Biology College

economic importance of mosquito larva list five​

Answers

Answer 1

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

Related Questions

Dehydration synthesis leads to WICH formation

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Answer:

Polymers...........

Explanation:

Two different enzymes catalyze the same reaction and both exhibit the same Vmax. When Enzyme A was run with a 40 uM substrate, the initial rate (Vo) was 10 uM/min, and when g it was run with a 4 mM substrate, the Vo was 20 uM/min. Estimate the approximate Vmax and Km of Enzyme A. When Enzyme B was run with 80 uM substrate, the initial rate (Vo) was 10 uM per minute. Estimate the Km of Enzyme B.

Answers

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

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