theoretical genetics
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Theoretical Genetics. Stephen Taylor. http://sciencevideos.wordpress.com. T his image shows a pair of homologous chromosomes. Name and annotate the labeled features. . Definitions. http://sciencevideos.wordpress.com. T his image shows a pair of homologous chromosomes. - PowerPoint PPT PresentationTRANSCRIPT
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.
Definitions
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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
Making Babies
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Activity from:http://www.nclark.net/Genetics
1. Count the chromosomes in your envelope - there should be 46 in total.
2. Shuffle the chromosomes, so that they are well mixed up. Which aspects of meiosis and sexual reproduction give genetic variation?
3. Now arrange them in a karyotype (don't turn them over - leave them as they were).
4. What is the gender of your baby? Explain how gender is inherited in humans.
• Crossing-over in prophase I • Random orientation in metaphase I and II• Random fertilisation
Making Babies
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 5
Activity from:http://www.nclark.net/Genetics
• Crossing-over in prophase I • Random orientation in metaphase I and II• Random fertilisation
List all the traits in a table. Use the key above to determine the genotypes and phenotypes of your offspring. Draw a picture of your beautiful child’s face!
HL identify traits which are polygenic, involve gene interactions and some which are linked.
Explain this
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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
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“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
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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
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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.
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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.
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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Career-related Case Study
From the Times Higher Education Supplement – “So Last Century”http://www.timeshighereducation.co.uk/story.asp?sectioncode=26&storycode=415941&c=2
“According to the US Bureau of Labor Statistics, the graduate of today will change career four to six times in a lifetime. By one estimate, 65 per cent of the jobs that will be available upon college graduation for students now entering high school (that's eight years from now) do not yet exist. Consider the new interdisciplinary field of genetic counselling, which combines biological science with social work and ethics - it was ranked as one of the "top 10" career choices of 2010 because it offered far more openings than could be filled by qualified applicants.”
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 21
Career-related Case Study
From the Times Higher Education Supplement – “So Last Century”http://www.timeshighereducation.co.uk/story.asp?sectioncode=26&storycode=415941&c=2
“According to the US Bureau of Labor Statistics, the graduate of today will change career four to six times in a lifetime. By one estimate, 65 per cent of the jobs that will be available upon college graduation for students now entering high school (that's eight years from now) do not yet exist. Consider the new interdisciplinary field of genetic counselling, which combines biological science with social work and ethics - it was ranked as one of the "top 10" career choices of 2010 because it offered far more openings than could be filled by qualified applicants.”
You are a genetic counselor. A couple walk into your clinic and are concerned about their pregnancy. They each have one parent who is affected by phenylketonuria (PKU) and one parent who has no family history. Explain PKU and its inheritance to them. Deduce the chance of having a child with PKU and how it can be tested and treated.
Use the following tools in your explanations:• Pedigree chart• Punnet grid• Diagrams
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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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
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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
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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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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
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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: :
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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
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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:
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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
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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:
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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:
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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
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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 blindnessXN Xn XN Y
Normal 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 blindnessXN Xn XN Y
Normal maleCarrier female X
XN
Xn
XN YXN XN
XN Xn
XN YXn Y
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 blindnessXN Xn XN Y
Normal maleCarrier female X
XN
Xn
XN YXN XN
XN Xn
XN YXn Y
F1
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
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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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 58
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 59
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 62
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 63
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 64
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%
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 69
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 70
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 71
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 72
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
http://sciencevideos.wordpress.com 4.3 Theoretical Genetics 73
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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