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bio final answers - Biology 41 FINAL EXAM ANSWER SHEET...

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1 Biology 41 FINAL EXAM December 14, 2005. ANSWER SHEET Question 1. (Total = 10 points) Tomatoes can be tall or dwarf, smooth or peachy, and round or oblate. When a single tall, smooth and round plant was crossed to a group of pure-breeding dwarf, peachy and oblate plants, the following progeny were obtained: 155 tall, smooth, round 33 tall, smooth, oblate 11 tall, peach, oblate 2 tall, peach, round 158 dwarf, peach, oblate 29 dwarf, peach, round 12 dwarf, smooth round 0 dwarf, smooth oblate (Total= 400) a. What would have been the results if the same individual tall, smooth and round plant had been crossed to another group of pure-breeding tall, smooth, round plants? (2 pts) You would get all tall, smooth, and round plants since these traits are dominant and a pure-breeding tall, smooth, round plant would be homozygous for all traits. You know that they are dominant because the original dwarf, peachy, oblate tomato was pure-breeding. If these traits were dominant, all the progeny would have been dwarf, peachy, and oblate. b. If the single individual tall, smooth and round plant used in these crosses was the progeny of a cross between two individuals from pure-breeding lines, what were the phenotypes of the pure-breeding parents? (3 pts) The parents must have been tall, smooth, and round X dwarf, peachy, and oblate. Since you get all categories of progeny, you know that the individual tall, smooth, and round plant used is a heterozygote for all traits. Thus, it must have received a its dominant copies from one pure-breeding parent and its recessive copies from the other pure-breeding parent . c. Draw as precise a genetic map as possible for the three genes. Call the genes “size”, “skin texture” and “shape”. (5 pts) size -- 6.25 m.u. -- skin texture -- 16 m.u. – shape size-skin texture = ((11+12+2+0)/400) X 100 = 6.25 m.u. or centimorgans skin texture-shape = ((33+29+2+0)/400) X 100 = 16.0 m.u. or centimorgans size-shape = ((33+11+29+12+(2X2)+(0X2)/400) X 100 = 22.25 m.u. or centimorgans
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2 Question 2. (Total= 7 points) Marsupial wolves were long thought to be extinct. However, a geneticist recently found a population of the animals in a remote valley on the island of Tasmania. One difference amongst the wolves was that some ran while others hopped. A series of crosses were done in order to try to understand the genetic basis of this behavior. The following results were obtained when single individuals were crossed. Parents Progeny hopping running running X running 23 74 genoypes:____ Rr, Rr (1 pt)__________________________________ hopping X running 48 55 genoypes:_____ rr, Rr (1 pt)_________________________________ running X running 0 53 genoypes:_______ RR, RR or RR, Rr (2 pts)_______________________________ hopping X running 0 41 genoypes :_____ _ rr, RR (1 pt)________________________________ hopping X hopping 51 0 genoypes:_______ rr, rr (1 pt) _______________________________ a. Describe the genetics of these traits. Be sure to name each gene you use in your description and indicate which alleles(s) are recessive and dominant. One autosomally inherited gene: R dominant, r recessive (1 pt for identifying that this is a single autosomal dominant/recessive gene. Any gene names are acceptable)b. Indicate as much as possible about the genotype of the parental wolves in each cross by writing them in below the parental phenotypes above.
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