Graphs are _____ of a relationship

What is the function of DNA and where is it found in a eukaryote cell

swat32

Answer:

The function of DNA is it has the instructions on how to recreate itself and how to do basic functions in cells and in life. It is found in the nucleus of a eukaryote

Explanation:

Hope this helps and have a great day!

7 0

3 years ago

Tissues and organs worksheet answers describe the basic components of a tissue

Kazeer [188]

The most common bone disease is

osteomyelitis
osteoporosis
osteosarcoma
osteomalacia
rickets

4 0

3 years ago

Analysis Questions:

Drupady [299]

1. List five observations Darwin made about the Amblyrhynchus lizards.

a. They measure about 2-3 feet long.
b. They are cold-blooded.
c. They feed on crab and sea lion feces.
d. They are the only marine iguanas in the world.
e. Their nostrils are connected to salt glands that allow them to expel excess sea salt.

2. Describe two examples of structural adaptations that Darwin observed.

a. Darwin discovered that finches in the Galapagos have thirteen different kinds of beaks.
b. Darwin observed two kinds of giant tortoises that are distinguished by the shapes of their carapaces (dome-shaped and saddle-shaped).

3. Describe two behavioral adaptations that Darwin observed.

a. The marine iguanas have developed several ways of coping with the cold. At night, they huddle together for warmth.
b. In the face of famine, the marine iguanas have adapted by having the ability to shrink their body size. They are able to grow back to their original size once the food is supply increases.

4. Explain how these observations influenced his theory of evolution?

These observations led Darwin to believe that if a certain trait is more advantageous in a specific environment, then the carriers of that trait are more likely to survive into adulthood and to generate offspring than those who do not have that trait. Eventually, the advantageous traits for a species will proliferate. Darwin later called this as "survival of the fittest".

6 0

3 years ago

Read 2 more answers

Phagocytosis is to eating as pinocytosis is to

VladimirAG [237]

Phagocytosis is to eating as pinocytosis is to drinking.

3 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

3 years ago