

Not Affected

deceased



2. Alice


4. Britney

Xh Y

XH Xh



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Key: female male

affected

Not Affected

deceased



2. Alice


4. Britney

Xh Y

XH Xh

XH Y or Xh Y



53



Key: female male

affected

Not Affected

deceased



2. Alice


4. Britney

Xh Y

XH Xh

XH Y or Xh Y

XH XH or XH Xh



54

Pedigree Chart PracticeKey: female male

affected

Not Affected

deceased

Dominant or Recessive? Autosomal or Sex-linked?


55


affected

Not Affected

deceased

Dominant or Recessive?Dominant. A and B are both affected but have produced unaffected (D & F). Therefore A and B must have been carrying recessive healthy alleles.

If it were recessive, it would need to be homozygous to be expressed in A & B – and then all offspring would be homozygous recessive.

Autosomal or Sex-linked?


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affected

Not Affected

deceased

Dominant or Recessive?Dominant. A and B are both affected but have produced unaffected (D & F). Therefore A and B must have been carrying recessive healthy alleles.

If it were recessive, it would need to be homozygous to be expressed in A & B – and then all offspring would be homozygous recessive.

Autosomal or Sex-linked?Autosomal. Male C can only pass on one X chromosome. If it were carried on X, daughter H would be affected by the dominant allele.

Tip: Don’t get hung up on the number of individuals with each phenotype – each reproductive event is a matter of chance. Instead focus on possible and impossible genotypes. Draw out the punnett squares if needed.


57

Super Evil Past Paper Question

Key: female male

affected

Not Affected

deceased

In this pedigree chart for hemophilia, what is the chance that offspring ? will be affected?

A. 0%

B. 12.5%

C. 25%

D. 50%


58


Key: female male

affected

Not Affected

deceased


A. 0%

B. 12.5%

C. 25%

D. 50%



59


Key: female male

affected

Not Affected

deceased


A. 0%

B. 12.5%

C. 25%

D. 50%


What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y

XH

Y


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Key: female male

affected

Not Affected

deceased


A. 0%

B. 12.5%

C. 25%

D. 50%


What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH YThere is an equal chance of F being XH XH or XH Xh

So:

XH XH XH Xh

XH

Y


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Key: female male

affected

Not Affected

deceased


A. 0%

B. 12.5%

C. 25%

D. 50%



So:

XH XH XH Xh

XH XH XH XH XH XH XH XH Xh

Y XH Y XH Y XH Y Xh Y


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Key: female male

affected

Not Affected

deceased


A. 0%

B. 12.5%

C. 25%

D. 50%



So:

XH XH XH Xh

XH XH XH XH XH XH XH XH Xh

Y XH Y XH Y XH Y Xh Y

So there is a 1 in 8 (12.5%) chance of the offspring being affected!


Cystic Fibrosis is a recessive, autosomal disorder.

If you have two unaffected parents, and a child with cystic fibrosis, what must their genotype be for the trait?

Cystic Fibrosis affects these protein channels all over your body!

Lungs, Pancreas, Sweat Glands

This reduces their capacity to deal with the build up of mucus.

Check out all the mucus build up!

This mucus eventually builds up and creates a great environment for bacteria to grow.

Most people die of infections of Pseudomonas aeruginosa (Option D)

http://learn.genetics.utah.edu/content/disorders/singlegene/cf/

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