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Colt1911 [192]
2 years ago
14

In what ways do organelles make a cell more efficient

Biology
2 answers:
zloy xaker [14]2 years ago
6 0
Carry on different on jobs
yKpoI14uk [10]2 years ago
5 0
They carry on different on jobs to help the cell work. For example, mitochondria produces energy from the oxidation of glucose.
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What do scientists think the first organisms on Earth were?
Setler [38]

Answer:

Ameboas

Explanation:

4 0
3 years ago
Identify the features that distinguish animals from organisms in other multicellular kingdoms.
Nata [24]
The features that <span>distinguish animals from organisms in other multicellular kingdoms are that a</span>nimals are ingestive heterotrophs.
7 0
2 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.  

Remark  

Sheryl's Avatar  

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
3 years ago
Describe what happens during translation
masha68 [24]

What happens is that the RNA<span> is synthesized from the template and RNA polymerase moves along the DNA, elongating the RNA transcript molecule; an enzyme forms the hydrogen bonds between the bases of the DNA strand and the complementary bases of the RNA molecule found in the nucleus.</span>

<span>Hope this helped !!</span>

5 0
3 years ago
The map of a walking trail is drawn on a coordinate grid with three points of interest. The trail starts at R(−2, 4) and goes to
Zinaida [17]

Answer:

The total length will be "10 units". A further explanation is given below.

Explanation:

The given points are:

R(−2, 4)

S(3, 4)

T(3, −1)

On applying the distance formula,

⇒ RS=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

⇒       =\sqrt{(3-(-2))^2+(4-4)^2}

⇒       =\sqrt{(3+2)^2+(0)^2}

⇒       =\sqrt{25}

⇒       =5 \ units

and,

⇒ ST=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

⇒       =\sqrt{(3-3)^2+(-1-4)^2}

⇒       =\sqrt{0+(-5)^2}

⇒       =5 \ units

Now,

The total trail length will be:

= RS+ST

= 5+5

= 10 \ units

7 0
3 years ago
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