BIOLOGY COLLEGE

Use the template of a replication fork to draw arrows that represent both continuous and discontinuous DNA synthesis. Draw one long arrow to show continuous DNA synthesis and three arrows to show discontinuous DNA synthesis. The arrows should point in the direction of nucleotide addition, and the three arrows showing discontinuous synthesis should be numbered 1, 2, and 3, in the order of fragment synthesis.

Answers

Answer 1
Answer:

Answer: seen below

Explanation:

Deoxyribonucleic acid (DNA) synthesis is the biological process by which DNA molecule is created. In the cell, each of the two strands of the DNA molecule acts as a template for the synthesis of a complementary strand. This strands are separated during replication (DNA replication is the biological process of producing two identical replicas of DNA from one original DNA molecule). DNA polymerase synthesizes the new strands by adding nucleotides that complement each (template) strand.

Deoxyribonucleic acid (DNA) synthesis is continuous because extension of a single ribonucleic acid (RNA) primer occurs without interruption into the replication fork as it continues to open to the right as seen on the diagram below. On the upper lagging strand which is the discontinuous synthesis, here synthesis is discontinuous, since new RNA primers must be added as opening of the replication fork continues to expose new template.

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At time 0s, a population has 2 members. At time 2s, there are 4 members. At time 6s, there are 16 members. What type of growth is being experienced by this population?

Answers

Exponential growth is seen

Given that initial population, at time 0 members =2

i.e. P0 =2

P(2) = 4

In 2 minutes it became 4 times

And in 6 minutes it became 16 members

P(0) =2

P(2) =4

P(6) =16

i.e. every 2 minutes the population doubles

Hence exponential growth of population doubling in 2 minutes

P(t) = 2^{\frac{t}2}

will be the function representing the population at time t.
Exponential Growth....members are increasing to its double value within every 2s.

) You treat some cells with a proteolytic enzyme that is too large to penetrate the cell membrane (Set 1). Another group of cells is made permeable before treatment with the enzyme (Set 2). A third set of cells was not treated with the enzyme at all (controls). Proteins are then extracted from the three different sets of cells and applied to an SDS-PAGE gel. Protein W migrates to the same distance on a gel of proteins from Set 1 and Set 2; Protein W migrates a shorter distance on gels of proteins extracted from the control group than on gels of proteins extracted from Set 1 and Set 2 treated cells. Protein X migrates to the same distance on a gel of proteins from control cells and the gels of the proteins from Set 1 and Set 2. Protein Y migrates a longer distance when extracted from Set 1 cells than does protein Y in the controls; Protein Y moves an even larger distance in the gel of the extracts from Set 2. Protein Z migrates the same distance on gels of proteins from the controls and the proteins extracted from Set 1, but it migrates a longer distance in extracts from Set 2 cells. Which protein is exposed only on the exterior of the cell?

Answers

Answer:

Proten W

Explanation:

SDS-PAGE gel is a method used for the separation of proteins in which proteins are separated based on their length (smaller proteins move faster through the gel, due to less resistance).

When treated with proteolytic enzymes, proteins are cut and become short fragment. This means that the fragments formed after the use of proteolytic enzymes, will move faster and thus, migrate a longer distance. Proteolytic enzymes in Set 2 cells will act only on plasma membrane proteins (because they cannot penetrate), while in Set 2 they will act on both, plasma membrane and interior proteins. Control group will have only the large fragments (not treated with enzyme).

Protein W travels the same distance on a gel of proteins from Set 1 and Set 2, but different than control group. It means that the proteolytic enzyme worked the same on Set 1 and Set 2.

PLEASE HELP WITH THIS ONE QUESTION What statement about the inheritance of mitochondrial DNA is true?

A) The sperm cell donates most of the material in the cytoplasm to the zygote, including mitochondrial DNA.

B) The egg and sperm cells contribute equal amounts of mitochondrial DNA to the zygote.

C) The egg cell donates most of the material in the cytoplasm to the zygote, including mitochondrial DNA.

D) The sperm cell donates most of the DNA in the nucleus to the zygote, including mitochondrial DNA.

Answers

Answer:

The answer should be C

Explanation:

The egg contributes all the cytoplasm and organelles to the zygote. The sperm only contributes one set of chromosomes.

The typical sperm midpiece contains approximately 50-75 mitochondria with one copy of mtDNA in each. This represents an 8- to 10-fold decline in copy number during spermiogenesis.

Produce three small cells thatMeiosis and unequal cytoplasmic division in seed plant
disintegrate and one large megaspore.
O a gametophytes
O b. pollen sacs
O c pollen grains
O d. microspores
o e ovules
Meiosis and unequal cytoplasmic division in seed plant produce three small cells that disintegrate and one large megastore

Answers

The answer is e. ovules

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B. Lipid
C. Protein
D. Nucleic Acid

Answers

Answer:

Carbohydrates - monosaccharides

Lipids - fatty acids, glycerol

Protein - amino acids

Nucleic acid - nucleotides

Answer: Carbohydrates: molecules composed of sugar monomers. They are necessary for energy storage. ...

Lipids: water-insoluble molecules that can be classified as fats , phospholipids, waxes, and steroids. ...

Proteins: biomolecules capable of forming complex structures. ...

Nucleic Acids: molecules consisting of nucleotide monomers linked together to form polynucleotide chains. ...

Explanation: I hope this helps.

An individual has a genotype of IBi. What is their blood type?Group of answer choices

Type O

Type BO

Type AB

Type B

Answers

the answer is 4) type b
Type B

Explanation: Blood Type B shows complete dominance over Blood Type O (i) - not codominance.
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