In which taxonomic levels will u find both the ringtail and the human

Glucose provides energy for cells. Different cells have different mechanisms for glucose intake. Intestinal cells contain protei

Vlada [557]

Answer:

D

Explanation:

Intestinal cell takes in glucose against Concentration gradient (uses energy in the form of ATP) to transport glucose into cell.

In blood cell, uptake of glucose occurs down concentration gradient. Hence, its facilitated diffusion (passive transport)

Glucose is a polar molecules due to presence of polar hydrophilic - OH group, hence carrier proteins are needed to provide a hydrophilic channel for glucose to enter the cell via facilitated diffusion.

4 0

4 years ago

Consider two different traits in mice affecting tail length and fur color. A two-factor cross was made involving true-breeding m

Stells [14]

Answer:

See the answer below

Explanation:

Null hypothesis: The two genes are independently assorting and therefore, are not linked to each other.

Let A represents the allele for tail length and B for fur color. Long-tail A is dominant over short tail a while brown fur B is dominant over white fur b.

For the first cross involving true-breeding mice:

AABB x aabb

F1 AaBb (long tails and brown fur)

For the second cross:

AaBb x aabb

4 AaBb - Long-tail, brown fur

4 Aabb - Long tail, white fur

4 aaBb - short tail, brown fur

4 aabb - short tail, white fur

Since the phenotypic ratio from the cross is 1:1:1:1, if the null hypothesis was to be true, it means that the expected phenotype ratio should be 1:1:1:1.

In order to test this hypothesis, we use Chi-square:

phenotype O E X^2

Long-tail, brown fur 118 97.5 $\frac{(118-97.5)^2}{97.5}$ = 4.31

Long tail, white fur 77 97.5 4.31

short tail, brown fur 81 97.5 2.79

short tail, white fur 114 97.5 2.79

Total 14.2

Degree of freedom = 4 - 1 = 3

Critical Chi-square value = 7.815

The calculated Chi-square value is more than the critical value, hence, the null hypothesis is rejected.

8 0

3 years ago

Centrioles are found in animal cells t or f

vekshin1

True, centrioles can be found in animal cells

6 0

4 years ago

What are two important fuels that comes out of the oil refining process?

goldfiish [28.3K]

Many fuels can come out of the process, such as gasoline, kerosene, diesel fuel, liquified petroleum gas, and jet fuel.

6 0

3 years ago

A recessive allele on the X chromosome is responsible for red-green color blindness in humans. A woman with normal vision whose

Igoryamba

Answer:

In a cross of a homozygous prevailing (AA) individual and a homozygous recessive(AA) person,

a. on the off chance that they have a kid, what is the likelihood that the kid will be heterozygous?

If you cross somebody who is homozygous predominant with one who is homozygous latent, the entirety of the posterity will be heterozygous (AA). Accordingly, there is a 100% possibility of delivering a posterity that is heterozygous.

b. what is the likelihood that the youngster will be homozygous latent?

There is no way of creating a kid that is homozygous passive. The entirety of the posterity will be heterozygous. In this way, the likelihood is zero.

c. in the event that they have two kids, what is the likelihood of the first being homozygous prevailing and the second being homozygous passive?

Again, this can't be determined, on the grounds that there is zero chance of delivering a posterity that is homozygous prevailing or homozygous passive. The likelihood is zero. In any case, for no reason in particular, suppose that the punnet square uncovered that there was 1/4 possibility of creating a kid that is homozygous passive. To decide the likelihood that the subsequent youngster will likewise be homozygous passive, you complete this straightforward duplication:

1/4 x 1/4 = 1/16

In this manner, there would be a 1/16 possibility of the subsequent kid being homozygous passive. In the event that you needed to discover the likelihood of the third kid being homozygous passive, you would do the accompanying:

1/4 x 1/4 x 1/4 = 1/64

Thus, there would be a 1/64 possibility of the third kid being homozygous latent. Once more, I was simply giving in model, yet it doesn't have any significant bearing right now, the entirety of the posterity will be heterozygous.

Another question...A latent allele on the X chromosome is answerable for red-green visual weakness in people. A typical lady whose father is visually challenged weds a partially blind man. What is the likelihood that this couples child will be visually challenged?

XX = female

XY = male

Let C = typical vision (predominant)

Let c = red-green visual weakness (latent)

A typical lady what father's identity is' partially blind must be a transporter, or heterozygous. This implies she acquired the ordinary allele from her mom and the visually challenged allele from her dad.

Genotypes:

Ordinary lady - Xc

Partially blind man - Xc Y

On the off chance that these two mate, here are the accompanying prospects:

half of the female posterity will be bearers with ordinary vision (Xc)

half of the female posterity will be homozygous passive and partially blind (Xc)

half of the male posterity will have typical vision (XC Y)

half of the male posterity will be visually challenged (Xc Y)

In this manner, the likelihood that the couple's child will be partially blind is half, or 1/2.

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Sheryl addressed this Was this answer accommodating?

XX= lady, XY=man

Alleles:

XC=normal; Xc=colorblind

Typical Genotypes:

XC (typical lady)

Xc (typical lady, yet bearer)

Xc (visually challenged lady)

XC Y (typical man)

Xc Y (visually challenged man)

Lady has typical vision, yet her dad is visually challenged. In this way, she needed to get XC from her mother (must have one, she's ordinary) and Xc from her father (it was everything he could give her): Xc

Man is visually challenged: Xc Y

Xc Y

XC Y

Xc Y

6 0

4 years ago