Jenna Rdz Lab 3 - Meiosis - Recombination and mapping - Worksheet key (S23) (1)

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Dec 6, 2023

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Lab 3 – Meiosis/Recombination Worksheet part 1 You will be performing simulated crosses between Drosophila (fruit flies), to determine which genes are linked on the same chromosomes. Drosophila have 4 different chromosomes, but the traits we’re looking at will be located on 3 of those chromosomes. There are 8 traits involved: Yellow body, Black body, Ebony body (Select in Body Color) Purple eyes, Sepia eyes, White eyes (Select in Eye Color) Curved wings (Select in Wing Shape) Crossveinless (Select in Wing Veins) The gene associated with ‘Yellow body’ is on chromosome 1 and has two of the other traits linked to it. Black body is on chromosome 2 and has two traits linked to it. Ebony body is on chromosome 3 and has one trait linked to it. Go to https://www.sciencecourseware.org/FlyLabJS/ to perform the crosses. In each case, the P1 generation will be a wild-type female and a male with two recessive mutant traits. 1) Design – select a female with all wild-type for mating 2) Design male fly with the two mutant traits in question for each cross (see table below) and select for mating. 3) Mate flies 4) All offspring should be wild-type (heterozygous). Now select the female from the F1 generation and select the homozygous recessive mutant from the P1 generation to cross. Mate these flies. 5) Both males and females should show all four possible combinations of wild-type/mutant phenotypes: wild-type/wild-type, wild-type/mutant, mutant/wild-type, mutant/mutant. Click on the ‘Analyze’ tab. 6) You are now going to test if the observed results fit with an expected test cross ratio (1:1:1:1). Click on ‘ignore sex of flies’, then click ‘include a test of hypothesis’. 7) In each box under the ‘Hypothesis’ column, enter a ‘1’ to test the 1:1:1:1 ratio. Click on ‘test your hypothesis’. 8) The chi-square value will be compared to a critical value of 7.815 (p=0.05, df=3). If the ‘Chi- squared test statistic’ is below the critical value, you do not reject the 1:1:1:1 ratio and you assume the genes are ‘unlinked’, i.e. on different chromosomes or distant on the same chromosome. 9) If the ‘Chi-squared test statistic’ is above the critical value, you reject the 1:1:1:1 ratio and you assume the genes are ‘linked’, i.e. on the same chromosome. To obtain the recombination frequency, you add the proportion of the recombinant types (wild-type/mutant, mutant/wild- type – these should be the two middle rows in the table) and multiple by 100. 10) Determine which 2 traits involve genes linked to ‘Yellow Body’ on chromosome 1. White Eyes & Crossveinless

11) Determine which 2 traits involve genes linked to ‘Black Body’ on chromosome 2. Curved wings & Purple Eyes 12) Determine which trait involves a gene linked to ‘Sepia eyes’ on chromosome 3. Ebony Body 13) Summarize your results in the table below. Mutant types Chi-square value (test 1:1:1:1 ratio) Reject Or Not reject Reject = genes are linked Not reject = genes are not linked Recombination Frequency (proportion of the rare recombinant types*) Yellow body / white eyes 925.80 Reject 1.53 Yellow body / curved wings 3.67 Not Reject 51.73 Yellow body / sepia eyes 0.57 Not Reject 50.63 Yellow body / purple eyes 3.80 Not Reject 52.21 Yellow body / crossveinless 645.63 Reject 10.08 Black body / white eyes 1.29 Not Reject 49.02 Black body / curved wings 333.61 Reject 20.05 Black body / sepia eyes 0.73 Not Reject 51.22 Black body / purple eyes 767.30 Reject 6.12 Black body / crossveinless 3.24 Not Reject 52.14 Ebony body / white eyes 1.32 Not Reject 50.82 Ebony body / curved wings 4.67 Not Reject 51.2 Ebony body / sepia eyes 155.56 Reject 30.27 Ebony body / purple eyes 0.69 Not Reject 51.2 Ebony body / crossveinless 4.94 Not Reject 47.01 1pt for each row with correctly identified linked genes and recombination frequency. (5pts) 0.5pt for each row where the genes are correctly identified as being unlinked. (5pts) (+ 1pt for participation and completion) *Recombination Frequency. The output from the simulation gives the proportion of each phenotype produced by the test cross. Take the two middle rows (i.e. the wild-type/mutant and mutant/wild-type phenotypes), add their proportions together and multiple by 100 to obtain the recombination frequency. E.g. Proportion of two middle rows (should be the two rarest classes) 0.006 + 0.004 = 0.01. 0.01 x 100 = 1% A recombination frequency of 1% is equivalent to 1 map unit apart between the two genes on the same chromosome.

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