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an eltner form a mustache (M) or beard (m). M IS domina RM Rm rM rm RRmm Rrmm Rm Rrmm RRMM RrMm RrMm rm rrMm rrmm From the cross, the phenotypes should be in the ratio: 3:3:1:1 Total observed number- 48

an eltner form a mustache (M) or beard (m). M IS domina RM Rm rM rm RRmm Rrmm Rm Rrmm RRMM RrMm RrMm rm rrMm rrmm From the cross, the phenotypes should be in the ratio: 3:3:1:1 Total observed number- 48

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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Please solve question 5 and letter B

The pads on cats' feet are either made of rubber (R) or metal (r). R is dominant. Their whiskers can either form a mustache (M) or beard (m). M is dominant. RM Rm rM rm RRMM RrMm Rm RRmm Rrmm RrMm Rrmm rm Irmm From the cross, the phenotypes should be in the ratio: 3:3:1:1 Total observed number= 48 Cat RrMm mates with cat Rrmm (note that the parents' genotypes are NOT identical), and they have kittens: What fraction do you "еxpect?" What # of Here is what you kittens do you "еxpect?"* |3/8 x48=D18 3/8x 48 = 18 "observe." Rubber feet/mustache 3/8 14 Rubber feet/beard 3/8 15 Metal feet/mustache 1/8 178x 48 = 6 10 Metal feet/ beard 1/8 x 48 = 6 Hint: The number you expect is based on how many total kittens there were; you'll have to add up all the actual 1/8 kittens in the next column yourself because I didn't do it for you. a) (4) Fill out the blank columns (2 points each column). b) (3) What is the Chi Square formula? X={(Observed Valve-Expected Value) Expected Value c) (2) how many degrees of freedom in this case? df=n(Number of classes)-1 df=4-1 df=3 (5) Perform a Chi-square test on these data, to see if there is likelihood that the R and M genes are linked. What is your resulting Chi² number? (SHOW WORK, of course) Probabilities d) (3) Using the table below, what is your interpretation and why? Explain how you used the table. df 0. 95 0.90 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001 1 0.004 0.016 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83 2. 0.10 0.21 0.71 1.39 2.41 3.22 4.61 5.99 9.21 13.82 3 0.35 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 16.27 4 0.71 1.06 2.20 3.36 4.88 5.90 7.78 9.49 13.28 18.47 1.15 1.61 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52 1.64 2.20 3.83 5.35 7.23 8.56 10.65 12.59 16.81 22.46 7 2.17 2.83 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32 2.73 3.49 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.13 3.33 4.17 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88 10 3.94 4.87 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59 11 4.58 5.58 8.15 10.34 12.90 14.63 17.28 19.68 24.73 31.26 12 5.23 6.30 9.03 11.34 14.01 15.81 18.55 21.03 26.22 32.91 13 5.89 7.04 9.93 12.34 15.12 16.99 19.81 22.36 27.69 34.53 14 6.57 7.79 10.82 13.34 16.22 18.15 21.06 23.69 29.14 36.12 15 7.26 8.55 11.72 14.34 17.32 19.31 22.31 25.00 30.58 37.70 20 10.85 12.44 16.27 19.34 22.78 25.04 28.41 31.41 37.57 45.32 25 14.61 16.47 20.87 24.34 28.17 30.68 34.38 37.65 44.31 52.62 30 18.49 20.60 25.51 29.34 33.53 36.25 40.26 43.77 50.89 59.70 50 34.76 37.69 44.31 49.34 54.72 58.16 63.17 67.51 76.15 86.66 Fail to reject | Reject at 9.05 level
Transcribed Image Text:The pads on cats' feet are either made of rubber (R) or metal (r). R is dominant. Their whiskers can either form a mustache (M) or beard (m). M is dominant. RM Rm rM rm RRMM RrMm Rm RRmm Rrmm RrMm Rrmm rm Irmm From the cross, the phenotypes should be in the ratio: 3:3:1:1 Total observed number= 48 Cat RrMm mates with cat Rrmm (note that the parents' genotypes are NOT identical), and they have kittens: What fraction do you "еxpect?" What # of Here is what you kittens do you "еxpect?"* |3/8 x48=D18 3/8x 48 = 18 "observe." Rubber feet/mustache 3/8 14 Rubber feet/beard 3/8 15 Metal feet/mustache 1/8 178x 48 = 6 10 Metal feet/ beard 1/8 x 48 = 6 Hint: The number you expect is based on how many total kittens there were; you'll have to add up all the actual 1/8 kittens in the next column yourself because I didn't do it for you. a) (4) Fill out the blank columns (2 points each column). b) (3) What is the Chi Square formula? X={(Observed Valve-Expected Value) Expected Value c) (2) how many degrees of freedom in this case? df=n(Number of classes)-1 df=4-1 df=3 (5) Perform a Chi-square test on these data, to see if there is likelihood that the R and M genes are linked. What is your resulting Chi² number? (SHOW WORK, of course) Probabilities d) (3) Using the table below, what is your interpretation and why? Explain how you used the table. df 0. 95 0.90 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001 1 0.004 0.016 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83 2. 0.10 0.21 0.71 1.39 2.41 3.22 4.61 5.99 9.21 13.82 3 0.35 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 16.27 4 0.71 1.06 2.20 3.36 4.88 5.90 7.78 9.49 13.28 18.47 1.15 1.61 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52 1.64 2.20 3.83 5.35 7.23 8.56 10.65 12.59 16.81 22.46 7 2.17 2.83 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32 2.73 3.49 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.13 3.33 4.17 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88 10 3.94 4.87 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59 11 4.58 5.58 8.15 10.34 12.90 14.63 17.28 19.68 24.73 31.26 12 5.23 6.30 9.03 11.34 14.01 15.81 18.55 21.03 26.22 32.91 13 5.89 7.04 9.93 12.34 15.12 16.99 19.81 22.36 27.69 34.53 14 6.57 7.79 10.82 13.34 16.22 18.15 21.06 23.69 29.14 36.12 15 7.26 8.55 11.72 14.34 17.32 19.31 22.31 25.00 30.58 37.70 20 10.85 12.44 16.27 19.34 22.78 25.04 28.41 31.41 37.57 45.32 25 14.61 16.47 20.87 24.34 28.17 30.68 34.38 37.65 44.31 52.62 30 18.49 20.60 25.51 29.34 33.53 36.25 40.26 43.77 50.89 59.70 50 34.76 37.69 44.31 49.34 54.72 58.16 63.17 67.51 76.15 86.66 Fail to reject | Reject at 9.05 level
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