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elena-14-01-66 [18.8K]
3 years ago
7

If a gene has only one allele, how many different traits can the allele produce?

Biology
1 answer:
salantis [7]3 years ago
3 0
One allele can only produce one trait
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What is the role of the nervous system in digestion
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Answer: motility, ion transport associated with secretion and absorption, and gastrointestinal blood flow.

Explanation:

The nervous system exerts a profound influence on all digestive processes, namely motility, ion transport associated with secretion and absorption, and gastrointestinal blood flow. The magnitude and complexity of the enteric nervous system is immense - it contains as many neurons as the spinal cord.

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Which body segment of an insect contains the heart and most of the digestive and excretory systems?
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The organs are in the abdomen
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Which of these types of weathering requires the presence of water?
serg [7]

Answer:

A. Abrasion

Explanation:

<em>"In abrasion, one rock bumps against another rock. Gravity causes abrasion as a rock tumbles down a mountainside or cliff. Moving water causes abrasion as particles in the water collide and bump against one another."</em>

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8 0
3 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
When ice is placed in warm water it melts. The order or structure of the molecules decreases. This is a natural process explaine
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Answer:

What is the question you’re asking

Explanation:

7 0
3 years ago
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