1. Define the following terms and
utilize these terms in an appropriate context:
Chiasma
chromosome theory
complete linkage
crossing over
double crossover
linkage
linkage group
coupling
repulsion
cis/trans
parental group
crossover group
2. Demonstrate ways to indicate a genotype involving linked genes.
3. Using chromosome diagrams and
gene symbols (genotypes), show how the offspring
resulting from a test
cross of individuals heterozygous for two traits could be mainly of two
phenotypic classes
(demonstrating linkage but no crossovers).
4. Explain how it could be shown
that the results of the cross referred to in objective #3 were
not caused by any affinity
of dominant gennes for each other, or recessive genes for each other.
5. Explain a crossover event and illustrate the resulting offspring of such an event.
6. Explain why more crossing over
occurs between two distantly linked genes than between
two genes that are
very close together on the same chromosome.
7. Explain why double-crossover events
are expected in lower frequencies than single-crossover
events.
8. In general terms, give the relationship
between the spacing of genes on a chromosome and
the crossover
frequency; and state specifically the relationship between crossover
frequency
and the map distance
between genes.
9. Explain and diagram what is meant by coupling phase and repulsion phase (cis and trans).
10. Describe how you could go about
determining the number of linkage groups for any species,
and the linkage group for any gene.
11. If given the map distances between
a series of genes obtained from two-point map crosses,
determine the order of the genes on a chromosome.
12. Considering three linked genes,
and given the numbers of individuals in each phenotype
which result from a test cross, determine the non-crossover classes, single
crossover
classes and double crossover classes, the position of the gene loci with
respect to each
other, and the map distances apart.
13. Give a possible explanation of
why crossingover seems to occur so precisely between
genes rather than within genes.
Resources: Text Chapter 8,
Cartoon Guide pgs. 69-78
Genetic Linkage and Chromosome Mapping
1903- Sutton, boveri (cytology
+ genetics)- organisms have more "unit factors" than
chromosomes.
Soon after ----> discovered
that certain genes were not transmitted according to law of
independent assortment.
-seemed to be inherited together, transmitted as single unit.
**Organisms have thousands of genes controlling
all aspects of its life, therefore, each
chromosome must carry
large numbers of genes (Chromosome theory)
Genes that are part of the same chromosome are said to be Linked and demonstrate Linkage
Chromosomes, not gene, unit of transmission and subject to independent assortment.
Modifications: Crossing over, provides potential for variation, diversity, survivability
Mendel worked w/ 7 genes, peas have seven
chromosomes, he did 21 different crosses.
Why didn't he find
linkage? Lucky??
Bateson, Punnett -----> Coupling
Example w/ unlinked genes (dihybrid cross)
Sepia/ Curved
X
Heterozygous normal
sese cc
+se +c
Gametes se c ++; +c; se+; se c
Offspring
+se+c
+se cc
sese +c
sese cc
normal
normal curved sepia normal
sepia curved
1: 1: 1: 1
Example w/ linked genes (**Note notation differences**)
G- grey body g- black body R- red r- purple
P generation grey/red X black/purple
G R g r
g r g r
(Gg Rr) (gg rr)
Gametes G R g r g r
Offspring G R or g r
g r g r
Grey/red black/purple Expected 1:1 ratio
However!!!!!
Actual results
black/purple 47%
grey/ red 47%
Parental chromosomal types (non recombinant)
black/ red 3%
grey/ purple 3%
Where did these come from? (non parental chromosomal
types, recombinant)
Crossing Over- during meiosis
chromosomes replicate forming sister chromatids;
homologous chromsosmes
pair, crossover occurs between non-sister chromatids;
chromosomes break and
rejoin
from previous example: black/red; grey/
purple resulted from cross overs in heterozygous parent
Alleles of douple heterozygotes can appear
in two positions:
Coupling
Phase: dominants one one chromosome, recessives on other
in our example: G R Parental non crossovers G R, g r
g r
Recombinant crossovers G r, g R
Least Frequent
Repulsion
Phase: dominant of one locus and recessive allele of other locus
on
same chromosome
In our example: G r
g R
Parental (Non crossovers) G r and g R
Recombinant (crossovers) G R and g r
**will make difference in results, parental types are most frequent
Summary: Recombinant
groups due to crossing over
1. Certain genes assort at random
2. other genes do not segragate randomly, but are linked, these groups
are transmitted
together
3. chromosomes occur in pairs (diploids)
4. linked grnes do not always stay together but are often exchanged (crossing
over)
5. chromosomes are seen to form chiasmata and exchange parts; chiasmata
and
crossover happen with similar frequency.
Using crossover frequencies we can form a linkage map
Map unit = 1% recombination
-frequency of crossove is governed by distance between genes
*as distance increases, frequency of crossover increases
*short distance between linked genes, fewer crossovers, fewer recombinants
Linkage groups = # of chromosomes in haploid set; Humans, therefore, have 23 linkage groups
Mapping based upon crossover frequencies-
Sturtevant
-studied yellow, white,
miniature traits
-during meiosis limited number of crossovers occur
-each recombinant event
occurs randomly, therefore, the closer two loci reside, less likely
crossover will cocur between them.
Limits of Recombination
if 2 genes are located so far apart that the prob. of a crossover occurring
is 100%
then 50% of gemetes will be parental (noncrossover) and 50% will be recombinant
(crossover). Ex. coin, there is a 100% prob that it will land flat, therefore,
we expect a
1:1 ratio of heads to tails.
Genetic Mapping
2 point cross- easiest way to detect crossovers is through test cross progeny
Suppose dihybrid in coupling phase A C X a c
a c a c
Progeny 37% AC
37% ac (parental, non crossovers)
13% Ac
13% aC (recombinants, crossover)
26% of all gametes were crossover types, therefore, distance between
A and C = 26 map units
Ex. X-linked traits
Red- w+ (+) white- w normal wing- m+ (+) miniature- m
P
w +
X
+ m
w +
Y
white eyed normal wing female red eyed miniature male
F1 w + X w +
+ m
Y
F2 (males) w + 226
+ m 202 66.5% parental ( if no crossover you expect 100%)
+ + 114
w m 102 33.5% recombinant
****during meiosis of F1 mother
w + ---------> w m + +
+ m
3 Point Test Cross
Introduction of double crossover groups
DCO's usually do not
occur between genes less than 5 mu's apart. For genes further
apart it is advisable
to use third marker to detect DCO's
For successful cross:
1. genotype of organism producing crossover gametes must be heterozygous
2. phenotype must reflect genotype
3. large #'s must be produced to recover crossover classes.
Assumptions: -parental types (non co's) will be
largest group
-double co's will be smallest
Trihybrid testcross A B C X a b c
a b c a b c
Progeny A B C 36%
a b c 36%
72% parental, noncrossover types
A b c 9%
a B C 9%
18% single crossovers between A and B
(region I)
A B c 4%
a b C 4%
8% single crossovers between B and C
(region II)
A b C 1%
a B c 1% 2% double crossovers
To find distance A-B must count all crossovers that occurred in Region I
CO Region I = 18% + 2% or 20% or 20 map units between A and B
To find distance B-C must count all crossovers that occurred in Region II
CO Region II = 8% + 2% or 10% or 10 map units between B and C
Therefore, 30 map units between A and C
Determining Gene Order
1. Assume any of three possibilities w/ 3 point cross
2. Observe double crossover groups
-if you started with coupling phase then only one gene will be changed
and
therefore must be in middle
Ex. y - w - ec; y - ec - w; w - y - ec
assume original parent producing crossover was
w y ec
if this order correct then DCO's would be w y+ ec, or w+ y ec+
w+ y+ ec+
After fertilization it was found that the DCO's were w+ y ec and w y+ ec+
Notice these do not correspond to our predicted
**Following a double
crossover event, allele representing the middle gene will find
itself present with the outside or flanking alleles present on the opposite
parental gamete
Therefore, our proper order ----> y+ w+ ec+
y w ec
Recombinant frequencies aid in determining
order
assume- if co frequencies
between a and b is two times greater than between b and c
then a and b must be
twice as far apart on the chromosome that b and c
Chickens Br (brown eye) Li (light down) S (silver plummage)
10% recombinant freq between Br and Li
26% recombinant freq between Br and S
16% recombinant freq between S and Li
Gene order Br
Li
S
Ex. Drosophila dumpy wing (dp), clot eye (cl), apterous wing (ap) linked on chromosome #2
map distances: dp - ap 42
dp - cl 3
ap - cl 39
dp
cl
ap
Ex. in a testcross
a b c
X
a b c
A B C
a b c
the rarest classes were abC and ABc
What is the correct sequence?