I. Allelic, Genic, and Environmental Interactions

1
I. Allelic, Genic, and Environmental
Interactions II. Sex Determination and Sex
Linkage III. Linkage - Overview Linkage is
a pattern of correlated inheritance between
traits governed by genes on the same chromosome.
Because the genes are part of the same physical
entity, they are inherited together rather than
independently.
INDEPENDENT ASSORTMENT (IA)
LINKED
a
A
AB
ab
B
b
ab
AB
a
A
a
A
b
B
B
b
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III. Linkage - Overview Linkage is a
pattern of correlated inheritance between traits
governed by genes on the same chromosome.
Because the genes are part of the same physical
entity, they are inherited together rather than
independently. Only crossing-over can cause
them to be inherited in new combinations.
Cross-over products
3
III. Linkage A. Complete Linkage - if
genes are immediate neighbors, they are almost
never separated by crossing over and are always
inherited together. The pattern mimics that of a
single gene.
AABB
aabb
AB
ab
X
ab
AB
4
III. Linkage A. Complete Linkage - if
genes are immediate neighbors, they are almost
never separated by crossing over and are always
inherited together. The pattern mimics that of a
single gene.
AABB
aabb
AB
ab
X
ab
AB
Gametes
Double Heterozygote in F1 no difference in
phenotypic ratios compared to IA
F1
5
III. Linkage A. Complete Linkage - if
genes are immediate neighbors, they are almost
never separated by crossing over and are always
inherited together. The pattern mimics that of a
single gene.
11 ratio Aa 11 ratio Bb 11 ratio ABab
Test cross
X
Phenotype
Gametes
AaBb
AB 50
aabb
Ab 50
Ab
Aabb
Ab
If it was I.A., also
aB
aaBb
aB
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III. Linkage A. Complete Linkage B.
Incomplete Linkage
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III. Linkage A. Complete Linkage B.
Incomplete Linkage - Crossing over in a
region is rare
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III. Linkage A. Complete Linkage B.
Incomplete Linkage - Crossing over in a
region is rare - Crossing over events increase
as the distance between genes increases
MORE LIKELY IN HERE
LESS LIKELY IN HERE
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III. Linkage A. Complete Linkage B.
Incomplete Linkage - Crossing over in a
region is rare - Crossing over events increase
as the distance between genes increases - So,
the frequency of crossing over (CO) gametes can
be used as an index of distance between genes!
(Thus, genes can be mapped through crosses)
FEWER CO GAMETES Ab, aB
MORE CO GAMETES bC, Bc
10
III. Linkage A. Complete Linkage B.
Incomplete Linkage - Crossing over in a
region is rare - Crossing over events increase
as the distance between genes increases - So,
the frequency of crossing over (CO) gametes can
be used as an index of distance between genes!
(Thus, genes can be mapped through
crosses) - How can we measure the frequency of
recombinant (cross-over) gametes? Is there a
type of cross where we can see the frequency of
different types of gametes produced by the
heterozygote as they are expressed as the
phenotypes of the offspring?
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TEST CROSS !!!
III. Linkage A. Complete Linkage B.
Incomplete Linkage
b
b
A
a
12
TEST CROSS !!!
III. Linkage A. Complete Linkage B.
Incomplete Linkage - So, since crossing-over
is rare (in a particular region), most of the
time it WONT occur and the homologous
chromosomes will be passed to gametes with these
genes in their original combinationthese gametes
are the parental types and they should be the
most common types of gametes produced.
b
b
a
A
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TEST CROSS !!!
III. Linkage A. Complete Linkage B.
Incomplete Linkage - Sometimes, crossing
over WILL occur between these loci creating new
combinations of genes This produces the
recombinant types
b
b
a
A
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TEST CROSS !!!
III. Linkage A. Complete Linkage B.
Incomplete Linkage As the other parent only
contributed recessive alleles (whether crossing
over occurs or not)
b
b
a
A
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TEST CROSS !!!
III. Linkage A. Complete Linkage B.
Incomplete Linkage As the other parent only
contributed recessive alleles (whether crossing
over occurs or not) SO the phenotype of the
offspring is determined by the gamete received
from the heterozygote
b
b
a
A
gamete genotype phenotype
ab aabb ab
ab AaBb AB
ab aaBb aB
ab Aabb Ab
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TEST CROSS !!!
III. Linkage A. Complete Linkage B.
Incomplete Linkage
b
b
a
A
ALTERNATIVES IA LINKAGE
gamete genotype phenotype
ab aabb ab
ab AaBb AB
ab aaBb aB
ab Aabb Ab
LOTS of PARENTALS
FREQUENCIES EQUAL TO PRODUCT OF INDEPENDENT
PROBABILITIES
FEWER COS
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• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• Example
• How do we discriminate between these two
  alternatives?
• Conduct a Chi-Square Test of Independence

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
18

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• Example
• How do we discriminate between these two
  alternatives?
• Conduct a Chi-Square Test of Independence
• - Compare the observed results with what you
  would expect if the genes assorted independently

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
19

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• Example
• How do we discriminate between these two
  alternatives?
• Conduct a Chi-Square Test of Independence
• - Compare the observed results with what you
  would expect if the genes assorted independently

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
The frequency of AB should f(A) x f(B) x N
55/100 x 51/100 x 100 28
20

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• Example
• How do we discriminate between these two
  alternatives?
• Conduct a Chi-Square Test of Independence
• - Compare the observed results with what you
  would expect if the genes assorted independently

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
The frequency of AB should f(A) x f(B) x N
55/100 x 51/100 x 100 28 The frequency of
Ab should f(A) x f(B) x N 55/100 x 49/100
x 100 27 The frequency of aB should f(a) x
f(B) x N 45/100 x 51/100 x 100 23 The
frequency of ab should f(a) x f(b) x N
45/100 x 49/100 x 100 22
21

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• Example
• How do we discriminate between these two
  alternatives?
• Conduct a Chi-Square Test of Independence
• This is fairly easy to do by creating a
  contingency table

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B b Row Total
A 43 12
a 8 37
Col. Total
22

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• Example
• How do we discriminate between these two
  alternatives?
• Conduct a Chi-Square Test of Independence
• This is fairly easy to do by creating a
  contingency table
• Add across and down
• This gives the totals for each trait
  independently.

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B b Row Total
A 43 12 55
a 8 37 45
Col. Total 51 49 100
23

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• Example
• How do we discriminate between these two
  alternatives?
• Conduct a Chi-Square Test of Independence
• This is fairly easy to do by creating a
  contingency table
• Then, to calculate an expected value based on
  independent assortment (for AB, for example),
  you multiple Row Total x Column Total and
  divide by Grand Total.
• 55 x 51 / 100 28

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B Exp. b Row Total
A 43 28 12 55
a 8 37 45
Col. Total 51 49 100
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Example Repeat to
calculate the other expected values (This is
just an easy way to set it up and do the
calculations, but you should appreciate it is the
same as the product rule F(A) x f(B) x N
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total 51 49 100
55 X 51 x 100 55x51 100 100
100
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Compare our observed
results with what we would expect if the genes
assort independently.
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total 51 49 100
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Compare our observed
results with what we would expect if the genes
assort independently. . If our results are
close to the expectations, then they support the
hypothesis of independence.
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total 51 49 100
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Compare our observed
results with what we would expect if the genes
assort independently. . If our results are
close to the expectations, then they support the
hypothesis of independence. If they are far
apart from the expected results, then they refute
that hypothesis and support the alternative
linkage.
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total 51 49 100
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Typically, we reject the
hypothesis of independent assortment (and accept
the hypothesis of linkage) if our observed
results are so different from expectations that
independently assorting genes would only produce
results as unusual as ours less than 5 of the
time
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total 51 49 100
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B Exp. b Exp. Row Total
A 43 28 12 27 55
a 8 23 37 22 45
Col. Total 51 49 100
III. Linkage A. Complete Linkage B.
Incomplete Linkage We determine this
probability with a Chi-Square Test of
Independence.
Phenotype Obs Exp (o-e) (o-e)2/e
AB 43 28 15 8.04
Ab 12 27 -15 8.33
aB 8 23 -15 9.78
ab 37 22 15 10.23
X2 36.38
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 First, we
determine the degrees of freedom (r-1)(c-1)
1
B b Row Total
A 43 12 55
a 8 37 45
Col. Total 51 49 100
31
III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 First, we
determine the degrees of freedom (r-1)(c-1)
1 Now, we read across the first row in the
table, corresponding to df 1. The column
headings are the probability that a number in
that column (diff. between observed and expected)
would occur by chance. In our case, it is the
probability that our hypothesis of independent
assortment (expected values are based on that
hypothesis) is true. (and the diff is just due to
chance).
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 Note that
larger values have a lower probability of
occurring by chance This should make sense, and
the value increases as the difference between
observed and expected values increases.
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 So, for
instance, a value of 2.71 will occur by chance
10 of the time.
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 So, for
instance, a value of 2.71 will occur by chance
10 of the time. But a value of 6.63 will only
occur 1 of the time... (if the hypothesis is
true and this deviation between observed and
expected values is only due to chance).
35
III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 So, for
instance, a value of 2.71 will occur by chance
10 of the time. For us, we are interested in
the 5 level. The table value is 3.84. Our
calculated value is much greater than this so
the chance that independently assorting genes
would yield our results is WAY LESS THAN 5. Our
results are REALLY UNUSUAL for independently
assorting genes.
36.38
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 Our results
are REALLY UNUSUAL for independently assorting
genes. So, either our results are wrong, or the
hypothesis of independent assortment is wrong.
If you did a good experiment, then you should
have confidence in your results reject the
hypothesis of IA and conclude the alternative
the genes are LINKED.
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III. Linkage A. Complete Linkage B.
Incomplete Linkage Our X2 36.38 Our results
are REALLY UNUSUAL for independently assorting
genes. So, either our results are wrong, or the
hypothesis of independent assortment is wrong.
If you did a good experiment, then you should
have confidence in your results reject the
hypothesis of IA and conclude the alternative
the genes are LINKED. Of course, there is still
that 5 probability that chance is responsible
for the deviation you observed, and your
conclusion is WRONG.
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III. Linkage A. Complete Linkage B.
Incomplete Linkage OK so we conclude the
genes are linked NOW WHAT?
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
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III. Linkage A. Complete Linkage B.
Incomplete Linkage OK so we conclude the
genes are linked NOW WHAT? We map the genes
using the knowledge that crossing-over is rare,
and the frequency of crossing-over correlates
with the distance between genes.
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
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III. Linkage A. Complete Linkage B.
Incomplete Linkage OK so we conclude the
genes are linked NOW WHAT? 1. Crossing-over is
rare so the RARE COMBINATIONS must be the
products of Crossing-Over. The OTHERS, the MOST
COMMON products, represent the PARENTAL
TYPES.
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
41

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• OK so we conclude the genes are linked NOW
  WHAT?
• Crossing-over is rare so the RARE COMBINATIONS
  must be the products of Crossing-Over. The
  OTHERS, the MOST COMMON products, represent the
  PARENTAL TYPES.
• This tells us the original arrangement of
  alleles in the heterozygous parent

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
42

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• OK so we conclude the genes are linked NOW
  WHAT?
• Crossing-over is rare so the RARE COMBINATIONS
  must be the products of Crossing-Over. The
  OTHERS, the MOST COMMON products, represent the
  PARENTAL TYPES.
• This tells us the original arrangement of
  alleles in the heterozygous parent
• Segregation without crossing over produces lots
  of AB and ab gametes.

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
43

• III. Linkage
• A. Complete Linkage
• B. Incomplete Linkage
• OK so we conclude the genes are linked NOW
  WHAT?
• Crossing-over is rare so the RARE COMBINATIONS
  must be the products of Crossing-Over. The
  OTHERS, the MOST COMMON products, represent the
  PARENTAL TYPES.
• 2. The frequency of crossing-over is used as an
  index of the distance between genes
• The other progeny are the products of crossing
  over, and they occurred 20 times in 100 progeny,
  for a frequency of 0.2. Multiply that by 100 to
  free the decimal, and this becomes 20 map units
  (CentiMorgans).

AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
20 map units
44
III. Linkage A. Complete Linkage B.
Incomplete Linkage So, we used the chi-square
test to determine whether genes are on the
chromosome. Then, we mapped the distance between
genes on the chromosome all by looking at the
frequency of phenotypes in the offspring of a
test cross.
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
20 map units
45
III. Linkage A. Complete Linkage B.
Incomplete Linkage So, we used the chi-square
test to determine whether genes are on the
chromosome. Then, we mapped the distance between
genes on the chromosome all by looking at the
frequency of phenotypes in the offspring of a
test cross. LIMITATION Genes that are 50 map
units apart (or more) will recombine 50 of the
time. Producing parentals and recombinants in
equal frequency just like independent assortment
predicts!!!
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
20 map units
46
III. Linkage A. Complete Linkage B.
Incomplete Linkage So, we used the chi-square
test to determine whether genes are on the
chromosome. Then, we mapped the distance between
genes on the chromosome all by looking at the
frequency of phenotypes in the offspring of a
test cross. LIMITATION Genes that are 50 map
units apart (or more) will recombine 50 of the
time. Producing parentals and recombinants in
equal frequency just like independent assortment
predicts!!! So, genes that are gt 50 mu apart
assort independently (even though they are on the
same chromosome).
AaBb x aabb
Offspring Number
AB 43
Ab 12
aB 8
ab 37
20 map units
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III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping Suppose you conduct two-point mapping
and find A - C are 23 centiMorgans apart C
B are 10 centiMorgans apart
A
C
B
OR
A
C
B
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III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping
Three Point Test Cross AaBbCc x
aabbcc Phenotypic Ratio ABC 25 ABc
3 Abc 42 AbC 85 aBC 79 aBc
39 abc 27 abC 5
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III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping - combine complementary sets
Three Point Test Cross AaBbCc x
aabbcc Phenotypic Ratio ABC 25 ABc
3 Abc 42 AbC 85 aBC 79 aBc
39 abc 27 abC 5
ABC 25 abc 27 52
ABc 3 abC 5 8
Abc 42 aBC 39 81
AbC 85 aBc 79 164
50
III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping - combine complementary sets
Three Point Test Cross AaBbCc x
aabbcc Phenotypic Ratio ABC 25 ABc
3 Abc 42 AbC 85 aBC 79 aBc
39 abc 27 abC 5
ABC 25 abc 27 52
ABc 3 abC 5 8
Abc 42 aBC 39 81
AbC 85 aBc 79 164
MOST ABUNDANT ARE PARENTAL TYPES
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III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping - combine complementary sets
Three Point Test Cross AaBbCc x
aabbcc Phenotypic Ratio ABC 25 ABc
3 Abc 42 AbC 85 aBC 79 aBc
39 abc 27 abC 5
ABC 25 abc 27 52
ABc 3 abC 5 8
LEAST ABUNDANT ARE DOUBLE CROSSOVERS
D
E
F
Abc 42 aBC 39 81
AbC 85 aBc 79 164
d
e
f
MOST ABUNDANT ARE PARENTAL TYPES
DCOs very rare, and gene in middle switches
chromosomes
52
III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping - combine complementary sets
Three Point Test Cross AaBbCc x aabbcc
Compare Parentals and DCOs to determine which
gene is in the middle Parentals AbC
aBc DCOs abC ABc B and C
alleles stay together A switches and so is in
middle. REMAKE YOUR DRAWING TO DEPICT ALLELES AND
ORDER
ABC 25 abc 27 52
ABc 3 abC 5 8
LEAST ABUNDANT ARE DOUBLE CROSSOVERS
b
A
C
Abc 42 aBC 39 81
AbC 85 aBc 79 164
B
a
c
MOST ABUNDANT ARE PARENTAL TYPES
53
III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping - combine complementary sets
Three Point Test Cross AaBbCc x aabbcc
So, a single cross-over (SCO) between the B and
A locus will produce the gametes/phenotypes BAC
and bac find their frequency 52 plus add
DCOs (8) 60/305 0.197 x 100 19.7 cM
ABC 25 abc 27 52
ABc 3 abC 5 8
LEAST ABUNDANT ARE DOUBLE CROSSOVERS
b
A
C
Abc 42 aBC 39 81
AbC 85 aBc 79 164
B
a
c
MOST ABUNDANT ARE PARENTAL TYPES
19.7 cM
54
III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping - combine complementary sets
Three Point Test Cross AaBbCc x aabbcc
81 8 89/306 0.292 x 100 29.2
ABC 25 abc 27 52
ABc 3 abC 5 8
LEAST ABUNDANT ARE DOUBLE CROSSOVERS
b
A
C
Abc 42 aBC 39 81
AbC 85 aBc 79 164
B
a
c
MOST ABUNDANT ARE PARENTAL TYPES
19.7 cM
29.2 cM
55
III. Linkage A. Complete Linkage B.
Incomplete Linkage C. Three-point
Mapping - combine complementary sets
Three Point Test Cross AaBbCc x aabbcc
Coefficient of Coincidence are SCOs
occurring independently? If so, then DCOs
product of SCOs (0.197) x (0.292) x 305
17 We only observed 8. C.O.C. obs/exp 8/17
0.47 Interference 1 c.o.c. 0.53
ABC 25 abc 27 52
ABc 3 abC 5 8
LEAST ABUNDANT ARE DOUBLE CROSSOVERS
b
A
C
Abc 42 aBC 39 81
AbC 85 aBc 79 164
B
a
c
MOST ABUNDANT ARE PARENTAL TYPES
19.7 cM
29.2 cM
56
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