Fruit fly genetics simulation:
An assignment worth 50 points plus an extra credit 5pts.
------------SEE FLOW CHART HANDOUT TO HELP YOU WITH THIS.
1. Go to http://biologylab.awlonline.com
2. Click on “Fly lab”
3. Log in using username "lima101" (ending in one zero one) and the password
"hypothesis2"
4. Select fruit fly mutant types and, by experimental matings, determine the
nature of each mutation. Here are the five possible types of mutations:
- dominant and not sex linked
- recessive and not sex linked
- dominant and sex linked
- recessive and sex linked
- lethal when homozygous
Number your paper 1-10. After each number put the name of a mutation and what
type that mutation is from the above 5 possible types. You will, thus, select 10
fruit fly mutations. Then show the appropriate genetic notation and final punnett
square from your experimental matings that fully demonstrate why the mutation is
what you say it is. Clearly indicate the phenotype of flies represented by each
box of your punnett square.
Only use X and Y chromosomes in your genetic notation and punnett squares if you
are trying to show that the mutation is sex linked. Otherwise ignore sex and
don’t use X and Y chromosomes.
2.5 points for each correct answer. 2.5 additional points for the correct genetic
notation and punnett squares supporting your correct answer.
A bonus of five points will be awarded if you can completely correctly identify
one each of all of the above five types of mutations.
Do not use the mutations White Eyes or Dumpy Wings because there were done for you
in class today.
If F1 are
1/2 mutant & 1/2 normal with no sex
bias, then homozygous lethal mutant is
possible
To verify this cross
two mutants.
Mutant X Mutant
If offspring are
2/3 mutant and
1/3 normal
with no sex bias
then mutant is
homozygous lethal
If F1 are
1/2 normal females and
1/2 mutant males
then mutant is sex linked and recessive
Start Here
Normal male X Mutant female
If F1 offspring are all normal with no
sex bias, then mutant may be recessive.
To verify this cross two F1.
F1 normal X F1 normal
If F2 are
3/4 normal & 1/4
mutant with no sex
bias, then mutant is
recessive
and not sex linked
If F1 are all mutant, then mutant may be
dominant. To verify, cross two F1.
F1 mutant X F1 mutant
If F2 are 3/4 mutant and 1/4 normal with no sex bias,
then mutant is dominant and not sex linked.
If F2 are
3/4 mutant, including both males and females, and
1/4 normal males with no normal females
then mutant is sex linked and dominant.