AP Lab 7 Genetics of Organisms

1
AP Lab 7Genetics of Organisms

• Virtual Fly Lab

2
Instructions

• Instructions
• 1. Find the Fly Lab Manual in your box of
  materials use the access code found in the manual
  to register for the lab at www.biologylabsonline.c
  om
• 2. Read through the introduction and objectives
  for the lab.
• 3. Then click START LAB

3
Your Choice

• Monohybrid cross
• Create a monohybrid cross between a wild-type
  female and male fly with sepia-colored eyes.
  Sepia is a nonlethal recessive allele located on
  Chromosome III.

• Dihybrid Cross
• Design a dihybrid cross by selecting and crossing
  an ebony body female fly with a male fly that has
  the vestigial mutation for wing size.

Sex-linked cross Design and perform the following
crosses to examine the inheritance of sex-linked
alleles in Drosophila. Cross a female fly with a
tan body with a wild-type male.
4
Now What?

• Based on what you know about the principles of
  Mendelian genetics, predict the phenotypic ratio
  that you would expect to see for the F1 offspring
  of this cross and describe the phenotype of each
  fly.
• (In other words, use a punnett square)
• Select 10,000 flies as the number of offspring
  for this mating. Click on the Mate button.

• Scroll down to see the wild type offspring.
  Record this on your data sheet. These offspring
  are the F1 generation.
• Note The actual number of F1 offspring created
  by FlyLab does not exactly equal the 10,000
• offspring that you selected. This difference
  represents the experimental error introduced by
  the FlyLab.
• Are the phenotypes of the F1 offspring what you
  would have predicted for this cross? Why or why
  not?

5
Next Step

• To set-up the F2 generation, click Select under
  the F1 female and Click Select under the F1
  male. And mate
• them.
• Record F2 offspring use the scroll buttons to
  view all the phenotypes of the F2 offspring.
• Are phenotypes of the F2 offspring what you would
  have predicted for this cross? Why or why not?

• From pg. 85 of AP Biology Lab Manual
• Statistics can be used to determine if
  differences among groups are significant or
  simply the result of predictable error. The
  statistical test most frequently used to
  determine whether data obtained experimentally
  provide a good fit or approximation to the
  expected or theoretical data is the
• chi square test.

6
kahy-skwair

• Determine the number of degrees of freedom (df)
  based on the number of different phenotypes minus
  1.
• The null hypothesis states that there is no
  statistically significant difference between the
  observed and expected data.
• If our cross resulted in a phenotype ratio that
  was expected, we would seek to accept our null
  hypothesis.
• If our
• actual data do not fit the expected ratios, then
  we would reject the null hypothesis.
• The minimum probability for rejecting a null
  hypothesis is 0.05, so use this row in the
  chi-square table provided
• in your manual. These results are said to be
  significant at a probability of p0.05, which
  means that only 5 of the
• time would you expect to see similar data if the
  null hypothesis was correct.
• Fill in the chi-square chart on your data sheet
  with the data for the F2 cross.

7
Chi What?

• Lets do a sample analysis.

8
Suppose you preformed a simple monohybrid cross
between two individuals that were heterozygous
for the trait of interest.

• Results of a monohybrid cross between two
  heterozygotes for the 'a' gene.

Our expected numbers due to our prediction
through a punnett square
25 AA 50 Aa 25 aa
10 AA 75 Aa 15 aa
The phenotypic ratio 85 of the A type and 15 of
the a-type (homozygous recessive). In a
monohybrid cross between two heterozygotes,
however, we would have predicted a 31 ratio of
phenotypes. In other words, we would have
expected to get 75 A-type and 25 a-type. Are our
results different?
9
Calculate the chi square statistic x2 by
completing the following steps 1. For each
observed number in the table subtract the
corresponding expected number (O E). 2. Square
the difference (O E)2 . 3. Divide the
squares obtained for each cell in the table by
the expected number for that cell (O - E)2 / E
. 4. Sum all the values for (O - E)2 / E. This
is the chi square statistic.
10
We now have our chi square statistic (x2 5.33),
our predetermined alpha level of significalnce
(0.05), and our degrees of freedom (df 1).
Entering the Chi square distribution table (pg.
86) with 1 degree of freedom and reading along
the row we find our value of x2 5.33) lies
between 3.841 and 5.412. The corresponding
probability is 0.05ltPlt0.02. This is smaller than
the conventionally accepted significance level of
0.05 or 5, so the null hypothesis that the two
distributions are the same is rejected. In other
words, when the computed x2 statistic exceeds the
critical value in the table for a 0.05
probability level, then we can reject the null
hypothesis of equal distributions. Since our x2
statistic (5.33) exceeded the critical value for
0.05 probability level (3.841) we can reject the
null hypothesis that the observed values of our
cross are the same as the theoretical
distribution of a 31 ratio.
11
You Can Do It!