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Biology A level - Patterns of Inheritance Cheat Sheet by

This is an OCR A Gateway spec A level Biology cheat sheet for Chapter 20 module 6. Specification reference: 6.1.2

Key words for topic

Genotype
Genetics makeup of an organisms - the alleles present.
Phenotype
The expression of genes (also affected by the enviro­nment).
Homozygous
A pair of homologous chromo­somes carrying same allele for a gene (e.g. AA, aa).)
Hetero­zygous
A pair of homologous chromo­somes carrying different alleles for a gene (e.g. Aa).
Recessive allele
Allele only expressed in the absence of dominant alleles
Dominant allele
Allele always expressed in phenotype.
Codominant
Multiple alleles are equally dominant and expressed in the phenotype.
Sex-li­nkage
A gene with a locus in the X chromo­some.
Autosomal linkage
Genes in the same chromosome (not sex chromo­some). An autosomal chromosome is any chromosome other than sex chromo­somes.
Epistasis
When one gene modifies the expression of a different gene.
Monohybrid
Genetics inheri­tance cross of trait determined by one gene.
Dihybrid
Genetics inheri­tance cross of trait determined by two gene.
Gene pool
All alleles of all genes within popula­tion.
Allele frequency
Proportion of an allele in a gene pool.
Some of these terms will be further explained in this cheat sheet.

Notation systems for topic

Type of inheri­tance
Coding
Monohybrid
Single letter capital / lower case e.g. A, a
Codomi­nance
Geneallele e.g. CWCR, IAIB
Multiple alleles for one gene (more than two)
Geneallele e.g. IAIO
Sex linkage
Chromosomeallele e.g. XRXr, XrY
Autosomal linkage and epistasis
Single letters e.g. Aa Bb
 

Monohybrid crosses

Worked example: Cystic fibrosis is caused by a recessive allele, what is the probab­ility that two carrier parents will have a child with cystic fibrosis?
Parent genotypes ("both carrie­rs"): Ff x Ff
Each gamete will carry either the F or f allele and can fuse with gametes containing either F or f alleles too because parents are hetero­zygous.
 
F
f
F
FF (no CF)
Ff (no CF)
f
Ff (no CF)
ff (CF)
Therefore probab­ility of child with cystic fibrosis = 1/4 = 25%.

Codominant inheri­tance

Worked example: A flower can be three colours: white, red or pink. The alleles for white and red are codominant to form pink flowers. If two pink flowers reproduce, what is the probab­ility of forming a red offspring flower?
Pink flowers must have the alleles both for red and white as they are codominant to produce pink. They are therefore hetero­zygous.
Parent genotype: CRCW x CRCW
 
CR
CW
CR
CRCR (red)
CRCW (pink)
CW
CRCW (pink)
CWCW (white)
Therefore probab­ility of forming a red offspring is 1/4 = 25%.

Multiple alleles inheri­tance

Key example: Blood groups
There are three alleles for blood group: A, B and O, repres­ented as IA, IB and IO.
Both IA and IB are codominant to form phenotype group AB, and IO is recessive.
Worked example: parents with blood group AB and O reproduce. WHat so the probab­ility that they will produce an offspring with blood group A?
Parent genotype: IAIB x IOIO
 
IA
IB
IO
IAIO (group A)
IBIO (group B)
IO
IAIO (group A)
IBIO (group B)
Therfore probab­ility is 2/4 = 50%.

Sex-li­nkage

Sex linked alleles only occur in X chromo­somes as Y chromo­somes contain less genetic inform­ation (too small to carry more). This makes males more likely to carry recessive sex-linked disorders as their homologous Y chromosome cannot contain the dominant allele.
Worked example: Colour blindness is caused by a recessive sex-linked allele only found in the X chromo­some. If a colour blind male reproduces with a hetero­zygous female, what is the probab­ility that they would have a colour blind child?
Parent genotype: XBXb x XbY
 
XB
Xb
Xb
XBXb
XbXb
Y
XBY
XbY
The probab­ility of a colour blind child is therefore 2/4 = 50%.
Be sure to read the question carefully. Some exam questions will choose animals where the female has XY chromo­somes and the male has XX chromo­somes which can throw you off.

Epistasis

One gene affects the expression of another, therefore multiple genes will be at play here.
(Common) worked example: Labradors have two genes which will affect their fur colour. The first gene controls which colour is expressed. If allele B is expressed, the dog will be black. If allele b is expressed, the dog will be brown. The second gene codes for pigment produc­tion. If allele E is expressed, pigment will be produced. If allele e is expressed, no pigment will be produced and the dog will be yellow. Parents hetero­zygous for both genes reproduce.
Parent genotype: Bb Ee x Bb Ee
Parent possible gametes: BE, Be, bE, be x BE, Be, bE, be
 
BE
Be
bE
be
BE
BBEE
BBEe
BbEE
BbEe
Be
BBEe
BBee
BbEe
Bbee
bE
BbEe
BbEe
bbEE
bbEe
be
BbEe
Bbee
bbEe
bbee
Offspring phenot­ypes:
Black : Brown : Yellow
9:3:4

Dihybrid crosses

Two genes considered at the same time.
Key example: Mendel's peas
Peas can be two different colours, yellow or green. They can also be either round or wrinkled. Roundness is a dominant allele (R) and yellow colour is also dominant (Y). Show the propor­tions of phenotypes for the offsprings of two hetero­zygous peas.
Parent genotype: RrYy x RrYy
Gametes: RY, Ry, rY, ry x RY, Ry, rY, ry
 
RY
Ry
rY
ry
RY
RRYY
RRYy
RrYY
RrYy
Ry
RRYy
RRyy
RrYy
Rryy
rY
RrYy
RrYy
rrYY
rrYy
ry
RrYy
Rryy
rrYy
rryy
Offspring phenot­ypes:
Round and yellow : Wrinkled and yellow : Round and green : Wrinkled and green
9:3:3:1
This ratio will always be present in dihybrid hetero­zygous crosses.
This proportion is only true IF:
- There is no autosomal linkage.
- There is no crossing over during meiosis.
- There are no mutations.
- There is no sexual selection (e.g. black rabbits only mate with other black rabbits).
- There is no epistasis.

Autosomal linkage and crossing over

When genes are linked, it means they occur on the same chromo­some.
For example, R and Y are on one chromosome and r and you on the other.
This means it is not possible to get all the gametes predicted in the previous box.
R and y cannot form a gamete because R and Y are always inherited together.
In the example we used, the only possible gametes for hetero­zygous parents if the genes are linked are:
RY, ry x RY, ry
 
RY
ry
RY
RRYY
RrYy
ry
RrYy
rryy
The ratio therefore changes from 9:3:3:1 to 3:1 (in this example.)
Crossing over (meiosis)
It is, however, possible for observ­ations to show more than just two phenot­ypes, or unexpected propor­tions of these phenot­ypes.
This is because of crossing over in meiosis. Homologous chromo­somes can overlap and swap parts of non-siter chroma­tids. This can form new gametes.
In our example, part of the chromosome containing the Y and another chromatid containing the y allele could swap, creating the gametes Ry and rY. These are called recomb­inant genes.
Genes are less likely to split if they are close together, because there is less space for a chiasmata to form between them so they are less likely to be separated and will be inherited together (linked).

Autosomal linkage

Note the genes occuring on the same chromosome but far apart as marked as not linked because they are likely to be separated by a chiasmata during meiosis.
 

Chi-sq­uared

Statis­tical test used to calculate whether what we expect is different from what we actually observe.
(O-E)2/E
Degrees of freedom = number of categories - 1
Worked example: Corn can be yellow (Y) or purple (y). Hetero­zygous cross - Mendelian genetics expected:
 
Y
y
Y
YY
Yy
y
Yy
yy
yellow : purple = 3:1
A student observed 21 yellow and 13 purple kernels. Is this signif­icantly different from expect­ation?
Null hypoth­esis: no signif­icant difference between expected and observed colour of corn.
1. Make expected a proportion
21+13 = 34
3+1 = 4
34/4 = 8.5
3x8.5 = 25.5
1x8.5 = 8.5
2. Observed - Expected
21-25.5 = -4.5
13-8.5 = 4.5
3. Square answers
-4.52 = 20.25
4.52 = 20.25
4. Divide by expected value
20.25/25.5 = 0.794
20.25/8.5 = 2.382
5. Add up values
Chi squared = 3.176
6. Calculate degrees of freedom and compare calculated value to critical value table
2-1 = 1
critical value = 3.841
Because our calculated value is lower than the critical value, there is more than 5% probab­ility that any differ­ences is due to chance. This means results are not signif­icantly different and we accept the null hypoth­esis.
So the colour observed matched the Mendelian genetics we originally calcul­ated.
If the observed data does not match the expected data, this means that genes are linked and therefore not subject to indepe­ndent assort­ment.

The Hardy-­Wei­nberg principle

Mathem­atical model used to determine the frequency of alleles.
p2 + 2pq + q2 = 1
p2 = frequency of homozygous dominant genotype (AA)
2pq = frequency of hetero­zygous genotype (Aa)
q2 = frequency of homozygous recessive genotype (aa)
p + q = 1
p = frequency of dominant allele
q = frequency of
Worked example: Haemoc­hro­matosis is caused by a recessive allele. In one country, every 1 in 400 people have haemoc­hro­mat­osis. What percentage of the population is a carrier for the haemoc­hro­matosis recessive gene?
The data given to us is the proportion of homozygous recessive indivi­duals, so q2.
Therefore, q2 = 1/400 = 0.0025.
1. Use the inform­ation given to work out other values.
If we know q2, then we can find q then p.
√q2 = q
√0.0025 = q = 0.05
p+q = 1, therefore p + 0.05 = 1
1-0.05 = p = 0.95
2. Substitute for answer
Carriers = hetero­zygous, so 2pq
2x0.95­x0.05 = 0.095
0.095x100 = 9.5%
The Hardy-­Wei­nberg principle does make assump­tions which could make it inappr­opriate to use for some contexts. It relies on:
- A large popula­tion.
- Random mating.
- No natural selection.
- No migration / gene flow.
- No mutations.

Types of selection

Speciation

Creation of new species. A group gets reprod­uct­ively isolated from the population (cannot breed together) and develop differ­ences in gene pools.
Allopatric speciation
Become geogra­phi­cally isolated (e.g. new mountain range). Evolve separately through natural selection, accumulate mutations.
Sympatric speciation
Occupy same area but behaviours change so don't breed together.

Genetic drift

Change in allele frequency in popula­tion. Impacts are greater in smaller popula­tions.
Factors that can lead to genetic drift are:
Genetic bottle­necks
Event (e.g. natural disaster, overfi­shi­ng...) kills off most of a popula­tion, leaving a few survivors behind (small gene pool).
Founder effect
A few indivi­duals (small gene pool) first colonise an area, isolated from original popula­tion. Can even make rare homozy­gotic recessive phenotypes more frequent.

Natural v. Artificial selection

Natural selection
Evolution. Variety in phenotypes due to genetic bad enviro­nmental factors.
Artificial selection
Humans select desirable features and breed those indivi­duals together.
 
Ethical concern: due to selected features, some animals (e.g. pugs) will have medical issues.

Genetic banks

Gene banks store DNA from plants or animals.
Selective breeding often involves inbree­ding, so gene banks can be used to reduce this and increase genetic diversity of a species.
                   
 

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