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ipn [44]
3 years ago
5

The Mono Lake ecosystem is being destroyed by dropping water levels and increasing salinity because ______.

Biology
2 answers:
Yuki888 [10]3 years ago
7 0
C) Snow melt is diverted to other areas
Alik [6]3 years ago
3 0

snowmelt is diverted to other areas

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Carica Papaya

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A population continues at a stable size for many years. Suddenly, in a single season, the population size drops by half. Is the
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What are the two reactants and two products for the process of photosynthesis?
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The two reactants are carbon dioxide and water.
The two products are oxygen and glucose.

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describe a situation in nature where genetic drift might be a more important influence on evolution than natural selection.
inna [77]

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Example of genetic drift

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a population of rabbits with alleles B and b, both alleles are present in equal frequencies p = 0.5 and q = 0.5 if 10 parents reproduce the probability of having an offspring with alleles B or b is 0.5; however, by chance, a slight difference in the offspring allele frequency might occur due ...

4 0
2 years ago
Calculate the final concentration of BSA in the problems below using the formula
Rudiy27

Answer:

A. C_2=1.5\frac{mg}{mL}

B. C_2=0.075\frac{mg}{mL}

C. C_2=0.01\frac{mg}{mL}

D. C_2=0.001\frac{mg}{mL}

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

C_2=\frac{C_1V_1}{V_2}

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}

Best regards.

7 0
3 years ago
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