LORD OF THE FLIES: Virtual Fruit Fly Genetics Lab, part 2
Di-Hybrid Crosses – you will be looking at 2 traits at the same time. Your goal will be to determine whether two genes are on the same chromosome or not, and how far apart they are. Before you set out to solve the problem, let’s do a warm up cross.
Warm up Cross. Cross a wild type female+/+ with a
male – design – Bristles -spineless(SS) and Body color – black(BL).
Based on your F1 results make a claim about how the black body and spineless bristles are inherited (e.g. autosomal, dominant, etc.). Justify your claim.
What are the genotypes of the F1 flies?
Using 2X2 punnett squares and the multiplication rule calculate the probability/proportions of the possible phenotypes of the F2 flies from a F1xF1 cross.
Body Color:
Bristles:
Phenotypes Probability:
+/+ _______ +/BL _________SS/+ __________ SS/BL ________
Cross the F1 female to the F1 male. And click on the Analyze tab. Check Ignore sex of Flies box only if you are certain none of the traits are sex linked. Check the box Include a test of hypothesis and fill out the expected values based on your answers above. Don’t forget to look at the total number of your flies (different each time you do a cross) in the table and use it. Make sure your expected numbers add up to the total number of flies, you can round to the nearest tenth. Record your results below.
Degrees of freedom _______________________ Chi-Square Value ______________
Look at the table and make a statement about the Null hypothesis
Next -the problem you are going to solve. Where are these genes located, different chromosomes or same chromosome. If they are on the same chromosome, how far apart are they?
Your job is to create a linkage map of the following traits, all of which are autosomal recessive
purple eye color
vestigial wing shape
spineless bristles
black body color
To do this you will need to cross a double mutant fly with a wild type fly,
then create an F2 generation from the offspring.
The F2 generation numbers will be tested using Chi-Square for the Null hypothesis – no difference from expected mendelian ratios if the genes are inherited independently.
If you reject the Null hypothesis, you will need to identify the recombinant offspring and calculate recombination frequencies, aka map distances.
Note: You must cross every trait with every other trait – this is 6 crosses in total.
But, here is the good news, you have already performed one of the crosses, your warm up cross. Simply transfer the data to the first table.
The 1st column of the table is filled out for you.
Cross #1: _____Black Body/Spineless Bristles____
P generation parents:_____________________________
F2 Generation Results:
Phenotype Ratio expected if genes are not linked Ratio expected if genes are linked Observed frequencies
(proportion column)
+, + 9/16= 0.56 +, _BL_ 3/16= 0.19 _SS_, +3/16=0.19 _BL_, _SS__ 1/16= 0.06 Are these two genes on the same chromosome? ______
If yes, what % of the offspring are recombinant? ____________
If no, support your answer with a Chi Square test
X2 = ________________________ Df = _______________________
Critical X2 value on table for p=0.05 = _________________________
Cross #2: P generation parents:_____________________________
F2 Generation Results:
Phenotype Ratio expected if genes are not linked Ratio expected if genes are linked Observed frequencies
(proportion column)
+, + +, _____ _____, + _____, _____ Are these two genes on the same chromosome? _____________
If yes, what % of the offspring are recombinant? ____________
If no, support your answer with a Chi Square test
X2 = ________________________ Df = _______________________
Critical X2 value on table for p=0.05 ______________
Cross #3: P generation parents:_____________________________
F2 Generation Results:
Phenotype Ratio expected if genes are not linked Ratio expected if genes are linked Observed frequencies
(proportion column)
+, + +, _____ _____, + _____, _____ Are these two genes on the same chromosome? _____________
If yes, what % of the offspring are recombinant? ____________
If no, support your answer with a Chi Square test
X2 = ________________________ Df = _______________________
Critical X2 value on table for p=0.05 ______________
Cross #4: P generation parents:_____________________________
F2 Generation Results:
Phenotype Ratio expected if genes are not linked Ratio expected if genes are linked Observed frequencies
(proportion column)
+, + +, _____ _____, + _____, _____ Are these two genes on the same chromosome? _____________
If yes, what % of the offspring are recombinant? ____________
If no, support your answer with a Chi Square test
X2 = ________________________ Df = _______________________
Critical X2 value on table for p=0.05 ______________
Cross #5: P generation parents:_____________________________
F2 Generation Results:
Phenotype Ratio expected if genes are not linked Ratio expected if genes are linked Observed frequencies
(proportion column)
+, + +, _____ _____, + _____, _____ Are these two genes on the same chromosome? _____________
If yes, what % of the offspring are recombinant? ____________
If no, support your answer with a Chi Square test.
X2 = ________________________ Df = _______________________
Critical X2 value on table for p=0.05 ______________
Cross #6: P generation parents:_____________________________
F2 Generation Results:
Phenotype Ratio expected if genes are not linked Ratio expected if genes are linked Observed frequencies
(proportion column)
+, + +, _____ _____, + _____, _____ Are these two genes on the same chromosome? _____________
If yes, what % of the offspring are recombinant? ____________
If no, support your answer with a Chi Square test
X2 = ________________________ Df = _______________________
Critical X2 value on table for p=0.05 ______________
Summarize your results in the table to help you construct a map of the 4 traits. Use N/A for map units if you think 2 genes are NOT on the same chromosome.
Gene/map units Bl SS VG PR
BL SS VG PR Using the table construct a map of the 4 traits. You can draw by hand and REPLACE the image. See sample map. In the sample map 4 traits a,b,c,d are placed on 2 different chromosomes (lines). The distance between genes on the same chromosome are map units (m.u.) = recombination frequency. Note, because recombination frequencies vary and are not 100% precise, the distance between d and b and between b and c may not add up precisely to the distance between d and c, but should be adding up – approximately/roughly.
REPLACE your drawing here.