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Midterm 1
• Mean = 74.6%
> 100: | |
> 90: | | | | | | | | | |
> 80: | | | | | | | | | | | | |
> 70: | | | | | | | | | | | | |
> 60: | | | | | | | | | | | |
> 50: | | | | | | | | | | |
> 40: |
Extra Credit, To Date: = 18 pts
Please see me if you have
questions about arithmetic, or
obvious misgrades.
If you want to argue a point, I
reserve the right to re-grade
the entire test.
Midterm 2
• Lectures,
• Assignments,
• Through Chapter 5, Chromosomal Mutations.
Linkage
• Genes located on the same chromosome do
not recombine,
– unless crossing over occurs,
• The recombination frequency gives an
estimate of the distance between the genes.
Recombination Frequencies
• Genes that are adjacent have a recombination frequency
near 0%,
• Genes that are very far apart on a chromosome have a
recombination frequency of 50%,
• The relative distance between linked genes influences the
amount of recombination observed.
Linkage Ratio
E
x
p
e
r
i
m
e
n
t
a
l
l
y
(How do you determine it?)
P GGWW x ggww
Testcross F1: GgWw x ggww
GW
Gw
gW
gw
?
?
?
?
recombinant
total progeny
=
Linkage Ratio
Linkage Ratio
(If Sorting Independently)
P GGWW x ggww
Testcross F1: GgWw x ggww
GW
Gw
gW
gw
50
50
50
50
50 (Gw) + 50 (gW)
200 (all classes)
=
.5
Linkage Ratio Units
% = mu (map units)
- or % = cm (centimorgan)
Fly Crosses
(white eyes, minature, yellow body)
• In a white eyes x miniature cross, 900 of the 2,441
progeny were recombinant, yielding a map
distance of 36.9 mu,
• In a separate white eyes x yellow body cross, 11 of
2,205 progeny were recombinant, yielding a map
distance of 0.5 mu,
• When a miniature x yellow body cross was
performed, 650 of 1706 flies were recombinant,
yielding a map distance of 38 mu.
Simple Mapping
• white eyes x miniature = 36.9 mu,
• white eyes x yellow body = 0.5 mu,
• miniature x yellow body = 38 mu,
0.5 mu
36.9 mu
y w
m
38 mu
Three Point Testcross
Triple Heterozygous
(AaBbCc )
x
Triple Homozygous Recessive
(aabbcc)
Three Point Mapping Requirements
• The genotype of the organism producing the gametes must
be heterozygous at all three loci,
• You have to be able to deduce the genotype of the gamete
by looking at the phenotype of the offspring,
• You must look at enough offspring so that all crossover
classes are represented.
w
g
d
Representing linked genes...
P
W
w
G
g
D
d
= WwGgDd
d
d
= wwggdd
x
Testcross
w
w
g
g
w
g
d
Representing linked genes...
P
+
w
+
g
+
d
= WwGgDd
d
d
= wwggdd
x
Testcross
w
w
g
g
Phenotypic Classes
GWgg
Ddd
Ddd
D-
G-
dd
W-G-DW-G-dd
W-gg-D
W-gg-dd
wwG-DwwG-dd
ww
gg
Ddd
wwggDwwggdd
#
Crossovers
W-G-D-
179
0
wwggdd
173
0
W-G-dd
46
1
wwggD-
52
1
wwG-D-
22
1
W-gg-dd
22
1
W-gg-D
2
2
wwG-dd
4
2
W
G
D
w
g
d
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
II
I
W
G
D
w
g
d
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Total =
500
I
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
W
G
D
w
g
d
Region I:
46 + 52 + 2 + 4
500
= 20.8 mu
x 100
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Total =
500
II
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
20.8 mu
W
G
D
w
g
d
Region II:
22 + 22 + 2 + 4
500
= 10.0 mu
x 100
10.0 mu
W
w
20.8 mu
G
D
g
d
0.1 x 0.208 = 0.0208
NO GOOD!
W-gg-D
wwG-dd
Total =
2
4
500
Recombinants,
double crossover
6/500 = 0.012
Interference
…the affect a crossing over event has on a second
crossing over event in an adjacent region of the
chromatid,
– (positive) interference: decreases the probability of a
second crossing over,
• most common in eukaryotes,
– negative interference: increases the probability of a
second crossing over.
Gene Order in Three Point Crosses
• Find either double cross-over phenotype, based on
the recombination frequencies,
• Two parental alleles, and one cross over allele will
be present,
• The cross over allele fits in the middle...
#
A-B-C-
2001
aabbcc
1786
A-B-cc
46
aabbC-
52
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
Which one is the odd
one?
II
I
A
C
B
a
c
b
#
A-B-C-
2001
aabbcc
1786
Region I
990 + 887 + 46 + 52
A-B-cc
46
6951
aabbC-
52
= 28.4 mu
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
x 100
I
A
C
B
a
c
b
#
A-B-C-
2001
aabbcc
1786
Region II
600 + 589 + 46 + 52
A-B-cc
46
6951
aabbC-
52
= 18.5 mu
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
18.5
II mu
x 100
28.4 mu
A
C
B
a
c
b
Master
• Problems 1, 2
• Questions 4.1 - 4.4, 4.6 - 4.16, 4.19 - 4.20
Quantitative Traits
Optional; Klug and Cummings (at the library),
– Chapter 5, pp. 115 - 120, Insights and Solutions #1 (pp.
130), Questions 1-6 (pp. 131-132)
Reading Assignments
Read 15.5
Available as a PDF, online.
Will answer question on Monday.
Quantitative Traits
…traits that show a continuous variation in
phenotype over a range,
…often result from multiple genes and are
further termed polygenic traits.
The Club Footed Boy
•Heart disease
•Spina bifida
•Neural tube disorders
•Diabetes
•Etc.
Ribera, 1642
Environmental Factors
Optional: Chapter 15
Discontinuous Traits
• Discontinuous Traits,
– exhibit only a few distinct phenotypes and can
be described in a qualitative manner,
• Mendel and others worked with “inbred”
lines,
Father:
DD BB cc aa EE FF GG
Mother:
dd BB cc aa EE FF GG
Continuous Traits
• “Outbred” transmission,
Father:
Mother:
dd BB CC AA EE FF GG
DD bb CC aa EE ff gg
• Continuous Traits,
– Display a spectrum of phenotypes, and must be
described in quantitative terms,
F3 #1: Cross Two Individuals With 10 Cm Ears.
F3 #2: Cross Two Individuals With 17 Cm Ears.
Polygenic Traits and Mendel
• It is possible to provide a Mendelian explanation
for continuous variation by considering numbers
of genes contributing to a phenotype,
– the more genes, the more phenotypic categories,
– the more categories, the more the variation seems
continuous.
Melanin Pigmentation
Calculating the Number of Genes
• If you know the frequency of either extreme
trait, then the number of genes (n) can be
calculated using the formulae…
1
n
4
= ratio of F2 individuals expressing
either extreme phenotype.
1/4 show the extreme traits,
1
4n
n=1
=
1
41
1/16 show either extreme traits,
1
4n
n=2
=
1 =
16
1
42
Continuous Variation
• Two or more genes that influence the same phenotype,
often in an additive way,
– additive allele: contributes a set amount to the phenotype,
– non-additive allele: does not contribute to the phenotype.
• The affect of each allele on the phenotype is relatively
small, and roughly equivalent,
• Substantial variation is observed when multiple genes
control a single trait,
• Must be studied in large populations.
•
•
•
•
Inbred strain #1, mean height = 24 cm,
Inbred strain #2, mean height = 24 cm,
F1 mean height = 24 cm,
F2, 12 - 36 cm range, mean = 24 cm,
– 4/1000 are 12 cm,
– 4/1000 are 36 cm,
smallest class
largest class
a. What mode of inheritance?
b. How many gene pairs?
Quantitative.
1/4n = 1/250, n = 4
4/1000 = 1/250 are the extreme class,
solve 1/4n = ratio of extreme class
42 = 16, 43 = 64, 44 = 256, etc.
•
•
•
•
Inbred strain #1, mean height = 24 cm,
Inbred strain #2, mean height = 24 cm,
F1 mean height = 24 cm,
F2, 12 - 36 cm range, mean = 24 cm,
– 4/1000 are 12 cm,
– 4/1000 are 36 cm,
smallest class
largest class
c. How much does each allele contribute?
range: largest (36 cm) - smallest (12 cm) = 24 cm
alleles: 4 genes, 8 potential additive alleles
24 cm / 8 additive alleles = 3 cm / additive allelle
•
•
•
•
Inbred strain #1, mean height = 24 cm,
Inbred strain #2, mean height = 24 cm,
F1 mean height = 24 cm,
F2, 12 - 36 cm range, mean = 24 cm,
– 4/1000 are 12 cm,
– 4/1000 are 36 cm,
smallest class
largest class
d. Indicate one possible set of P1, F1.
Base Height is 12 cm, so each P1, F1 has 4 additive alleles.
P1
AABBccdd x aabbCCDD
F1
AaBbCcDd x AaBbCcDd