what will most likely happen if a population is large and no migration, mutation, or environmental change occurs?
1. natural selection will increase
2. nonrandom mating will start to occur
3. the rate of evolution will increase
4. gene frequencies will remain constant

Answer:

a) k_m = 4.08 uM

  V_{max} = 20.07 uM/min

b) k_m = 8.16 uM

Explanation:

Given that:

For Enzyme A:

the substrate concentration [S] = 40 uM

the initial velocity rate v = 10 uM/min

when it was 4mM, v = 20 uM/min

i.e.

at 4mM = 4000 uM;

Using Michealis -menten equation;

when v = 10

[tex]V = \dfrac{V_{max}[S]}{k_m+[S]}[/tex]

∴

[tex]10 = \dfrac{V_{max}\times 40}{k_m + 40}[/tex]

[tex]10 (k_m + 40) = V_{max}40[/tex]

[tex]40V_{max} -10k_m = 400 --- (1)[/tex]

when v= 20

[tex]20= \dfrac{V_{max}\times 4000}{k_m + 4000}[/tex]

[tex]20 (k_m + 4000) = V_{max}4000[/tex]

[tex]4000V_{max} -20k_m = 8000 --- (2)[/tex]

equating equation (1) and (2):

[tex]40V_{max} -10k_m = 400 --- (1)[/tex]

[tex]4000V_{max} -20k_m = 8000 --- (2)[/tex]

let multiply equation (1) by 100 and equation (2) by 1

4000V_{max} - 1000K_m = 4000

4000V_{max} - 20 k_m = 8000    

  0        -980k_m = 4000

k_m = 4000/-980

k_m = 4.08 uM

replacing the value of k_m into equation (1)

40{V_max } - 10(4.08) = 400

40{V_max } - 40.8 = 400

40{V_max } = 400 +  40.8

40{V_max } = 440.8

V_{max} = 440.8/40

V_{max} = 11.02 uM/min

b)

Since V_{max} of A ie equivalent to that of B; then:

V_{max} of B = 11.02 uM/min

Here;

[S] = 80  uM

V = 10 uM/min

∴

[tex]10 = \dfrac{11.02 \times 80}{k_m + 80 }[/tex]

10(k_m +80) = 881.6

10k_m = 881.6  - 800

10k_m = 81.6

k_m = 81.6/10

k_m = 8.16 uM

Explanation:

economic importance of mosquito larva are

Mosquito larvae grow by consuming microorganisms such as algae and microbes that decompose decaying plant material. Larval mosquitoes contribute to aquatic food chains by serving as food sources for many predators, including fish and birds.i hope it will help you