Solving Genetics Probability Problems

In probability method, we have to decide the basic terms of probability.. The main aim of genetics probability subject is to give the fundamental properties the importance of their use with some basic examples of genetics. In a genetics probability, the basic term and properties is not easier to understand but work with sums we will start to get comfortable with in genetics probability.


Basic Definitions of genetics probability:

In a genetic probability method, an testing method is continuously repeated for infinite numbers of time and it happened by the expected probability in the genetics

Genetics probability solving problems:

Problem 1:

To solve consider a locus with two chromosome, C and c. If the frequency of CC is 0.64, what is the frequency of A under Hardy-Weinberg?

Solution:

Solving methods of genetic probability,

Under H-W, if q = freq(C) , then q2 = freq(CC), hence q2 =0.64 or q = 0.8.

Problem 2:

To solve if the genotypes CC, Cc, and cc have frequencies 0.7, 0.49, and 0.49 (respectively), what are q = freq (C)? r = freq (c)? After a single generation of random mating, what is the expected frequency of CC? Of Cc? Of cc?

Solution:

Solving methods of genetic probability,

p = freq (CC) + (1/2) freq (Cc) = 0.7 + (1/2)(0.49) = 0.945

q = 1-p = 0.945

freq (CC) = p2 = 0.9452 = 0.8930

freq (Cc) = 2qr = 2*0.945*0.8930 =1.6877

freq (cc) = r2= 0.89302= 0.797449

Problem 3:

To solve the suppose 60 out of 850 women’s are green eyes. What is frequency of green eyes? If a random women’s is chosen, what is the probability they are a green eyes?

Solution:

Solving methods of genetic probability,

Freq (Redheads) = 60/850 = 0.070 or 7.05 percent

Probability of a Redhead = 30/850, or 7.05 percent

Problem 4:

Solve the genotype cc is lethal and yet population has an equilibrium frequency for c of 30.  Here the Cc is the fitness and its value is one.  Find the CC genotype fitness?

Recall if the genotypes CC: Cc: cc have fitness 1-s : 1 : 1-t, then the equilibrium frequency of A is s/(r+s).

Here, s= 1 (as cc is lethal), so that freq(C) = 1- freq(c) =0.6 = 1/(1+s)

Solving gives 1+s = 1/0.8, or s = 1/0.8 -1 = 3.5

Hence fitness of CC = 1-s = 1-3.5=-2.5.

Learn Derivative at a Point

Learn derivative at a point involves the process solving derivative problems at an exam point view. The rate of change of the given function is determined with the help of calculus. To find the rate of change the calculus is divided into two types such as differential calculus and integral calculus. At exam point of view the derivative is easily carried out by differentiating the given function with respect to input variable. The following are the solved example problems in derivatives at exam point of view for learn.

Learn derivative at a point example problems:


Example 1:

Find the derivative for the given differential function.

f(e) = 3e 3 -4 e 4  - 5e

Solution:

The given function is

f(e) = 3e 3 -4 e 4  - 5e

The above function is differentiated with respect to e to find the derivative

f '(e) = 3(3e 2 )-4(4e 3  ) - 5

By solving above terms

f '(e) = 9e 2 -  8e 3 - 5

Example 2:

Compute the derivative for the given differential function.

f(e) = 6e6 - 5 e5 - 4 e4 - e

Solution:

The given equation is

f(e) = 6e6 - 5 e5 - 4 e4 - e

The above function is differentiated with respect to e to find the derivative

f '(e) =  6(6e 5)  -5 (5 e4 ) -4(4 e3) - 1

By solving above terms

f '(e) =  36e 5  - 25 e4  - 16 e3 - 1

Example 3:

Compute the derivative for the given differential function.

f(e) = 2e 2 -4 e 4  - 15

Solution:

The given function is

f(e) = 2e 2 -4e 4  - 15

The above function is differentiated with respect to e to find the derivative

f '(e) = 2(2e  )-4(4 e 3 ) - 0

By solving above terms

f '(e) = 4e -16e3

Example 4:

Compute the derivative for the given differential function.

f(e) = 5e5 -4e 4 -3e 3  - 2

Solution:

The given function is

f(e) = 5e5 -4e 4 -3e 3  - 2

The above function is differentiated with respect to e to find the derivative

f '(e) = 5(5e 4 )-4(4e 3 ) -3( 3e 2) -0

By solving above terms

f '(e) = 25e 4 -16e 3  -9 e 2


Learn derivative at a point practice problems:


1) Compute the derivative for the given differential function.

f(e) = 2e 3 -3 e 4  - 4 e 5

Answer: f '(e) = 6e 2 -12 e3 - 20 e 4

2) Compute the derivative for the given differential function.

f(e) = e 3-e5 - 4 e 6

Answer: f '(e) = 3e2 - 5e4 - 24 e 5

Linear Algebra Test

Linear function is a function with polynomial degree 1. This linear function matches a dependent variable with independent variable and maintains a relation in a simpler way. The linear algebra is the important method which is used in the linear function, variable function, linear equation, vector, matrix etc. This linear function is used in the high school and college students are using the linear function by using the linear algebra.

Sample Problem for linear algebra:


f is a linear function. Value of x and f(x) are given in the table below; complete the table using linear algebra

X                           f(x)

-1                          11

1                           –

–                           1

6                          -12

7                         --

–                          -35

Solution:

f is a linear function in linear algebra whose formula has the form to test is

f(x) = a x + b

where a and b are constants to be test and found. Note that 2 ordered pairs (-2,18) and (5,-19) are given in the table. These are two ordered pairs to test are used to write a system of linear equations as follows

11 = - 1 a + b and -12 = 6 a + b

Solve the above system to obtain a = - 3 and b = 2 and write the formula for function f as follows

f(x) = - 3 x + 2

We now use the formula for the f to find f(x) given x or find x given f(x).

for x = 0 , f(0) = -3(0) + 2 = 2

for f(x) = 1 , 1 = -3 x + 2 which gives x = 1/3

for x = 7 , f(7) = -3(7) + 2 = - 19

for f(x) = - 30 , -30 = -3 x + 2 which gives x = 32/3

We now put the values of the calculated above in the answer.

X                           f(x)

-1                          11

1                           2

1/3                         1

6                         -12

7                         -19

32/3                      -35


Practice problem:


Problem 1:
f is a linear function in linear algebra. Value of x and f(x) to test are given in the table below; complete the table.

X                              f(x)

-5                             15

1                               –

–                              1

10                         -14

8                             --

–                           -35


Problem 2:
f is a linear function. Value of x and f(x) to test are given in the table below; complete the table.

X                               f(x)

-6                             18

1                              –

–                             1

9                           -20

10                             --

–                             -40