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k0ka [10]
3 years ago
7

What are each of these blood vessels responsible for when it comes to transporting blood in the body?

Biology
1 answer:
Nitella [24]3 years ago
8 0
Is veins in charge of transporting the blood
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Can you answer these all right? About: cell facts.
Karo-lina-s [1.5K]

Answer:

1. True

2. True

3. True

4. False

5. False

6. True

7. False

8. False

7 0
3 years ago
Why is carbon so important for living things
Blizzard [7]

Answer:

Organic compounds make up the cells and other structures of organisms and carry out life processes. Carbon is the main element in organic compounds, so carbon is essential to life on Earth. Without carbon, life as we know it could not exist.

Explanation:

8 0
3 years ago
Read 2 more answers
I need help please!
swat32
Okay so have the chart here to show you H is dominant and h is recessive

8 0
3 years ago
In what type of relationship do fig trees and fig wasps engage
Sergio039 [100]

Answer: Symbiotic Relationship

Both fig trees and fig wasps are important for each others survival.

Explanation: The fig provides a joke for the wasps and the wasps provide pollen that the fig needs for  the fruit to ripen.

It basically begins when a female fig wasp enters the fig and lays eggs ,they pollinate the fig with pollen that they had collected in their pouches after mating ,when the wasp enters the fig its wings and parts of antennae may get broken because of the small opening in the fruit ,the wasp completes it's task and never leave the fruit after.

4 0
3 years ago
Suppose that we are dealing with four genes, each gene consisting of a dominant allele (capital letter) and a recessive allele (
Sauron [17]

Answer: The probability of obtaining an individual who is CcmmLLPp 1/32.

Explanation: This can be achieved by crossing similar genes to obtain individual probability as shown in the attached image.

When you cross Cc in the first genotype with CC in the second genotype, the following probabilities will be obtained;

P (CC) = ¾, P(Cc) = ¼.

Similarly, crossing Mm with mm, we get;

P (Mm) = ½, P (mm) = ½

Crossing Ll with Ll, we get

P (LL) = ¼, P (Ll) = ½, P (ll) = ¼

Crossing PP with pp, we get

P (Pp) = 1

Therefore, the probability of individual with genotype CcmmLLPp will be;

P (Cc) x P (mm) x P (LL) x P (Pp)

= ¼ x ½ x ¼ x 1

= 1/32

3 0
3 years ago
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