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Number 183
Variations from expected Mendelian
Monohybrid Ratios
Gregor Mendel demonstrated 3 key phenotypic ratios when he
carried out a series of experiments investigating monohybrid
crosses in garden peas. All Biologists need to know these ratios
and how to use them to solve problems. However there are a few
important situations where the usual outcomes do not arise.
Example 2
•
•
•
•
Exam Hint: You must know the 3 key ratios and how to draw a
genetic diagram to demonstrate genotypic and phenotypic
outcomes.
One parent is heterozygous (Tall genotype)
Half the gametes will be T (dominant) and half will be t (recessive)
One parent is homozygous recessive (short genotype)
This parent can only produce gametes with the recessive allele
(t)
Male Gametes
T
The three key ratios
From the phenotypic ratio of the progeny, you can work out the
genotypes of the parents
•
•
•
•
•
1:0 – Either: at least one parent is homozygous dominant; or:
both parents are homozygous recessive
1:1 – One parent is heterozygous and one is homozygous
recessive
3:1 – Both parents are heterozygous
•
tt
Short
Example 3
•
•
One parent is homozygous dominant
This parent can only produce gametes with the dominant allele
(T)
All progeny will have the dominant phenotype – the genotype
of the other parent is irrelevant
Both parents are heterozygous (Tall genotype)
These parents both produce gametes with either the dominant
allele (T) or the recessive allele (t)
Female Gametes
Male Gametes
T
t
T
TT
Tall
t
Female Gametes
T
t
T
TT
Tall
Tt
Tall
T
TT
Tall
Tt
Tall
tt
Short
Exam Hint: You may be asked how to find the genotype of an
individual with the dominant phenotype. This is the Test Cross.
If, when the individual is crossed with a homozygous recessive,
all the offspring show the dominant phenotype then the
individual is homozygous dominant (Ratio 1:0); if the ratio is
1:1 then the parent is heterozygous.
Always use the same letter for a gene – usually the first letter of
the dominant allele or first letter of the characteristic
Use UPPER CASE for dominant alleles and lower case for
recessive alleles
E.g. Mendel’s first experiment was on the height of pea plants –
start by demonstrating that you understand this convention:
“let T represent Tall and t represent short”
Male Gametes
Tt
Tall
Phenotypic Ratio = 1 Tall:1 Short
(1/2 Tall:1/2 Short)
Example 1
•
•
Tt
Tall
t
Genetic Diagrams
•
Female Gametes
t
t
Tt
Tall
Tt
Tall
tt
Short
Phenotypic Ratio = 3 Tall:1 Short
(3/4 Tall:1/4 Short)
When will these expected ratios not occur?
These ratios will always apply to monohybrid crosses except in one
of the following situations:
1. A Mutation occurs
2. There are more than 2 alleles (Multiple Alleles), in this case two
or more of the alleles are usually equally dominant (CoDominance)
3. The gene has its locus on the sex chromosomes (Sex Linkage)
4. One genotype results in failure of the progeny to survive to
birth (Lethal Gene)
Phenotypic Ratio = All tall = 1:0
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Mutation
Sex Linkage
This is an unpredictable change to the genome of the individual. It
could be that just one of the alleles is changed by as little as one base
in the DNA sequence. An example of a Gene Mutation such as this is
Sickle Cell Anaemia. Alternatively, there may be a change to a
chromosome so that a whole gene is missing or has multiple versions.
This is a Chromosome Mutation. For Example Down’s Syndrome,
where the person has THREE of chromosome number 21.
Sex linked genes have a locus on the sex chromosomes. The much
smaller Y chromosome has many loci missing. Males only have 1
allele for the gene – on their one X chromosome. This one allele
automatically expresses itself in the phenotype so recessive sex
linked alleles affect males far more than females. Two major sex
linked genes are those for HAEMOPHILIA and RED-GREEN
COLOUR BLINDNESS.
In genetic diagrams, use X and Y to represent the sex chromosomes
and show the gene as superscript.
Multiple alleles and Co-Dominance
Each individual only receives one allele from each parent and clearly
only possesses two alleles for each gene (hence diploid). If the
gene has more than two alleles then clearly no individual can have
them all in their genotype. The gene is said to have Multiple Alleles
if there are more than two. At the same time the presence of more
than two alleles affects how the gene expresses itself. Usually two
or more alleles are neither dominant nor recessive in relation to
each other – they are said to be Co-Dominant. They may both be
dominant over the other alleles or both recessive in relation to
another allele.
Example 6
Consider HAEMOPHILIA with a heterozygous mother (XHXh) and
a healthy father (XHY)
Female Gametes
Male Gametes
XH
Xh
XH
Healthy
Female
Y
Example 4
One parent is group A and is heterozygous (dominant A,
recessive O) - IA Io
The other parent is group B and is heterozygous - IB Io
Male Gametes
IA
Female Gametes
IB
Io
IA IB
IA Io
AB
Io
IB Io
B
Io Io
O
IA
Female Gametes
IB
Io
IA Io
IA Io
A
IB
IB Io
B
Haemophiliac
Male
Either the homozygous dominant or the homozygous recessive
condition results in failure of the offspring to develop properly so
they do not appear in the progeny, affecting the expected ratio. An
example of this is in Manx cats. These cats have a dominant allele
that causes them to have a much-shortened tail. However, if the
dominant allele is inherited from both parents the condition is lethal.
Hence all surviving short-tailed Manx cats are heterozygous, but
the phenotypic ratio of “Manx” to “normal-tailed” cats is not the
expected 3:1, but 2:1 (See example 3 above).
One parent is group AB and is co-dominant (dominant A,
dominant B) - IA IB
The other parent is group O and is homozygous recessive - Io Io
Male Gametes
Xh Y
Lethal Gene
Example 5
•
Healthy
Female (carrier)
Exam Hint: You must know how to use Superscript in genetic
diagrams to demonstrate how multiple alleles and sex linkage
affect inheritance.
A
Offspring could be any blood group, with an equal chance (1/4) of
each occurring
•
XH Xh
All daughters are healthy – though half are carriers. Half of the
sons suffer from Haemophilia.
Unfortunately, the allele controls production of “Factor VIII”, one
of the factors in the blood clotting process. This means that the
blood of Haemophiliacs will not clot, they suffer frequent bouts of
serious internal bleeding (haemorrhages) and until recently did not
survive beyond childhood. The inheritance of haemophilia can be
observed in the British Royal Family. Queen Victoria was a carrier
for the condition and passed the recessive allele on to several of
her 9 children (probability of half of them!) and it continued through
European Royalty from there, though not in the current British Royal
Family. Try to find a pedigree diagram of Queen Victoria’s
descendants and predict the genotype of each individual.
When constructing genetic diagrams a few new “rules” apply:
• There are 3 alleles (“multiple alleles”) so the usual method of
denoting a gene using UPPER or lower case cannot apply.
• Use a letter for the gene and superscript for the alleles e.g. IA
•
XH Y
Healthy
Male
An example of this is the ABO blood-grouping gene in humans,
where the alleles for antigen A and antigen B are co-dominant over
the allele for no antigen (O).
•
XH XH
Example 7
A
Male Gametes
M
IB Io
B
Female Gametes
M
m
MM
Mm
Manx
Lethal
m
Offspring all have different blood groups from parents – ½ A and ½
B.
A person with phenotype AB cannot pass this on to their children,
they can only pass on one of the alleles.
Mm
mm
Manx
Phenotypic Ratio = 2 Manx:1 Normal length tail
(2/3 Manx:1/3 Normal)
2
Normal
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Practice Questions
1. A homozygous tall pea plant is crossed with a heterozygous tall plant. All the offspring are tall. Use a genetic diagram to explain this,
with the letter T representing the gene.
2. Explain how you would find which of the offspring are homozygous and which are heterozygous.
3. Use a genetic diagram to explain the phenotypic ratio when a person with blood group AB has children with a group A person if a) the
second parent is homozygous; b) is heterozygous
4. Use a genetic diagram to explain the phenotypic ratio when a colour-blind man has children with a woman who is a carrier for the
condition.
Answers
1.
Male Gametes
T
T
Female Gametes
T
t
TT
Tall
Tt
TT
Tall
Tt
Tall
Tall
Phenotypic Ratio = All tall = 1:0; all progeny must receive a dominant allele from their homozygous parent.
2. Use a test cross. Breed the plants with a short (homozygous recessive) plant. Progeny from the homozygous tall plants will all be tall
(ratio 1:0); half of those from the heterozygous parent will be tall and half will be short (ratio 1:1)
3. (a)
Male Gametes
Female Gametes
IA
IB
IA
IA IA
A
IA IA
A
IB
IB IA
AB
IB IA
B
Phenotypic ratio = 1 A: 1 AB – half the children have group A and half have group AB
(b)
Male Gametes
Female Gametes
I0
IA
IA
IA IA
A
IA I0
A
IB
IB IA
AB
I B I0
B
The heterozygous parent must have genotype IA Io, the phenotypic ratio = 2 A: 1 AB: 1 B (½ AB, ¼ A, ¼ B)
4. Let XH represent the healthy allele, Xh the colour-blind allele, and Y the male sex chromosome without the gene locus. (XC and Xc are
not recommended as the letter C is the same in upper and lower case)
Female Gametes
Male Gametes
XH
Xh
Xh
XH Xh
Healthy
Female (carrier)
Y
XH Y
Healthy
Male
Xh Xh
Colour blind
Female
Xh Y
Colour blind
Male
Phenotypic ratio = 1 healthy male: 1 colour blind male: 1 healthy female: 1 colour blind female
(¼ healthy male: ¼ colour blind male: ¼ healthy female: ¼ colour blind female)
Acknowledgements:
This Factsheet was researched and written by AndyPeters.
Curriculum Press, Bank House, 105 King Street, Wellington, Shropshire, TF1 1NU.
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