Homework Assignment #2 Answers

1. a) P(P) = 3/4
   a) P(p) = 1/4
   a) N = 10
   a)
   
   a) r = # purple flowered plants in family of size 10
   

Purple	White	Probability
10	0	[3628800/(3628800)(1)][(3/4)(1/4)]	= 0.056000
9	1	[3628800/(362880)(1)][(3/4)(1/4)]	= 0.188000
8	2	[3628800/(40320)(2)][(3/4)(1/4)]	= 0.282000
7	3	[3628800/(5040)(6)][(3/4)(1/4)]	= 0.250000
6	4	[3628800/(720)(24)][(3/4)(1/4)]	= 0.146000
5	5	[3628800/(120)(120)]\(3/4)(1/4)]	= 0.058000
4	6	[3628800/(24)(720)][(3/4)(1/4)]	= 0.016000
3	7	[3628800/(6)(5040)][(3/4)(1/4)]	= 0.000100
2	8	[3628800/(2)(40320)][(3/4)(1/4)]	= 0.000040
1	9	[3628800/(1)(362880)][(3/4)(1/4)]	= 0.000030
0	10	[3628800/(1)(3628800)][(3/4)(1/4)]	= 0.000001
			1.0

1. b) P (at least one purple plant in family size eight)
   
   b) = 1 - P(eight white plants)
   
   b) == 1 -1 (1)(0.0000153)
   b)                                 wq  = 1 - 0.0000153
   b)                                 wq  = 0.999985
   
   
2. Test Ho:1:1 ratio for the A locus
   n = 49
   
   

class	Aabb	aabb
observed	27.0	22.0
expected	24.5	24.5
deviation	2.5	-2.5


X = S(| O - E| - 1/2) = (2.5 - 0.5) + (2.5 - 0.5)
                 E                  24.5            24.5
X = S(| O - E| - 1/2) = 0.163 + 0.163
X = S(| O - E| - 1/2) = 0.326

Critical X for 1df, a = 0.05 is 3.84


Test Ho:1:1 ratio for the B locus
n = 257



class	aaBb	aabb
observed	120	137.0
expected	128.5	128.5
deviation	-8.5	8.5


X = S (O-E) = (-8.5) + (8.5) = 0.562 + 0.562
            E     w 128.5     128.5

                                            = 1.125


Test Ho:15:1 ratio
n = 300



class	A_B_	aabb
observed	275.00	25.0
expected	281.25	18.75
deviation	-6.5	6.25


X = S(|O - E|- 1/2) = (|-6.25| -0.5)+(6.25 - 0.5)
                 E                     281.25           18.75

                           W= 0.1176 + 1.763
                           W= 1.88