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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.

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Theoretical Genetics


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theoretical genetics

Theoretical Genetics

Stephen Taylor

4.3 Theoretical Genetics

http://sciencevideos.wordpress.com

definitions

This image shows a pair of homologous chromosomes.

Name and annotate the labeled features.

Definitions

4.3 Theoretical Genetics

http://sciencevideos.wordpress.com

definitions1

This image shows a pair of homologous chromosomes.

Name and annotate the labeled features.

Genotype

The combination of alleles

of a gene carried by an organism

Homozygous dominant

Having two copies of the same dominant allele

Phenotype

The expression of alleles

of a gene carried by an organism

Homozygous recessive

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

Definitions

Centromere

Joins chromatids in cell division

Codominant

Pairs of alleles which are both expressed when present.

Alleles

Different versions of a gene

Dominant alleles = capital letter

Recessive alleles = lower-case letter

Heterozygous

Having two different alleles.

The dominant allele is expressed.

Gene loci

Specific positions of genes on a chromosome

Carrier

Heterozygous carrier of a

recessive disease-causing allele

4.3 Theoretical Genetics

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making babies

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.

Making Babies

  • Crossing-over in prophase I
  • Random orientation in metaphase I and II
  • Random fertilisation

Activity from:

http://www.nclark.net/Genetics

4.3 Theoretical Genetics

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making babies1

Making Babies

  • 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 interactionsand some which are linked.

Activity from:

http://www.nclark.net/Genetics

4.3 Theoretical Genetics

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explain this

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

Explain this

Mendel from:

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

4.3 Theoretical Genetics

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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

Mendel from:

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

4.3 Theoretical Genetics

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segregation1

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

F0

Genotype:

Y y

Y y

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

Segregation

Y or y

Y or y

Gametes:

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

Mendel from:

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

4.3 Theoretical Genetics

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mono hybrid cross

Crossing a singletrait.

F0

Genotype:

Y y

Y 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).

Monohybrid Cross

Y or y

Y or y

Gametes:

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

Mendel from:

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

4.3 Theoretical Genetics

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mono hybrid cross1

Crossing a singletrait.

F0

Genotype:

Y y

Y 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).

Monohybrid Cross

Y or y

Y or y

Gametes:

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

Mendel from:

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

4.3 Theoretical Genetics

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mono hybrid cross2

Crossing a singletrait.

F0

Genotype:

Y y

Y 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.

Monohybrid Cross

Y or y

Y or y

Gametes:

Punnet Grid:

Yy

yy

Yy

YY

F1

Genotypes:

Phenotypes:

3 : 1

Phenotype ratio:

Mendel from:

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

4.3 Theoretical Genetics

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mono hybrid cross3

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

F0

Key to alleles:

Y = yellow

y = green

Phenotype:

Genotype:

Monohybrid Cross

Homozygous recessive

Homozygous recessive

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

4.3 Theoretical Genetics

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mono hybrid cross4

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

F0

Key to alleles:

Y = yellow

y = green

Phenotype:

Genotype:

yy

y y

Monohybrid Cross

Homozygous recessive

Homozygous recessive

Punnet Grid:

yy

yy

yy

yy

F1

Genotypes:

Phenotypes:

All green

Phenotype ratio:

4.3 Theoretical Genetics

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mono hybrid cross5

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

F0

Key to alleles:

Y = yellow

y = green

Phenotype:

Genotype:

Monohybrid Cross

Homozygous recessive

Heterozygous

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

4.3 Theoretical Genetics

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mono hybrid cross6

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

F0

Key to alleles:

Y = yellow

y = green

Phenotype:

Genotype:

yy

Yy

Monohybrid Cross

Homozygous recessive

Heterozygous

Punnet Grid:

yy

yy

Yy

Yy

F1

Genotypes:

Phenotypes:

1 : 1

Phenotype ratio:

4.3 Theoretical Genetics

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mono hybrid cross7

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

F0

Key to alleles:

Y = yellow

y = green

Phenotype:

Genotype:

Monohybrid Cross

Homozygous dominant

Heterozygous

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

4.3 Theoretical Genetics

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mono hybrid cross8

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

F0

Key to alleles:

Y = yellow

y = green

Phenotype:

Genotype:

Y Y

Yy

Monohybrid Cross

Homozygous dominant

Heterozygous

Punnet Grid:

Yy

Yy

YY

YY

F1

Genotypes:

Phenotypes:

All yellow

Phenotype ratio:

4.3 Theoretical Genetics

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test cross

Used to determine the genotype of an unknown individual.

The unknown is crossed with a known homozygous recessive.

F0

Key to alleles:

R = Red flower

r = white

Phenotype:

R ?

Genotype:

r r

Test Cross

unknown

Homozygous recessive

Possible outcomes:

F1

Phenotypes:

Unknown parent = RR

Unknown parent = Rr

4.3 Theoretical Genetics

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test cross1

Used to determine the genotype of an unknown individual.

The unknown is crossed with a known homozygous recessive.

F0

Key to alleles:

R = Red flower

r = white

Phenotype:

R ?

Genotype:

r r

Test Cross

unknown

Homozygous recessive

Possible outcomes:

F1

Some white, some red

All red

Phenotypes:

Unknown parent = RR

Unknown parent = Rr

4.3 Theoretical Genetics

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career related case study

“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.”

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

4.3 Theoretical Genetics

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career related case study1

“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.”

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

  • 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

4.3 Theoretical Genetics

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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.

I

Phenylketonuria (PKU)

II

A

B

III

?

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

4.3 Theoretical Genetics

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phenylketonuria pku1

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.

I

Phenylketonuria (PKU)

II

A

B

III

?

  • 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.

4.3 Theoretical Genetics

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phenylketonuria pku2

Clinical example.

A mis-sensemutation 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.

Phenylketonuria (PKU)

Diagnosis- blood test taken at 6-7 days after birth

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

The Boy with PKU ideo clip from:

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

4.3 Theoretical Genetics

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phenylketonuria pku3

Clinical example.

A recessivemis-sensemutation 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?

Phenylketonuria (PKU)

Chromosome 12 from:

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

Diagnosis- blood test taken at 6-7 days after birth

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

4.3 Theoretical Genetics

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phenylketonuria pku4

Clinical example.

A recessivemis-sensemutation 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

Phenylketonuria (PKU)

Chromosome 12 from:

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

Diagnosis- blood test taken at 6-7 days after birth

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

4.3 Theoretical Genetics

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phenylketonuria pku5

Clinical example.

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

F0

Key to alleles:

T = Normal enzyme

t = faulty enzyme

Phenotype:

carrier

carrier

T t

T t

Phenylketonuria (PKU)

Genotype:

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

4.3 Theoretical Genetics

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phenylketonuria pku6

Clinical example.

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

F0

Key to alleles:

T = Normal enzyme

t = faulty enzyme

Phenotype:

carrier

carrier

T t

T t

Phenylketonuria (PKU)

Genotype:

Punnet Grid:

Tt

tt

Tt

TT

F1

Genotypes:

Phenotypes:

Normal enzyme

PKU

Therefore 25% chance of a child with PKU

3 : 1

Phenotype ratio:

4.3 Theoretical Genetics

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pedigree charts

Key to alleles:

T= Has enzyme

t= no enzyme

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.

Pedigree Charts

Looks like

4.3 Theoretical Genetics

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pedigree charts1

Key to alleles:

T= Has enzyme

t= no enzyme

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.

Pedigree Charts

Looks like

4.3 Theoretical Genetics

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pedigree charts2

Key to alleles:

T= Has enzyme

t= no enzyme

Individuals D and $ are planning to have another child.

Calculate the chances of the child having PKU.

Pedigree Charts

$

Looks like

Genotypes:

D =

$ =

Phenotype ratio

Therefore

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pedigree charts3

Key to alleles:

T= Has enzyme

t= no enzyme

Individuals D and $ are planning to have another child.

Calculate the chances of the child having PKU.

Pedigree Charts

$

Looks like

Genotypes:

D = Tt (carrier)

$ = tt (affected)

Phenotype ratio

1 : 1 Normal : PKU

Therefore 50% chance of a child with PKU

4.3 Theoretical Genetics

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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 present

IA= type A anitgens present

IB = type B antigens present

Codominance

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more about blood typing

A Nobel breakthrough in medicine.

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.

More about blood typing

Blood typing game from Nobel.org:

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

Images and more information from:

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

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sickle cell

Another example of codominance.

Remember the notation used: superscripts represent codominant alleles.

In codominance, heterozygous individuals have amixed phenotype.

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

Complete the table for these individuals:

Sickle Cell

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sickle cell1

Another example of codominance.

Remember the notation used: superscripts represent codominant alleles.

In codominance, heterozygous individuals have amixed phenotype.

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

Complete the table for these individuals:

Sickle Cell

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sickle cell2

Key to alleles:

HbA = Normal Hb

HbS= Sickle cell

Another example of codominance.

Predict the phenotype ratio in this cross:

F0

Phenotype:

carrier

affected

Genotype:

Sickle Cell

Punnet Grid:

F1

Genotypes:

Phenotypes:

:

Phenotype ratio:

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

4.3 Theoretical Genetics

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sickle cell3

Key to alleles:

HbA = Normal Hb

HbS= Sickle cell

Another example of codominance.

Predict the phenotype ratio in this cross:

F0

Phenotype:

carrier

affected

HbAHbs

HbSHbs

Genotype:

Sickle Cell

Punnet Grid:

HbAHbS&HbSHbS

F1

Genotypes:

Carrier & Sickle cell

Phenotypes:

1 : 1

Phenotype ratio:

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

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sickle cell4

Key to alleles:

HbA = Normal Hb

HbS= Sickle cell

Another example of codominance.

Predict the phenotype ratio in this cross:

F0

Phenotype:

carrier

carrier

Genotype:

Sickle Cell

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

4.3 Theoretical Genetics

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sickle cell5

Key to alleles:

HbA = Normal Hb

HbS= Sickle cell

Another example of codominance.

Predict the phenotype ratio in this cross:

F0

Phenotype:

carrier

carrier

HbAHbS

HbAHbS

Genotype:

Sickle Cell

Punnet Grid:

HbAHb& 2 HbAHbS&HbSHbS

F1

Genotypes:

Unaffected & Carrier & Sickle cell

Phenotypes:

1: 2 : 1

Phenotype ratio:

Therefore 25% chance of a child with sickle cell disease.

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sickle cell6

Key to alleles:

HbA = Normal Hb

HbS= Sickle cell

Another example of codominance.

Predict the phenotype ratio in this cross:

F0

Phenotype:

carrier

unknown

HbAHbS

Genotype:

Sickle Cell

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

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sickle cell7

Key to alleles:

HbA = Normal Hb

HbS= Sickle cell

Another example of codominance.

Predict the phenotype ratio in this cross:

F0

Phenotype:

carrier

unknown

HbAHbS

HbAHbAorHbAHbS

Genotype:

Sickle Cell

Punnet Grid:

F1

Genotypes:

Phenotypes:

Phenotype ratio:

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sickle cell8

Key to alleles:

HbA = Normal Hb

HbS= Sickle cell

Another example of codominance.

Predict the phenotype ratio in this cross:

F0

Phenotype:

carrier

unknown

HbAHbS

HbAHbAorHbAHbS

Genotype:

Sickle Cell

Punnet Grid:

3 HbAHbA& 4 HbAHbS&1HbSHbS

F1

Genotypes:

3 Unaffected & 4 Carrier & 1 Sickle cell

Phenotypes:

3 : 4 : 1

Phenotype ratio:

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

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sex determination

It’s all about X and Y…

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.

Sex Determination

Karyotype of a human male, showing X and Y chromosomes:

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

SRY

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.

Chromosome images from Wikipedia:

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

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sex determination1

It’s all about X and Y…

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.

Sex Determination

Segregation of the sex chromosomes in meiosis.

SRY gene determines maleness.

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

Chromosome images from Wikipedia:

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

Therefore there is an even chance*

of the offspring being male or female.

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

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sex determination2

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 Syndrome

Monosomy of X, leads to short stature, female children.

XXX Syndrome:

Fertile females. Some X-carrying gametes can be produced.

XXY: KlinefelterSyndrome:

Males with enhanced female characteristics

Sex Determination

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.

The sex chromosomes are non-homologous.

There are many genes on the X-chromosome

which 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.

Non-homologous region

Sex Linkage

Non-homologous region

Examples of sex-linked genetic disorders:

- haemophilia

- colour blindness

X and Y SEM from

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

Chromosome images from Wikipedia:

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

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sex linkage1

X and Y chromosomes are non-homologous.

What number do you see?

Sex Linkage

Chromosome images from Wikipedia:

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

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sex linkage2

X and Y chromosomes are non-homologous.

What number do you see?

5 = normal vision

2 = red/green colour blindness

Sex Linkage

Chromosome images from Wikipedia:

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

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sex linkage3

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.

Sex Linkage

XNXN

XNY

no allele carried, none written

Normal female

Normal male

Key to alleles:

N = normal vision

n = red/green colour

blindness

Xq28

XnY

XnXn

Affected female

Affected male

XNXn

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

Carrier female

Chromosome images from Wikipedia:

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

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sex linkage4

X and Y chromosomes are non-homologous.

Key to alleles:

N = normal vision

n = red/green colour

blindness

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

F0

XNXn

XNY

Genotype:

X

Phenotype:

Carrier female

Normal male

Sex Linkage

Punnet Grid:

F1

Chromosome images from Wikipedia:

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

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sex linkage5

X and Y chromosomes are non-homologous.

Key to alleles:

N = normal vision

n = red/green colour

blindness

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

F0

XNXn

XNY

Genotype:

X

Phenotype:

Carrier female

Normal male

Sex Linkage

XN

Y

Punnet Grid:

XNXN

XN

XNY

F1

XnY

XNXn

  • Xn

Chromosome images from Wikipedia:

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

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sex linkage6

X and Y chromosomes are non-homologous.

Key to alleles:

N = normal vision

n = red/green colour

blindness

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

F0

XNXn

XNY

Genotype:

X

Phenotype:

Carrier female

Normal male

Sex Linkage

XN

Y

Punnet Grid:

XNXN

XN

XNY

Normal male

Normal female

F1

XnY

XNXn

  • Xn

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?

Chromosome images from Wikipedia:

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

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red green colour blindness

How does it work?

The OPN1MW and OPN1LWgenes 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.

Red-GreenColour Blindness

Xq28

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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slide55

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.

Hemophilia

XHXH

XH Y

no allele carried, none written

Normal female

Normal male

Key to alleles:

XH = healthy clotting factors

Xh= no clotting factor

Xh Y

XhXh

Affected female

Affected male

XHXh

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

Heterozygous females are carriers.

Carrier female

Chromosome images from Wikipedia:

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

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hemophilia1

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

Chromosome images from Wikipedia:

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

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hemophilia2

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

Hemophilia

Chromosome images from Wikipedia:

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

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hemophilia3

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

Hemophilia

Royal Family Pedigree Chart from:

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

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hemophilia4

Pedigree chart practice

State the genotypes of the following family members:

Leopold

Alice

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

Britney

Hemophilia

Key to alleles:

H = healthy clotting factors

h= no clotting factor

Royal Family Pedigree Chart from:

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

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hemophilia5

Pedigree chart practice

State the genotypes of the following family members:

Leopold

Alice

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

Britney

  • Xh Y

Hemophilia

Key to alleles:

H = healthy clotting factors

h= no clotting factor

Royal Family Pedigree Chart from:

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

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hemophilia6

Pedigree chart practice

State the genotypes of the following family members:

Leopold

Alice

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

Britney

  • Xh Y

Hemophilia

  • XH Xh

Key to alleles:

H = healthy clotting factors

h= no clotting factor

Royal Family Pedigree Chart from:

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

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hemophilia7

Pedigree chart practice

State the genotypes of the following family members:

Leopold

Alice

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

Britney

  • Xh Y

Hemophilia

  • XH Xh

XH Y or XhY

Key to alleles:

H = healthy clotting factors

h= no clotting factor

Royal Family Pedigree Chart from:

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

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hemophilia8

Pedigree chart practice

State the genotypes of the following family members:

Leopold

Alice

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

Britney

  • Xh Y

Hemophilia

  • XH Xh

XH Y or XhY

  • XH XHorXH Xh

Key to alleles:

H = healthy clotting factors

h= no clotting factor

Royal Family Pedigree Chart from:

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

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pedigree chart practice

Pedigree Chart Practice

Dominant or Recessive?

Autosomal or Sex-linked?

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pedigree chart practice1

Pedigree Chart Practice

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 practice2

Pedigree Chart Practice

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

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

A. 0%

B. 12.5%

C. 25%

D. 50%

Super Evil Past Paper Question

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super evil past paper question1

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 factors

Xh= no clotting factor

Super Evil Past Paper Question

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super evil past paper question2

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 factors

Xh= no clotting factor

Super Evil Past Paper Question

What do we know?

A = XHY B = XHXh (because G = XhY) E = XHY

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super evil past paper question3

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 factors

Xh= no clotting factor

Super Evil Past Paper Question

What do we know?

A = XHY B = XHXh (because G = XhY) E = XHY

There is an equal chance of F being XHXHorXHXh

So:

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super evil past paper question4

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 factors

Xh= no clotting factor

Super Evil Past Paper Question

What do we know?

A = XHY B = XHXh (because G = XhY) E = XHY

There is an equal chance of F being XHXHorXHXh

So:

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super evil past paper question5

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 factors

Xh= no clotting factor

Super Evil Past Paper Question

What do we know?

A = XHY B = XHXh (because G = XhY) E = XHY

There is an equal chance of F being XHXHorXHXh

So:

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

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slide74

Whirling Gene activity from the awesome Learn.Genetics site:

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

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for more ib biology resources
For more IB Biology resources:

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