Topic guide 7.5: Patterns of inheritance

Unit 7: Molecular biology and genetics
.
75
Patterns of
inheritance
What are the inheritance patterns for the following genetic diseases: cystic
fibrosis, Huntington’s disease, sickle cell anaemia, phenylketonuria?
Can you think of any genetic diseases that are not inherited?
On successful completion of this topic you will:
•• be able to determine patterns of inheritance (LO4).
To achieve a Pass in this unit you will need to show that you can:
•• explain the terminology used in determining patterns of inheritance (4.1)
•• construct appropriate crosses from information provided to show
patterns of inheritance (4.2).
•• relate inherited diseases to patterns of inheritance (4.3).
1
Unit 7: Molecular biology and genetics
1 Some background science
Mendel, now considered the father of genetics, crossed true breeding tallstemmed pea plants with true breeding short-stemmed pea plants and all the
offspring were tall-stemmed. However, when he let this generation interbreed,
three-quarters of the offspring were tall and one-quarter short.
When Mendel presented his work it was ignored but it was later rediscovered and
its importance realised. The inheritance of some characteristics follows simple
Mendelian rules but scientists have found that many inheritance patterns follow
variations of these rules. Molecular genetics studies are now revealing many subtle
interactions between genes and between genes and environment that can affect
predicted Mendelian ratios.
2 Exemplar crosses
Key terms
Gene: A length of DNA that contains
a specific base pair sequence that
codes for one (or sometimes more)
polypeptide (protein). Some genes
code for mRNA and this acts to
regulate other genes.
Allele: A version of a gene. If the
original gene has undergone a
mutation then the nucleotide base
sequence is altered which may result
in a different polypeptide being
produced.
Multiple alleles: Within a population
there may be more than two alleles
of a particular gene, but in any
individual there will only be two
alleles present. There are three alleles
for the human ABO blood group, IA, IB
and Io, but each of us has only two of
those three.
Genotype: Describes an individual’s
genetic makeup, usually for a
particular characteristic, for example
IoIo denotes the genotype of someone
with blood group O.
Recessive: Characteristic that
is expressed in the phenotype
only if there is no allele for
the corresponding dominant
characteristic present in the
genotype. This allele for the
characteristic may also be described
as recessive.
7.5: Patterns of inheritance
There are many different inheritance patterns, including:
•• monohybrid
•• dihybrid
•• autosomal linkage
•• sex-linkage
•• pleiotropy
•• epistasis
•• polygenic inheritance.
Monohybrid inheritance
This is the inheritance pattern of just one characteristic, governed by one gene
(pair of alleles). A Punnett square can be used to predict the possible genotypes
of offspring of such a cross. There are conventions to be followed, as shown in
Figure 7.5.1.
Figure 7.5.1: Conventions
for genetic diagrams
and Punnett squares.
Mother
Parents’ phenotypes Blood group A X
IAIO
Parents’ genotypes
IA
Gametes
Father
X
Blood group B
IBIO
X
IO
X
IB
IO
Punnett square
male
gametes
IB
IO
female
gametes
IA
IAIB
blood group AB
IAIO
blood group A
IO
IBIO
blood group B
IOIO
blood group O
There are four possible phenotypes, group A, group AB, group B and group O. The
genotype for people with group O blood must be IoIo as it is a recessive trait.
2
Unit 7: Molecular biology and genetics
Multiple alleles
Link
See Unit 8: Pharmacological principles
of drug actions, Topic guide 8.6 to find
out more about antigens.
The letter I here stands for isoagglutinogen – a type of antigen found on the
surface membrane of red blood cells. The gene coding for the making of the
antigen is found on the long arm of chromosome 9. Within the human population,
or the gene pool, there are three different alleles of this gene but every person has
only two of them. IA and IB code for slightly different antigens and Io is so altered
that no antigens are produced.
Codominance
Key terms
Dominant: Characteristic expressed
in the phenotype even if the
individual is heterozygous and has
only one allele governing it in the
genotype, the other allele not being
expressed in the phenotype. This
allele for the characteristic may also
be described as dominant.
Phenotype: Observable
characteristic of an organism/
individual, for example, hair colour.
Codominance: Where two different
alleles of the same gene both
contribute to the phenotype. The
alleles IA and IB are codominant (both
will contribute to the phenotype if
both are present in the genotype) but
both are dominant to Io.
Heterozygous: Genotype of an
individual (known as a heterozygote)
who has two different alleles for a
particular gene.
Homozygous: Genotype of an
individual (known as a homozygote)
who has two of the same alleles for a
particular gene.
Meiosis: Type of cell division where
chromosome number is halved.
Used in many organisms to produce
gametes (eggs and sperms).
IA and IB are both dominant to Io (which is recessive), but if both are present in the
genotype each will contribute to the phenotype, so they are codominant. Hence
there is only one possible genotype, IAIB, for group AB blood.
However, people with group A or group B blood could be heterozygous or
homozygous.
Activity: Blood groups
1 List the possible genotypes for people with (i) group A blood and (ii) group B blood.
2 Use Punnett squares to predict the possible offspring genotypes and phenotypes from the
following parents:
(i) Mother with blood group AB and father with blood group O.
(ii)Mother with blood group B, whose own father was blood group A, and her partner with
blood group O.
3 Use the blood group kits and find out your own blood group.
Take it further: Blood groups
1 The A and B antigens on the surface of red blood cells are glycolipids (carbohydrates attached
to lipid molecules in the cell membrane). What makes them different from each other is the
type of sugar tacked on to the end of the carbohydrate molecule. Genes code for polypeptides
(proteins) and not directly for carbohydrates. Explain how the gene on chromosome 9
influences which type of antigen is produced on the surface of red blood cells.
2 Find out about the contribution of Austrian doctor, Karl Landsteiner, to safe blood transfusions.
Dihybrid inheritance
This is the inheritance pattern resulting from a dihybrid cross involving two
characteristics, each governed by a different gene. The genes are on separate
chromosomes so they can undergo independent assortment during meiosis.
Link
See Unit 14: Cell biology, Topic guide 14.4 for more details about meiosis.
7.5: Patterns of inheritance
3
Unit 7: Molecular biology and genetics
Gillian breeds horses and has a mare that is a black trotter. She pays a stud fee and takes the mare
to be mated with a stallion, who is also a black trotter. If both these horses are heterozygous at both
loci she can work out the chances of getting a foal that is also a black trotter.
Horse breeder
Black coat is dominant and chestnut coat is recessive. The gene responsible has two alleles B/b.
Trotting gait is dominant and pacing gait is recessive. The gene has two alleles T/t. If a homozygous
black trotting mare is bred with a homozygous chestnut pacing stallion, then all the foals will be black
trotters. However, if black trotters, heterozygous at both gene loci, are interbred, then, from several
matings, foals of the following phenotypes and in the following proportions should be obtained:
9 black trotters; 3 black pacers; 3 chestnut trotters; 1 chestnut pacer.
A Punnett square shows how this happens (see Figure 7.5.2).
Note that there are conventions to follow. Each gamete must have one allele for each characteristic.
When writing the genotypes of the diploid organisms keep the alleles of the same gene together
and when the genotype is heterozygous write the upper case letter first.
This shows the classic Mendelian dihybrid ratio of 9:3:3:1 of phenotypes in the F2 (second generation
offspring) generation.
Figure 7.5.2: Punnett
square showing possible
genotypes and phenotypes
of offspring resulting from
a cross between two horses,
both heterozygous for
colour and trotting gait.
Mare
X
Stallion
Parents’ phenotypes
black trotter
X
black trotter
Parents’ genotypes
BbTt
X
BbTt
male
gametes
BT
Bt
bT
bt
female
gametes
BT
BBTT
Black trotter
BBTt
Black trotter
BbTT
Black trotter
BbTt
Black trotter
Bt
BBTt
Black trotter
BBtt
Black pacer
BbTt
Black trotter
Bbtt
Black pacer
bT
BbTT
Black trotter
BbTt
Black trotter
bbTT
Chestnut trotter
bbTt
Chestnut trotter
bt
BbTt
Black trotter
Bbtt
Black pacer
bbTt
Chestnut trotter
bbtt
Chestnut pacer
Key
Black trotter (9/16)
Black pacer (3/16)
Chestnut trotter (3/16)
Chestnut pacer (1/16)
Key term
Locus (plural loci): Position on a
chromosome where a gene is located.
7.5: Patterns of inheritance
4
Unit 7: Molecular biology and genetics
Key term
Activity: Dihybrid crosses
Test cross: Used to determine the
genotype of an individual showing a
dominant trait. The individual is bred
with another individual showing the
recessive trait, whose genotype is
therefore known, and the phenotypes
of the offspring are observed.
In a certain breed of rabbit, black coat is dominant and brown coat is recessive. Short hair coat is
dominant and long hair coat is recessive. Use the symbols B/b and S/s for the two genes involved
and answer the following questions.
1
2
3
4
5
List all the possible genotypes for black, short-haired rabbits.
List all the possible genotypes for black, long-haired rabbits.
List all the possible genotypes for brown, short-haired rabbits.
List all the possible genotypes for brown, long-haired rabbits.
Use a Punnett square and predict the outcome (genotypes and phenotypic ratios of the
offspring) of several matings between two black, short-haired rabbits, heterozygous at both
gene loci.
6 Test cross: A rabbit breeder has a brown long-haired doe and a black short-haired buck. He
does not know the genotype of the buck. However, he knows that by doing test crosses
and observing the phenotypes, he may be able to tell the genotype. He allows these rabbits
to breed and produce several litters of kittens. Altogether there are 19 with black short hair;
21 with black long hair; 20 with brown short hair and 21 with brown long hair. What is the
genotype of the buck? Use a Punnett square to explain your answer.
Figure 7.5.3: Genotypes and phenotypes
for offspring resulting from a cross where
the genes are autosomally linked.
Female
Autosomal linkage
Male
X
Purple flowers, X Pink flowers,
long seed pods
short seed pods
Parents’
phenotypes
PP
LL
Parents’ genotypes
pp
ll
X
p
l
P
L
Gametes
Note that when genes
are on linked loci, we
write the two genes
one beneath the other
Pp
Ll
Offspring genotype
Purple flowered,
long seed pods
Offspring (F1)
phenotype
Cross between
members of
F1 generation
Pp
Ll
P
L
Gametes
Pp
Ll
X
p
l
P
L
X
p
l
Punnett square
male
gametes
P
L
p
l
P
L
PP
LL
Pp
Ll
p
l
Pp
Ll
pp
ll
female
gametes
Ratio of
F2 genotypes
Ratio of
F2 phenotypes
1
PP
LL
:
2 Pp
Ll
3 purple flowered
long seed pods
7.5: Patterns of inheritance
If two genes (each having two alleles) are on the
same autosome, they are inherited together. The
inheritance pattern is different from that of a dihybrid
cross (as shown in Figure 7.5.2) as the genes cannot
independently segregate.
•• In a species of flowering plant, one gene, P/p,
controls flower colour. Purple is dominant and
pink is recessive. Another gene, L/l, on the same
chromosome, controls seed pod length. Long is
dominant and short is recessive.
•• If two plants, both homozygous for both loci but
one being recessive for both features and the other
dominant, are cross-bred, all the offspring have
purple flowers and long seed pods.
•• If these offspring are allowed to interbreed, the
next generation does not show the classic 9:3:3:1
Mendelian dihybrid ratio. Instead it shows a 3:1 ratio
– that which would be expected in the F2 generation
from a monohybrid cross.
Figure 7.5.3 shows how this occurs.
:
:
1 pp
ll
Sometimes a few offspring show recombinant features,
such as purple flowered with short pods and pink
flowered with long pods. This shows that crossing over
happened in meiosis, between the two gene loci. The
greater the percentage of recombinants, the further
apart the two gene loci are on the chromosome.
1 pink flowered
short pods
5
Unit 7: Molecular biology and genetics
Key terms
Autosome: Chromosome that is not involved in determining the sex of an individual.
Humans have 23 pairs of chromosomes made up of 22 pairs of autosomes and one pair of sex
chromosomes.
Crossing over: Exchange of alleles between non-sister chromatids (replicated copies of
chromosomes of a homologous pair) during meiosis 1.
Sex chromosomes: The pair of chromosomes that determines whether an individual is male
(XY) or female (XX). These chromosomes may also contain genes for other characteristics that are
therefore sex-linked.
Figure 7.5.4: How crossing over during
meiosis produces genetic recombinants.
Ll
l
L
p
P
P
One pair of homologous chromosomes undergoing
crossing over during prophase 1 of meiosis, from an
individual heterozygous at both gene loci
p
L
L
l
l
P
P
P
p
L
L
l
l
P
p
P
p
Non-sister chromatids have swapped alleles
These chromatids contain new combinations of
alleles. If two gametes containing recombinants
combine at fertilisation, plants with pink flowers
and long seed pods or purple flowers and short
seed pods may appear in the F2 generation
7.5: Patterns of inheritance
6
Unit 7: Molecular biology and genetics
Sex-linkage
Key term
Sex-linked: Characteristic, not
directly connected with sex
determination, encoded by a gene
on a sex chromosome, for example,
genes for some blood clotting factors
are on the X chromosome.
If a gene for a characteristic that is nothing to do with sex determination is on the
X or Y chromosome, that characteristic is described as sex-linked.
Most sex-linked traits, such as haemophilia, Duchenne muscular dystrophy and
red-green colour blindness, are governed by a gene on the larger X chromosome,
which contains about 1300 genes. The small Y chromosome has about 40–50
protein-encoding genes, about half of which are male specific and needed for
aspects of male reproductive organ development. The SRY gene codes for SRY
protein, which activates a pathway causing certain embryonic tissue to develop
as testes, rather than ovaries. The Y chromosome passes from father to son. Sons
always inherit their X chromosome from their mother.
Because males have only one X chromosome, even if they inherit a recessive allele,
they will have the disorder because they have no other normal allele to produce
the required protein. Figure 7.5.5 shows the inheritance pattern for a sex-linked
disorder.
Figure 7.5.5: Inheritance pattern for
haemophilia A, a sex-linked characteristic.
Parental
phenotypes
Carrier mother X Normal father
XH Xh
Parental genotypes
Gametes
XH
XH Y
X
Xh
XH
Y
Offspring genotypes
male
gametes
XH
Y
XH
XH XH
Normal female
XH Y
Normal male
Xh
XH Xh
Carrier female
Xh Y
Haemophiliac male
female
gametes
H = allele for normal factor VIII
h = allele for non-functioning factor VIII
7.5: Patterns of inheritance
7
Unit 7: Molecular biology and genetics
Case study: Haemophilia A
Matthew has haemophilia A. This was diagnosed when he was about three years old, after having attended casualty departments for excessive
bruising whenever he fell over. At first the casualty staff suspected possible child abuse but, after some investigations, they concluded that this was
very unlikely and so they had a blood coagulation-time test and a blood cell count carried out. Levels of factor VIII in his blood were also measured.
This confirmed the diagnosis.
He inherited the condition from his mother who is a symptomless carrier. The mutation is to the large (186 kilo-base pairs including 26 exons
(expressed regions of DNA)) gene for a blood-clotting plasma protein, factor VIII, on the X chromosome. There are many different types of mutation
to this gene including large deletions, insertions and small deletions leading to frameshift mutations or nonsense mutations leading to formation
of truncated proteins.
Depending on the type of mutation, different patients suffer varying degrees of severity of the disease. Matthew has mild haemophilia as his blood
contains 10% normal level of factor VIII and his parents have learned how to manage his bleeding episodes. Surface cuts are not as dangerous as
knocks and bruises, because they can be seen and attended to, whereas knocks lead to internal bleeding and may damage joints. He has to go to
hospital if a large internal bleed is suspected. His dentist needs to know of his condition, as do his teachers.
Matthew would need a transfusion of blood plasma containing factor VIII if he were to have surgery or had a severe injury. Patients with severe
haemophilia are given this treatment prophylactically. All such blood products are obtained from donors and are heat-treated to remove viruses
such as HIV, and hepatitis A, B and C.
1 Why do Matthew’s dentist and teachers need to know of his medical condition?
2 Explain why Matthew may be given factor VIII as a treatment but boys with severe haemophilia are given factor VIII prophylactically.
3 Why, before the mid 1980s, do you think some haemophiliacs became HIV+?
Take it further: Baldness
Pattern baldness is a sex-influenced trait but it is NOT sex-linked. The gene, B/b, is on an autosome but its expression is influenced by the male
hormone testosterone. Males of genotype BB or Bb will become bald, whereas females of Bb will not be bald, but those of genotype BB will develop
baldness. The B allele is not a common mutation, so there are few females with pattern baldness. Even when they have the genotype BB their
baldness is less pronounced than in males of the same genotype and it develops much later in life.
•• How might taking certain steroids (that lead to greater production of testosterone) affect female body-builders of genotype BB?
Pleiotropy
Sometimes a single gene governs more than one phenotypic effect (trait), and the
characteristics are often very different. This phenomenon is called pleiotropy.
The gene in question codes for a product that is used by different types of cell in the
body or may be used as a cell signaller for various target cells.
The genetic disease phenylketonuria (PKU) is an autosomal recessive single gene defect
but the disease has different characteristics.
•• The gene codes for an enzyme, phenylalanine hydroxylase, which converts a dietary
amino acid, phenylalanine, into tyrosine, which can then be used to make the skin
and hair pigment, melanin. Homozygous recessive individuals have inherited two
mutated alleles and lack this enzyme so they cannot make tyrosine (and therefore
cannot make melanin) and are very fair/albino.
•• Because it is not converted to tyrosine, phenylalanine builds up in their blood and
saturates transporter proteins in the blood-brain barrier, preventing other amino
acids entering brain tissue and therefore reducing the formation of essential
neurotransmitters and leading to brain damage. It is also converted to other chemicals
such as phenylketone – which appears in the urine – hence the name of the condition.
Hence there are two distinct effects, in all people with PKU, from a mutation to one gene.
7.5: Patterns of inheritance
8
Unit 7: Molecular biology and genetics
3 The chi-squared (χ2) test
The chi-squared test is a statistical test to test predicted values based on the
inheritance theory we are using and find out if the difference between observed
categorical data and expected data is small enough to be due to chance. The
sample size must be relatively large with no zero values.
As with all statistical tests, the null hypothesis is tested and can be accepted or
rejected.
Example
Some mice have yellow fur and some have agouti (banded) fur. Agouti (grey) fur is the normal wild colour and the hairs are black at the tip
and base and yellow in the middle. The agouti gene (A) when switched on during the hair growth cycle produces the yellow band in the hair.
More than 25 mutated agouti alleles have been identified. Two notable alleles are lethal yellow, Ay, and viable yellow, Avy, on chromosome 2. The
mutated allele, a, never gets switched on and homozygotes (aa) have black fur.
When many pairs of heterozygous yellow-haired mice, genotype AAy, were interbred, 78 had yellow coats and 42 had agouti coats. The expected
ratio for a Mendelian monohybrid cross is 3:1 so out of 120 pups, we expect 90 to have yellow coats and 30 to have agouti coats.
The chi-squared test can tell us the probability of the difference between observed and expected ratio being due to chance. If it is less than 5%
(0.05) probability due to chance, we need to repeat the experiment and, if the data are always concordant, think about the inheritance pattern
and its underlying mechanism.
The null hypothesis is: there is no statistically significant difference between the observed and expected data; any difference is due to chance.
∑ (observed numbers [O] –E expected numbers [E])
2
χ2 =
Using the example above:
(O – E)2
E
Category
Observed (O)
Expected (E)
O–E
(O – E )
yellow fur
78
90
–12
144
144/90 = 1.6
grey fur
42
30
12
144
144/30 = 4.8
2
χ2 = 6.4
When comparing two categories of data, there is one degree
of freedom (number of categories −1). Using the χ2 distribution
table (Figure 7.5.6) we can see that χ2 is 6.4, which is greater
than the critical value at the 0.05 (5%) probability level. This
means that the difference is significant and less than 5% likely to
be due to chance so we reject the null hypothesis.
This means that with this type of monohybrid cross we do not
get a 3:1 ratio but a 2:1 ratio. What is the mechanism for this? It
is another example of pleiotropy. Mice of genotype AAy survive
but will develop insulin resistance, obesity and increased risk of
neoplasia (abnormal proliferation of cells that may be benign or
cancerous). Mice fetuses with genotype AyAy do not develop as
this is a lethal combination of alleles.
Number Degrees
of
of
classes freedom
2
2
1
0.00
0.10
0.45
1.32
2.71
3.84
5.41
6.64
3
2
0.02
0.58
1.39
2.77
4.61
5.99
7.82
9.21
4
3
0.12
1.21
2.37
4.11
6.25
7.82
9.84 11.34
5
4
0.30
1.92
3.36
5.39
7.78
9.49 11.67 13.28
6
5
0.55
2.67
4.35
6.63
9.24 11.07 13.39 15.09
Probability that
0.99 0.75 0.50 0.25 0.10 0.05
deviation is due to
(99%) (75%) (50%) (25%) (10%) (5%)
chance alone
Accept null hypothesis (any difference is due to chance
and not significant)
Figure 7.5.6: Part of a χ2
distribution table.
7.5: Patterns of inheritance
Critical value
of 2 0.05 p level; this is the level at which we
are 95% certain the result is not due to chance,
agreed on by statisticians as a cut-off point
0.02
(2%)
0.01
(1%)
Reject null
hypothesis; accept
experimental
hypothesis
(difference is
significant, not due
to chance)
9
Unit 7: Molecular biology and genetics
Leander is a health visitor and visits mothers with their newborn babies before the babies are a
week old. One of her jobs is to take blood from the baby’s heel and then carry out the Guthrie
test for PKU. Although this is a rare condition (about 1 in 16 000 births; first identified in 1934), its
effects are severe and irreversible and the Guthrie test, developed by a microbiologist in the early
1960s, is easy, simple and cheap so national screening has been carried out since 1966.
Health visitor
The blood is added to discs that are then placed on agar plates containing B. subtilis bacteria and
a growth inhibitor. If the baby’s blood has raised levels of phenylalanine, the bacteria will grow
and be visible. The amount of bacterial growth correlates to levels of phenylalanine in the blood.
Other tests, such as measuring the ratio of phenylalanine to tyrosine, can be carried out using mass
spectrometry to confirm the diagnosis.
Leander then has to explain to the parents of affected babies that a diet restricted in phenylalanine
and enhanced with tyrosine can prevent mental retardation. The child’s diet must contain very low
amounts of meat, chicken, cheese, dairy and legumes but a formula feed, first developed in 1951
by a German doctor, Professor Horst Bickel, containing all other essential amino acids is given as
well. She also tells them that diet drinks, sweetened with aspartame, should be avoided because
aspartame contains phenylalanine.
Leander informs the parents that their child’s blood levels of phenylalanine will need to be
regularly monitored and that there is now medication available – a drug called sapropterin
dihydrochloride, available since 2007, to reduce raised phenylalanine levels in patients as a result
of illness or fever. However, this drug does have some side effects.
During her studies, Leander read a book by Pearl Buck, published in 1950, called The Child Who
Never Grew, about Pearl’s daughter Carol who had undiagnosed, and therefore untreated, PKU.
Phenylketonuria was first named in 1934 by a Norwegian doctor, Ivar Asbjørn Følling.
One of Leander’s colleagues, a nurse, has looked after a young man with severe mental retardation
due to PKU. When he was diagnosed, soon after birth, the parents were advised of the condition
but did not follow the dietary treatment advice because they assumed that, if a condition was
genetic, then diet would not make any difference.
Epistasis
This is interaction between two genes where one gene masks the expression of
the other. (Epistasis is from a Greek word meaning ‘stoppage’).
The A and B antigens for blood groups are glycolipids. There is a gene on
chromosome 9 which codes for an enzyme that catalyses a stage in the pathway to
synthesise these molecules. However, there first has to be a precursor substance,
H, synthesised. This is coded for by another gene, H/h.
The recessive allele h results from a very rare mutation. A woman in Bombay, who
was genetically group B (she had one parent with blood group AB and one child
with blood group B), was phenotypically blood group O – her red blood cells did
not have any antigens on them.
She had the genotype hh and did not make the precursor substance H so her
B antigens could not be produced.
This is an example of epistasis as the h alleles prevent the expression of the
B allele. This particular example is known as the Bombay phenotype.
7.5: Patterns of inheritance
10
Unit 7: Molecular biology and genetics
Activity: More examples of epistasis
1 Show why having one parent with group AB blood meant that the woman with the Bombay
phenotype could not have been genetically blood group O.
2 How did the fact that her child was blood group B help to confirm her genotype for blood
group?
3 In mice the wild coat colour is agouti – it looks grey but each hair has black and yellow bands.
Agouti, A, is dominant to black, a. At a separate locus a gene, B/b, codes for an enzyme to
change a precursor substance to the black pigment. Individuals with the bb genotype are
albino, regardless of their A/a genotype, as they cannot produce black pigment and cannot
convert this to agouti. Use a Punnett square to predict the ratios of offspring phenotypes from
a cross between two individuals heterozygous at both gene loci (AaBb).
4 Two strains of white-flowered sweet peas, genotypes AAbb and aaBB, when crossed, produce
plants with purple flowers. The gene A/a converts a colourless precursor substance to a
colourless intermediate substance. Gene B/b converts the intermediate substance to a final
product (purple). One dominant allele of each gene must be present for purple flowers to
develop. Genotypes with aa or bb produce white flowers. Use a Punnett square to predict the
phenotypic ratios of offspring from a cross between two individuals of genotype AaBb.
Polygenic inheritance
From a monohybrid inheritance (one gene with two alleles) two distinct
phenotypes are produced (for example, Mendel’s tall- and short-stemmed
pea plants). There are no intermediate phenotypes and this is discontinuous
variation.
Where one gene and three alleles are involved, four distinct phenotypes are
produced. In dihybrid inheritance, four distinct phenotypes are produced.
Some characteristics are governed by more than two genes (each with alleles) and
each of the alleles makes a contribution to the characteristic. They have an additive
effect. They produce much phenotypic variation (continuous variation) and the
phenotypes are also able to be influenced by the environment.
An example is wheat with red grain (AABB) being crossed with wheat with white
grain (aabb). All the offspring (AaBb) have pink grains – intermediate between red
and white. When these offspring are interbred, they produce 1/16 red, 4/16 pale
red, 6/16 pink, 4/16 pale pink and 1/16 white. Note that there are five phenotypes
and the distinction between them is more ‘blurred’.
This is because each allele, A and B, is governing the making of a certain amount
of red pigment. The alleles a and b cause no pigment to be made. If there are four
dominant alleles present a large amount of red pigment is made. Three dominant
alleles causes less red pigment to be made, two even less, one very little and no
dominant alleles = no red pigment.
Human polygenic traits include height, skin colour and intelligence and, for all of
these, many genes are involved, each exerting a small additive effect.
7.5: Patterns of inheritance
11
Unit 7: Molecular biology and genetics
Activity: Polygenic inheritance
1 Use a Punnett square to explain how the cross between wheat plants, both having the genotype
AaBb, produces the five phenotypes. Show the phenotype distribution in graphical form.
2 In humans, two genes, D/d and E/e, at different loci, contribute to eye colour by controlling the
deposition of melanin in the front layer of the iris.
Genotypes: DDEE DdEE, DDEe DDee, DdEe, ddEE Ddee, ddEe ddee
Number of dominant
alleles in genotype 4
3
2
1
0
Phenotype: dark brown/black medium brown light brown deep blue/green pale blue
Use a Punnett square to calculate the frequency (percentage) of phenotypes from a cross
between two individuals both with the genotype DdEe. Plot these frequencies using a bar graph.
3 In a species of maize plant, three genes, A/a, B/b and C/c, contribute to length of corn cobs.
Dominant alleles each contribute 4 cm and recessive alleles contribute 2 cm.
(a)What is the potential length of the corn cobs of plants with the following genotypes?
AABBCC AaBBCC AabbCc AaBbCc AaBBCc
(b)Two batches of maize plants of genotype AABBCC were grown in the same type of soil
containing the same concentration of minerals. They were given the same volume of water
every day. One (batch X) was grown at 20 °C with high light intensity and the other
(batch Y) at 15 °C with lower light intensity. Those in batch X had a greater mean cob length
than those in batch Y. Suggest reasons for the difference in cob lengths.
Checklist
At the end of this topic guide you should be familiar with the following ideas:
 some phenotypic characteristics show straightforward Mendelian recessive or dominant
inheritance patterns
 monohybrid inheritance looks at the inheritance of one gene governing one characteristic
 some genes have more than two alleles in the population gene pool, although any individual
only has two alleles
 when genes are linked on the same autosome they cannot segregate at meiosis so the
expected dihybrid ratio is not seen
 recombinant phenotypes appear if there has been crossing over between these linked gene loci
 genes for traits, other than sex-determination, that are on the X or Y chromosome are
described as sex-linked
 some genes exert more than one phenotypic effect – this is called pleiotropy
 some genes interact and one affects the expression of other genes – this is called epistasis
 some characteristics are determined by many genes – this is polygenic inheritance
 the chi-squared test is used by geneticists to analyse data from crosses and determine if any
differences between their observations and expected ratios are significant or due to chance
 if the differences are significant they need to re-examine their inheritance pattern model.
Acknowledgements
The publisher would like to thank the following for their kind permission to reproduce their photographs:
Corbis: MedicalRF.com
All other images © Pearson Education
We are grateful to the following for permission to reproduce copyright material:
Figure 7.5.5: Inheritance pattern for haemophilia A, a sex-linked characteristic, from Fig 1 from
Pearson A2 Bio for OCR page 124. Used with permission of Pearson Education Ltd; Figure 7.5.6: Part
of a χ2 distribution table, from Fig 2 from Pearson A2 Bio for OCR page 135. Used with permission
of Pearson Education Ltd.
Every effort has been made to trace the copyright holders and we apologise in advance for any
unintentional omissions. We would be pleased to insert the appropriate acknowledgement in any
subsequent edition of this publication.
7.5: Patterns of inheritance
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