theoretical genetics stephen taylor 4.3 theoretical genetics1

Theoretical GeneticsStephen Taylor

http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 1

http://sciencevideos.wordpress.com/


http://sciencevideos.wordpress.com/about/creative-commons/

Definitions


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


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


“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 fertilisation, some offspring peas are homozygous recessive – they express a green colour.

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



“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.

gametesPunnet Grid:

Genotypes:

Phenotypes:

Phenotype ratio:






Monohybrid Cross



F0

F1

Genotype: Y y Y y



Fertilisation results in diploid zygotes.

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

gametesPunnet Grid:

Genotypes:

Phenotypes:

Phenotype ratio:

Crossing a single trait.







F0

F1

Genotype: Y y Y y





gametes Y yY YY Yy

y Yy yy

Punnet Grid:

Genotypes:

Phenotypes:

Phenotype ratio:

Monohybrid Cross Crossing a single trait.







F0

F1

Genotype: Y y Y y





Ratios are written in the simplest mathematical form.

gametes Y yY YY Yy

y Yy yy

Punnet Grid:

YY Yy Yy yyGenotypes:

Phenotypes:

Phenotype ratio: 3 : 1

Monohybrid Cross Crossing a single trait.






F0

F1

Genotype:

gametesPunnet Grid:

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





F0

F1

Genotype: y y y y

gametes y yy yy yy

y yy yy

Punnet Grid:

yy yy yy yyGenotypes:

Phenotypes:

Phenotype ratio: All green


Homozygous recessiveHomozygous recessive






F0

F1

Genotype:

gametesPunnet Grid:

Genotypes:

Phenotypes:

Phenotype ratio:


Phenotype:

HeterozygousHomozygous recessive

Key to alleles:Y = yellowy = green





F0

F1

Genotype: y y Y y

gametes Y yy Yy yy

y Yy yy

Punnet Grid:

Yy Yy yy yyGenotypes:

Phenotypes:

Phenotype ratio: 1 : 1


HeterozygousHomozygous recessive






F0

F1

Genotype:

gametesPunnet Grid:

Genotypes:

Phenotypes:

Phenotype ratio:


Phenotype:

HeterozygousHomozygous dominant

Key to alleles:Y = yellowy = green





F0

F1

Genotype: Y Y Y y

gametes Y yY YY Yy

Y YY Yy

Punnet Grid:

YY YY Yy YyGenotypes:

Phenotypes:

Phenotype ratio: All yellow


HeterozygousHomozygous dominant






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





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




Dihybrid Crosses

http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 17

Consider two traits, each carried on separate chromsomes (the genes are unlinked).

Key to alleles:Y = yellowy = greenS = smooths = rough

In this example of Lathyrus odoratus (sweet pea), we consider two traits: pea colour and pea surface.

What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci?

F1

gametesPunnet Grid:

F0Genotype:

Phenotype:Heterozygous at both loci Heterozygous at both loci




Dihybrid Crosses






F1

gametesPunnet Grid:

F0Genotype: SsYy SsYy


Smooth, yellow Smooth, yellow




Dihybrid Crosses






F1

gametes SY Sy sY sySY SSYY SSYy SsYY SsYy

Sy SSYy SSyy SsYy Ssyy

sY SsYY SsYy ssYY ssYy

sy SsYy Ssyy ssYy ssyy

Punnet Grid:







Dihybrid Crosses






F1

gametes SY Sy sY sySY SSYY SSYy SsYY SsYy




Punnet Grid:




Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green




Dihybrid Crosses





Calculate the predicted phenotype ratio for:

F1

Punnet Grid:

F0Genotype:

Phenotype:Heterozygous at both loci Heterozygous for S, homozygous dominant for Y

Phenotypes:




Dihybrid Crosses






F1

Punnet Grid:

F0Genotype: SsYy SsYY



Phenotypes:




Dihybrid Crosses






F1

gametes SY sYSY

Sy

sY

sy

Punnet Grid:




Phenotypes:




Dihybrid Crosses






F1

gametes SY sYSY SSYY SsYY

Sy SSYy SsYy

sY SsYY ssYY

sy SsYy ssYy

Punnet Grid:




Phenotypes:




Dihybrid Crosses






F1

gametes SY sYSY SSYY SsYY

Sy SSYy SsYy

sY SsYY ssYY

sy SsYy ssYy

Punnet Grid:




Phenotypes: 3 Smooth, yellow : 1 Rough, yellow Present the ratio in the simplest mathematical form.

6 Smooth, yellow : 2 Rough, yellow




Dihybrid Crosses


Common expected ratios of dihybrid crosses.

SY Sy sY sy

SY SSYY SSYy SsYY SsYy




Heterozygous at both loci


SsYySsYy

9 : 3 : 3 : 1

SY sY

SY SSYY SsYY

Sy SSYy SsYy

sY SsYY ssYY

sy SsYy ssYy


Heterozygous at one locus, homozygous dominant at the other

SsYySsYy

3 : 2

Sy sy

SY SSYy SsYy

Sy SSyy Ssyy

sY SsYy ssYy

sy Ssyy ssyy


SsyySsYy

4 : 3 : 1

Heterozygous/Homozygous recessive

ssYYSSyy = All SsYy

ssYySsyy = 1 : 1 : 1 : 1

ssyySSYY = all SyYy




Dihybrid Crosses





A rough yellow pea is test crossed to determine its genotype.

Punnet Grid:

F0Genotype:

Phenotype: Rough, yellow

F1Phenotypes:




Dihybrid Crosses






Punnet Grid:

F0Genotype: ssYy


F1

gametes sY sy

Phenotypes:




Dihybrid Crosses






F1

gametes sY sy sY sYPunnet Grid:

F0Genotype: ssYy or ssYY


Phenotypes:




Dihybrid Crosses






F1

gametes sY sy sY sYAll sy

Punnet Grid:

F0Genotype: ssYy or ssYY ssyy


Phenotypes:

Remember: A test cross is the unknown with a known homozygous recessive.




Dihybrid Crosses






F1

gametes sY sy sY sYAll sy ssYy ssyy ssYy ssYy

Punnet Grid:



Phenotypes:




Dihybrid Crosses






F1

gametes sY sy sY sYAll sy ssYy ssyy ssYy ssYy

Punnet Grid:



Phenotypes:

Some green peas will be present in the offspring if the unknown parent

genotype is ssYy.

No green peas will be present in the offspring if the unknown parent genotype is ssYY.




Dihybrid Crosses





A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring.

F1

Punnet Grid:

F0Genotype:

Phenotype: Smooth, green

Phenotypes:




Dihybrid Crosses






F1

gametes

All sy

Punnet Grid:

F0Genotype: ssyy


Phenotypes:




Dihybrid Crosses






F1

gametes Sy SyAll sy Ssyy Ssyy

Punnet Grid:

F0Genotype: SSyy ssyy


Phenotypes:




Dihybrid Crosses






F1

gametes Sy Sy Sy syAll sy Ssyy Ssyy Ssyy ssyy

Punnet Grid:

F0Genotype: SSyy or Ssyy ssyy


Phenotypes:




Dihybrid Crosses






F1

gametes Sy Sy Sy syAll sy Ssyy Ssyy Ssyy ssyy

Punnet Grid:

F0Genotype: SSyy or Ssyy ssyy


Phenotypes:

No rough peas will be present in the offspring if the unknown parent

genotype is SSyy.

The presence of rough green peas in the offspring means that the unknown genotype must be Ssyy.

The expected ratio in this cross is 3 smooth green : 1 rough green. This is not the same as the outcome. Remember that each reproduction event is chance and the sample size is very small. With a much larger sample size, the outcome would be closer to the expected ratio, simply due to probability.





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 anitgens presentIB = type B antigens present





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 recognises '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://nobelprize.org/educational/medicine/landsteiner/readmore.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://nobelprize.org/educational/medicine/landsteiner/readmore.html


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?






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






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

gametesPunnet Grid:

Genotypes:

Phenotypes:

F0Genotype:

Phenotype: carrier affected

Phenotype ratio: :









F1

gametes HbS HbS

HbA HbAHbS HbAHbS

HbS HbSHbS HbSHbS

Punnet Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA Hbs HbS Hbs

Phenotype: carrier affected

Phenotype ratio:

Carrier & Sickle cell

1 : 1

HbAHbS & HbSHbS








F1

gametesPunnet Grid:

Genotypes:

Phenotypes:

F0Genotype:

Phenotype: carrier carrier

Phenotype ratio:









F1

gametes HbA HbS

HbA HbAHbA HbAHbS

HbS HbAHbS HbSHbS

Punnet Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA HbS HbA HbS


Phenotype ratio:

Unaffected & Carrier & Sickle cell

1: 2 : 1

HbAHb & 2 HbAHbS & HbSHbS








F1

gametes

HbA

HbS

Punnet Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA HbS

Phenotype: carrier unknown

Phenotype ratio:








F1

gametes HbA HbA HbA HbS

HbA

HbS

Punnet Grid:

Genotypes:

Phenotypes:

F0Genotype: HbA HbS HbA HbA or HbA HbS


Phenotype ratio:








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

Punnet 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





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


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










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


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: - haemophilia- colour 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




What number do you see?






What number do you see?

5 = normal vision2 = red/green colour blindness






How is colour-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 colour-blindness.

Xq28Key to alleles:N = normal visionn = red/green colour 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:






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

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

XN Xn XN YNormal maleCarrier female X

F1

Punnet Grid:

F0 Genotype:

Phenotype:









XN

Xn

XN YXN XN

XN Xn

XN Y

Xn YF1

Punnet Grid:

F0 Genotype:

Phenotype:









XN

Xn

XN YXN XN

XN Xn

XN Y

Xn YF1

Punnet 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 colour-blind male and a homozygous normal female?b. a normal male and a colour-blind female?





Red-Green Colour 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 colour 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.





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

http://illinoisreview.typepad.com/illinoisreview/2015/02/myths-and-facts-about-parcc-in-illinois.html



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

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

key female male

affected

Not Affected

deceased


Phenylketonuria (PKU) Clinical example.

Pedigree 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.

?Is PKU dominant or recessive? How do you know?• •

A B

I

II

III




key female male

affected

Not Affected

deceased



Pedigree 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.

?Is PKU dominant or recessive? How do you know?• Recessive • Unaffected mother in Gen I has produced

affected II A. Mother must have been a carrier.

A B

I

II

III






A mis-sense mutation in the gene that produces tyrosine hydroxylase means that phenylalanine cannot be converted to tyrosine in the body - so it builds up.

This results in brain developmental problems and seizures. It is progressive, so it must be diagnosed and treated early.

Dairy, breastmilk, meat, nuts and aspartame must be avoided, as they are rich in phenylalanine.

The Boy with PKU ideo clip from:http://www.youtube.com/watch?v=KUJVujhHxPQ

Diagnosis- blood test taken at 6-7 days after birthhttp://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/




http://www.youtube.com/watch?v=KUJVujhHxPQ

http://www.youtube.com/watch?v=KUJVujhHxPQ

http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/




Chromosome 12 from:http://commons.wikimedia.org/wiki/File:Chromosome_12.svg


A recessive mis-sense mutation in the gene that produces tyrosine hydroxylase means that phenylalanine cannot be converted to tyrosine in the body - so it builds up.

Genetics review: 1. What is a missense mutation?

2. Is this disorder autosomal or sex-linked?

3. What is the locus of the tyrosine hydroxlase gene?




http://commons.wikimedia.org/wiki/File:Chromosome_12.svg





Chromosome 12 from:http://commons.wikimedia.org/wiki/File:Chromosome_12.svg


A recessive mis-sense mutation in the gene that produces tyrosine hydroxylase means that phenylalanine cannot be converted to tyrosine in the body - so it builds up.

Genetics review: 1. What is a missense mutation?It is a base-substitution mutation where the change in a single base results in a different amino acid being produced in the polypeptide.

2. Is this disorder autosomal or sex-linked?Autosomal – chromosome 12

3. What is the locus of the tyrosine hydroxlase gene? 12q22 - 24









What is the probability of two parents who are both carriers of the recessive allele producing children affected by PKU?

Key to alleles:T = Normal enzymet = faulty enzyme

F1

gametes T tT

t

Punnet Grid:

Genotypes:

Phenotypes:

F0Genotype: T t T t


Phenotype ratio:






What is the probability of two parents who are both carriers of the recessive allele producing children affected by PKU?

Key to alleles:T = Normal enzymet = faulty enzyme

Therefore 25% chance of a child with PKU

F1

gametes T tT TT Tt

t Tt tt

Punnet Grid:

TT Tt Tt ttGenotypes:

Phenotypes:

F0Genotype: T t T t


Phenotype ratio:

Normal enzyme PKU

3 : 1





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





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





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





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





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







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?







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




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








Key: female male

affected

Not Affected

deceased



2. Alice


4. Britney

Xh Y








Key: female male

affected

Not Affected

deceased



2. Alice


4. Britney

Xh Y

XH Xh








Key: female male

affected

Not Affected

deceased



2. Alice


4. Britney

Xh Y

XH Xh

XH Y or Xh Y








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






Pedigree Chart PracticeKey: female male

affected

Not Affected

deceased

Dominant or Recessive? Autosomal or Sex-linked?






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?






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 punnet grids if needed.





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%






Key: female male

affected

Not Affected

deceased


A. 0%

B. 12.5%

C. 25%

D. 50%







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






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






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






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!





Whirling Gene activity from the awesome Learn.Genetics site:http://learn.genetics.utah.edu/archive/pedigree/mapgene.html



http://learn.genetics.utah.edu/archive/pedigree/mapgene.html



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