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MidtermIF08key - BIS 101-003(Engebrecht Fall 2008 Name Last...

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BIS 101-003 (Engebrecht) Name: ______________________________ Fall 2008 Last, First ID #:_______________________________ BIS 101-003 - MIDTERM IA This exam has a total of 100 points. READ THE QUESTIONS CAREFULLY BEFORE YOU ANSWER. SHOW ALL WORK TO GET FULL CREDIT. Please sign your name on the score sheet. This exam has eight pages including this cover page and the score sheet HONOR CODE: My signature below affirms that I wrote this exam in the spirit of the honor system of the University of California, Davis. I neither received nor furnished any “help” during the exam, nor did I use any unauthorized references. Signature: ______________________________________________________ Good luck Authorization for Public Distribution of Graded Examination I, ______________________________________, ID# _______________________ authorize that after this exam is graded, it can be placed in a holding bin, labeled BIS 101-021/Engebrecht, on the first floor of Briggs Hall.
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1. (20pts) You are a Drosophila geneticist and notice two different fly variants in your bottles: one fly has yellow-eyes and the other fly has stubby wings. You are interested in determining the nature of these characteristics. You propagate the flies to obtain true breeding lines. You then cross a true breeding yellow-eyed straight winged female to a true breeding red-eyed stubby winged male. You find that all the F1 flies have red-eyes and straight wings. a. (4pts) Define alleles of these characters and show genotype of parents (P) and offspring (F1). y+=red; y=yellow; st+=straight; st=stubby (other allele designations are fine, as long as consistent with dominance and recessive relationship of indicated phenotypes) (1pt) P: yyst+st+ x y+y+stst (1pt for each correct parent) F1: y+yst+st (1pt) b. (4pts) You perform the reciprocal cross (red-eyed stubby winged females x yellow-eyed straight winged males) and obtain red-eyed straight winged females and red-eyed stubby winged males. Please draw cross and explain this result. P: y+y+XstXst x yyXst+Y (2pts) F1: y+yXstXst+ and y+yXstY (2pts) Because there is a difference in the behavior of the reciprocal cross, the stubby wing must reside on the X chromosome. c. (4pts) The F1s from b above are crossed to obtain F2s. What proportion of the F2 red-eyed straight winged flies are true breeding? There are 6 red-eyed straight winged flies in the F2, of those only 1 is true breeding or homozygous (y+y+Xst+Y). Therefore 1/6. (4pts) [2pts for 6/16 or 3/8] d. (4pts) What proportion of the F2 yellow-eyed stubby winged flies are true breeding (from the same cross)? There are 2 yellow-eyed stubby winged flies in the F2 and both are
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MidtermIF08key - BIS 101-003(Engebrecht Fall 2008 Name Last...

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