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2 years ago

Name the

1 answer:

3 0

carbonic anhydrases (or carbonate dehydratases) form a family of enzymes that catalyze ... Essentially an increase in carbon dioxide results in lowered blood pH which lowers ... The α-CAs are found in humans.

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If humans had been present to build log structures during the Carboniferous period (though they were not!), which plant types wo

aliina [53]

Answer:

The plant types suitable during that period would be ferns, horsetails, and lycopods.

Explanation:

A geologic period and system, which covers 60 million years from the termination of the Devonian period to the start of the Permian period is known as the Carboniferous period. During the Carboniferous period, the plant life was luxuriant and extensive, mainly at the time of Pennsylvanian. It comprises ferns and fernlike trees, that is, the giant horsetails known as club mosses, calamites, or lycopods like Sigillaria and Lepidodendron.

Thus, if humans had been present to construct log structures during the Carboniferous period, then they would have used logs of ferns, horsetails, and lycopods for construction work.

8 0

3 years ago

How can organisms be so unique if the nucleotides making up dna are all the same between organisms?

sergejj [24]

Answer:

All organisms are made up of the four nucleotide bases of the DNA. Yet still all organisms are different from one another due to the sequence of the arrangement of these nucleotides. The pattern of arrangement of the nucleotides determines which organisms will be more similar and which will be more different from one another. The pattern of arrangement leads to the formation of genetic code which will differ in organisms. Hence, all organisms are made of the 4 nucleotides but differ due to the pattern of arrangement of the nucleotides.

3 0

2 years ago

Which of the following is true ?

yulyashka [42]

Water is important to some organism.... i think

6 0

3 years ago

Read 2 more answers

2. Dominant trait: cleft chin (C) Mother’s gametes: Cc

andre [41]

.2. Offspring Genotypes will be Cc or cc.

Offspring phenotypes : Cleft chin or no cleft chin.

% chance child will have cleft chin: 50%

3. % chance child will have arched feet: 25%

4. % chance child will have blonde hair: 50%

5. % chance child will have normal vision: 25%

Explanation:

CASE 1 :

Dominant trait: cleft chin (C)

Recessive trait: lacks cleft chin (c)

Father’s gametes: cc

Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents= $\frac{1}{2}C + \frac{1}{2} c$ ........(i)

Gametes of cc parents = $\frac{1}{2}c + \frac{1}{2}c$ .........(ii)

Combining (i) and (ii) we get,

$\frac{1}{2} Cc + \frac{1}{2} cc$

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin = $\frac{0.5}{1}$ ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

A from mother, a from father =Aa

a from mother, A from Father = Aa

a from mother, a from father = aa

Gametes of Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

Gametes of other Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

..................................................................................

$\frac{1}{4} AA + \frac{1}{4} Aa$

+ $\frac{1}{4} Aa$ + $\frac{1}{4} aa$

..........................................................................................

$\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa$

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = $\frac{0.25}{1}$ × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive (bb)

Father’s gametes: Heterozygous (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in half Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous (Ff)

Father’s gametes: Heterozygous (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF then phenotype farsightedness

Genotype Ff then phenotype farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

3 0

3 years ago

If a forest experiences a fire, which adaptation would most likely help an organism survive?

kirill115 [55]

Answer:

cellulose

Explanation:

5 0

2 years ago