P1251328618TyqZV

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VII. Gene Linkage and Gene Mapping When two or
more pairs of genes are transmitted, they do not
always follow the independent assortment rule.
Instead, they segregate together as if they were
linked to each other. Punnet 1905 Purple
flower/long pollen X Red /round pollen F2
phenotypes 12 purple and long 1 purple
and round 1 red and long 3 red and round
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The above data revealed 1). Purple and
long are dominant traits, red and round
are recessive traits. 2). The purple and long
genes seem to stick together and red and round
genes stick together. This is the earliest
evidence for gene linkage.
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Independent assortment the two genes are located
on two different chromosomes.
A-
B-
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Gene linkage - alleles of two different genes
segregate together during meiosis. For two
genes to be linked, they have to be located on
the same chromosome.
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When two parental individuals mate AA BB x aa
bb, there can be two outcomes for F1 Cis
configuration F1 genotype AB ab. (A and B are
linked and so are a b AB or ab segregate
together.)
Trans configuration F1 genotype as Ab aB The
two dominant genes separate from each other and
so do two recessive genes, Ab aB recombination
occur.
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• Linked genes do not always show complete linkage.
  
• Instead, crosses between two such genes always
• occur.
• Recombination
• Reshuffling of genes.
• When an organism makes gametes with combination
• different from its parents, we say there is gene
  
• recombination.

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A dihybrid cross with parental genotypes AA
BB x aa bb P1 gametes AB ab if F1
gametes AB ab (Cis, showing linkage
between A/B and a/b) or F1
gametes Ab aB (trans arrangement)
recombination occurred
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Morgan Allele recombination is due to physical
exchange between two homologous chromosomes
during meiosis - meiotic crossing over. AB
and ab Ab and aB If no C.O. between AB or ab,
F1 will produce only AB and ab gametes, like
parental, ...
Crossing over
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Consider a dihybrid cross of corn W_ colored
kernels ww white kernels S_ full kernels ss
shrunken kernels wwSS (white/full) x WWss
(colored/shrunken) Parental gametes wS and
Ws F1 genotype WwSs (colored and full) If
genes are not linked, F1 gametes would be WS,
Ws, wS, ws (1111 ratio)
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There is 50 recombination (the maximum rate
of recombination) because two new gametes, WS,
ws formed. 2/4 50. If W-s or w-S are linked,
the F1 gametes would be mostly Ws and wS, same
as those of parental. no recombination. How
do we see the genetic composition in a gamete?
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Testcross F1 cross with a homologous
recessive ? ? x wwss (white and shrunken)
F2 phenotypes should tell us what F1 gametes were
made, because F2 phenotype is determined by
F1(??). If we see F2 Phenotype F2 genotype
F1 Gametes ratio Colored, shrunken
Wwss Ws (parental) 48.4 White,
full wwSs wS (parental)
48.4 Colored, full WwSs WS
(recombinant) 1.6 White, shrunken wwss
ws (recombinant) 1.6
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Results Four phenotypes - there were 4 types
of F1 gametes, The phenotype ration is 48.4
48.4 1.6 1.6 Interpretation Most gametes
were non recombinants - the two genes are
linked. But there was low rate of C.O.
resulting 3.2 of gametes with gene
recombination. Conclussion Partial linkage
between genes W and S.
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In AB x ab cross if there is low frequency of
C.O. between the two linked genes, most F1
gametes will be AB and ab, only a few are
recombinants, Ab and aB - partial linkage or
incomplete linkage.
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Linkage detected by testcross
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Crossing over - Reciprocal exchange of
chromosome segments. A rare event which
provides chance for genetic variation.
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Advantage of sexual reproduction it provides
two kinds of recombination a, by independent
assortment for unlinked genes b, by C.O.
for linked genes. Recombination and
independent assortment ensure the inheritance
and yet diversity of a population.
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The role of distance The frequency of C.O. is
directly proportional to the distance between
the two loci - the interlock distance. If the
two linked genes are far apart, the chance of
C.O. is increased.
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The correlation between distance and the
frequency of C.O. can be used as the basis for
chromosome mapping. The frequency of C.O.
products (phenotypes) can be use as a direct
index of distance between two loci.
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Consider dihybrid cross AABB x aabb F1
gametes AB 38 Ab 12 aB 16 ab
29 Which ones are C.O. products
Ab and aB. The frequency of c.o. is 12 16 28
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To calculate the relative distance The
frequency of the recombinants / tall
frequency Ab aB 28, 28 / 95 (total ) x
100 0.337 (33.7). Converted to map units
33.7 mu. In the corn exp (1.61.6) /100
3.2 3.2 mu between W and s.
Note, map unit is a derived (or a predicted)
distance, not a real distance that can be
measured.
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One mu represents 1 recombinants in all
offspring. It can also be described as centi
Morgan (cM). A small mu value means the two
loci are very close to each other, and C.O.
would be difficult between them. What is the
largest mu value?
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If the two linked genes are very far apart, the
rate of recombination will be close to 50 (but
never exceeds 50).
Maximum recombination results in a 1111 ratio
of the four types of F1 gametes. Two
recombinants verses two parental 50, same as
the case of independent assortment of two gene.
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The study of single C.O. between two linked genes
provides the basis to determine the distance
between them (two point mapping). When three
or more linked genes are investigated
simultaneously, it is possible to determine the
sequence of the three.
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Multiple Exchanges and 3 point mapping It is
possible that more than one crossing overs occur
between non-sister chromatids. By studying a
double exchange, the sequence of three linked
genes can be mapped.
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Consider genes a, b, and c being three linked
genes. To map them, (1) do a - b testcross to
determine ab distance, (2) do b and c testcross
to determine bc distance. (3) do a - c
distance. Exp. A-b 5, b-c 10, a-c
13 What is the order? A b c ? or b c a ?
or a-c-b?
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The sequence of the three genes a-b-c does not
exactly fit but quite close. Why? The sum of two
smaller distances is always greater than the
apparent a c distance. Because in a double
C.O., the second one may cancel the effect of the
first C.O., and makes the apparent distance
smaller than real. So, the sum of a-b and b-c
will provides more accurate distance between a
and c.
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An alternative method trihybrid testcross
Textbook p154-p155.
AaBbCc (F1) x aabbcc 370 offspring Eight
possible F1 gametes (obtained from
testcross) ABC 103 abc 112 aBC 6 Abc
4 AbC 43 aBc 57 ABc 22 abC 23
1). Determine crossing over products aBC and
Abc, the least frequent forms.
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2). Determine loci order DCO products are aBC
and Abc. The gene that has been switched in DCO
is the gene in the middle A - Sequence is CAB
or BAC.
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3) Determine distance by the frequency of
recombination. What to look at? ABC
103 abc 112 aBC 6 Abc 4 AbC 43 aBc
57 ABc 22 abC 23
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Distance between A and B a, aBC and Abc (DCO)
6 4 10 b, AbC and aBc (single CO between
a and b) 43 57 100 100 10 110,
110/370 (total) 0.27 x 100 27 mu
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Distance between a and c Single CO (ABc and
abC) 22 23 45, DCO (aBC and Abc) 6 4
10 (10 45)/370 0.149 x 100
14.9mu Distance between B and C 15 27 42
(mu) Or 10 45 100 155. 155/370 0.42
42 mu.
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Yeast crossing over Yeast and fungi spent most
of their life cycle as haploids, although
diploids do form during fertilization. Fungi
or yeast reproduce by producing spores.
Sexual reproduction two cells from different
strains fuse to form a heterozygous diploid
cell. Their nuclei also fuse and enter meiosis.

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In yeast and fungi, meiosis occurs in a sack-like
structure - ascus (asci). Within each ascus,
a meiotic cell can develop into 4 haploid cells
called spores tetrad. Tetrad 8
ascospores The arrangement of ascospores
reflects the genotypes of parentals. - Ordered
tetrad ...
mitosis
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One fertilized cell can produce 8 ascospores
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In the early stage of meiosis, crossing over
between chromatids may occur.
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By crossing two fungus strains and observing
the arrangement of ascospores in different
colors one can test gene linkage and do gene
mapping.
Before test cross, phenotypes of the parental
must be known.
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For exp., gene b codes for black color (wild
type), t for tan. When two strains of
Sodaria are crossed No crossover the pattern
of ascospores in the ascus would be bbbb tttt,
same as the parental.
Crossover between strands 2 and 3 ascospore
arrangement will be bb tt bb tt Crossover is
between strand 1 and 3 tt bbbb tt .
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To calculate the distance between the gene and
the centromere Data must be collected from a
large number of asci. The distance between gene
b and the centromere 1/2(recombinant asci) /
total asci scored
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If we get 65 wild type and 70 recombinants The
distance between gene a and the centromere is
(1/2)(70)/ 135 0.259 25.9 mu
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Somatic-cell hybridization is an important
technique in creating human chromosome maps
Somatic-cell hybridization Barsky (1960)
found that cells from different organisms can
also fuse together. The cell fusion produces a
cell type called hetero- karyon, which contains
two nuclei in one cell.
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With somatic-cell hybridization technique, it is
possible to fuse human and mouse cells. After
cell culture, the nuclei will also fused to form
cells called synkaryons. After many
generations, these synkaryonic cells randomly
lose some chromosomes, mostly from one parental
cell line.
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In the case of the human-mouse hybrid,
most human chromosomes are lost. The cells
will stabilize when one or few human chromosomes
left. This phenomenon was utilized to map
human genes on different chromosomes.
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One can test the protein products (for exp.,
enzymes) that these cells make, and correlate
the protein products to the human chromosomes
that are present in the synkaryonic cell
line. Ideally, a panel of 23 hybrid cell lines,
each with only one unique human chromosome,
would allow the immediate assignment of a
particular chromosome that is responsible for
certain gene products
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However, in reality, a fusion cell line usually
contains several human chromosomes. The
correlation of the presence or absence of each
chromosome with the presence or absence of each
gene product is called synteny testing.
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For exp., The chart on p172 listed 23 human
chromosomes verses 8 cell lines. A gene product
TK is analyzed. Cell line A is TK positive and
contains chromosomes 1,2,3,4,5,10. and 17.
Line B is also TK and contains chromosomes
1,3,4, 6, and 17. Line C is TK and contains
1,2,3,4,9,10,11 and 17
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Chromosome 17 is always present in TK cell
lines and is always absent in TK cell lines.
This correlation allows you to conclude that
TK gene resides on chromosome 17.
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Solve problems p139 9, 20 p182- 183 6, 8, 9,
10, 11,