Answer:
A)100mL B)50mL C)The second option D)Hypoosmotic Environment
Explanation:
The average Na concentration in the seas and oceans of the world is around 3,5% which mean that in 100 ml of sea water, there is around 3,5 grams of Na.
The weight of one mol of NaCl is 58,44 grams. For 3,5 grams of NaCl, we get 3,5/58,44 = 0,060 mol of NaCl which is 0,060x1000 = 60 mmol/100ml. According to this and the information given in the question about the secretion of the salt glands', if the average sodium concentration is 600mmol/L, we have 60*10 = 600mmol/L so it would take 100 mililiters of water to excrete.
If the average Na concentration of the salt gland's secretion were 300 mmol/L, only 50 mililiters of water would be needed to excrete the same sodium load.
The second option of secretion is hyperosmotic to seawater because the concentration is higher.
Osmoregulation is the process of balancing the amount of water and salt between the body of the organism and its surrounding environment. For salt glands to be advantageous for osmoregulation, they need to be in a hypoosmotic environment.
I hope this answer helps.
The DNA replication products visualized during the sanger method of DNA sequencing are observed in which nucleotides are added.
Sanger sequencing is based on the process of DNA replication. A scientist creates a copy of his DNA strand. Then observe which nucleotides have been added. This way you can see the sequence of nucleotides. A laser excites the fluorescent labels in each band and a computer detects the resulting light.
Sanger sequencing produces extension products of various lengths ending in dideoxynucleotides at the 3' ends. Extension products are separated by capillary electrophoresis or CE. Molecules are injected by an electric current into a long glass capillary filled with gel polymer. Selective incorporation of chain-terminating dideoxynucleotides by DNA polymerases during in vitro DNA replication.
Learn more about DNA replication here:-brainly.com/question/21265857
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QUESTION--> <span>The nucleolus is where _______ are constructed.
</span>ANSWER--> C) ribosomes
Answer:
- Calcium binds to troponin C
- Troponin T moves tropomyosin and unblocks the binding sites
- Myosin heads join to the actin forming cross-bridges
- ATP turns into ADP and inorganic phosphate and releases energy
- The energy is used to impulse myofilaments slide producing a power stroke
- ADP is released and a new ATP joins the myosin heads and breaks the bindings to the actin filament
- ATP splits into ADP and phosphate, and the energy produced is accumulated in the myosin heads, starting a new cycle
- Z-bands are pulled toward each other, shortening the sarcomere and the I-band, producing muscle fiber contraction.
Explanation:
In rest, the tropomyosin inhibits the attraction strengths between myosin and actin filaments. Contraction initiates when an action potential depolarizes the inner portion of the muscle fiber. Calcium channels activate in the T tubules membrane, releasing <u>calcium into the sarcolemma.</u> At this point, tropomyosin is obstructing binding sites for myosin on the thin filament. When calcium binds to troponin C, troponin T alters the tropomyosin position by moving it and unblocking the binding sites. Myosin heads join to the uncovered actin-binding points forming cross-bridges, and while doing so, ATP turns into ADP and inorganic phosphate, which is released. Myofilaments slide impulsed by chemical energy collected in myosin heads, producing a power stroke. The power stroke initiates when the myosin cross-bridge binds to actin. As they slide, ADP molecules are released. A new ATP links to myosin heads and breaks the bindings to the actin filament. Then ATP splits into ADP and phosphate, and the energy produced is accumulated in the myosin heads, which starts a new binding cycle to actin. Finally, Z-bands are pulled toward each other, shortening the sarcomere and the I-band, producing muscle fiber contraction.
