Lab 5: Genetic Maps Based On Recombination Gene Linkage - PowerPoint PPT Presentation

1 / 24
About This Presentation
Title:

Lab 5: Genetic Maps Based On Recombination Gene Linkage

Description:

long arm (q) chromosome 9. centromere. Individual is ... Lab Exercises. Work problems and answer questions in the lab manual. Also, F1 flies have emerged! ... – PowerPoint PPT presentation

Number of Views:421
Avg rating:3.0/5.0
Slides: 25
Provided by: dh480
Category:

less

Transcript and Presenter's Notes

Title: Lab 5: Genetic Maps Based On Recombination Gene Linkage


1
Lab 5 Genetic Maps Based On RecombinationGene
Linkage
  • For two or more genes, the Mendelian principle of
    independent assortment (considered previously)
    applies only when genes are on different
    chromosomes

2
Lab 5 Genetic Maps Based On RecombinationGene
Linkage
  • However an individual chromosome may carry many
    genes, and we can also follow the inheritance of
    alleles of two or more genes found on the same
    chromosome.
  • A chromosome can be defined genetically as a
    linkage group, or a group of genes that are
    physically linked (joined together)
  • Alleles of genes that are linked will tend to be
    inherited together

3
Lab 5 Genetic Maps Based On RecombinationGene
Linkage and Recombination
  • However, new combinations of linked alleles can
    be produced by recombination
  • Recombination is defined here as the physical
    exchange of genetic material between homologous
    chromosomes
  • Recombination (also known as crossing over) is a
    normal process that occurs during meiosis
  • Recombination events are detected only in the
    progeny of an individual

4
Consider a real example of two genes found on
human chromosome 9 One gene is responsible for a
genetic disorder known as Fanconis anemia This
is a are genetic disorder that results in a
deficiency of all bone marrow elements including
red blood cells, white blood cells, and
platelets. This leads to heart, kidney, and/or
skeletal abnormalities.
5
Fanconi's Anemia There are different subtypes of
this disorder resulting from mutations of
different genes. One gene causing Fanconis
anemia type C) is located on chromosome 9
(9q22.3). The mutant allele can be designated
fac Exhibits an autosomal recessive pattern of
inheritance patients with this disorder are
homozygotes (fac//fac)
Fanconis anemia gene (locus) long
arm (q) chromosome 9
fac
centromere
Chromosome 9
Individual is homozygous both homologs carry
this allele
fac
6
Another gene found on chromosome 9 is the ABO
gene (determines ABO blood type) A and B
alleles are both dominant, o is recessive Found
at 9q34.1 (more distal on chromosome 9) Imagine
that you diagnose a patient with Fanconis
anemia, and you also determine that they have
blood type O (homozygous for the o allele)
In this case, the fac allele and the o allele
are linked
7
Will the fac allele will always be found linked
to the o allele? Imagine another person that
carries the Fanconis anemia allele as a
heterozygote. Their genotype fac// (
normal allele) This person also has blood type
B, and their chromosomes could look like this

B
During gamete formation (meiosis), homologous
chromosomes align and each chromosome is
duplicated Recombination events are detected as
the products of meiosis
8
Case 1 No recombination
fac o
B
In this case, all of the chromosomes are
nonrecombinants
Case 2 Recombination
fac o
B
In this case, two new recombinant types are
produced in addition to the nonrecombinant types
9
Summary For recombination cases involving two
genes Two of the chromosomes are nonrecombinant
types
fac o
B
And two are recombinant types
10
Recombination events for three genes on the same
chromosome Categories produced 1. Two
nonrecombinant types 2. Two recombinant types
reflecting cross-over events between the first
two genes (region I)
11
3. Two recombinant types reflecting cross over
events between the second and the third gene
(region II)
I
II
a
b c
4. Two recombinant types reflecting double
cross over events (cross-overs in both regions I
and II)
12
  • Recombination can also be used to determine the
  • relative location of genes on a chromosome
  • In general, recombination will occur more
    frequently
  • between genes that are farther apart on a
    chromosome
  • The amount (frequency) of recombination between
    genes
  • reflects their relative distance apart on the
    chromosome
  • Recombination frequencies can also be converted
    to
  • distances to construct a recombination map

13
Sample dataset F2 progeny of a cross in
Drosophila involving three mutations a, b, c (
symbol wild type)
Phenotype Number of Individuals (no
mutations) 362 b, , 4 c ,, 90 b, c
, 39 a, b, c (all three mutations) 396 a, c
, 2 a,, 34 a, b, 100 Total 102
7
14
To construct a map you must first determine which
individuals correspond to each of the different
categories (nonrecombinant vs. each of the
different recombinant types) The nonrecombinant
types are generally the most frequent types
(largest numbers of individuals) The double
recombinant types are generally the least
frequent (smallest numbers)
15
The most abundant phenotypes are wildtype (
) 362 a, b, c 396 These should be the
nonrecombinant types
The least abundant phenotypes are b, , 4 a,
c, 2 These should be the double recombinant
types
16
Next, determine a gene order that will give you
the correct classes of double recombinant
types Hint Start with a heterozygous individual
and draw out the double cross over events (see
lab manual) Dont assume that it will always be
a, b, c You must know the correct gene order
to determine region I and region II
17
Determine the distances between the first two
genes based on the frequency of recombination
events (also include the double
recombinants) Convert the frequency to a
percentage. Each 1 recombination frequency
1 cM (cM centiMorgan, a unit of map distance)
18
a
Crossover between a and b
b c
Phenotype a, , 34
b, c, 39 double recombinants b, ,
4 a, c, 2 Total 79
79 / 1027 0.077 x 100 7.7
or 7.7 cM
19
Repeat this process for crossovers in region
II Draw a linear map (e.g., if gene order is
a-b-c)
a b
c
7.7 cM
19.1 cM
20
Coefficient of coincidence
  • The observed frequency of double recombinants may
    not always
  • coincide with the expected frequency of double
    recombinants
  • The coefficient of coincidence quantifies the
    difference
  • between the observed and expected frequencies of
    double
  • recombination events

Observed frequency of double
recombinants Expected frequency of
double recombinants
cc
Observed freq of dbl recombinants/total
number offspring
Expected freq (recomb. freq. a-b) (recomb.
freq. b-c)
21
Ideally, the cc1 In reality it may be deviate
from 1 in either direction (positive or negative).
Any deviation from a value of 1 indicates some
type of interference. This is quantified as
Interference (I) 1 - coefficient of coincidence
22
Lab 3. Linkage and RecombinationLab Exercises
  • Work problems and answer questions in the lab
    manual
  • Also, F1 flies have emerged!
  • Anesthetize flies as before, one vial at a time
  • Count and classify all flies according to
    phenotypes
  • (Obtain both individual and class data counts for
    each mutation)
  • Compare your F1 results to your predicted Punnett
    squares to indicate a mode of inheritance for
    each trait

23
Lab 3 Linkage and RecombinationLab summary
  • Purpose
  • Materials and Methods
  • Results and preliminary conclusions
  • Paper problems Show all calculations and answer
    all related questions thoroughly
  • Drosophila F1 results
  • Compare results for EACH mutation to F1 Punnett
    square predictions. State how you think each is
    inherited

24
  • Some additional aspects of linkage and
    recombination
  • 1. Recombination events can only be detected in
    the progeny of heterozygotes. The
    nonrecombinant progeny are also sometimes
    referred to as the parental types. How can you
    use this to work backwards and predict the
    genotypes of the parents used to produce the
    heterozygotes?
  • In animal breeding experiments, selecting for one
    characteristic (for example short hair) may
    result in other characteristics being selected
    for as well (for example brown eyes). What
    topic covered today helps to explain this
    phenomenon?
  • 3. A gene responsible for one form of the disease
    ALS is very closely linked to the ABO gene.
    Could you use the results of ABO testing to
    predict the likelihood of developing ALS?
Write a Comment
User Comments (0)
About PowerShow.com