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svp [43]
3 years ago
11

What do bacteria have in common with cells of other living organisms?

Biology
2 answers:
ELEN [110]3 years ago
7 0
D is your correct answer
geniusboy [140]3 years ago
4 0
The answer would be D. Contain cytoplasm.

bacteria(prokaryotic cells) have a cell membrane, cytoplasm, DNA, ribosomes but lack all membrane bound organelles that eukaryotic cells have.
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Plants use glucose to synthesize cellulose. Which statement accurately describes this process?
Sergeeva-Olga [200]

Answer:

C

Explanation:

Sugar, which is a product of the photosynthetic process of green plants can be converted to glucose (a simple carbohydrate). Enormous amount of these glucose molecules can be linked together to form a complex carbohydrate called cellulose, which is a polysaccharide because it is made up of a long chain of glucose molecules. The cellulose is ultimately used to build the cell wall of plants.

Anabolism is the production of a complex molecule by a living organism from a much simpler one. Since, cellulose is produced by the building up of glucose molecules, the process can therefore, said to be anabolic

6 0
2 years ago
Forces that act in pairs are said to be _____________.
dimulka [17.4K]
The answer is A, stable.
7 0
3 years ago
Ocean acidification can lead to increased beach erosion because of which of its effects?
Salsk061 [2.6K]

Answer:

I feel like it's D

Explanation:

Because Coral can't survive at 2C so this mean that theres nothing to stop curants hitting the shores and so ye that or it's A hope this helps :3

4 0
3 years ago
Black coat color in horses is conferred by the presence of a dominant B allele. Chestnut horses are genotype bb. A population of
AysviL [449]

Answer:

Frequency of B allele is 0.6681

Explanation:

If p represents the frequency of dominant allele and q represents the frequency of  recessive allele, according to Hardy-Weinberg equilibrium:

p + q = 1

p² + 2pq + q² = 1

where p² = frequency of homozygous dominant genotype

q² = frequency of homozygous recessive genotype

2pq = frequency of heterozygous genotype

Given that number of recessive chestnut horse = 28

Total horses = 226 + 28 = 254

frequency of b² genotype = 28/254 = 0.1102

frequency of recessive b allele = √0.1102 = 0.3319

So, frequency of B allele =

1 - 0.3319 = 0.6681

Hence frequency of B allele is 0.6681

7 0
3 years ago
Nucleotide sequence comparisons are possible because all organisms share what
slamgirl [31]
Same base pairs !! Adenine;Guanine;Cytosine;Thymine !!
3 0
3 years ago
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