Thursday, December 10, 2009

Fruit Fly Genetics

An interactive website on fruit flies:
http://www.sciencecourseware.org/vcise/drosophila/Drosophila.php?guestaccess=1

Order two sets of flies:
1 - just noting specific eye color for the parents (this will be a monohybrid cross)
2 - order and note a wild type for eye color and wing shape and a mutant for both eye color and wing shape.

Example data for the dihybrid cross:
White eyes noted by “W” & vestigial wings by “vg”both are mutations and recessive
•Parental cross =
–wild type (X+X+++) female
–and XWYvgvg male
•F1
–607 wild type female
–603 wild type males
•F2
–458 wild type female
–225 wild type males
–151 red eyes and vestigial wings female
–88 red eye and vestigial wings male
–215 white eyes and normal wings male
–67 white eyes and vestigial wings male


RECORD all data for parents, F1 and F2 generation....

The Analysis?

Chi Square Analysis

•If the chi square value is higher than the critical value, the null hypothesis is rejected. If probability is lower than critical probability (we’re told is .05) the null hypothesis is also rejected.
•The probabilities given on the table is
•The probability the observed and expected results being EXACTLY the same.

Lab #7 Genetics of Organisms Post Lab Analysis worth 50 points

Date

Observations/ Tables

Questions

Total pts possible:

50 pts total

Week 1

Draw the possible parental flies and label their phenotypic characteristics.

We probably never got to “observe” these, only their offspring (the F1 generation)

What does a Wild-type fly look like?

5pts

Week 2

Table 7.1 notations of F1 sex & corresponding mutations.

Label your data found in each vial separately.

Read through #1-5 and consider implications of data thus far…

4pts

Week 3

N/A

Week 4

Table 7.2 notations of F2 sex & corresponding mutations.

#1-7 AND discussion questions #1-3

Table 7.2 4pts

#1-7 – 21pts

#1-3 – 6 pts

Statistical Analysis

Example #1 (table 7.3 & 7.4)

Practice problems

a-c and (i) & (ii)

10pts

Conclusion Questions #1-7 (28 points)

  1. Description of the mutations possible for each vial.

Vial 1 – (eyes and sex-linked)

Vial 2 – eyes and wings (dominantly inherited and linked traits)

  1. Provide 2 null hypothesis (If…., then….)

A null hypothesis is what you expected (based on the punnett square)

  1. Punnett Squares: Your expected data for BOTH vials and BOTH generations

Vial 1 – Eyes only (sex-linked)

What must the parents’ genotypes have been? Use the F1 offspring:

Vial 2 – Eyes and wings (linked; therefore, a dihybrid cross)

Predict parents to get F1 offspring inside Use F1 as parents & F2 offspring inside

  1. Table

Vial and generation

Phenotypic ratio

Genotypic ratio

Vial 1 – F1

Vial 1 – F2

Vial 2 – F1

Vial 2 – F2

  1. Do the actual (observed) deviated from the expected (punnett squares)?

(for both vial 1 & 2)

6. Add an extra column to include both vial 1 and 2

  1. Do the full chi square analysis for BOTH vials. Show all your work.
    1. Degrees of freedom for sex-linked eyes – (n-1)
    2. Degrees of freedom for linked eyes and wings – (n-1)

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