Transcript lecture_07(LP)

Announcements
-First midterm exam will be in this room on Friday
(4-25) from 10:30AM-12:20PM
-Exam will cover material presented in lecture and
quiz section through the end of last week,
however…
-this weeks material will reinforce some of
the previous concepts you have learned
-this weeks material WILL be covered on
the next midterm exam
Leo Pallanck’s office hours: Friday afternoons by
appointment ([email protected])
What this course is about From Lecture 1
Inheritance
how is genetic information read within cells?
how are traits transmitted to progeny?
how are unique physical traits determined by genes?
Mutant analysis
how are biological processes studied by analyzing
Today
mutants?
Genomics
what can we learn by studying whole genomes?
Throughout the quarter . . .
how and why are model organisms used in genetics?
how does that information apply to humans?
More
to
come!
Mutant analysis (AKA Genetic Analysis)
The use of mutants to understand how a
biological process normally works*
Very powerful - can be used to study metabolic
pathways, animal development, neurobiology,
cell division, etc.
A simple analogy…
*See the Salvation of Doug article at the following site:
http://bio.research.ucsc.edu/people/sullivan/savedoug.html
Analysis of pizza synthesis
Analagous
to genes
Analysis of pizza synthesis
Analagous to a mutation
Analagous
to genes
The mutant
phenotype
No red sauce?!
Mutant analysis involves model organisms
What is a model organism?
A species that one can experiment with to
ask a biological question
Why bother with model organisms?
- All organisms are related at the molecular level
- Not always possible to do experiments on the
organism you want
- If the basic biology is similar, it may make
sense to study a simple organism rather than a
complex one
Which of these cameras do you think would be
easier to understand?
IMAX 3-D camera
Box camera
Features of a good model organism
• Short generation time
• Small, easy to maintain
• Large numbers of progeny
• Well-studied life cycle, biology
• Appropriate
for theorganism—the
question at hand
• Mendel
used a model
garden pea
- relatively short generation time—one per year
- lots of progeny per cross
- self-pollination and out-crossing possible
Telomeres
- true-breeding varieties readily available from
local merchant
96 million telomeres per cell!
Some commonly used model organisms
- Bacteria — Escherichia coli
- Budding yeast — Saccharomyces
cerevisiae
- Fruit fly — Drosophila melanogaster
QuickTime™ and a
TIFF (U ncompressed) decompressor
are needed to see t his picture.
- Nematode — Caenorhabditis elegans
- Mouse — Mus musculus
Quiz Section this week:
Complementation analysis of yeast mutants
An introduction to yeast . . .
Yeast as a model “genetic” organism
Budding yeast—a single-celled fungus that divides by budding
Yeast cells can exist as haploids…
Mutagenesis is easier in single-cell organisms with haploid
lifestyles
Haploid life cycle:
mitosis
The haploid
life cycle (1n)
cytokinesis
mutation
The life cycle of “budding” yeast
Yeast cells can also exist as diploids…
1n
a
haploid
a
haploid
mating
The haploid
life cycle (1n)
diploid
zygote
The life cycle of “budding” yeast (cont)
Mendelian
segregation
occurs here
A tetrad with 4
haploid spores
meiosis
(“gametes”)
2 a cells
1n
2 a cells
a/a diploid
life cycle
(2n)
Conducting a mutant analysis with yeast
Case study: analyzing the adenine biosynthetic
pathway by generating and studying “ade” mutants
Wild-type yeast can survive on ammonia, a few vitamins, a few
mineral salts, some trace elements and sugar…
They synthesize everything else they need, including adenine
What genes does yeast
need to synthesize adenine?
(Why might we care about adenine?)
Identifying yeast mutants that require adenine
Treat wt haploid
cells with a
mutagen:
plate
cells
Adeninerequiring
colonies
(ade
mutants)
“complete”
plate
-adenine
plate
m3
m2
“Replica-plating”
m1
sterile
piece of
velvet
Are the adenine-requiring mutants recessive?
That is, are they LOF mutations? Why do we care?
“a” mating type
m1
“a” mating type
wild-type
Genotypes:
ade
ADE
diploids
“complete”
plate
What do you conclude?
replica-plate
using velvet
-adenine plate
ADE is dominant over ade
Are all of the mutations in one gene?
Are m1 and m2 alleles of the same gene?
What would you predict if…
• only one enzyme is needed for synthesis of adenine?
All mutants would be alleles of the same gene.
• many enzymes are needed for synthesis of adenine?
Different genes might be mutant.
How to find out how many different genes we have
mutated?
Do complementation test to ask:
are the mutations alleles of the same
gene or of different genes?
Performing a complementation test
“a” mating type
“a” mating type
m1
m2
diploids
replica-plate
“complete”
4
-adenine
Do m1 and m2 complement, or fail to complement?
Are m1 and m2 alleles of the same gene, or alleles of
different genes?
Complementation tests with ade mutants
What do you conclude from the pair-wise crosses shown below?
x m
m 1o
1
m
2
m
3
m
4
m
5
m
6
m
7
m
2
+
m
3
+
m
4
+
m
5
o
m
6
+
m
7
o
o
Conclusion?
m1, m5, m7 are
mutations in one
gene
o
o
o
o
o
o = no growth on -ade
+ = growth on -ade
Complementation tests with ade mutants
What do you conclude from the pair-wise crosses shown below?
Four complementation groups
x m
m 1o
1
m
2
m
3
m
4
m
5
m
6
m
7
Usually means four genes
Conclusion?
m
m
2
m
3
m
4
m
5
m
6
o
+
o
+
+
+
o
+
+
+
+
m2, m4 are in
one gene
o
+
+
+
m3
o
+
o
m6
o
+
+
+
+
o
+
7
o
o
m1, m5, m7 are
mutations in one
gene
o = no growth on -ade
+ = growth on -ade
Practice Question
Yeast cells can normally grow on a sugar called galactose as the sole carbon source.
Seven mutant “a” haploid yeast strains have been isolated that are unable to grow
on galactose (“gal”) plates.
Six of these mutant strains were each cross-stamped on a gal plate with a wild type
“a” strain. The resulting pattern of growth on the gal plates is depicted below
(shading = growth). In all plates, the wild type strain is in the horizontal streak.
On the leftmost plate, mark the location of the a/a diploid with a circle.
What is the mode of inheritance of mutant phenotype in mutants 1-6? How can you
tell?
Diploids grow on gal plate… so, wild type is dominant
Practice question (continued)
Each of the seven “a” mutant strains was cross-stamped on gal plates against “a”
versions of the seven mutants. The results are depicted below:
Mutant 1
Mutant
1
Mutant
2
Mutant
3
Mutant
4
Mutant 2
Mutant 3
Mutant 4
Mutant 5
Mutant 6
Mutant 7
m1, m2, m5
m3, m6
m4
Mutant
5
Mutant
6
Mutant
7
Looking just at mutants 1–6 for now… group these six mutants by
complementation group.
Practice question (continued)
Now consider mutant 7. What is surprising about the result in the complementation
table?
Fails to complement any of the others… how could it be an
allele of 3 different genes?
Mutant 7 was cross-stamped on gal plate with wild type as you saw with the other
six mutants earlier:
What do you conclude about the mode of inheritance of
mutant 7? How does that help you explain the
complementation test result for mutant 7?
Complementation test fails with a dominant mutation…
heterozygote will always show the mutant phenotype
What can you conclude about how many genes are represented in this collection of
seven mutants?
At least 3 genes (can’t tell about m7)
Complementation is relevant to humans
Family A
Family B
aaBB
= deaf
AAbb
AaBb
Within each family, does deafness look like it’s dominant or
recessive?
recessive
Assign genotypes (A, B, etc.) to the deaf individuals in these
pedigrees.
Complementation is relevant to humans
Niemann Pick Type C disease (NPC): a recessive human
neurodegenerative disease resulting in premature death
Cellular cholesterol accumulation accompanies the disease (can
be detected using a chemical called ‘filipin’ which fluoresces
upon binding cholesterol)
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
QuickTime™ and a
TIFF (Uncompressed) decompressor
TIFF (Uncompressed) decompressor
are needed to see this picture.
are needed to see this picture.
normal
NPC
NPC1:NPC2 NPC1:NPC3
Do NPC1 & NPC2 complement? Do NPC1 & NPC3 complement?
Practice Question
Hearing mice… independently assorting genes A and B, both
needed for hearing:
a, b: complete LOF, recessive
AaBb
x
AaBb
WITHOUT drawing a Punnett square, predict the progeny
phenotypes and proportions with respect to hearing ability.
using mutants to order the steps in a pathway
..
.
HC
N
COOCH
C
H2 N
(AIR)
For example,
this molecule
accumulates in an
ade13 mutant.
CH2
N
HC
R
-OOC
H2 N
R
C
N
CH
C
(SAICAR)
N
(CAIR)
C
H2 N
CH
C
H
N
COO-
N
C
O
ADE13
encodes the enzyme that
carries out the next step
N
R
..
O
.
C
C
HN
N
CH
HC
C
N
NH
Adenine
The Yeast Adenine Biosynthetic Pathway
ADE4 ADE5*
A
B
ADE8 ADE6 ADE7 ADE2 ADE1
X
D
E
F
Y
G
C
ADE13
ADE3*
H
red
ADE17
pigment
ADE16
I
ADE17
ADE16
J
ADE12
K
ADE13
AMP
A second phenotype of some ade mutants…
Mutagenize:
plate
cells
complete
plate
Replica-plate
-adenine
plate
Some of the adeninerequiring mutants are red!
Are the red ade mutations recessive?
m8
wild-type
Genotypes:
ade
complete plate
White color
is dominant
replica-plate
using velvet
ADE
-adenine plate
Ability to make
adenine is dominant
Hypothesis: Two phenotypes/one LOF mutation
How can one LOF mutation generate two very different
phenotypes?
some
intermediate
“X”
X
X
ADE1
ade1
another
adenine
intermediate
“Y”
LOF
mutation
adenine
Y
UNK1*
red
pigment
*not a real gene name!
This gene has not yet
been identified
Suppose we isolate LOTS of independent red
mutants:
Are all red mutants defective in the SAME
GENE?
How to tell?
Complementation tests of red mutants
m9
m10
diploids
mutations fail to complement =
same gene
m9
m11
white
diploids
mutations complement =
different genes
All pairwise combinations reveal two complementation groups.
Must modify the hypothesis.
Modified hypothesis for red phenotype
X
X
X
ADE2
Y
ade2
ADE1
adenine
ADE1
Y
adenine
“UNK1”
red
pigment
But how do mutations in ADE1 result in a build-up of X?
Modified hypothesis for red phenotype
X
X
“UNK1”
ADE2
Y
ade1
Y
adenine
red
pigment
Mutations in either ADE1 or ADE2 lead to a defect
in adenine biosynthesis and lead to the build-up of
intermediates in the pathway.
Excess “X” is converted to a red pigment.
Test your understanding
ADE7
X
X
“UNK1”
red
pigment
ade2
Y
Y
ade1
ADE13
adenine
1. Phenotype of ade1 ade2 double mutation?
Same as ade2 single mutation! Red and adenine-requiring.
2. Phenotype of ade2 ade7 double mutation?
Same as ade7 single mutation! White and adenine-requiring.
3. Phenotype of ade2 ade13 double mutation?
Same as ade2 single mutation! Red and adenine-requiring.
4. Phenotype of unk1?
White and able to grow on -ade plates.
Practice Questions
ADE3
X
X
ADE2
Y
Y
ADE1
adenine
“UNK1”
red
pigment
1A. A MATa ade2 ADE3 mutant was mated to a MATa ADE2 ade3 mutant to create
a diploid. What are the phenotypes of the three strains? Assume all other genes
are wild type.
strain:
MATa ade2 ADE3
MATa ADE2 ade3
Diploid
color on
complete
plate?
growth on
-adenine plate?
(yes or no)
Practice Questions
ADE3
X
X
ADE2
Y
Y
ADE1
adenine
“UNK1”
red
pigment
1A. A MATa ade2 ADE3 mutant was mated to a MAT ADE2 ade3 mutant to create
a diploid. What are the phenotypes of the three strains? Assume all other genes
are wild type.
strain:
color on
complete
plate?
growth on
-adenine plate?
(yes or no)
MATa ade2 ADE3
red
no
MATa ADE2 ade3
white
no
Diploid
white
yes
1B. ADE2 and ADE3 assort independently. Draw the chromosomes at metaphase of
meiosis I such that the two WILD TYPE alleles face the same pole. Place a
crossover on the other chromosome arm relative to the ADE2 and ADE3 genes.
A
ADE2
ADE3
B
C
ADE2
ADE3
ade2
ade3
D
ade2
ade3
1C. Recall that each tetrad contains
the products of a single meiosis.
Predict the genotypes and growth
properties of each spore resulting
from this meiosis.
Spore
A
B
C
D
complete genotype?
grow without adenine?
yes
yes
no
no
ADE2 ADE3
ADE2 ADE3
ade2 ade3
ade2 ade3
1D. Analysis of many tetrads demonstrates that three types are found, depending
on the behavior of the chromosomes in meiosis. Which tetrad best fits the meiosis
you just drew? Letter the spores below to match the genotypes in your table.
Tetrad on complete plates
#1
#2
#3
A
C
red
B
D
1E. Now draw the chromosomes at metaphase of meiosis I such that one wild type
and one mutant allele face each pole. Place a crossover on the other chromosome
arm relative to the Adenine genes.
A
ADE2
ade3
B
ADE2
ade3
1F. Predict the genotypes and growth Spore complete genotype?
properties of each spore resulting
A
from this meiosis.
B
grow without adenine?
C
D
1G. Which tetrad best fits the meiosis you just drew? Letter the spores below to
match the genotypes in your table.
Tetrad on complete plates
#1
#2
red
#3
1E. Now draw the chromosomes at metaphase of meiosis I such that one wild type
and one mutant allele face each pole. Place a crossover on the other chromosome
arm relative to the Adenine genes.
A
ADE2
ade3
C
B
ADE2
ade3
ade2
ADE3
D
ade2
ADE3
1F. Predict the genotypes and growth Spore complete genotype?
properties of each spore resulting
ADE2 ade3
A
from this meiosis.
ade2 ADE3
B
C
D
ADE2 ade3
ade2 ADE3
grow without adenine?
no
no
no
no
1G. Which tetrad best fits the meiosis you just drew? Letter the spores below to
match the genotypes in your table.
Tetrad on complete plates
#1
A
C
B
D
#2
red
#3
1H. Now draw the chromosomes at metaphase of meiosis I such that one wild type
and one mutant allele face each pole. On one chromosome, place a crossover on
the other chromosome arm relative to the Adenine gene. On the other
chromosome, place a crossover BETWEEN the centromere and the Adenine gene.
1I. Predict the genotypes and growth Spore complete genotype?
properties of each spore resulting
A
from this meiosis.
B
grow without adenine?
C
D
1J. Which tetrad best fits the meiosis you just drew? Letter the spores below to
match the genotypes in your table.
Tetrad on complete plates
#1
#2
red
#3