List three examples of single celled eukaryotes.

Biology

1 answer:

VMariaS [17]3 years ago

6 0

Answer:

Explanation:

Algae, phytoplankton and zooplankton

You might be interested in

A nurse assisting a nursing student with medications asks the student to describe how penicillins (PCNs) work to treat bacterial

Mnenie [13.5K]

Penicillins disrupts bacterial cell wall synthesis.

<h3>How does penicillin affect bacterial cell walls?</h3>

Penicillin kills bacteria by inhibiting the proteins which cross-link peptidoglycans in the cell wall .
When a bacterium divides in the presence of penicillin, it cannot fill in the “holes” left in its cell wall.
β-Lactam antibiotics, including penicillins, cephalosporins, monobactams, and carbapenems, are distinguished by a lactam ring in their molecular structure and act by inhibiting the synthesis of the peptidoglycan layer of bacterial cell walls.
Penicillins work by bursting the cell wall of bacteria. Drugs in the penicillin class work by indirectly bursting bacterial cell walls.
They do this by acting directly on peptidoglycans, which play an essential structural role in bacterial cells.

To learn more about Penicillin from the given link

brainly.com/question/11849121

#SPJ4

5 0

1 year ago

A substance that speeds up the rate of a chemical reaction is called

maw [93]

Answer:

It is called An enzyme

6 0

2 years ago

Glucose-6-phosphate dehydrogenase deficiency (G6PD) is inherited as an X-linked recessive gene 43) in humans. A woman whose fath

LiRa [457]

Answer:

50% or 1/2. The result remains unchanged if the husband were to have G6PD.

Explanation:

For X-linked recessive inheritance, a female (XX) needs two recessive alleles to be affected while a male needs only one (XY). It is hypothetically assumed that the Y chromosome does not carry any trait.

Assuming the allele for the disease is represented by g, a woman whose father suffered from G6PD is a carrier for the disease with genotype $X^GX^g$ . A normal man will have the genotype $X^GY$ . When the 2 marries:

$X^GX^g$ x $X^GY$ = $X^GX^G (normal female), X^GY (normal male), X^GX^g (carrier female), X^gY (affected male)$ It thus means that 50% or 1/2 of their sons will be expected to have G6PD.

Now, assuming the husband has G6PD, the mating becomes:

$X^GX^g$ x $X^gY$ = $X^GX^g (carrier female), X^GY (normal male), X^gX^g (affected female), X^gY (affected male)$ 50% or 1/2 of their sons is still expected to have G6PD. The ratio remains unchanged.