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AleksandrR [38]
4 years ago
7

True or False. An action potential can exist when both sides of the cell membrane have the same charge.

Biology
2 answers:
aleksandr82 [10.1K]4 years ago
8 0

Answer:

That answer would be True

Damm [24]4 years ago
5 0

The answer for that is true’

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A woman has her personal genome analyzed for the BRCA1 mutation after learning that her father is heterozygous and carries one m
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a. 50%

Explanation:

<em>BRCA1 genes is a tumor supressor gene whose harmful mutation can cause hereditary breast ovarian cancer syndrome in both males and females. The BRCA actually stands for breast cancer.</em>

<em>Mutation in BRCA1 gene is heritable and every progeny of a carrier parent irrespective of the sex has a 50% chance of inheriting the trait from such parent, be it from the mother or the father.</em>

Hence, the correct option is a.

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A structure that helps prevent food from entering the pharynx prematurely is the
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Uvula

Explanation:

The uvula is a fleshy structure found at the back of the soft palate in the mouth. It is the structure seen hanging at the back of the throat when someone opens his/her mouth and views in the mirror.

<em>The uvula is made up of flexible tissues with the ability to produce saliva. During eating or swallowing of food, the uvula along with the soft palate move to seal off the pharynx in order to prevent food materials from entering the nasal passage.</em>

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4 years ago
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What vibrates as an electromagnetic wave passes by?
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Prominences in your own words
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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
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