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JulsSmile [24]
3 years ago
15

How would a small, uncharged lipid move across the cell membrane?

Biology
1 answer:
ElenaW [278]3 years ago
5 0

Answer:

Through simple diffusion, down the concentration gradient.

Explanation:

The phospholipids of the membrane are amphipathic with hydrophillic heads and hydrophobic tails. Other polar molecules cannot go through this hydrophobic interior. Since small uncharged lipids are non polar and hydrohobic, they are able to go through the membrane without the help of transport proteins. Therefore, the last two options can be ruled out because facilitated diffusion includes the use of a protein. Diffusion involves molecules moving down the concentration gradient so the second option can be ruled out.

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The upper limb is supplied by a main arterial vessel called the ______ artery and the lower limb's main arterial vessel is calle
OlgaM077 [116]

The <u>brachial artery</u> is located in each upper limb, therefore it supplies the upper limb and the lower limb's main arterial vessel is called the <u>femoral artery</u>.

<h3>What is brachial artery?</h3>

It is the main blood vessel that is responsible for transporting oxygenated blood to supplies the upper limb from where the arm begins until it divides into two branches (radial and ulnar), and culminates in the hand and fingers.

<h3>What is femoral artery?</h3>

The femoral artery is located in each lower limb, specifically in the anterior and internal part of the thigh, to later become the popliteal and tibial arteries, it transports the blood with oxygen that comes from the left ventricle.

Therefore, we can conclude that the femoral artery is a pathway that belongs to the vascular system, which supplies the lower limb's, and the brachial artery is a prominent blood vessel that is responsible for irrigating the tissues located in the upper limb.

Learn more about arteries here: brainly.com/question/14821054

3 0
2 years ago
Can you help me with both
Ahat [919]
Decisions, studied I think
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3 years ago
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Dna is the instructions manual for a living thing. Each time one of your cells divide , your dna is copied. That way, each new c
sergey [27]

Answer:

DNA is the hereditary material in humans and almost all other ... Human DNA consists of about 3 billion bases, and more than 99 ... for building and maintaining an organism, similar to the way in which ... is critical when cells divide because each new cell needs to have an ... Thank you for your feedback!

Explanation:

Dna la me la traga fuerte mi monda

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3 years ago
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In humans, excess blood glucose is stored in the liver and in muscle tissue in the form of glycogen. Glycogen is a long chain of
serious [3.7K]

Answer:

Polysaccharide

Explanation:

These are large molecules which are formed from the union of many monosaccharides units through condensation. Excess glucose in the body is stored in form of glycogen and can be hydrolyzed when glucose levels go down. Glucagon stimulates liver cells to break down glycogen into glucose.

3 0
3 years ago
Duchenne muscular dystrophy is an X-linked recessive disease. A phenotypically normal couple wants to start a family. The woman’
Andreyy89

Answer:

\frac{1}{8}

Explanation:

From the question: Duchenne muscular dystrophy is an X-linked recessive disease.

Now for an X-linked recessive disorder to be affected by a male individual; it only requires the presence of only one copy of the recessive allele of the disease to be present BUT in a female individual. both copies of the recessive allele must be present.

Let the X-linked recessive disease(i.e Duchenne muscular dystrophy) be = (ⁿ)

However, we are told that the couples are normal and are unaffected. Therefore;

the male partner (XY) will be:   X^NY

the female partner (XY) will be: X^NX^N

Similarly, the question proceeds by telling us that: the woman's brother has the disease.

Definitely, it's possible that this unaffected woman is a carrier of the disease.

So, if she is a carrier; we have her traits to be: X^NX^n

NOW, if a cross exist between these couples; we have

X^NY     ×        X^NX^n

                     X^N                         Y

X^N                X^NX^N                  X^NY

X^n                 X^NX^n                   X^nY

So, we have offspring as follows:

X^NX^N  = normal unaffected female  

X^NY     = normal unaffected male

X^NX^n   = female carrier for the Duchenne muscular dystrophy disease

X^nY      = affected male with the Duchenne muscular dystrophy disease

So, the probability of the child to be affected with the disease = \frac{1}{4}

Also, the probability that the first child will be a male or a female = \frac{1}{2}

∴

the  probability that the couple’s first child will be affected = \frac{1}{4}*\frac{1}{2}

=\frac{1}{8}

4 0
3 years ago
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