Question

In: Biology

A female with Muppetrus bristle mates with a male with Rheingoldenbach body color, producing the following...

 A female with Muppetrus bristle mates with a male with Rheingoldenbach body color,
producing the following counts:

F1 Generation
Males
WT-WT   Disease-WT      WT-Disease      Disease-Disease
260     252             0               0

Females
WT-WT   Disease-WT      WT-Disease      Disease-Disease
258     255             0               0

I have determined that at locus 1, the mode of inheritance is sex linked and at locus 2, the mode of inheritance is autosomal recessive.

Could someone help me determine the expected number of individuals that will have the following phenotypes, and fill out the E column.

Chi Square test:

Gender

Two locus phenotype

Observed (o)

Expected (E)

(o-E)^2 /E

Female

WT/ Disease

0

Disease/Disease

0

Disease/ WT

255

WT/WT

258

Male

Disease/WT

252

WT/disease

0

WT/WT

260

Disease/Disease

0

Total

1025

DF=

P value=

Solutions

Expert Solution

Expected value ( Total of observed values/total number):

               ( 1025/8 )

  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125

Observed - expected

  • -128.125
  • -128.125
  • 126.875
  • 129.875
  • 123.875
  • -128.125
  • 131.875
  • -128.125

( Observed - Expected )2

  • 16,416.015625
  • 16,416.015625
  • 16,097.265625
  • 16,867.515625
  • 15,345.015625
  • 16,416.015625
  • 17,391.015625
  • 16,416.015625

( Observed - Expected )2/exp

  • 128.125
  • 128.125
  • 125.637
  • 131.648
  • 119.765
  • 128.125
  • 135.734
  • 128.125

Total = 1,025.284

DF ( degree of freedom ) = N-1

     N=8

DF = N-1 = 8-1 = 7

P value = It depends upon hypotheseis

HO: u = 0

Ha: u is not equal to zero

P value depends on the Z table

You can see the result of p value on the Z table. You have a degree of freedom 7 and at the point of 7 you can check the p value.Expected value ( Total of observed values/total number):

               ( 1025/8 )

  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125
  • 128.125

Observed - expected

  • -128.125
  • -128.125
  • 126.875
  • 129.875
  • 123.875
  • -128.125
  • 131.875
  • -128.125

( Observed - Expected )2

  • 16,416.015625
  • 16,416.015625
  • 16,097.265625
  • 16,867.515625
  • 15,345.015625
  • 16,416.015625
  • 17,391.015625
  • 16,416.015625

( Observed - Expected )2/exp

  • 128.125
  • 128.125
  • 125.637
  • 131.648
  • 119.765
  • 128.125
  • 135.734
  • 128.125

Total = 1,025.284

DF ( degree of freedom ) = N-1

     N=8

DF = N-1 = 8-1 = 7

P value = It depends upon hypotheseis

HO: u = 0

Ha: u is not equal to zero

P value depends on the Z table

You can see the result of p value on the Z table. You have a degree of freedom 7 and at the point of 7 you can check the p value.

gladiator answered 1 year ago
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