Download Drosophila lab recitation notes

Recitation 1 (8/27)
♦ Pass out lab manual
♦ Welcome. Explanation of the recitation and lab structure and how this fits into the new
curriculum. Explanation of purpose of recitation and objectives of lab.
♦ Roll.
♦ Who's teaching? When? - Hauber (T 2:00-5:00), Dorn (W 1:30-4:30), Coltman (R 2:005:00), Shuh (F 12:30-3:30). If student anticipates missing scheduled lab, they should check
the schedule and see if they can attend one of the other labs. Inform the instructor (and
regular instructor).
♦ Review syllabus including deadlines, lab quizzes, final recitation grand quiz. Explain
grading.
♦ This week:
Genetic analysis in Drosophila
Ø Bring lab manual to lab and next week to recitation.
Ø Prepare to take notes in recitation and lab - ask lab instructor about lab notebook
preference.
Ø Students will work with a partner on this lab, which carries on through the next few
weeks.
Ø This week in lab, a culture of wild-type flies will be to be dispensed to each pair of
students in order for them to learn about fruit fly life cycle, how to anaesthetize flies,
manipulate anaesthetized flies, prepare and label culture vials, sex flies with dissecting
scope.
Ø Lab partners will share data but submit separate, independent lab reports.
Single-gene inheritance in Drosophila
Traits involve:
Eye color, eye shape, hair morphology (head and thorax), body color, wing size, wing shape
Wild-type phenotype:
Brick-red, oval eyes; normal spines and bristles; gray body; normal wing (see p. 4 in handout)
Mutant phenotype (e.g.):
Purple eyes
Purple and wild-type (red) are alleles for the eye color gene.
The expression of every trait is determined by a pair of genes, one gene from each parent.
Homozygous means the pair of genes are the same (two wild-type or two purple)
Heterozygous means the pair of genes are different (one wild-type, one purple)
Genotype refers to the genetic constitution of the pair of genes for a given trait in an individual.
P1 cross - explain
homozygous wild-type female X
F1 = ?
homozygous purple male
P1: a+a+bb
x
mutant b, wild-type a
F1:
aab+b+
mutant a, wild-type b
wild-type for both traits
Therefore the a and b mutations are recessive and autosomal and
the F1 genotype would be a+ab+b
F1:
males: mutant b, wild-type a
females: wild-type for both traits
Therefore the a and b mutations are sex-linked and the male F1
genotype is
a+ b
a+ b
The female genotype is
a
b+
Are the mutants sex-linked recessive or sex-linked dominant?
How can you tell?
If autosomal, use F2 results to determine if genes are
independently assorting or linked
F1: a+ab+b x a+ab+b - assume that the F1's were all wild-type
for both traits.
F2: Four possible phenotypes:
a+_b+_
a+_bb
aab+_
aabb
What is the expected ratio of the F2 phenotypes?
9:3:3:1? What does this mean? How is this obtained?
What if the gene loci are linked?
a+ b
F1:
a+ b
X
a
b+
a
b+
-all F1s wild-type
Note: no crossing over in male fruit flies
F2:
a+_b+_
a+_bb
aab+_
aabb
What is the expected ratio of the F2 phenotypes?
How does one determine whether their data are supporting
independent assortment or linkage?
Chi-square test - Statistical test for goodness-of-fit
If you want to investigate whether the distribution of F2
phenotypes that you observed in your results is different from the
expected distribution if the genes were independently assorting?
(or if the genes were linked), then you perform a chi-square (χ²)
test. χ² tests can only be carried out on actual numbers, not
percentages or proportions.
The χ² test is carried out in the following steps :
Subtract each expected number (E) from the
observed number (O) for each phenotype
Square the difference
Divide the squared differences by the expected
number for that phenotype
The calculated χ² value is the (O - E)² / E
summed over all phenotypes
Compare the calculated χ² value with the
values in the χ² table (p. 19 in lab manual)
To determine which row to use in the table,
you must calculate your degrees of freedom
(df)
Degrees of freedom equals the number of
classes (i.e., phenotypes) minus one.
O-E
(O - E)²
(O - E)² / E
(O - E)² / E
df =
# of classes - 1
The χ² probability (p) is determined by finding where your
calculated χ² value falls in the table and finding the
corresponding probability in the top row of the table (0.9 to
0.001) (try this website if the table is confusing
http://www.cyber.vt.edu/mbo/commres/lectures/statinf/chiprob.h
tml). Your hypothesis is that there will be no difference between
the expected distribution of phenotypes and the observed
distribution. If your χ² probability is < 0.05 (5%), then you
reject your hypothesis. If your χ² probability is > 0.05, then you
fail to reject your hypothesis.
*probability that the deviation between the observed and
expected distribution is due to chance.
Sample F2 data:
Hypothesis : ?
phenotype
a+_b+_
a+_bb
aab+_
aabb
Totals
obs.
49
13
10
3
75
exp.
o-e
(o-e)2
(o-e)2/e
o-e
(o-e)2
(o-e)2/e
d.f. = ?
χ² value = ?
reject or fail to reject?
Alternate hypothesis: ?
phenotype
a+_b+_
a+_bb
aab+_
aabb
Totals
obs.
49
13
10
3
75
d.f. = ?
χ² value = ?
reject or fail to reject?
exp.
If genes are sex-linked, you already know they are linked,
the question you waqnt to answer is how close or how far
apart from each other are they on the X chromosome.
a b+
P1:
X
a
F1:
a+ b
b+
a+ b
a
b+
X
a
F2: a+ b+
a+ b
a b+
ab
b+
co
par. or nco
par. or nco
co
map distance = # of c.o. types (males) x 100
total # of males