3.4 Inheritance
Essential idea: The inheritance of genes follows patterns.
The patterns that genes and the phenotypes they generate can be
mapped using pedigree charts. The image show a small section of a
pedigree chart that maps the inheritance of hair colour in an
extended family over several generations. Analysis of pedigree
charts enables us to the nature of the inheritance; controlled by
dominant or recessive alleles? linked to the sex chromosomes?
controlled by multiple genes or a single gene?
By Chris Paine
http://www.bioknowledgy.info/
http://www.indiana.edu/~oso/lessons/Genetics/RealColors.html
Understandings
Statement
3.4.U1
3.4.U2
3.4.U3
3.4.U4
3.4.U5
3.4.U6
3.4.U7
3.4.U8
3.4.U9
Guidance
Mendel discovered the principles of inheritance with
experiments in which large numbers of pea plants were
crossed.
Gametes are haploid so contain only one allele of each
gene.
The two alleles of each gene separate into different
haploid daughter nuclei during meiosis.
Fusion of gametes results in diploid zygotes with two
alleles of each gene that may be the same allele or
different alleles.
Dominant alleles mask the effects of recessive alleles
but co-dominant alleles have joint effects.
Many genetic diseases in humans are due to recessive
alleles of autosomal genes, although some genetic
diseases are due to dominant or co-dominant alleles.
Some genetic diseases are sex-linked. The pattern of Alleles carried on X chromosomes should be
inheritance is different with sex-linked genes due to
shown as superscript letters on an upper case X,
their location on sex chromosomes.
such as Xh.
Many genetic diseases have been identified in humans
but most are very rare.
Radiation and mutagenic chemicals increase the
mutation rate and can cause genetic diseases and
cancer.
Applications and Skills
Statement
Guidance
3.4.A1
Inheritance of ABO blood groups.
The expected notation for ABO blood group
alleles: O = i, A=IA, B = IB.
3.4.A2
Red-green colour blindness and hemophilia as
examples of sex-linked inheritance.
Inheritance of cystic fibrosis and Huntington’s disease.
Consequences of radiation after nuclear bombing of
Hiroshima and accident at Chernobyl.
Construction of Punnett grids for predicting the
outcomes of monohybrid genetic crosses.
Comparison of predicted and actual outcomes of
genetic crosses using real data.
Analysis of pedigree charts to deduce the pattern of
inheritance of genetic diseases.
3.4.A3
3.4.A4
3.4.S1
3.4.S2
3.4.S3
3.4.U1 Mendel discovered the principles of inheritance with experiments in which large numbers of pea
plants were crossed.
Mendel’s principles of inheritance
Johann Gregor Mendel
(1822-1884)
Learn about Mendel and his work by using the weblinks
Gregor Mendel: Great Minds by SciShow
Because of his work
with pea plants Mendel
is considered the father
of modern genetics.
https://youtu.be/GTiOETaZg4w?list=PLC31B0C382
F9585D6
Gregor Mendel and pea plants
https://www.dnalc.org/view/16002-GregorMendel-and-pea-plants.html
He planted 1000s of
seeds per trial and
carried out many trials
to be sure of his results.
Biologica: Mendel’s Peas
His published work
(1865) is now
considered important,
but at the time was
ignored for 30 years.
http://biologica.concord.org/webtest1/
web_labs_mendels_peas.htm
https://upload.wikimedia.org/wikipedia/commons/3/3d/Gregor_Mendel_oval.jpg
Nature of science: Making quantitative measurements with replicates to ensure reliability. Mendel’s genetic
crosses with pea plants generated numerical data. (3.2)
To use statistical tests correctly and reach valid conclusions samples of
quantitative data has to be sufficiently large
First to develop theory scientists must make
deductions and test hypotheses: both processes
rely on quantitative data.
Secondly It is not enough to just have numerical
data, the sample size must be sufficiently large
to be judged reliable.
In smaller samples
anomalous values
are more likely to
skew the
calculated mean
and standard
deviation
*The standard deviation of the population is constant: (small) samples
have a higher standard deviation than the population the sample
comes from.
Larger samples give smaller
standard deviation*, this in
turn makes it easier to find a
statistically significant result at
a higher confidence level
The sample size required
varies:
• The larger the natural
variation the larger
the sample
• Depends on the type
of statistical test used
http://www.conceptstew.co.uk/PAGES/nsamplesize.html
Nature of science: Making quantitative measurements with replicates to ensure reliability. Mendel’s genetic
crosses with pea plants generated numerical data. (3.2)
To use statistical tests correctly and reach valid conclusions samples of
quantitative data has to be sufficiently large
First to develop theory scientists must make
deductions and test hypotheses: both processes
rely on quantitative data.
Secondly It is not enough to just have numerical
data, the sample size must be sufficiently large
to be judged reliable.
In smaller samples
anomalous values
are more likely to
skew the
calculated mean
and standard
deviation
*The standard deviation of the population is not affected, remember
that (small) samples have a higher standard deviation than the
population the sample comes from.
Larger samples give smaller
standard deviation*, this in
turn makes it easier to find a
statistically significant result at
a higher confidence level
The sample size required
depends on:
• The larger the natural
variation the larger
the sample
• Type of statistical test
used
http://www.conceptstew.co.uk/PAGES/nsamplesize.html
Definitions
This image shows a pair of homologous chromosomes.
Name and annotate the labeled features.
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
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
Review: 3.3.U2 The halving of the chromosome number allows a sexual life cycle with fusion of gametes.
Many eukaryotes reproduce by sexual reproduction. Even organisms capable of
asexual reproduction will reproduce sexually as well. Sexual reproduction involves
fertilisation, the fusion of gametes (sex cells), one from each parent.
Because fertilisation
involves the fusion of
gametes the number of
chromosomes in the next
generation is doubled.
To prevent a doubling of
chromosomes in each
generation a halving
mechanism is needed
during the life cycle.
To compensate for
the chromosome
doubling during
fertilisation gametes
undergo meiosis,
which halves the
chromosomes
present in gametes
compared to the
parent.
http://www.biologycorner.com/resources/diploid_life_cycle.gif
3.4.U3 The two alleles of each gene separate into different haploid daughter nuclei during meiosis. AND 3.4.U2 Gametes are
haploid so contain only one allele of each gene. AND 3.4.U4 Fusion of gametes results in diploid zygotes with two alleles of
each gene that may be the same allele or different alleles.
Because fertilisation
involves the fusion of
gametes the number
of chromosomes is
doubled. The diploid
organism also now
contains two alleles
for each gene locus.
The alleles present at
a gene locus maybe
similar or different.
Meiosis, along with
halving the
chromosomes
present in gametes
also reduces the
number of alleles
of each gene locus
from two to one.
http://www.biologycorner.com/resources/diploid_life_cycle.gif
3.4.A1 Inheritance of ABO blood groups.
The ABO blood type classification system uses the presence or absence of
certain antigen on red blood cells to categorize blood into four types.
Distinct molecules called agglutinogens (a type of antigen) are attached to the surface of red
blood cells. There are two different types of agglutinogens, type "A" and type "B”.
http://www.ib.bioninja.com.au/_Media/abo_blood_groups_med.jpeg
http://www.anatomybox.com/tag/erythrocytes/
3.4.A1 Inheritance of ABO blood groups.
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 no
antigens against which the recipient immune system can
react. Type AB is the universal recipient, as the blood has
no antibodies which will react to AB antigens.
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/
3.4.A1 Inheritance of ABO blood groups.
The ABO blood type is controlled by a single gene, the ABO gene. This gene has
three different alleles:
i O allele (no anitgen is produced)
IA A allele (type “A” anitgen is produced)
IB B allele (type “B” anitgen is produced)
A
I
Allele variant
Gene (lower case for ‘recessive’ alleles)
Diploid cells possess two alleles therefore the possible
genotype and phenotype combinations are:
Genotype
(allele combination)
Antigen production
Phenotype
(characteristic expressed)
ii
No antigen produced
Blood type O
IAIA and IAi
Type “A” anitgen produced
Blood type A
IBIB and IBi
Type “B” anitgen produced
Blood type B
IAIB
Both type “A” and “B” anitgens produced
Blood type AB
http://www.anatomybox.com/tag/erythrocytes/
3.4.U5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects.
Type “A” allele present and blood
type is A therefore the type “A”
allele is dominant to type “O”
Dominant alleles have the same effect on
the phenotype whether it is present in the
homozygous or heterozygous state
A
I i
Recessive alleles only have an effect on the
phenotype when present in the homozygous
state
Codominant alleles are pairs of different
alleles that both affect the phenotype when
present in a heterozygote
Type “O” allele present and blood
type is not O therefore the type “O”
allele is recessive to type “A”
A
B
I I
Type “A” and “B” alleles are
present and blood type is AB
therefore type “A”and “B” alleles
are codominant
http://www.anatomybox.com/tag/erythrocytes/
3.4.U5 Dominant alleles mask the effects of recessive alleles but co-dominant alleles have joint effects.
Type “A” allele present and blood
type is A therefore the type “A”
allele is dominant to type “O”
Dominant alleles have the same effect on
the phenotype whether it is present in the
homozygous or heterozygous state
A
I i
Recessive alleles only have an effect on the
phenotype when present in the homozygous
state
Codominant alleles are pairs of different
alleles that both affect the phenotype when
present in a heterozygote
Type “O” allele present and blood
type is not O therefore the type “O”
allele is recessive to type “A”
A
B
I I
Type “A” and “B” alleles are
present and blood type is AB
therefore type “A”and “B” alleles
are codominant
http://www.anatomybox.com/tag/erythrocytes/
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
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
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
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
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
Segregation
“alleles of each gene separate into different
gametes when the individual produces gametes”
F0
F1
Key to alleles:
Y = yellow
y = green
Genotype:
Yy
Yy
Gametes:
Y or y
Y or y
Punnet Grid:
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
Alleles segregate during
meiosis (anaphase I) and end
up in different haploid gametes.
Simplified notation of using
upper case for dominant
and lower case for
recessive is acceptable in
the case of two alleles
without co-dominance.
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
Monohybrid Cross
Crossing a single trait.
F0
Key to alleles:
Y = yellow
y = green
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:
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
Monohybrid Cross
Crossing a single trait.
F0
Genotype:
Yy
Yy
Gametes:
Y or y
Y or y
Punnet Grid:
F1
Key to alleles:
Y = yellow
y = green
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:
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
Monohybrid Cross
Crossing a single trait.
F0
Genotype:
Yy
Yy
Gametes:
Y or y
Y or y
Punnet Grid:
F1
Key to alleles:
Y = yellow
y = green
gametes
Genotypes:
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:
3:1
Mendel from:
http://history.nih.gov/exhibits/nirenberg/popup_htm/01_mendel.htm
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
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
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
Homozygous recessive
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
Monohybrid Cross
F0
Phenotype:
Genotype:
yy
yy
Homozygous recessive
Punnet Grid:
F1
What is the expected ratio of phenotypes
in this monohybrid cross?
Homozygous recessive
gametes
y
y
y
yy
yy
y
yy
yy
Genotypes:
yy
yy
yy
Phenotypes:
Phenotype ratio:
All green
yy
Key to alleles:
Y = yellow
y = green
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
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
gametes
Genotypes:
Phenotypes:
Phenotype ratio:
Heterozygous
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
Monohybrid Cross
F0
What is the expected ratio of phenotypes
in this monohybrid cross?
Phenotype:
Genotype:
yy
Yy
Homozygous recessive
Punnet Grid:
F1
Heterozygous
gametes
Y
y
y
Yy
yy
y
Yy
yy
Genotypes:
Yy
Yy
Phenotypes:
Phenotype ratio:
1:1
yy
yy
Key to alleles:
Y = yellow
y = green
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
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
Genotypes:
Phenotypes:
Phenotype ratio:
gametes
Heterozygous
3.4.S1 Construction of Punnett grids for predicting the outcomes of monohybrid genetic crosses.
Monohybrid Cross
F0
What is the expected ratio of phenotypes
in this monohybrid cross?
Phenotype:
Genotype:
YY
Yy
Homozygous dominant
Punnet Grid:
F1
Heterozygous
gametes
Y
y
Y
YY
Yy
Y
YY
Yy
Genotypes:
YY
YY
Yy
Phenotypes:
Phenotype ratio:
All yellow
Yy
Key to alleles:
Y = yellow
y = green
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
Cat genetics – collecting real world data
Do the inherited traits match what we know about cat genes?
http://www.g3journal.org/content/4/10/1881/F4.large.jpg
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
Cat genetics – collecting real world data
Locus
Phenotype and genotypes guide
Cats surveyed
1
1. Sex
2
…
12
Male (XY) / Female (XX)
record genotype
2. Hair length
short (ll) / long (LL, Ll)
record phenotype
3. Dominant white
completely white (WW, Ww) / some colour
(ww)
record phenotype - stop here if white is recorded
4. Piebald
No white (ss) / Some White (Ss) / Mostly
White (SS)
record genotype, some is < 50%, more is > 50%
5. Pigment density
dense - black, brown or orange (DD, Dd) /
diffuse - gray, light brown or cream (dd)
record phenotype
5. Orange
orange or cream (XOXO, XOY) / orange and
black or cream and grey (XOXo) / black or
grey (XoXo, XoY)
record genotype appropriate for the sex
Cat images: https://www.petfinder.com/
Source: http://udel.edu/~mcdonald/mythintro.html
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
Cat genetics – collecting real world data
1. Use the slide (previously) or first hand data collected from a family of cats
2. Survey the cats by completing the table (on the prior slide) – make sure you
indicate which cats are the parents and which are the offspring
It is important you do not create results, if you cannot determine a characteristic learn the box empty or
write unknown
3. From the parental background on the genetic traits construct an expected
frequency table
4. Test the expected frequencies against those observed using the Chi-Squared test
(see the following slides
n.b. especially with small sample sizes it is not always the case that observed
data will support theory. Outcomes are individual events and independent of the
collective probability, but the larger the sample the more likely the general
outcome will follow the theoretical expectation.
Cat images: https://www.petfinder.com/
Source: http://udel.edu/~mcdonald/mythintro.html
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
Cat genetics – collecting real world data
How well does the piebald frequency in the offspring match the predictions made from our
knowledge of the genes?
F0
Phenotype:
Genotype:
Punnet Grid:
F1
Genotypes:
Phenotypes:
Phenotype ratio:
Some white
No white
Ss
ss
gametes
s
s
S
Ss
Ss
s
ss
ss
Ss
Ss
Some white
1:1
ss
Key to alleles:
S = White
s = no white
Key to genotypes:
ss = no white
Ss = some white
SS = mostly white
ss
No white
Expected ratios
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
Cat genetics – collecting real world data
How well does the piebald frequency in the offspring match the predictions made from our
knowledge of the genes?
genotype
SS
(mostly white)
Ss
(some white)
ss
(no white)
observed
0
df
critical values
at 5%
1
3.84
2
5.99
3
7.82
4
9.49
5
11.07
expected
0
3
2
1
2
N = number of classes
Degrees of freedom (df) = N – 1
=3–1
=2
Chi-squared value =
Chi-squared value < critical value therefore we
support the hypothesis of piebald coat colour
= (0 – 0)2 + (3 – 2)2 + (1 – 2)2
0
2
2
=1
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
Chi-squared Test and genetics
More help and examples using the Chi-squared test:
http://www.slideshare.net/gurustip/the-chisquared-test
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
Test Cross
F0
Used to determine the genotype of an unknown individual.
The unknown is crossed with a known homozygous recessive.
Phenotype:
Genotype:
R?
unknown
r r
Key to alleles:
R = Red flower
r = white
Homozygous recessive
Possible outcomes:
F1
Phenotypes:
Unknown parent = RR
gametes
Unknown parent = Rr
gametes
3.4.S2 Comparison of predicted and actual outcomes of genetic crosses using real data.
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
3.4.U8 Many genetic diseases have been identified in humans but most are very rare.
3,358 genes with a phenotype-causing
mutation (OMIM, March 19, 2015)
Estimated total of 20,000-25,000 genes that
are expressed as proteins (International
Human Genome Sequencing Consortium,
2004).
Although it is impossible to give a single value
estimate we can say that genetic diseases are
very rare.
The number of genes present in the human
genome along with the fact that most
conditions are autosomal recessive: it is
unlikely that one parent will have a mutation
on a disease related gene, but the probability
that both parents have a mutation on the
same gene is very small.
http://learn.genetics.utah.edu/content/history/geneticrisk/
For example: Phenylketonuria (PKU) is a rare metabolic disorder that can be destructive to
the nervous system, causing intellectual disability. About 1 out of every 15,000 babies is born
with PKU. (Source: http://learn.genetics.utah.edu/content/disorders/singlegene/pku/)
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
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
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
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 cystic fibrosis (CF) and
one parent who has no family history. Explain CF and its inheritance to them. Deduce
the chance of having a child with CF and how it can be tested and treated.
Use the following tools in your explanations:
• Pedigree chart
• Punnet grid
• Diagrams
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Cystic Fibrosis (CF)
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 CF dominant or recessive? How do you know?
•
•
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Cystic Fibrosis (CF)
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.
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Cystic Fibrosis (CF)
Clinical example.
A mutation in the CFTR gene causes secretions (e.g. mucus, sweat and digestive juices) which are
usually thin instead become thick.
Instead of acting as a lubricant, the secretions block tubes, ducts and passageways, especially in the
lungs and pancreas. Despite therapeutic care lung problems in most CF sufferers leads to a early death
(life expectancy is between 35 and 50 years).
Diagnosis- blood test taken at 6-7 days after birth
http://www.flickr.com/photos/ozewiezewozewiezewallakristallix/263
2833781/
https://youtu.be/-a-WHZoTX0E
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Cystic Fibrosis (CF)
Clinical example.
What is the probability of two parents who are both carriers of the recessive allele producing
children affected by CF?
F0
Phenotype:
carrier
Genotype:
Tt
Punnet Grid:
gametes
carrier
Tt
T
T
t
F1
Genotypes:
Phenotypes:
Phenotype ratio:
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
t
Key to alleles:
T = Normal allele
t = mutated allele
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Cystic Fibrosis (CF)
Clinical example.
What is the probability of two parents who are both carriers of the recessive allele producing
children affected by CF?
F0
F1
Phenotype:
carrier
Genotype:
Tt
Punnet Grid:
gametes
T
t
T
TT
Tt
t
Tt
tt
Genotypes:
Phenotypes:
TT
carrier
Tt
Tt
Tt
Normal
Phenotype ratio:
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
Key to alleles:
T = Normal allele
t = mutated allele
tt
CF
3:1
Therefore 25% chance
of a child with CF
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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 = Normal allele
t = mutated allele
Key:
female
affected
Not
Affected
deceased
Looks like
Deduce the genotypes
of these individuals:
A&B
Genotype
Reason
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
C
D
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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 = Normal allele
t = mutated allele
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
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
Pedigree Charts
Key to alleles:
T = Normal allele
t = mutated allele
Individuals D and $ are planning to have another child.
Calculate the chances of the child having CF.
Key:
female
affected
Not
Affected
$
deceased
Looks like
Genotypes:
D=
Gametes
Phenotype ratio
$=
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
Therefore
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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 CF.
Key:
female
affected
$
Not
Affected
deceased
Looks like
Genotypes:
D = Tt (carrier)
$ = tt (affected)
T
t
t
Tt
tt
t
Tt
tt
Gametes
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
Phenotype ratio
1 : 1 Normal : CF
Therefore 50% chance of a
child with CF
male
Review: 3.1.A1 The causes of sickle cell anemia, including a base substitution mutation, a change to the base
sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin.
Review: 3.1.A1 The causes of sickle cell anemia, including a base substitution mutation, a change to the base
sequence of mRNA transcribed from it and a change to the sequence of a polypeptide in hemoglobin.
https://youtu.be/1fN7rOwDyMQ
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
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
Phenotype
Malaria
protection?
Homozygous HbA
Heterozygous
Homozygous HbS
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
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
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
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:
:
Therefore 50% chance of a
child with sickle cell disease.
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
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
Genotypes:
Phenotypes:
Phenotype ratio:
gametes
affected
HbS Hbs
HbS
HbS
HbA
HbAHbS HbAHbS
HbS
HbSHbS HbSHbS
HbAHbS & HbSHbS
Carrier & Sickle cell
1:1
Therefore 50% chance of a
child with sickle cell disease.
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
Punnet Grid:
F1
Genotypes:
Phenotypes:
Phenotype ratio:
gametes
carrier
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
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
Genotypes:
Phenotypes:
Phenotype ratio:
gametes
carrier
HbA HbS
HbA
HbS
HbA
HbAHbA HbAHbS
HbS
HbAHbS HbSHbS
HbAHb & 2 HbAHbS & HbSHbS
Unaffected & Carrier & Sickle cell
1: 2 : 1
Therefore 25% chance of a
child with sickle cell disease.
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA HbS
Punnet Grid:
gametes
HbA
HbS
F1
Genotypes:
Phenotypes:
Phenotype ratio:
unknown
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA HbS
Punnet Grid:
gametes
HbA
HbS
F1
Genotypes:
Phenotypes:
Phenotype ratio:
Key to alleles:
HbA = Normal Hb
HbS = Sickle cell
unknown
HbA HbA or HbA HbS
HbA
HbA
HbA
HbS
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some
genetic diseases are due to dominant or co-dominant alleles.
Sickle Cell
Another example of codominance.
Predict the phenotype ratio in this cross:
F0
Phenotype:
carrier
Genotype:
HbA HbS
Punnet Grid:
F1
Genotypes:
Phenotypes:
Phenotype ratio:
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
3 Unaffected & 4 Carrier & 1 Sickle cell
3:4:1
Therefore 12.5% chance of a
child with sickle cell disease.
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Huntington's Disease (HD) is a brain
disorder that affects a person's ability to
think, talk, and move. HD is caused by a
mutation in a gene on chromosome 4.
Genetics review:
1. Is this a dominant or recessive
condition?
2. Is this disorder autosomal or sex-linked
3. Produce a punnett square to explain the
inheritance pattern in the diagram.
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Huntington's Disease (HD) is a brain
disorder that affects a person's ability to
think, talk, and move. HD is caused by a
mutation in a gene on chromosome 4.
Genetics review:
1. Is this a dominant or recessive
condition?
Dominant – individuals are affected with
only a single mutated allele.
2. Is this disorder autosomal or sex-linked
Autosomal – chromosome 4
3. Produce a punnett grid to explain the
inheritance pattern in the diagram.
[Next slide]
http://learn.genetics.utah.edu/content/disorders/singlegene/hunt/
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Huntington's Disease (HD)
Clinical example.
What is the probability of an unaffected mother and a heterozygous affected father (for HD)
producing children affected by HD?
F0
Phenotype:
normal
Genotype:
tt
Punnet Grid:
gametes
affected
Tt
T
t
t
F1
Genotypes:
Phenotypes:
Phenotype ratio:
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
t
Key to alleles:
T = mutated allele
t = normal gene
3.4.U6 Many genetic diseases in humans are due to recessive alleles of autosomal genes, although some genetic diseases
are due to dominant or co-dominant alleles. AND 3.4.A3 Inheritance of cystic fibrosis and Huntington’s disease.
Huntington's Disease (HD)
Clinical example.
What is the probability of an unaffected mother and a heterozygous affected father (for HD)
producing children affected by HD?
F0
F1
Phenotype:
normal
affected
Genotype:
tt
Punnet Grid:
gametes
T
t
t
Tt
tt
t
Tt
tt
Genotypes:
Phenotypes:
Phenotype ratio:
Tt
Tt
Tt
HD
tt
tt
Normal
1:1
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
Key to alleles:
T = mutated allele
t = normal gene
Therefore 50% chance
of a child with HD
3.4.U7 Some genetic diseases are sex-linked. The pattern of inheritance is different with sex-linked genes
due to their location on sex chromosomes.
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. Alleles in this regions are expressed whether
they are dominant or recessive, as there is no
alternate allele carried on the Y chromosome.
Therefore sex-linked genetic disorders are
more common in males.
Non-homologous
region
Examples of sex-linked genetic disorders:
- haemophilia
- colour blindness
Edited from: http://www.slideshare.net/gurustip/theoretical-genetics
http://www.angleseybonesetters.co.uk/bones_DNA.html
http://en.wikipedia.org/wiki/Y_chromosome
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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
Xn
Xn
XN
Xn
Affected female
Carrier female
XN
Y
Normal male
Xn
Y
Affected male
no allele carried, none written
Key to alleles:
N = normal vision
n = red/green colour
blindness
Xq28
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
Human females can be homozygous or
heterozygous with respect to sex-linked genes.
Heterozygous females are carriers.
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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:
XN
Xn
Carrier female
X
XN
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
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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:
XN
X
Carrier female
N
X
Punnet Grid:
F1
Xn
XN
Y
Key to alleles:
N = normal vision
n = red/green colour
blindness
Normal male
Y
XN XN XN XN Y
Xn
XN
Xn
n
X
Y
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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?
There is a 1 in 4 (25%)
chance of an affected child.
Chromosome images from Wikipedia:
http://en.wikipedia.org/wiki/Y_chromosome
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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
3.4.A2 Red-green colour blindness and hemophilia as examples of sex-linked inheritance.
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
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
Hemophilia
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
This pedigree chart of the English Royal Family gives us a
picture of the inheritance of this X-linked disorder.
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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:
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
Not
Affected
deceased
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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:
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
Not
Affected
deceased
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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:
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
Not
Affected
deceased
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
Hemophilia
Pedigree chart practice
State the genotypes of the following family members:
1. Leopold
Xh Y
2. Alice
X H Xh
3. Bob was killed in a tragic croquet accident before
his phenotype was determined.
XH Y or Xh Y
4. Britney
Key:
affected
Key to alleles:
H = healthy clotting factors
h = no clotting factor
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
Not
Affected
deceased
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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
XH XH or XH Xh
Royal Family Pedigree Chart from:
http://www.sciencecases.org/hemo/hemo.asp
Key:
affected
Not
Affected
deceased
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
Pedigree Chart Practice
Key:
affected
Not
Affected
deceased
Dominant or Recessive?
Autosomal or Sex-linked?
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
Pedigree Chart Practice
Key:
affected
Not
Affected
deceased
Dominant or Recessive?
Dominant.
A and B are both affected but have produced
unaffected (D & F). Therefore A and B must have
been carrying recessive healthy alleles.
If it were recessive, it would need to be
homozygous to be expressed in A & B – and then
all offspring would be homozygous recessive.
Autosomal or Sex-linked?
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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
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.
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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:
affected
Not
Affected
deceased
female
male
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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:
affected
Not
Affected
deceased
female
male
Key to alleles:
XH = healthy clotting factors
Xh = no clotting factor
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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%
Key:
female
male
What do we know?
A = XH Y B = XH Xh (because G = Xh Y) E = XH Y
affected
Not
Affected
XH
deceased
Y
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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%
Key:
affected
Not
Affected
female
male
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
deceased
Y
XH
XH
Xh
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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%
Key:
affected
Not
Affected
female
male
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
XH Xh
Y
XH Y
XH Y
XH Y
Xh Y
deceased
3.4.S3 Analysis of pedigree charts to deduce the pattern of inheritance of genetic diseases.
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%
Key:
affected
Not
Affected
female
male
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 X H
XH 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!
3.4.U9 Radiation and mutagenic chemicals increase the mutation rate and can cause genetic diseases and cancer.
A mutation is a change in an organisms genetic code.
A Gene mutation is a change in the nucleotide sequence of a
section of DNA coding for a particular feature
Mutagens are agents that cause gene
mutations such as:
• chemicals that cause mutations (like
some found in tobacco smoke) are
referred to as carcinogens
• high energy radiation such as X-rays
• ultraviolet light
• Some viruses*
Alleles of a gene are similar, but have variations in the base
sequence. New alleles are created by gene mutation.
*Though an important source of mutation it is not a
focus of this syllabus point.
https://commons.wikimedia.org/wiki/File:Papierosa_1_ubt_0069.jpeg
Review: 1.6.U6 Mutagens, oncogenes and metastasis are involved in the development of primary
and secondary tumours.
If a mutation occurs in an oncogenes it can become cancerous. In normal cells
oncogenes control of the cell cycle and cell division.
mutation in a oncogene
malfunction in the control
of the cell cycle
uncontrolled cell division
tumour formation
http://en.wikipedia.org/wiki/Oncogene#mediaviewer/File:Oncogenes_illustration.jpg
3.4.U9 Radiation and mutagenic chemicals increase the mutation rate and can cause genetic diseases and cancer.
A mutation is a change in an organisms genetic code.
A Gene mutation is a change in the nucleotide sequence of a
section of DNA coding for a particular feature
Mutations can be classed as being beneficial, neutral (due to the
degenerate nature of DNA) or harmful. Most mutations are neutral
or harmful.
Mutations that occur in body (somatic cells) remain within
the organism. Mutations that occur in gametes can be
inherited by offspring: this is how genetic diseases arise.
http://www.nature.com/scitable/topicpage/rare-genetic-disorders-learning-about-genetic-disease-979
3.4.A4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
accident at Chernobyl nuclear power station
Radioactive isotopes released into the environment
exposing humans and other organisms to potentially
dangerous levels of radiation.
nuclear bombing of Hiroshima
http://i.telegraph.co.uk/multimedia/archive/02446/hiroshimabomb_2446747b.jpg
https://upload.wikimedia.org/wikipedia/commons/1/16/VOA_Mark
osian_-_Chernobyl02.jpg
3.4.A4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
nuclear bombing of Hiroshima
• Elevated rate of Leukemia (with the
greatest impact in children and young
adults)
• Elevated rates of other cancers
• No evidence of stillbirth or mutations in
the children of those exposed to radiation
https://upload.wikimedia.org/wikipedia/commons/f/f6/Hiroshima_girl.jpg
https://upload.wikimedia.org/wikipedia/commons/e/e9/The_patient%27s_skin_is_burned_in_a_pattern_corresponding_to_the_dark_porti
ons_of_a_kimono_-_NARA_-_519686.jpg
http://i.telegraph.co.uk/multimedia/archive/02446/hiroshima-bomb_2446747b.jpg
3.4.A4 Consequences of radiation after nuclear bombing of Hiroshima and accident at Chernobyl.
accident at Chernobyl nuclear power station
• A large area of pine forest downwind of
the reactor turned brown and died.
• Horses and cattle near the plant died
from radiation damage to their thyroid
glands.
• Bioaccumulation of radioactive caesium
in fish (Scandinavia and Germany) and
lamb (Wales) - contaminated meat was
banned from sale for years afterward.
• Drinking water (and milk) contaminated with radioactive iodine - at least 6,000
thyroid cancer attributed to radioactive iodine.
• No clear evidence to support an increase in the rate of leukemia other cancers – in
part due to the widely dispersed variable radiation and measures taken in
European populations.
https://upload.wikimedia.org/wikipedia/commons/1/16/VOA_Markosian_-_Chernobyl02.jpg
http://i.guim.co.uk/img/static/sys-images/Guardian/Pix/pictures/2014/6/27/1403890449199/933cb303-bf75-4e9a-8b0b-806bbfa6a37b2060x1373.jpeg?w=620&q=85&auto=format&sharp=10&s=abe2802021d01fe090859454e9020a44
Whirling Gene activity from the awesome Learn.Genetics site:
http://learn.genetics.utah.edu/archive/pedigree/mapgene.html
More practise questions for inheritance – the best
way to learn genetic theory is by practise.
Excellent problems and tutorials
by the biology project
Sex linked inheritance problems:
http://www.biology.arizona.edu/mendelian_genetics/problem_sets/sex_linked_inheritance/sex_linked
_inheritance.html
Monohybrid inheritance problems:
http://www.biology.arizona.edu/mendelian_genetics/problem_sets/monohybrid_cross/monohybrid_cr
oss.html
Bibliography / Acknowledgments
Bob Smullen