Dissociate in water because water is a polar molecule
Answer: the cfu/g Gram-negative bacteria in the fecal sample is C = 3.0 × 10^3
Explanation:
We know that; Gram negative bacteria looks pale reddish in color under a light microscope from Gram staining.
therefore
There are 30 red bacterial colonies counted.
1 mL of from tube 1 was removed and added to tube with 99 mL saline (tube 2) dilution is 1/100.
transferred volume into the plate is 1 mL.
Now, we have to determine the cfu/g Gram-negative bacteria in the fecal sample
Formula to calculate CFU/g bacteria in fecal sample is expressed as;
C = n/(s×d )
where C is concentration (CFU/g)
, n is number of colonies
, s is volume transferred to plate
, d is dilution factor.
so we substitute
C = 30 / ((1/100) × 1)
C = 30 / 0.01
C = 3000
C = 3.0 × 10^3
THERFERE, the cfu/g Gram-negative bacteria in the fecal sample is C = 3.0 × 10^3
The frequency <em>p</em> of the yellow (A) allele is <em>p</em>= 0.3
The frequency <em>q</em> of the blue (a) allele is <em>q= </em><em>0.7</em>
Hardy–Weinberg equilibrium, states that allele and genotype frequencies in a population will remain constant from generation to generation. Equilibrium is reached in the absence of selection, mutation, genetic drift and other forces and allele frequencies p and q are constant between generations. In the simplest case of a single locus with two alleles denoted A and a with frequencies f(A) = p and f(a) = q, the expected genotype frequencies under random mating are f(AA) = p² for the AA homozygotes, f(aa) = q² for the aa homozygotes, and f(Aa) = 2pq for the heterozygotes.
p²+2*p*q+q²= 1 p+q= 1 q= 1-p
yellow (p²)= 9%= 0.09 p= √0.09= 0.3
green (2*p*q)= 42%= 0.42
blue (q²)=49%= 0.49 q=1-0.3= 0.7 <em>or</em> q= √0.49= 0.7