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Theoretical GeneticsStephen Taylor
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 1
Definitions
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 2
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 3
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
Segregation
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 4
“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).
Mendel from:http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
Segregation
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 5
Mendel from:http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
“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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 6
Mendel from:http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
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.
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.
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 7
Mendel from:http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
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.
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).
gametes Y yY YY Yy
y Yy yy
Punnet Grid:
Genotypes:
Phenotypes:
Phenotype ratio:
Monohybrid Cross Crossing a single trait.
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 8
Mendel from:http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
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.
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).
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.
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 9
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 10
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
Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross?
Homozygous recessiveHomozygous recessive
Phenotype:Key to alleles:Y = yellowy = green
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 11
F0
F1
Genotype:
gametesPunnet Grid:
Genotypes:
Phenotypes:
Phenotype ratio:
Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross?
Phenotype:
HeterozygousHomozygous recessive
Key to alleles:Y = yellowy = green
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 12
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
Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross?
HeterozygousHomozygous recessive
Phenotype:Key to alleles:Y = yellowy = green
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 13
F0
F1
Genotype:
gametesPunnet Grid:
Genotypes:
Phenotypes:
Phenotype ratio:
Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross?
Phenotype:
HeterozygousHomozygous dominant
Key to alleles:Y = yellowy = green
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 14
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
Monohybrid Cross What is the expected ratio of phenotypes in this monohybrid cross?
HeterozygousHomozygous dominant
Phenotype:Key to alleles:Y = yellowy = green
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 15
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 16
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
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 18
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: SsYy SsYy
Phenotype:Heterozygous at both loci Heterozygous at both loci
Smooth, yellow Smooth, yellow
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 19
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
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:
F0Genotype: SsYy SsYy
Phenotype:Heterozygous at both loci Heterozygous at both loci
Smooth, yellow Smooth, yellow
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 20
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
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:
F0Genotype: SsYy SsYy
Phenotype:Heterozygous at both loci Heterozygous at both loci
Smooth, yellow Smooth, yellow
Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 21
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.
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
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 22
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.
Calculate the predicted phenotype ratio for:
F1
Punnet Grid:
F0Genotype: SsYy SsYY
Phenotype:Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 23
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.
Calculate the predicted phenotype ratio for:
F1
gametes SY sYSY
Sy
sY
sy
Punnet Grid:
F0Genotype: SsYy SsYY
Phenotype:Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 24
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.
Calculate the predicted phenotype ratio for:
F1
gametes SY sYSY SSYY SsYY
Sy SSYy SsYy
sY SsYY ssYY
sy SsYy ssYy
Punnet Grid:
F0Genotype: SsYy SsYY
Phenotype:Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 25
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.
Calculate the predicted phenotype ratio for:
F1
gametes SY sYSY SSYY SsYY
Sy SSYy SsYy
sY SsYY ssYY
sy SsYy ssYy
Punnet Grid:
F0Genotype: SsYy SsYY
Phenotype:Heterozygous at both loci Heterozygous for S, homozygous dominant for Y
Smooth, yellow Smooth, yellow
Phenotypes: 3 Smooth, yellow : 1 Rough, yellow Present the ratio in the simplest mathematical form.
6 Smooth, yellow : 2 Rough, yellow
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 26
Common expected ratios of dihybrid crosses.
SY Sy sY sy
SY SSYY SSYy SsYY SsYy
Sy SSYy SSyy SsYy Ssyy
sY SsYY SsYy ssYY ssYy
sy SsYy Ssyy ssYy ssyy
Heterozygous at both loci
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 both loci
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
Heterozygous at both loci
SsyySsYy
4 : 3 : 1
Heterozygous/Homozygous recessive
ssYYSSyy = All SsYy
ssYySsyy = 1 : 1 : 1 : 1
ssyySSYY = all SyYy
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 27
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.
A rough yellow pea is test crossed to determine its genotype.
Punnet Grid:
F0Genotype:
Phenotype: Rough, yellow
F1Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 28
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.
A rough yellow pea is test crossed to determine its genotype.
Punnet Grid:
F0Genotype: ssYy
Phenotype: Rough, yellow
F1
gametes sY sy
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 29
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.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sYPunnet Grid:
F0Genotype: ssYy or ssYY
Phenotype: Rough, yellow
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 30
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.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sYAll sy
Punnet Grid:
F0Genotype: ssYy or ssYY ssyy
Phenotype: Rough, yellow
Phenotypes:
Remember: A test cross is the unknown with a known homozygous recessive.
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 31
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.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sYAll sy ssYy ssyy ssYy ssYy
Punnet Grid:
F0Genotype: ssYy or ssYY ssyy
Phenotype: Rough, yellow
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 32
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.
A rough yellow pea is test crossed to determine its genotype.
F1
gametes sY sy sY sYAll sy ssYy ssyy ssYy ssYy
Punnet Grid:
F0Genotype: ssYy or ssYY ssyy
Phenotype: Rough, yellow
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
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 33
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.
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
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 34
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.
A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring.
F1
gametes
All sy
Punnet Grid:
F0Genotype: ssyy
Phenotype: Smooth, green
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 35
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.
A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring.
F1
gametes Sy SyAll sy Ssyy Ssyy
Punnet Grid:
F0Genotype: SSyy ssyy
Phenotype: Smooth, green
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 36
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.
A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring.
F1
gametes Sy Sy Sy syAll sy Ssyy Ssyy Ssyy ssyy
Punnet Grid:
F0Genotype: SSyy or Ssyy ssyy
Phenotype: Smooth, green
Phenotypes:
Dihybrid Crosses
http://sciencevideos.wordpress.com 10.2 Dihybrid Crosses & Gene Linkage 37
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.
A smooth green pea is test crossed. Deduce the genotype. Smooth green = nine offspring. Rough green = one offspring.
F1
gametes Sy Sy Sy syAll sy Ssyy Ssyy Ssyy ssyy
Punnet Grid:
F0Genotype: SSyy or Ssyy ssyy
Phenotype: Smooth, green
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.
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 38
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 39
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://sciencevideos.wordpress.com 4.3 Theoretical Genetics 40
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?
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 41
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 HbA HbA HbA HbS HbS HbS
Description Homozygous HbA Heterozygous Homozygous HbS
Phenotype normal carrier Sickle cell disease
Malaria protection? No Yes Yes
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 42
Sickle Cell Another example of codominance.
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: :
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 43
Sickle Cell Another example of codominance.
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
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 44
Sickle Cell Another example of codominance.
Predict the phenotype ratio in this cross:
Key to alleles:HbA = Normal HbHbS = Sickle cell
F1
gametesPunnet Grid:
Genotypes:
Phenotypes:
F0Genotype:
Phenotype: carrier carrier
Phenotype ratio:
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 45
Sickle Cell Another example of codominance.
Predict the phenotype ratio in this cross:
Key to alleles:HbA = Normal HbHbS = Sickle cell
Therefore 25% chance of a child with sickle cell disease.
F1
gametes HbA HbS
HbA HbAHbA HbAHbS
HbS HbAHbS HbSHbS
Punnet Grid:
Genotypes:
Phenotypes:
F0Genotype: HbA HbS HbA HbS
Phenotype: carrier carrier
Phenotype ratio:
Unaffected & Carrier & Sickle cell
1: 2 : 1
HbAHb & 2 HbAHbS & HbSHbS
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 46
Sickle Cell Another example of codominance.
Predict the phenotype ratio in this cross:
Key to alleles:HbA = Normal HbHbS = Sickle cell
F1
gametes
HbA
HbS
Punnet Grid:
Genotypes:
Phenotypes:
F0Genotype: HbA HbS
Phenotype: carrier unknown
Phenotype ratio:
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 47
Sickle Cell Another example of codominance.
Predict the phenotype ratio in this cross:
Key to alleles:HbA = Normal HbHbS = Sickle cell
F1
gametes HbA HbA HbA HbS
HbA
HbS
Punnet Grid:
Genotypes:
Phenotypes:
F0Genotype: HbA HbS HbA HbA or HbA HbS
Phenotype: carrier unknown
Phenotype ratio:
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 48
Sickle Cell Another example of codominance.
Predict the phenotype ratio in this cross:
Key to alleles:HbA = Normal HbHbS = Sickle cell
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: carrier unknown
Phenotype ratio:
3 Unaffected & 4 Carrier & 1 Sickle cell
3 : 4 : 1
3 HbAHbA & 4 HbAHbS & 1 HbSHbS
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 49
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
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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
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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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
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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: - haemophilia- colour blindness
X and Y SEM fromhttp://www.angleseybonesetters.co.uk/bones_DNA.html
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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Sex Linkage X and Y chromosomes are non-homologous.
What number do you see?
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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Sex Linkage X and Y chromosomes are non-homologous.
What number do you see?
5 = normal vision2 = red/green colour blindness
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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Sex Linkage X and Y chromosomes are non-homologous.
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:
http://en.wikipedia.org/wiki/Y_chromosome
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Sex Linkage X and Y chromosomes are non-homologous.
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:
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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Sex Linkage X and Y chromosomes are non-homologous.
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
XN
Xn
XN YXN XN
XN Xn
XN Y
Xn YF1
Punnet Grid:
F0 Genotype:
Phenotype:
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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Sex Linkage X and Y chromosomes are non-homologous.
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
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
Normal female Normal male
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?
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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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.
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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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
key female male
affected
Not Affected
deceased
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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
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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?• Recessive • Unaffected mother in Gen I has produced
affected II A. Mother must have been a carrier.
A B
I
II
III
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Phenylketonuria (PKU) Clinical example.
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/
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Phenylketonuria (PKU) Clinical example.
Chromosome 12 from:http://commons.wikimedia.org/wiki/File:Chromosome_12.svg
Diagnosis- blood test taken at 6-7 days after birthhttp://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/
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?
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Phenylketonuria (PKU) Clinical example.
Chromosome 12 from:http://commons.wikimedia.org/wiki/File:Chromosome_12.svg
Diagnosis- blood test taken at 6-7 days after birthhttp://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/
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
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Phenylketonuria (PKU) Clinical example.
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: carrier carrier
Phenotype ratio:
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Phenylketonuria (PKU) Clinical example.
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: carrier carrier
Phenotype ratio:
Normal enzyme PKU
3 : 1
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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
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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
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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
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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
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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
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.
Key to alleles:XH = healthy clotting factorsXh = no clotting factor
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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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?
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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Hemophilia results from a lack of clotting factors. These are globular proteins, which act as enzymes in the clotting pathway.
Chromosome images from Wikipedia:http://en.wikipedia.org/wiki/Y_chromosome
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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
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Hemophilia Pedigree chart practice
Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp
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
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Hemophilia Pedigree chart practice
Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp
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
Xh Y
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Hemophilia Pedigree chart practice
Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp
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
Xh Y
XH Xh
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Hemophilia Pedigree chart practice
Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp
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
Xh Y
XH Xh
XH Y or Xh Y
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Hemophilia Pedigree chart practice
Royal Family Pedigree Chart from: http://www.sciencecases.org/hemo/hemo.asp
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
Xh Y
XH Xh
XH Y or Xh Y
XH XH or XH Xh
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Pedigree Chart PracticeKey: female male
affected
Not Affected
deceased
Dominant or Recessive? Autosomal or Sex-linked?
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Pedigree Chart PracticeKey: female male
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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Pedigree Chart PracticeKey: female male
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.
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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%
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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 to alleles:XH = healthy clotting factorsXh = no clotting factor
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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 to alleles:XH = healthy clotting factorsXh = no clotting factor
What do we know? A = XH Y B = XH Xh (because G = Xh Y) E = XH Y
XH
Y
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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 to alleles:XH = healthy clotting factorsXh = no clotting factor
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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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 to alleles:XH = healthy clotting factorsXh = no clotting factor
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 XH XH XH XH XH XH XH Xh
Y XH Y XH Y XH Y Xh Y
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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 to alleles:XH = healthy clotting factorsXh = no clotting factor
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 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!
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Whirling Gene activity from the awesome Learn.Genetics site:http://learn.genetics.utah.edu/archive/pedigree/mapgene.html
For more IB Biology resources:http://sciencevideos.wordpress.com
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