University:
Marymont University
Course:
BIOL 425 | Genetics
Academic year:

2021
Views:

367

Pages:

36
Author:

Dixie Summers

Numeric Response Practice (AP Review Notebook) Science Practices 1. A plant geneticist is investing the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Citrullus lanatus). Explosive rind is recessive and causes watermelons to burst when cut. Non-bitter watermelons are a result of the recessive genotype (susu). The geneticist wishes to determine if the genes assort independently. She performs a testcross between a bitter/nonexplosive hybrid and a plant homozygous recessive for both traits. The following offspring are produced: BN bitter/non-explosive-88 E Ju su BE bitter/explosive-68 NN non-bitter/non-explosive - 62 e Eeee SUSUSU SVIN NE non-bitter/explosive-81 e Eeee SS Calculate the chi-squared value for the null hypothesis that the two genes assort independently. Give your answer to the nearest hundredth. phenotype o 0 e o-e (0-0)2 o-ele BN 88 74.15 13,25 175,563 2,349 BE 68 74.75 -6.25 45.563 0.610 NN 62 74.75 -12.75 162.563 2.175 NE 81 74.75 6.25 39.063 0.523 299 X²= 5.657 = 5.66 2. Results of a Drosophila mating between Fi flies resulted in 58 flies showing red eyes (a dominant trait) and 42 flies showing sepia eyes (recessive). Calculate the Chi-square value for the hypothesis that both Fi flies were heterozygous for eye color. Give your answer to the nearest tenth. Rr RRRR Red rRrrr & sepia phenotype 이 e o-e (ef (o-ele Red 58 75 -17 289 3.853 42 25 17- 289 11.56 sepia 100 2 X²=15.413 = 15.4Rr XX R XY 3. A hetero red eyed female was crossed with a red eyed male. The results are shown below. Red eyes are sex-linked dominant to white, determine the chi-square value. Round to the nearest. hundredth Phenotype #flies observed Red Eyes 134 White Eyes 66 R XY XXXXRY Red & white XXXXY (oe) (o-e) (o-e)le phenotype Ω e Red 134 150 -16 256 white 66 50 16 256 1.707 5,12 200 X=6.827= 6.83 4. In corn (Zea mays), purple kernels (R) are dominant to yellow kernels (r). Cobs from the offspring of a cross between a purple plant and yellow plant were used in a lab. A student counts 329 purple and 299 yellow kernels on one cob. Rr Calculate the chi-squared value for the null hypothesis that the purple parent was heterozygous for purple kernels. Give your answer to the nearest hundredth. Rr purple ½ yellow phenotype 이 e (o-e) (o-e)² 2 (o-e)le purple 329 314 15 225 0.717 yellow 299 314 -15 225 0.717 628 x=1.434=1.43 32. What is the water potential for a solution that is 0.1 M? The solution is in an open container. The equation for water potential is: water potential (4) pressure potential (P) + solute potential () The equation for solute potential is: Y=-iCRT Ys=-0.24 where: C-molar concentration i ionization constant (assume that it is 1) R-pressure constant (R-0.00831 liter MPa/mole K) This is different than Ap formula sheet Ttemperature in Kelvin (assume a room temperature of 293 K) 4-(1)(0.1M) (0.00831 LMP4/molk) (293K) = 0.243 33. Four blocks of phenolphthalein agar are placed in a vinegar solution. The pl indicator solution changes to pink when in contact with an acidic solution, Block A: 2 cm x 4 cm x 4 cm Surface Area: 64 Volume: 32 Surface Area/Volume: 2 SA=2(2×4)+2(2x4)+2(4x4) = 64 V= 2x4x4 = 32 #=64/32-2 SA 34. Four blocks of phenolphthalein agar are placed in a vinegar solution. The pH indicator solution changes to pink when in contact with an acidie solution, Block B: 2 cm x 8 cm x 4 cm Surface Area: 112 Volume: 64 Surface Area/Volume: 1.78 SA=2(2x8)+2(8x4)+2(2×4) = 32+64+ 16 112 V= 2x8x4 = 64 =112/64=1.75 35. Four blocks of phenolphthalein agar are placed in a vinegar solution. The pH indicator solution changes to pink when in contact with an acidic solution, Block C: 1 cm x 8 cm x 8 cm Surface Area: 160 Volume: 64 Surface Area/Volume: 2.5 SA2(1×8)+2(1×8)+2(848) 16+16+128 = 160 V=7x8x8 = 64 #=160/64=2.5 36. Four blocks of phenolphthalein agar are placed in a vinegar solution. The pH indicator solution changes to pink when in contact with an acidic solution, Block D: 1 cm x 1 cm x 64 cm Surface Area: 258 Volume: 64 Surface Area/Volume: 4.03 SA-2(1x1)+2(1x64)+2(1x64) =2+128+128 258 V= 1x1x64 =64 -258/64-4.03 SA V a. Which block would the vinegar solution penetrate most thoroughly into after ten minutes? A=2 B=1.75 C-2.5 D = 4.03 *D because it has greatest Salv * b. Which block would have the greatest volume of pink phenolphthalein at the end of ten minutes? V | A = 32 B=64 C=64 D=64 * A because it has smallest volume * Unit 3: Cell Energetics 37. The following chart shows the energy products produced in various stages of the breakdown of glucose. Process ATP Produced NADH/FADH2 Produced Glycolysis 2 2 NADH Pyruvate Oxidation (per molecule of pyruvate) INADH Citric Acid (Krebs) Cycle per molecule of pyruvate 1 3 NADH 1 FADH: Each molecule of NADH results in approximately 2.5 molecules of ATP, whereas cach molecule of FADH: results in approximately 1.5 molecules of ATP when these molecules are fed into the electron transport chain. What is the difference in the total number of ATP molecules produced between three molecules of glucose that undergo fermentation compared to three molecules of glucose that undergo aerobic respiration?

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