Download www.psd150.org

Theoretical Genetics
Stephen Taylor
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
1
Definitions
Genotype
The combination of alleles
of a gene carried by an organism
Phenotype
The expression of alleles
of a gene carried by an organism
Centromere
This image shows a pair of homologous chromosomes.
Name and annotate the labeled features.
Homozygous dominant
Having two copies of the same
dominant allele
Homozygous recessive
Having two copies of the same
recessive allele. Recessive alleles are
only expressed when homozygous.
Joins chromatids in cell division
Codominant
Alleles
Different versions of a gene
Dominant alleles = capital letter
Recessive alleles = lower-case letter
Carrier
Heterozygous carrier of a
recessive disease-causing allele
http://sciencevideos.wordpress.com
Pairs of alleles which are both
expressed when present.
Heterozygous
Having two different alleles.
The dominant allele is expressed.
Gene loci
Specific positions of genes on a
chromosome
4.3 Theoretical Genetics
2
Explain this
Mendel crossed some yellow peas with some yellow
peas. Most offspring were yellow but some were green!
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
3
Segregation
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.
“alleles of each gene separate into different
gametes when the individual produces gametes”
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
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
4
Segregation
“alleles of each gene separate into different
gametes when the individual produces gametes”
F0
F1
Genotype:
Yy
Yy
Gametes:
Y or y
Y or y
Punnet Grid:
gametes
Alleles segregate during
meiosis (anaphase I) and end
up in different haploid gametes.
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
4.3 Theoretical Genetics
5
Monohybrid Cross
Crossing a single trait.
F0
Genotype:
Yy
Yy
Gametes:
Y or y
Y or y
Punnet Grid:
gametes
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).
F1
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
4.3 Theoretical Genetics
6
Monohybrid Cross
Crossing a single trait.
F0
Genotype:
Yy
Yy
Gametes:
Y or y
Y or y
Punnet Grid:
F1
gametes
Y
y
Y
YY
Yy
y
Yy
yy
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).
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
4.3 Theoretical Genetics
7
Monohybrid Cross
Crossing a single trait.
F0
Genotype:
Yy
Yy
Gametes:
Y or y
Y or y
Punnet Grid:
F1
Genotypes:
gametes
Y
YY
Yy
y
Yy
yy
YY
Yy
Fertilisation results in diploid
zygotes.
y
Y
Yy
Alleles segregate during
meiosis (anaphase I) and end
up in different haploid gametes.
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).
yy
Ratios are written in the
simplest mathematical form.
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
3:1
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
4.3 Theoretical Genetics
8
Monohybrid Cross
F0
What is the expected ratio of phenotypes
in this monohybrid cross?
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
Homozygous recessive
Punnet Grid:
F1
Homozygous recessive
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
9
Monohybrid Cross
F0
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
yy
yy
Homozygous recessive
Punnet Grid:
F1
What is the expected ratio of phenotypes
in this monohybrid cross?
Genotypes:
Homozygous recessive
gametes
y
y
y
yy
yy
y
yy
yy
yy
yy
yy
yy
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
All green
4.3 Theoretical Genetics
10
Monohybrid Cross
F0
What is the expected ratio of phenotypes
in this monohybrid cross?
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
Homozygous recessive
Punnet Grid:
F1
Heterozygous
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
11
Monohybrid Cross
F0
Phenotype:
Genotype:
yy
Yy
Homozygous recessive
Punnet Grid:
F1
What is the expected ratio of phenotypes
in this monohybrid cross?
Genotypes:
Heterozygous
gametes
Y
y
y
Yy
yy
y
Yy
yy
Yy
Yy
Key to alleles:
Y = yellow
y = green
yy
yy
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
1:1
4.3 Theoretical Genetics
12
Monohybrid Cross
F0
What is the expected ratio of phenotypes
in this monohybrid cross?
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
Homozygous dominant
Punnet Grid:
F1
Heterozygous
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
13
Monohybrid Cross
F0
Phenotype:
Genotype:
YY
Yy
Homozygous dominant
Punnet Grid:
F1
What is the expected ratio of phenotypes
in this monohybrid cross?
Genotypes:
Heterozygous
gametes
Y
y
Y
YY
Yy
Y
YY
Yy
YY
YY
Key to alleles:
Y = yellow
y = green
Yy
Yy
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
All yellow
4.3 Theoretical Genetics
14
Test Cross
F0
Used to determine the genotype of an unknown individual.
The unknown is crossed with a known homozygous recessive.
Key to alleles:
R = Red flower
r = white
Phenotype:
R?
Genotype:
unknown
r r
Homozygous recessive
Possible outcomes:
F1
Phenotypes:
Unknown parent = RR
gametes
http://sciencevideos.wordpress.com
Unknown parent = Rr
gametes
4.3 Theoretical Genetics
15
Test Cross
F0
Used to determine the genotype of an unknown individual.
The unknown is crossed with a known homozygous recessive.
Key to alleles:
R = Red flower
r = white
Phenotype:
R?
Genotype:
unknown
r r
Homozygous recessive
Possible outcomes:
F1
All red
Phenotypes:
Unknown parent = RR
Some white, some red
Unknown parent = Rr
gametes
r
r
gametes
r
r
R
Rr
Rr
R
Rr
Rr
R
Rr
Rr
r
rr
rr
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
16
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F0
Y = yellow
y = green
S = smooth
s = rough
Phenotype:
Heterozygous at both loci
Heterozygous at both loci
Genotype:
Punnet Grid:
gametes
F1
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
17
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F0
Phenotype:
Genotype:
Punnet Grid:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous at both loci
SsYy
SsYy
gametes
F1
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
18
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F0
Phenotype:
Genotype:
Punnet Grid:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous at both loci
SsYy
SsYy
gametes
SY
Sy
sY
sy
SY
SSYY
SSYy
SsYY
SsYy
Sy
SSYy
SSyy
SsYy
Ssyy
sY
SsYY
SsYy
ssYY
ssYy
sy
SsYy
Ssyy
ssYy
ssyy
F1
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
19
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
What is the predicted phenotype ratio for a cross between
two pea plants which are heterozygous at both loci?
F0
Phenotype:
Genotype:
Punnet Grid:
F1
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous at both loci
SsYy
SsYy
gametes
SY
Sy
sY
sy
SY
SSYY
SSYy
SsYY
SsYy
Sy
SSYy
SSyy
SsYy
Ssyy
sY
SsYY
SsYy
ssYY
ssYy
sy
SsYy
Ssyy
ssYy
ssyy
Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
20
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
F0
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
Phenotype:
Heterozygous at both loci
Heterozygous for S, homozygous dominant for Y
Genotype:
Punnet Grid:
F1
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
21
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
Calculate the predicted phenotype ratio for:
F0
Phenotype:
Smooth, yellow
Heterozygous at both loci
Genotype:
SsYy
Smooth, yellow
Heterozygous for S, homozygous dominant for Y
SsYY
Punnet Grid:
F1
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
22
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
Calculate the predicted phenotype ratio for:
F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous for S, homozygous dominant for Y
Heterozygous at both loci
Genotype:
Punnet Grid:
SsYY
SsYy
gametes
SY
sY
SY
Sy
sY
sy
F1
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
23
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
Calculate the predicted phenotype ratio for:
F0
Phenotype:
Smooth, yellow
Heterozygous for S, homozygous dominant for Y
Heterozygous at both loci
Genotype:
Punnet Grid:
F1
Smooth, yellow
SsYY
SsYy
gametes
SY
sY
SY
SSYY
SsYY
Sy
SSYy
SsYy
sY
SsYY
ssYY
sy
SsYy
ssYy
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
24
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
Calculate the predicted phenotype ratio for:
F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous for S, homozygous dominant for Y
Heterozygous at both loci
Genotype:
Punnet Grid:
SsYY
SsYy
gametes
SY
sY
SY
SSYY
SsYY
Sy
SSYy
SsYy
sY
SsYY
ssYY
sy
SsYy
ssYy
6 Smooth, yellow : 2 Rough, yellow
F1
Phenotypes: 3 Smooth, yellow : 1 Rough, yellow
http://sciencevideos.wordpress.com
Present the ratio in the simplest
mathematical form.
10.2 Dihybrid Crosses & Gene Linkage
25
Dihybrid Crosses
Common expected ratios of dihybrid crosses.
SsYy
SsYy
SsYy
Heterozygous at
both loci
Heterozygous at
both loci
Heterozygous at
both loci
SsYy
Heterozygous at one locus,
homozygous dominant at the other
SY
sY
SY
SSYY
SsYY
Ssyy
Sy
SSYy
SsYy
ssYY
ssYy
sY
SsYY
ssYY
ssYy
ssyy
sy
SsYy
ssYy
SY
Sy
sY
sy
SY
SSYY
SSYy
SsYY
SsYy
Sy
SSYy
SSyy
SsYy
sY
SsYY
SsYy
sy
SsYy
Ssyy
3:2
9:3:3:1
Ssyy
SsYy
Heterozygous at
both loci
Heterozygous/
Homozygous recessive
Sy
sy
SY
SSYy
SsYy
Sy
SSyy
Ssyy
sY
SsYy
ssYy
sy
Ssyy
ssyy
SSyy
ssYY
= All SsYy
SSYY
ssyy
= all SyYy
Ssyy
ssYy
=1:1:1:1
4:3:1
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
26
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A rough yellow pea is test crossed to determine its genotype.
F0
Phenotype:
Rough, yellow
Genotype:
Punnet Grid:
F1
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
27
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A rough yellow pea is test crossed to determine its genotype.
F0
Phenotype:
Rough, yellow
ssYy
Genotype:
Punnet Grid:
F1
gametes
sY
sy
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
28
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A rough yellow pea is test crossed to determine its genotype.
F0
Phenotype:
Rough, yellow
ssYy or ssYY
Genotype:
Punnet Grid:
F1
gametes
sY
sy
sY
sY
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
29
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A rough yellow pea is test crossed to determine its genotype.
F0
Phenotype:
Rough, yellow
Punnet Grid:
ssYy or ssYY
ssyy
Genotype:
gametes
sY
sy
sY
sY
All sy
F1
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
30
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A rough yellow pea is test crossed to determine its genotype.
F0
Phenotype:
Rough, yellow
Punnet Grid:
F1
ssYy or ssYY
ssyy
Genotype:
gametes
sY
sy
sY
sY
All sy
ssYy
ssyy
ssYy
ssYy
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
31
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A rough yellow pea is test crossed to determine its genotype.
F0
Phenotype:
Rough, yellow
Punnet Grid:
F1
ssYy or ssYY
ssyy
Genotype:
gametes
sY
sy
sY
sY
All sy
ssYy
ssyy
ssYy
ssYy
Phenotypes:
Some green peas will be present in
the offspring if the unknown parent
genotype is ssYy.
http://sciencevideos.wordpress.com
No green peas will be present in the
offspring if the unknown parent
genotype is ssYY.
10.2 Dihybrid Crosses & Gene Linkage
32
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F0
Phenotype:
Smooth, green
Genotype:
Punnet Grid:
F1
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
33
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F0
Phenotype:
Smooth, green
ssyy
Genotype:
Punnet Grid:
gametes
All sy
F1
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
34
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F0
Phenotype:
Smooth, green
Punnet Grid:
F1
SSyy
ssyy
Genotype:
gametes
Sy
Sy
All sy
Ssyy
Ssyy
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
35
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F0
Phenotype:
Smooth, green
Punnet Grid:
F1
SSyy or Ssyy
ssyy
Genotype:
gametes
Sy
Sy
Sy
sy
All sy
Ssyy
Ssyy
Ssyy
ssyy
Phenotypes:
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
36
Dihybrid Crosses
Consider two traits, each carried on separate
chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
F0
Phenotype:
Smooth, green
Punnet Grid:
F1
SSyy or Ssyy
ssyy
Genotype:
gametes
Sy
Sy
Sy
sy
All sy
Ssyy
Ssyy
Ssyy
ssyy
Phenotypes:
No rough peas will be present in the
offspring if the unknown parent
genotype is SSyy.
The presence of rough green peas in the
offspring means that the unknown
genotype must be Ssyy.
The expected ratio in this cross is 3 smooth green : 1 rough green. This is not the same as the outcome.
Remember that each reproduction event is chance and the sample size is very small. With a much larger sample
size, the outcome would be closer to the expected ratio, simply due to probability.
http://sciencevideos.wordpress.com
10.2 Dihybrid Crosses & Gene Linkage
37
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.
Key to alleles:
If they are present, they are either type A, B or both.
i = no antigens present
IA = type A anitgens present
Where the genotype is heterozygous for IA and IB, both
IB = type B antigens present
are expressed. This is codominance.
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
38
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:
http://nobelprize.org/educational/medicine/landsteiner/readmore.html
http://sciencevideos.wordpress.com
Images and more information from:
http://learn.genetics.utah.edu/content/begin/traits/blood/
4.3 Theoretical Genetics
39
Sickle Cell
Another example of codominance.
Remember the notation used: superscripts
represent codominant alleles.
In codominance, heterozygous individuals have a
mixed phenotype.
The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia.
Complete the table for these individuals:
Genotype
Description
Homozygous HbA
Heterozygous
Homozygous HbS
Phenotype
Malaria
protection?
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
40
Sickle Cell
Another example of codominance.
Remember the notation used: superscripts
represent codominant alleles.
In codominance, heterozygous individuals have a
mixed phenotype.
The mixed phenotype gives protection against malaria, but does not exhibit full-blown sickle cell anemia.
Complete the table for these individuals:
Genotype
HbA HbA
HbA HbS
HbS HbS
Description
Homozygous HbA
Heterozygous
Homozygous HbS
Phenotype
normal
carrier
Sickle cell disease
Malaria
protection?
No
Yes
Yes
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
41
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
affected
Genotype:
Punnet Grid:
F1
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
:
Therefore 50% chance of a
child with sickle cell disease.
4.3 Theoretical Genetics
42
Sickle Cell
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA Hbs
Punnet Grid:
F1
gametes
affected
HbS Hbs
HbS
HbS
HbA
HbAHbS HbAHbS
HbS
HbSHbS HbSHbS
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
HbAHbS & HbSHbS
Carrier & Sickle cell
1:1
Therefore 50% chance of a
child with sickle cell disease.
4.3 Theoretical Genetics
43
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
carrier
Genotype:
Punnet Grid:
F1
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
44
Sickle Cell
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA HbS
Punnet Grid:
F1
gametes
carrier
HbA HbS
HbA
HbS
HbA
HbAHbA HbAHbS
HbS
HbAHbS HbSHbS
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
HbAHb & 2 HbAHbS & HbSHbS
Unaffected & Carrier & Sickle cell
1: 2 : 1
Therefore 25% chance of a
child with sickle cell disease.
4.3 Theoretical Genetics
45
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA HbS
Punnet Grid:
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
unknown
gametes
HbA
HbS
F1
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
46
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA HbS
Punnet Grid:
gametes
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
unknown
HbA HbA or HbA HbS
HbA
HbA
HbA
HbS
HbA
HbS
F1
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
47
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA HbS
Punnet Grid:
F1
Genotypes:
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
unknown
HbA HbA or HbA HbS
gametes
HbA
HbA
HbA
HbA
HbAHbA
HbAHbA HbAHbA
HbAHbS
HbS
HbAHbS
HbAHbS
HbSHbS
HbAHbS
HbS
3 HbAHbA & 4 HbAHbS & 1 HbSHbS
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
3 Unaffected & 4 Carrier & 1 Sickle cell
3:4:1
Therefore 12.5% chance of a
child with sickle cell disease.
4.3 Theoretical Genetics
48
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
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
49
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.
X
Y
X
XX
XY
X
XX
XY
gametes
Therefore there is an even chance*
of the offspring being male or female.
http://sciencevideos.wordpress.com
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
http://www.hhmi.org/biointeractive/gender/lectures.html
4.3 Theoretical Genetics
50
Sex Determination
Non-disjunction can lead to gender disorders.
XYY Syndrome:
XXY: Klinefelter Syndrome:
Fertile males, with increased risk of learning difficulties.
Some weak connections made to violent tendencies.
Males with enhanced female characteristics
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:
http://www.hhmi.org/biointeractive/gender/click.html
http://sciencevideos.wordpress.com
Image from NCBI:
http://www.ncbi.nlm.nih.gov/bookshelf/br.fcgi?book=mga&part=A1179
4.3 Theoretical Genetics
51
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.
Non-homologous
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
http://www.angleseybonesetters.co.uk/bones_DNA.html
http://sciencevideos.wordpress.com
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
4.3 Theoretical Genetics
52
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
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
53
Sex Linkage
X and Y chromosomes are non-homologous.
What number do you see?
5 = normal vision
2 = red/green colour blindness
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
54
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.
XN
XN
Normal female
n
X
n
X
N
X
n
X
Affected female
Carrier female
XN
Y
Normal male
n
X
Y
Affected male
no allele carried, none written
Key to alleles:
N = normal vision
n = red/green colour
Xq28
blindness
Human females can be homozygous or
heterozygous with respect to sex-linked genes.
Chromosome images from Wikipedia:
Heterozygous females are carriers.
http://en.wikipedia.org/wiki/Y_chromosome
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
55
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?
F0
Genotype:
Phenotype:
N
X
n
X
Carrier female
X
N
X
Y
Key to alleles:
N = normal vision
n = red/green colour
blindness
Normal male
Punnet Grid:
F1
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
56
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?
F0
Genotype:
Phenotype:
Punnet Grid:
F1
N
X
n
X
Carrier female
X
N
X
Y
Key to alleles:
N = normal vision
n = red/green colour
blindness
Normal male
XN
Y
N XN
N
X
X
XN Y
n
X
N
X
n
X
Xn Y
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
57
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?
F0
Genotype:
Phenotype:
Punnet Grid:
F1
N
X
n
X
Carrier female
X
N
X
Y
Key to alleles:
N = normal vision
n = red/green colour
blindness
Normal male
XN
Y
N XN
N
X
X
XN Y
Normal female
n
X
N
X
n
X
Carrier female
Normal male
Xn Y
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?
http://sciencevideos.wordpress.com
There is a 1 in 4 (25%)
chance of an affected child.
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
4.3 Theoretical Genetics
58
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
to the Y chromosome, red-green colour blindness is known as a sexlinked disorder. The male has no allele on the Y chromosome to
combat a recessive faulty allele on the X chromosome.
Xq28
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
59
http://illinoisreview.typepad.com/illinoisrevie
w/2015/02/myths-and-facts-about-parcc-inillinois.html
Colour Blind cartoon from:
http://www.almeidacartoons.com/Med_toons1.html
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
60
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
I
female
male
affected
Not
Affected
II
A
III
B
deceased
?
Is PKU dominant or recessive? How do you know?
•
•
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
61
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
I
female
male
affected
Not
Affected
II
A
III
B
deceased
?
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.
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
62
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
http://sciencevideos.wordpress.com
Diagnosis- blood test taken at 6-7 days after birth
http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/
4.3 Theoretical Genetics
63
Phenylketonuria (PKU)
Clinical example.
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
Diagnosis- blood test taken at 6-7 days after birth
http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
64
Phenylketonuria (PKU)
Clinical example.
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
Diagnosis- blood test taken at 6-7 days after birth
http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/2632833781/
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
65
Phenylketonuria (PKU)
Clinical example.
What is the probability of two parents who are both carriers of the recessive allele producing
children affected by PKU?
F0
Phenotype:
carrier
Genotype:
Tt
Punnet Grid:
gametes
carrier
Tt
T
Key to alleles:
T = Normal enzyme
t = faulty enzyme
t
T
t
F1
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
66
Phenylketonuria (PKU)
Clinical example.
What is the probability of two parents who are both carriers of the recessive allele producing
children affected by PKU?
F0
F1
Phenotype:
carrier
Genotype:
Tt
Punnet Grid:
gametes
T
t
T
TT
Tt
t
Tt
tt
Genotypes:
Phenotypes:
Phenotype ratio:
http://sciencevideos.wordpress.com
TT
carrier
Tt
Tt
Tt
Normal enzyme
Key to alleles:
T = Normal enzyme
t = faulty enzyme
tt
PKU
3:1
Therefore 25% chance
of a child with PKU
4.3 Theoretical Genetics
67
Pedigree Charts
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 to alleles:
T= Has enzyme
t = no enzyme
Key:
female
male
affected
Not
Affected
deceased
Looks like
Deduce the genotypes
of these individuals:
A&B
C
D
Genotype
Reason
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
68
Pedigree Charts
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 to alleles:
T= Has enzyme
t = no enzyme
Key:
female
male
affected
Not
Affected
deceased
Looks like
Deduce the genotypes
of these individuals:
A&B
C
D
Genotype
Both Tt
tt
Tt
Reason
Trait is recessive, as both
are 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
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
69
Pedigree Charts
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.
Key:
female
male
affected
Not
Affected
$
deceased
Looks like
Genotypes:
D=
Gametes
Phenotype ratio
Therefore
$=
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
70
Pedigree Charts
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.
Key:
female
male
affected
$
Not
Affected
deceased
Looks like
Genotypes:
D = Tt (carrier)
$ = tt (affected)
T
t
t
Tt
tt
t
Tt
tt
Gametes
http://sciencevideos.wordpress.com
Phenotype ratio
1 : 1 Normal : PKU
Therefore 50% chance of a
child with PKU
4.3 Theoretical Genetics
71
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 Xh
Affected female
XH
Xh
Carrier female
XH
Y
Normal male
Xh Y
no allele carried, none written
Key to alleles:
XH = healthy clotting factors
Xh = no clotting factor
Affected male
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
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
72
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
73
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
74
Hemophilia
This pedigree chart of the English Royal Family gives us a
picture of the inheritance of this X-linked disorder.
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
75
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.
4. Britney
Key:
female
male
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
http://sciencevideos.wordpress.com
Not
Affected
deceased
4.3 Theoretical Genetics
76
Hemophilia
Pedigree chart practice
State the genotypes of the following family members:
1. Leopold
Xh Y
2. Alice
3. Bob was killed in a tragic croquet accident before
his phenotype was determined.
4. Britney
Key:
female
male
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
http://sciencevideos.wordpress.com
Not
Affected
deceased
4.3 Theoretical Genetics
77
Hemophilia
Pedigree chart practice
State the genotypes of the following family members:
1. Leopold
Xh Y
2. Alice
XH Xh
3. Bob was killed in a tragic croquet accident before
his phenotype was determined.
4. Britney
Key:
female
male
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
http://sciencevideos.wordpress.com
Not
Affected
deceased
4.3 Theoretical Genetics
78
Hemophilia
Pedigree chart practice
State the genotypes of the following family members:
1. Leopold
Xh Y
2. Alice
XH Xh
3. Bob was killed in a tragic croquet accident before
his phenotype was determined.
XH Y or Xh Y
4. Britney
Key:
female
male
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
http://sciencevideos.wordpress.com
Not
Affected
deceased
4.3 Theoretical Genetics
79
Hemophilia
Pedigree chart practice
State the genotypes of the following family members:
1. Leopold
Xh Y
2. Alice
XH Xh
3. Bob was killed in a tragic croquet accident before
his phenotype was determined.
XH Y or Xh Y
4. Britney
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
http://sciencevideos.wordpress.com
XH XH or XH Xh
Key:
female
male
affected
Not
Affected
deceased
4.3 Theoretical Genetics
80
Pedigree Chart Practice
Key:
female
male
affected
Not
Affected
deceased
Dominant or Recessive?
http://sciencevideos.wordpress.com
Autosomal or Sex-linked?
4.3 Theoretical Genetics
81
Pedigree Chart Practice
Key:
female
male
affected
Not
Affected
deceased
Dominant or Recessive?
Dominant.
Autosomal or Sex-linked?
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.
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
82
Pedigree Chart Practice
Key:
female
male
affected
Not
Affected
deceased
Dominant or Recessive?
Dominant.
Autosomal or Sex-linked?
Autosomal.
A and B are both affected but have produced
unaffected (D & F). Therefore A and B must have
been carrying recessive healthy alleles.
Male C can only pass on one X chromosome. If it
were carried on X, daughter H would be affected
by the dominant allele.
If it were recessive, it would need to be
homozygous to be expressed in A & B – and then
all offspring would be homozygous recessive.
Tip: Don’t get hung up on the number of
http://sciencevideos.wordpress.com
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.
4.3 Theoretical Genetics
83
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%
female
Key:
male
affected
Not
Affected
deceased
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
84
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%
Key to alleles:
XH = healthy clotting factors
Xh = no clotting factor
D. 50%
female
Key:
male
affected
Not
Affected
deceased
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
85
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%
Key to alleles:
XH = healthy clotting factors
Xh = no clotting factor
D. 50%
female
Key:
male
What do we know?
A = XH Y B = XH Xh (because G = Xh Y) E = XH Y
affected
Not
Affected
XH
deceased
Y
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
86
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%
Key to alleles:
XH = healthy clotting factors
Xh = no clotting factor
D. 50%
female
Key:
male
affected
Not
Affected
What do we know?
A = XH Y B = XH Xh (because G = Xh Y) E = XH Y
There is an equal chance of F being XH XH or XH Xh
So:
XH
XH
XH
Xh
XH
deceased
Y
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
87
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%
Key to alleles:
XH = healthy clotting factors
Xh = no clotting factor
D. 50%
female
Key:
male
affected
Not
Affected
What do we know?
A = XH Y B = XH Xh (because G = Xh Y) E = XH Y
There is an equal chance of F being XH XH or XH Xh
So:
XH
XH
XH
Xh
XH
XH XH
XH XH
XH XH
X H Xh
Y
XH Y
XH Y
XH Y
Xh Y
deceased
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
88
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%
Key to alleles:
XH = healthy clotting factors
Xh = no clotting factor
D. 50%
female
Key:
male
affected
Not
Affected
What do we know?
A = XH Y B = XH Xh (because G = Xh Y) E = XH Y
There is an equal chance of F being XH XH or XH Xh
So:
XH
XH
XH
Xh
XH
XH XH
XH XH
XH XH
X H Xh
Y
XH Y
XH Y
XH Y
Xh Y
deceased
So there is a 1 in 8 (12.5%) chance of the offspring being affected!
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
89
Whirling Gene activity from the awesome Learn.Genetics site:
http://learn.genetics.utah.edu/archive/pedigree/mapgene.html
http://sciencevideos.wordpress.com
4.3 Theoretical Genetics
90
For more IB Biology resources:
http://sciencevideos.wordpress.com
This presentation is free to view. Please make a donation to one of my
chosen charities at Gifts4Good and I will send you the editable pptx file.
Click here for more information about Biology4Good charity donations.
This is a Creative Commons presentation. It may be linked and embedded but not sold or re-hosted.
4.3 Theoretical Genetics
91