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2 years ago

15

GLUTs are integral membrane proteins that assist in the facilitated diffusion of glucose into and out of cells. What reaction in

glycolysis prevents glucose from being transported back out of the cell

1 answer:

krok68 [10]2 years ago

3 0

Answer:

i dont know that one im sorry

Explanation:

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What is true for urochordates

Leno4ka [110]

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3 years ago

1. Han Solo and Leia fall madly in love and have one child (Ben). What is the chance that they will have a child who is strong w

Andreas93 [3]

1.
Han Solo genotype: ff
Offspring genotypes: Ff, ff
There is a 50% chance the child will have the force.

2.
Let f be webbed feet and F be non-webbed feet
Parental genotypes: ff ; FF
Possibility of webbed feet in offspring: 0%

3.
Yes, it is possible if both parents are heterozygous. 25% of offspring will be grey.

4.
Pheonotypes: Hairy and non-hairy fur
genotypes: HH, Hh, hh

5.
Parental genotypes: RR and RL
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6.
Genotypes: BW
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8.
None of his children will be bald.<span />

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3 years ago

1. What percentage of the weight of Earth's crust is made of silicon?

sladkih [1.3K]

Answer: 27.7 percent

Explanation:

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3 years ago

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Vesna [10]

50% because of the ee and ee

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2 years ago

Glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as an X-linked recessive gene 43) in humans. A woman whose fath

LiRa [457]

Answer:

50% or 1/2. The result remains unchanged if the husband were to have G6PD.

Explanation:

For X-linked recessive inheritance, a female (XX) needs two recessive alleles to be affected while a male needs only one (XY). It is hypothetically assumed that the Y chromosome does not carry any trait.

Assuming the allele for the disease is represented by g, a woman whose father suffered from G6PD is a carrier for the disease with genotype $X^GX^g$ . A normal man will have the genotype $X^GY$ . When the 2 marries:

$X^GX^g$ x $X^GY$ = $X^GX^G (normal female), X^GY (normal male), X^GX^g (carrier female), X^gY (affected male)$ It thus means that 50% or 1/2 of their sons will be expected to have G6PD.

Now, assuming the husband has G6PD, the mating becomes:

$X^GX^g$ x $X^gY$ = $X^GX^g (carrier female), X^GY (normal male), X^gX^g (affected female), X^gY (affected male)$ 50% or 1/2 of their sons is still expected to have G6PD. The ratio remains unchanged.

3 0

3 years ago