Theoretical Genetics

Theoretical Genetics
Stephen Taylor

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

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


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

Genotype
The combination of alleles Homozygous dominant
of a gene carried by an organism Having two copies of the same
dominant allele
Phenotype
The expression of alleles Homozygous recessive
of a gene carried by an organism Having two copies of the same
recessive allele. Recessive alleles are
Centromere only expressed when homozygous.
Joins chromatids in cell division
Codominant
Alleles Pairs of alleles which are both
Different versions of a gene expressed when present.
Dominant alleles = capital letter
Recessive alleles = lower-case letter
Heterozygous
Having two different alleles.
The dominant allele is expressed.

Carrier Gene loci
Heterozygous carrier of a Specific positions of genes on a
recessive disease-causing allele chromosome


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

3. Now arrange them in a karyotype (don't turn them
over - leave them as they were).
Activity from:
http://www.nclark.net/Genetics
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

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.
Activity from:
http://www.nclark.net/Genetics


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

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


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

The yellow parent peas must
Mendel did not know about
be heterozygous. The yellow
DNA, chromosomes or meiosis.
phenotype is expressed.
Through his experiments he did
Through meiosis and
work out that ‘heritable factors’
fertilisation, some offspring
(genes) were passed on and
peas are homozygous
that these could have different
recessive – they express a
versions (alleles).
green colour.

Mendel from:


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

F0
Genotype: Yy Yy Alleles segregate during
meiosis (anaphase I) and end
up in different haploid gametes.
Gametes: Y or y Y or y
Punnet Grid: gametes

Genotypes:
F1 Phenotypes:

Phenotype ratio: Mendel from:


Monohybrid Cross Crossing a single trait.

F0
Fertilisation results in diploid
Punnet Grid: gametes 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).

Genotypes:
F1 Phenotypes:




F0
Punnet Grid: gametes Y y zygotes.

Y YY Yy deduce the potential outcomes
y Yy yy expected ratio of phenotypes in

Genotypes:
F1 Phenotypes:




F0
Punnet Grid: gametes Y y zygotes.

Y YY Yy deduce the potential outcomes
y Yy yy expected ratio of phenotypes in

Genotypes: YY Yy Yy yy Ratios are written in the
F1 Phenotypes:
simplest mathematical form.

Phenotype ratio: 3:1 Mendel from:


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

F0 Phenotype:
Key to alleles:
Y = yellow
y = green
Genotype:

Homozygous recessive Homozygous recessive


Genotypes:
F1 Phenotypes:

Phenotype ratio:


F0 Phenotype:
Key to alleles:
Y = yellow
y = green
Genotype: yy yy
Homozygous recessive Homozygous recessive

Punnet Grid: gametes y y
y yy yy
y yy yy
Genotypes: yy yy yy yy
F1 Phenotypes:

Phenotype ratio: All green


F0 Phenotype:
Key to alleles:
Y = yellow
y = green
Genotype:

Homozygous recessive Heterozygous


Genotypes:
F1 Phenotypes:

Phenotype ratio:


F0 Phenotype:
Key to alleles:
Y = yellow
y = green
Genotype: yy Yy
Homozygous recessive Heterozygous

Punnet Grid: gametes Y y
y Yy yy
y Yy yy
Genotypes: Yy Yy yy yy
F1 Phenotypes:

Phenotype ratio: 1:1


F0 Phenotype:
Key to alleles:
Y = yellow
y = green
Genotype:

Homozygous dominant Heterozygous


Genotypes:
F1 Phenotypes:

Phenotype ratio:


F0 Phenotype:
Key to alleles:
Y = yellow
y = green
Genotype: YY Yy
Homozygous dominant Heterozygous

Punnet Grid: gametes Y y
Y YY Yy
Y YY Yy
Genotypes: YY YY Yy Yy
F1 Phenotypes:

Phenotype ratio: All yellow

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

F0 Phenotype:
Key to alleles:
R = Red flower
r = white
Genotype: R? r r
unknown Homozygous recessive
Possible outcomes:

F1 Phenotypes:
Unknown parent = RR Unknown parent = Rr

gametes gametes


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

F0 Phenotype:
Key to alleles:
R = Red flower
r = white
Genotype: R? r r
unknown Homozygous recessive
Possible outcomes:

F1 Phenotypes: All red Some white, some red
Unknown parent = RR Unknown parent = Rr

gametes r r gametes r r
R Rr Rr R Rr Rr
R Rr Rr r rr rr


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.”
From the Times Higher Education Supplement – “So Last Century”
http://www.timeshighereducation.co.uk/story.asp?sectioncode=26&storycode=415941&c=2


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.”
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


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.

key female male
I
affected

Not
II Affected

A B deceased

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



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.

key female male
I
affected

Not
II Affected

A B deceased

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.



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.

Diagnosis- blood test taken at 6-7 days after birth
The Boy with PKU ideo clip from: http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/
http://www.youtube.com/watch?v=KUJVujhHxPQ



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?

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

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

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



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

F0 Phenotype: carrier carrier
Key to alleles:
T = Normal enzyme
t = faulty enzyme
Genotype: Tt Tt

Punnet Grid: gametes T t
T
t
Genotypes:
F1 Phenotypes:
Phenotype ratio:



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

Key to alleles:
T = Normal enzyme
t = faulty enzyme
Genotype: Tt Tt

Punnet Grid: gametes T t
T TT Tt
t Tt tt
Genotypes: TT Tt Tt tt
F1 Phenotypes: Normal enzyme PKU
Therefore 25% chance
Phenotype ratio: 3:1 of a child with PKU


Pedigree Charts Key to alleles:
T= Has enzyme
Pedigree charts can be used to trace family histories and deduce t = no enzyme
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 genotypes
of these individuals: A&B C D
Genotype

Reason


T= Has enzyme
Pedigree charts can be used to trace family histories and deduce t = no enzyme
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 genotypes
of these individuals: A&B C D
Genotype Both Tt tt Tt
To have produced affected
Trait is recessive, as both Recessive traits only
child H, D must have inherited
Reason are normal, yet have produced
an affected child (C)
expressed when
homozygous.
a recessive allele from either A
or B


T= Has enzyme
Individuals D and $ are planning to have another child. t = no enzyme
Calculate the chances of the child having PKU.
Key: female male

affected

Not
$ Affected

deceased

Looks like

Genotypes:
D= Gametes
Phenotype ratio

$= Therefore


T= Has enzyme
Individuals D and $ are planning to have another child. t = no enzyme
Calculate the chances of the child having PKU.
Key: female male

affected

Not
$ Affected

deceased

Looks like

Genotypes:
D = Tt (carrier) Gametes T t
Phenotype ratio
t Tt tt 1 : 1 Normal : PKU
$ = tt (affected)
Therefore 50% chance of a
t Tt tt child with PKU


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. Key to alleles:
i = no antigens present
Where the genotype is heterozygous for IA and IB, both IA = type A anitgens present
are expressed. This is codominance. IB = type B antigens present


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.

Blood typing game from Nobel.org: Images and more information from:
http://nobelprize.org/educational/medicine/landsteiner/readmore.html http://learn.genetics.utah.edu/content/begin/traits/blood/


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?


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
No Yes Yes
protection?


Sickle Cell Another example of codominance. Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
Predict the phenotype ratio in this cross:

F0 Phenotype: carrier affected
Genotype:


Genotypes:
F1 Phenotypes:
Phenotype ratio: : Therefore 50% chance of a
child with sickle cell disease.


HbA = Normal Hb
HbS = Sickle cell

F0 Phenotype: carrier affected
Genotype: HbA Hbs HbS Hbs

Punnet Grid: gametes HbS HbS
HbA HbAHbS HbAHbS
HbS HbSHbS HbSHbS
Genotypes: HbAHbS & HbSHbS
F1 Phenotypes: Carrier & Sickle cell

Phenotype ratio: 1:1 Therefore 50% chance of a


HbA = Normal Hb
HbS = Sickle cell

Genotype:


Genotypes:
F1 Phenotypes:
Phenotype ratio:


HbA = Normal Hb
HbS = Sickle cell

Genotype: HbA HbS HbA HbS

Punnet Grid: gametes HbA HbS
HbA HbAHbA HbAHbS
HbS HbAHbS HbSHbS
Genotypes: HbAHb & 2 HbAHbS & HbSHbS
F1 Phenotypes: Unaffected & Carrier & Sickle cell

Phenotype ratio: 1: 2 : 1 Therefore 25% chance of a


HbA = Normal Hb
HbS = Sickle cell

F0 Phenotype: carrier unknown
Genotype: HbA HbS


HbA
HbS
Genotypes:
F1 Phenotypes:
Phenotype ratio:


HbA = Normal Hb
HbS = Sickle cell

Genotype: HbA HbS HbA HbA or HbA HbS

Punnet Grid: gametes HbA HbA HbA HbS
HbA
HbS
Genotypes:
F1 Phenotypes:
Phenotype ratio:


HbA = Normal Hb
HbS = Sickle cell

Genotype: HbA HbS HbA HbA or HbA HbS

Punnet Grid: gametes HbA HbA HbA HbS
HbA HbAHbA HbAHbA HbAHbA HbAHbS

HbS HbAHbS HbAHbS HbAHbS HbSHbS

Genotypes: 3 HbAHbA & 4 HbAHbS & 1 HbSHbS
F1 Phenotypes: 3 Unaffected & 4 Carrier & 1 Sickle cell

Phenotype ratio: 3:4:1 Therefore 12.5% chance of a


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.

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


Sex Determination 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.

Segregation of the sex chromosomes in meiosis.

SRY gene determines maleness.
gametes X Y
Find out more about its role and
X XX XY just why do men have nipples?

X XX XY Chromosome images from Wikipedia:

Therefore there is an even chance*
of the offspring being male or female. http://www.hhmi.org/biointeractive/gender/lectures.html


Sex Determination Non-disjunction can lead to gender disorders.

XYY Syndrome: XXY: Klinefelter Syndrome:
Fertile males, with increased risk of learning difficulties. Males with enhanced female characteristics
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.

Interactive from HHMI Biointeractive: Image from NCBI:
http://www.hhmi.org/biointeractive/gender/click.html http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179


Sex Linkage X and Y chromosomes are non-homologous.

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

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

Examples of sex-linked genetic disorders:
- haemophilia
- colour blindness
X and Y SEM from Chromosome images from Wikipedia:
http://www.angleseybonesetters.co.uk/bones_DNA.html http://en.wikipedia.org/wiki/Y_chromosome



What number do you see?




What number do you see?

5 = normal vision
2 = red/green colour blindness




How is colour-blindness inherited?

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

Normal vision is dominant over colour-blindness.

XN XN
Normal female
XN Y
Normal male
no allele carried, none written

Key to alleles:
n n n N = normal vision Xq28
X X
Affected female
X Y
Affected male
n = red/green colour
blindness

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



What chance of a colour-blind child in the cross between a Key to alleles:
normal male and a carrier mother? N = normal vision
F0 Genotype: XN Xn XN Y
blindness

Phenotype: Carrier female X Normal male

Punnet Grid:

F1




blindness


Punnet Grid:
XN Y
XN XN XN XN Y
F1
Xn XN Xn Xn Y




blindness


Punnet Grid:
XN Y
XN XN XN XN Y
Normal female Normal male
F1
Xn XN Xn
Carrier female
Xn Y
Affected male
There is a 1 in 4 (25%)
chance of an affected child.

What ratios would we expect in a cross between:
a. a colour-blind male and a homozygous normal female? Chromosome images from Wikipedia:
b. a normal male and a colour-blind female? http://en.wikipedia.org/wiki/Y_chromosome


Red-Green Colour Blindness How does it work?

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
Xq28
to the Y chromosome, red-green colour blindness is known as a sex-
linked disorder. The male has no allele on the Y chromosome to
combat a recessive faulty allele on the X chromosome.



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


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
Normal female
XH Y
Normal male
no allele carried, none written

Key to alleles:
XH = healthy clotting factors
Xh Xh
Affected female
Xh Y
Affected male
Xh = no clotting factor

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


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

Read/ research/ review:

How can gene transfer be used to treat
hemophiliacs?

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



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



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

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


Hemophilia Pedigree chart practice

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.

Key: female male
4. Britney
affected
Key to alleles:
Not
H = healthy clotting factors Affected
h = no clotting factor
deceased



1. Leopold
Xh Y
2. Alice


Key: female male
4. Britney
affected
Key to alleles:
Not
deceased



1. Leopold
Xh Y
2. Alice
XH Xh


Key: female male
4. Britney
affected
Key to alleles:
Not
deceased



1. Leopold
Xh Y
2. Alice
XH Xh

XH Y or Xh Y
Key: female male
4. Britney
affected
Key to alleles:
Not
deceased



1. Leopold
Xh Y
2. Alice
XH Xh

XH Y or Xh Y
Key: female male
4. Britney
XH XH or XH Xh affected
Key to alleles:
Not
deceased


Pedigree Chart Practice
Key: female male

affected

Not
Affected

deceased

Dominant or Recessive? Autosomal or Sex-linked?


Key: female male

affected

Not
Affected

deceased

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.


Key: female male

affected

Not
Affected

deceased

Dominant. Autosomal.
A and B are both affected but have produced Male C can only pass on one X chromosome. If it
unaffected (D & F). Therefore A and B must have were carried on X, daughter H would be affected
been carrying recessive healthy alleles. by the dominant allele.

If it were recessive, it would need to be Tip: Don’t get hung up on the number of
homozygous to be expressed in A & B – and then individuals with each phenotype – each
all offspring would be homozygous recessive. reproductive event is a matter of chance. Instead
focus on possible and impossible genotypes.
Draw out the punnet grids if needed.


Super Evil Past Paper Question
In this pedigree chart for hemophilia, what is
the chance that offspring ? will be affected?
A. 0%
B. 12.5%
C. 25%
D. 50%

Key: female male

affected

Not
Affected

deceased


A. 0%
Key to alleles:
B. 12.5% XH = healthy clotting factors
C. 25%
D. 50%

Key: female male

affected

Not
Affected

deceased


A. 0%
Key to alleles:
C. 25%
D. 50%
What do we know?
A = XH Y B = XH Xh (because G = Xh Y) E = XH Y
Key: female male

affected

Not
Affected
XH
deceased
Y


A. 0%
Key to alleles:
C. 25%
D. 50%
What do we know?
Key: female male There is an equal chance of F being XH XH or XH Xh
So:
affected

Not XH XH XH Xh
Affected
XH
deceased
Y


A. 0%
Key to alleles:
C. 25%
D. 50%
What do we know?
So:
affected

Not XH XH XH Xh
Affected
XH XH XH XH XH XH XH XH Xh
deceased
Y XH Y XH Y XH Y Xh Y


A. 0%
Key to alleles:
C. 25%
D. 50%
What do we know?
So:
affected

Not XH XH XH Xh
Affected
XH XH XH XH XH XH XH XH Xh
deceased
Y XH Y XH Y XH Y Xh Y

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


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


For more IB Biology resources:
http://sciencevideos.wordpress.com
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