A student carries out a genetics experiment with fruit flies to investigate the inheritance pattern of the white eye trait. The
student crosses a homozygous white-eyed female with a wild-
type male and records observations about the flies in the F, generation. The student plans to use the F, data to perform a chi-square goodness-of-fit test for a model based on an X-linked
recessive pattern of inheritance. The student will use one degree of freedom and a significance level of p = 0.05. The setup for the student's chi-square goodness-of-fit test is presented in
Table 1
Table 1. Setup for the student's chl-square goodness-of-fit test
Phenotype
Red-eyed female
White-eyed male
Observed
53
47
Expected
50
50
The student calculates a chi-square value of 0.36. Which of the following statements best completes the student's chi-square goodness-of-fittest?
type male and records observations about the flies in the F, generation. The student plans to use the F, data to perform a chi-square goodness-of-fit test for a model based on an X-linked
recessive pattern of inheritance. The student will use one degree of freedom and a significance level of p = 0.05. The setup for the student's chi-square goodness-of-fit test is presented in
Table 1
Table 1. Setup for the student's chl-square goodness-of-fit test
Phenotype
Red-eyed female
White-eyed male
Observed
53
47
Expected
50
50
The student calculates a chi-square value of 0.36. Which of the following statements best completes the student's chi-square goodness-of-fittest?
1 answer:
Full question attached
Answer:
c
Explanation:
To carry out the Chi squared test the following formula is used:
(O - E)² / E
Where O is the observed values and E is the expected values. The sum of this equation for each phenotype is the Chi squared value
<h3>Red eyed female</h3>Observed: 53
Expected: 50
O - E= -3
(O - E)² = 9
(O - E)² / E = 9/ 50 = 0.18
<h3>White Eyed Male</h3>Observed: 47
Expected: 50
O - E= 3
(O - E)² = 9
(O - E)² / E = 9/ 50 = 0.18
Sum = 0.36
The p value is dependent upon the number of degrees of freedom. The degrees of freedom is the number of categories in the experiment - 1. So here, there is one degree of freedom are there are two phenotypes being observed.
If we look up the critical values for each degree of freedom, we can see that for 1 degree of freedom the critical value is 3.841
Since the chi squared value is less than 3.84 (it is 0.36), we cannot reject the null hypothesis. So the answer is c
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Explanation:
The given data is as follows.
m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)
Initial temp. of block 1, = (100 + 273) K = 373 K
Initial temp. of block 2, = (0 + 273) K = 273 K
So, heat released by block 1 = heat gained by block 2
Convert temperature into kelvin as (50 + 273) K = 323 K.
Also, we know that the relation between enthalpy and temperature change is as follows.
=
= 1243550 J
or, = 1243.5 kJ
Now, calculate entropy change for block 1 as follows.
=
=
= -554.12 J/K
Now, entropy change for block 2 is as follows.
=
=
= 647.49 J/K
Hence, total entropy will be sum of entropy change of both the blocks.
= -554.12 J/K + 647.49 J/K
= 93.37 J/K
Thus, we can conclude that for the given reaction is 1243.5 kJ and
is 93.37 J/K.
Answer: The reaction is not at equilibrium and will proceed to make more products to reach equilibrium.
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as
K is the constant of a certain reaction when it is in equilibrium, while Q is the reaction quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
The expression for is written as:
Given : = 54.8
Thus as , the reaction will shift towards the right i.e. towards the product side.