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.
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.
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