Term
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Definition
When specific alleles of two genes are transmitted
together, the genes are said to be linked.
• Linkage defies Mendel’s law of independent
assortment. |
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Term
| The examples of the gene linkage |
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Definition
• Example of Hemophilia A and colorblindness: in a
pedigree, all males have either both of these X-linked
diseases, or neither disease. These two diseases are
always inherited together: the genes that transmit them
are linked.
• In contrast, in the second pedigree, males may be
affected by either hemophilia B or colorblindness
individually, both together, or neither disease. These
two disorders are not linked. |
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Term
Genetic Basis of Linkage.
Independent assortment of genes located on...results
from... |
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Definition
the same
chromosome ( in our example, X chromosome)
Recombination |
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Term
Genetic Basis of Linkage
Examples of...the male in generation I had....Mutant genes were located... |
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Definition
...hemophilia B and colorblindness,
...both diseases
...on the same X
chromosome. |
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Term
Genetic Basis of Linkage
-• As the mutant alleles traversed through the daughter
to the affected males grandchildren...
-offspring in generation III |
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Definition
-...there was
recombination between the two gene loci in the
female (they have paired X chromosomes) - had either one, both or
neither disease. |
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Term
Linkage: Fundamental Rules
• The farther apart the two genes are on the same
chromosome... |
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Definition
...the more likely they are to crossover
during recombination. |
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Term
Linkage, fundamental rules
• This implies that since all three genes: Hemophilia
A, Hemophilia B and Colorblindness... those for Hemophilia B and
colorblindness ... Hemophilia A and
colorblindness.
• Based on this interpretation, we can draw a rough
map of the position of these three genes on the X
chromosome. |
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Definition
... are on the X
chromosome,
...must be farther apart from each
other than are the genes for |
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Term
Recombination frequency
When genes assort independently there will be....and 50% with a.... |
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Definition
... 50% offspring
with the same gene combination as the parent (parental
type)... with a ‘mixed’ combination (recombinant
type). |
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Term
| Recombination frequency is 50% when there is... |
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Definition
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Term
| Recombination frequency 50% is the maximum recombination frequency possible because |
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Definition
...for independently assorting genes, there will always
be equal number of parental and recombinant types. |
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Term
Recombination frequency
When genes are not assorting independently... |
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Definition
recombination
frequency will be less than 50%. |
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Term
Recombination frequency
Geneticists use recombination frequency to gauge |
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Definition
the
distance between two genes and to map their location on the
chromosome in relation to one another. |
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Term
Calculation of Recombination Frequency
Example: Drosophila Eye color and body color genes located on the |
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Definition
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Term
Calculation of Recombination Frequency
Eye color:
– w+ =
– w = |
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Definition
=red eyes (wild type)
=white eyes (recessive) |
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Term
Calculation of Recombination Frequency
Body color:
– y+ =
– y = |
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Definition
= brown body (wild type)
=yellow body (recessive) |
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Term
Calculation of Recombination Frequency
Cross between homozygous white eyed, brown bodied female with
a red eyed, yellow bodied male.
If there is no linkage, F2 will have ...
• However, if the two genes are close to each other on the X
chromosome... |
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Definition
-equal proportions of all four
types of males.
-linkage prevents crossing-over, resulting in deviation
from the expected ratios: |
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Term
| Two genes located on the X chromosome Drosophilia: |
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Definition
• Drosophila wing size :
– m+ = normal wings (wild type)
– m = miniature wings (recessive)
• Drosophila eye color:
– w+ = red eye color (wild type)
– w = white eyes (recessive |
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Term
Drosophilia Cross between a female with normal wings and red eye
color and a male with miniature wings and white eyes. |
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Definition
• F1 = all normal wings and red eyes.
• F2 males = both wild type = 412; miniature wings and
white eyes = 389, normal wings and white eyes = 185,
miniature wings and red eyes = 206. |
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Term
Linkage in Autosomal Genes
The dihybrid ratio of...
• The deviation from this ratio may be because of...
• Because progeny receive two copies of both genes
from both parents...
it is harder to follow than...
• Test crosses are very useful to determine... |
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Definition
...9:3:3:1 does not always hold true.
...cotransmission (lack of independent assortment) of
these genes.
...Xlinked
traits (where one can look at the phenotype
of the male to determine the allele transmitted by
mom).
...recombination frequency between autosomal genes |
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Term
| Autosomal Gene linkage, drosophilia the phenotypes |
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Definition
Autosomal gene in Drosophila for body color:
– b+ = brown body (wild type)
– b = black body (recessive)
• Autosomal gene for wing shape:
– c+ = straight edge (wild type)
– c = curve edge (recessive) |
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Term
| Autosomal Gene linkage, drosophilia |
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Definition
A cross between a male homozygous for brown body and curve
edged wings and a female homozygous for black body and straight
edged wings.
• F1offspring are all double heterozygotes.
• A test cross of F1female with a double recessive male will give F2
with different proportions of each phenotype.
• Since the male will only contribute recessive alleles, the phenotype
of F2 reflects the alleles contributed by the F1 female. |
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Term
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Definition
To predict whether two genes are linked or not,
one only needs to determine whether the ratios of
F1 gametes formed adhere to the 1:1:1:1
prediction.
• If there are equal numbers of parental and
recombinant gametes, genes are not linked; if the
parental gametes outnumber recombinant gametes,
the genes are linked.
• This method is rather qualitative; experimental
variations may cause small deviations from
expected ratios. |
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Term
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Definition
This test determines whether the deviation
from expected ratios can be attributed to
experimental variation/chance, or truly
represents linkage.
• The test is designed to determine the
probability that the experimental deviation
from calculated (predicted) values is a
chance event.
• It is a ‘goodness of fit’ test between
observed and predicted values. |
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Term
| Basic requirements for χ2 test |
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Definition
• Use of real numbers rather than percentages or ratios.
The greater the sample size, the better chance that the
observed numbers should match predicted values.
• Test a preformed hypothesis. Then determine whether the
data agree with it or not.
• If one forms a hypothesis that two genes are linked, it is
not possible to predict expected ratios (since they will
vary according to distance between genes), it is the norm
to form a hypothesis that the genes are not linked (the
ratios now can be predicted based on Mendelian
inheritance pattern and the laws of probability).
• This hypothesis that the two genes under study are not
linked is called the null hypothesis. |
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Term
| Data Tabulation for chi-squared test |
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Definition
Data tabulation:
– What are the total # of offspring analyzed?
– How many different classes of offspring are
there?
– In each class, how many offspring are
observed? |
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Term
| The Chi-squared test, predictions |
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Definition
How many offspring would be expected in each
class if the hypothesis (no linkage) is correct?
– E.g.: in case of no linkage for a dihybrid cross of
the type AaBb x AaBb, there should be 4 classes
with the expected ratios being 9:3:3:1.
– E.g: in case of no linkage for a dihybrid cross of
the type AaBb x aabb (test cross), there should
be 4 classes, with the expected ratios being
1:1:1:1. |
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Term
| The chi-squared test, calculation |
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Definition
– For each class of offspring subtract the expected number
(E) from the observed number (O) to obtain the difference,
d.
– Square the result: d2.
– Divide this value by the expected number for that class:
d2/E.
– Do this for each class and add the numbers together: Σ
d2/E.
– This is the value for χ2.
χ2 = Σ d2/E
– Compute the degrees of freedom: df = (number of
phenotypic classes –1). |
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Term
Use the chi square value and the degree of freedom value
to determine ...
• The p value is the probability that ...
• The greater the p value...
• Typically, a p value of 0.05 or less is required ... |
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Definition
...the p value from the table.
...a deviation from the
predicted (expected) values occurred by chance.
... the stronger the case for the null
hypothesis to be true (no linkage).
...to reject
the null hypothesis, and prove linkage |
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Term
• E.G.: A p value of 0.05 indicates that...
A p value of 0.9 indicates that there
is a .... |
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Definition
... there is 5%
probability that the deviation from predicted values
occurred by chance
90% probability that the deviation from the predicted
values occurred by chance |
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Term
| Limitations of chi squared test. |
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Definition
The chi square test is not accurate for small sample
size.
• Subjectivity in determining the boundary of
significance (p value threshold is arbitrary)
• Does not provide information about the extent of
linkage, or relative positions of genes on the
chromosome. |
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Term
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Definition
χ2 = Σ (observed – expected)2
/number expected |
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Term
Physical Markers of Chi squared
-Cytologically visible abnormalities or signs on
chromosomes...
- |
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Definition
-... that allow tracking of the origin of
the chromosomal segment.
• Provided evidence that recombination results from
reciprocal exchange of parts between maternal and
paternal chromosomes.
• E.G.: drosophila X chromosome markers:
– physical markers: chromosomal abbarations.
– Genetic markers: carnation eyes and kidney shaped
eyes. |
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Term
Physical Markers of Chi- squared
Provided evidence that recombination results from
reciprocal exchange of parts between...
• E.G.: drosophila X chromosome markers:– physical markers:
– Genetic markers: |
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Definition
...maternal and
paternal chromosomes.
...chromosomal abbarations.
... carnation eyes and kidney shaped
eyes |
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Term
Gene Mapping
Recombination frequency is a measure of the...
• 1 RF = 1 centimorgan (cM) or 1 map unit (m.u).
• This is not the actual distance between the genes...
• By calculating the map units between two pairs of
genes at a time, one can construct a...
• Example: |
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Definition
...physical distance separating the two genes on the
same chromosome.
...only a relative value.
... genetic map of
that region of the chromosome.
...: genes on the Drosophila X
chromosome: y, w, v, m, r and v. |
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Term
| Limitations of two point crosses |
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Definition
• A pairwise comparison of two genes at a
time is time consuming.
• Linear arrangement is difficult when
distances are too close (e.g. w and m vs. y
and m being 32.8 and 34.3 map units
respectively).
• Actual distances do not always add up,
indicating a limit to the of accuracy of
resolution. |
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Term
| Three point crosses, why are they better? |
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Definition
Faster and more accurate way to map genes
• Simultaneous analysis of three markers
• Information on the position of three genes relative to
each other can be obtained from one mating rather
than two independent matings. |
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Term
| Three point crosses, example: |
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Definition
Example: Drosophila autosomal genes:
– vg = vestigial wings; vg+ = normal
– b = black body; b+ = normal body
– pr = purple eyes; pr+ = normal eyes
• Cross of pure breeding vestigial winged, black
bodied, purple eyed female to a pure breeding wild
type male: vgvg bb prpr x vg+vg+ b+b+ pr+pr+ |
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Term
| Data Analysis, steps one-three |
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Definition
There are eight gametes from the female F1
• Identify the parental and recombinant types for each
gene pair at a time.
• Calculate the percentage of recombinant types
(recombination frequency) for each gene pair. This is
the map distance between the two genes. |
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Term
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Definition
Now you can place the genes with the greatest
distance between them on the two ends and place the
left over gene in the middle.
• The orientation from left to right is purely arbitrary.
• Do the numbers add up ? |
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Term
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Definition
• Recombination is caused by formation of chiasmata
along the chromosome at multiple points.
• If the distance between two genes is large enough,
there can potentially be multiple chiasmata formation
between them; so there could be multiple crossovers. |
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Term
What would happen if there were two crossovers
between the two outside genes (in this case vg and b)? |
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Definition
• Answer: there would appear to be fewer
recombinants between the two genes: it would appear
as if the genes are closer; the calculated map distance
between these genes will be less than actual. |
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Term
| Correcting for double cross overs, three steps |
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Definition
Correcting for double crossovers
• Find out the recombinants arising from double
crossovers: those offspring showing relocation of
the middle gene with respect to the two outside
genes.
• Recalculate the distance between the two outside
genes, adding the offspring resulting from double
crossovers twice to the other numbers.
• Check if this new recombination frequency (map
distance) now adds up to the sum of the individual
map distances between the two other gene
combinations. |
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Term
Calculation of Expected Frequency of
Double Crossovers |
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Definition
• Using the law of the product, one can calculate the
expected probability of double crossovers.
• Check if this is the same as the observed frequency.
• Example: Consider the following data:
– Distance between a and b is 10 m.u.
– Distance between b and c is 20 m.u.
– Distance between a and c is 28 m.u.
• This indicates gene b is in the middle.
• The probability of a crossover between a and b
AND one between b and c = 0.1 x 0.2 = 0.02 or 2%. |
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Term
| Inference in double crossovers |
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Definition
When the observed frequency of double crossovers is
less than the expected frequency, there is some
Interference in double crossover events occurring.
• Example: our earlier three point cross.
– Expected double crossover frequency is 0.79%
– Observed frequency is 0.52%.
• Indicates some interference: once one crossover has
occurred, there is some inhibition for a second one to
occur in an adjacent location.
• The extent of interference is not uniform along the
length of the chromosome, some regions may show
more interference than others. |
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Term
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Definition
First calculate coefficient of coincidence,
which is the ratio between the actual
frequency of double crossovers and the
expected frequency.
• Interference = 1- coefficient of coincidence.
• For our example, coefficient of coincidence
= 0.52/0.79 = 0.66.
• Interference = 1-0.66 = 0.34. |
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Term
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Definition
Interference = 1: there are no double
crossovers (coefficient of coincidence = 0).
• Interference = 0: the expected and observed
number of double crossovers are the same.
(coefficient of coincidence = 1). |
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