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dybincka [34]
2 years ago
15

Drag the labels to their appropriate locations in the flowchart below, indicating the sequence of events in the production of fr

agment B. (Note that pol I stands for DNA polymerase I, and pol III stands for DNA polymerase III.)

Biology
1 answer:
matrenka [14]2 years ago
3 0

(Complete question attached)

Answer:

  1. Pol III binds to 3' end of primer B
  2. Poll III moves to 5' to 3',adding DNA nucleotides to primer B
  3. Pol I binds to 5' end primer A
  4. Pol I replaces primer A with DNA
  5. DNA ligase links fragments A and B

Explanation:

Both strands of parental DNA acts as a template for the synthesis of new DNA. The site of synthesis is called replication fork because the daughter strands look similar to <em>two-pronged fork.</em> The strands formed from Okazaki fragments(short sequences of DNA nucleotides) is called the <u>lagging strands,</u> which is synthesized in short fragments and in the opposite direction. While the strand that is synthesized continuously and in the same direction as the movement of the replication fork is called the <u>leading strand.</u> Both strands are synthesized in a 5'→3' direction. DNA ligase join these fragments together.

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To visualize the structure of DNA, which has a width of 10 nm, which microscope would be best
damaskus [11]

Answer:

Electron microscope

Explanation:

Instead of using light, it utilized a beam of electron for illumination. Because an electron beam has a shorter wavelength than light (up to 100, 000 times shorter), they have a higher resolution power and are therefore able to view sub-cellular structures including DNA.

5 0
2 years ago
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In response to seasonal changes in temperature, many organisms must alter the composition of their plasma membranes to maintain
aleksandrvk [35]

Answer:

C. a decrease in phospholipid fatty acid side chain length and a decrease in side chain saturation

Explanation:

Temperature is a factor that has a huge impact on cell membrane structure, more precisely its fluidity. So, for example, if temperature increases, the cell membrane becomes more fluid because the fatty acid tails of the phospholipids become less rigid.

5 0
2 years ago
Compare and contrast the differences between mature human eggs and sperm in the following questions. For each prompt, determine
Llana [10]

Answer:

QUESTION 1:

a. They undergo unequal cytoplasmic division during meiosis- eggs

b. Their chromosomes have undergone genetic recombination and independent assortment- both gamete types

c. They contain specialized secretory vesicles called cortical granules just under the plasma membrane- eggs

d. They are characterized by a flagellum. Most of these gametes die without ever completing meiosis- sperm

e. The acrosomal reaction occurs in these cells- sperm

f. They can be derived from any adult cell type- neither gametes

QUESTION 2:

c) The egg is surrounded by a mineralized shell added before the egg exits the body.

Explanation:

a) Female reproductive cells undergo meiosis to produce one large egg and three polar bodies. However, an unequal cytoplasmic division occurs in favour of the large egg.

b) Since both egg cell and sperm cell are produced via meiosis and meiotic division involves the processes of genetic recombination and independent assortment, hence, both gamete types have their chromosomes underg genetic recombination and independent assortment.

c) Egg cells contain specialized secretory vesicles called cortical granules just under the plasma membrane. This cortical granules help prevent more sperms from fertilizing the same egg after it has been fertilized.

d) Sperm cells possess flagellum for motility in the female reproductive tract. Most of the sperm cells die without ever completing meiosis.

e) Sperm cells possess an head-covering structure called ACROSOME. Acrosome reaction is the process undergone by sperm cells permeate the zona pellucida of the egg.

f) Gametes are specifically derived from adult reproductive cells and hence, NEITHER TYPE OF GAMETE can be derived from any adult cell type

QUESTION 2:

The egg in Sea urchins is surrounded by a mineralized shell added before the egg exits the body to the external environment. This does not occur in the humans because the egg does not leave the body but fertilized internally.

8 0
2 years ago
Determine the new concentration, in mg/mL, of the starch solution you diluted in the Erlenmeyer flask. Recall from the Procedure
Semmy [17]

Answer:

0.05 mg/mL  ( B )

Explanation:

Given data:

20 mg/ml starch

2% solution = 2g of solute is in 100g of solvent

<u>Determine the new concentration in mg/ml </u>

Dilution equation = C1V1 = C2V2

new concentration ; applying the dilution factor

dilution factor = 1 : 400 ; ( 2 /400 )g = 0.005 g of solute is present in every  100 mL

∴ new concentration = 0.00005 g / 1 mL * ( 1000 mg / 1g ) = 0.05 mg/mL

6 0
2 years ago
You have two pure substances that you cannot identify each sample is solid at room temperature describe at least five steps in t
lapo4ka [179]

Answer:

Explanation:

If we have two solid samples, in order to identify what they are a series of ordered steps have to be performed.

1) The first thing to do is to observe the sample. If there is color it <u>may indicate</u> the presence of certain anions: for example if the sample is purple, it can be because of the presence of the permanganate ion (MnO₄⁻), if it is yellow it can be chromate ion (CrO₄⁻), if it is orange it can be the dichromate Cr₂O₇²⁻), etcetera. If the color of the sample is white we have no indication whatsoever.

2) Then we can use certain reactants to precipitate the cations of the sample. For example, we can add first HCN 3N to our sample. If there is precipitation, it means that the cations Ag⁺ or Pb²⁺ are present. If not, there are other cations and we must use a different reactant to precipitate them.

3) We then add H₂S to the sample (not adding it per se, but generating it heating thioacetamide with water). If we see a black precipitate, it can be because of the cations Pb²⁺, Bi³⁺ or Cu²⁺. If we see a yellow precipitate, it corresponds to Cd²⁺. If we do not see a precipitate, we need to add other reactant.

4) We add NaOH to the sample. If we see precipitate, it can be because of the the ions Fe³⁺, Ni²⁺, Co²⁺ or Mn²⁺.

5) We observe the color of this precipitate. If it is brown is Fe(OH)₃, if it is green is Ni(OH)₂, if it is pink is Co(OH)₂, and if it is white is Mn(OH)₂.

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Sobre el mismo papel de filtro se añade NH3 2N. En el papel de filtro si existe Hg22+ y se forma una mancha blanca, gris o negro, que es una mezcla de HgClNH2 y Hg0. En la disolución se forman Ag(NH3)2+, que se puede identificar con KI dando un precipitado de AgI amarillo claro.

3 0
2 years ago
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