Tutor profile: Henry B.
A 1000L balloon is compressed to 500L at a constant temperature. If the initial pressure of the 1000L balloon is 5 atm, what is the final pressure of the compressed, 500L balloon?
This question involves Boyle's gas law: Pi*Vi=Pf*Vf (Pi: initial pressure, Pf: final pressure, Vi: initial volume, Vf: final volume) In order to solve this problem, we have to plug in the given pressures and volumes into the appropriate variables and solve for our unknown variable, Pf: Pi = 5 atm Pf = ? atm = x Vi = 1000 L Vf = 500 L => (5 atm)*(1000 L) = (x)*(500 L) (5000 atm*L)/(500 L) = x 10 atm = x Therefore, our final pressure will be 10 atm. Note: Once students are comfortable with these gas law equations, I encourage them use their knowledge of the mathematical relationships between these physical qualities when solving these problems. In the case of Boyle's law, pressure and temperature are inversely related, therefore since the balloon's volume decreases by a factor of two (1000L to 500L) , we know that its pressure will increase by a factor of two (5 atm to 10 atm). This approach can prove to be a powerful tool that students can use to solve or estimate their answers.
Given the following coding strand of DNA: ATGCCCACATACTATAATCGTAAATCGTGGGAAAGTACATGA A) Determine the mRNA sequence if this strand is transcribed. B) Using your answer from A) and an RNA codon table, determine the amino acid sequence of this transcript if it is translated into protein. (Use the appropriate three letter abbreviations for all amino acids.) C) How many basic acid amino acid residues are in this protein? Acidic? D) How many amino acid residues in this protein can be phosphorylated? D) Where are you more likely to find this protein in the cell: the cytoplasm or cell membrane? Explain.
Given the following coding strand of DNA: ATGCCCACATACTATAATCGTAAATCGTGGGAAAGTACATGA A) Determine the mRNA sequence, if this strand is transcribed. Since we were given the CODING strand of the DNA (as opposed to the non-coding strand), all we have to do to find the mRNA transcript is replace our thymines (T's) with uracils (U's): AUGCCCACAUACUAUAAUCGUAAAUCGUGGGAAAGUACAUGA B) Using your answer from A) and an RNA codon table, determine the amino acid sequence of this transcript if it is translated into protein. (Use the appropriate three letter abbreviations for all amino acids.) In order to "translate" our mRNA transcript into protein, we have to consider the three nucleotide long codons and the amino acids they code for: AUG CCC ACA UAC UAU AAU CGU AAA UCG UGG GAA AGU ACA UGA Met-Pro-Thr-Tyr-Tyr-Asn-Arg-Lys-Ser-Trp-Glu-Ser-Thr-STOP (Note: STOP does not represent an amino acid, rather it illustrates that UGA is a STOP codon that terminates translation.) C) How many basic acid amino acid residues are in this protein? Acidic? There are two basic amino acid residues—arginine (Arg) and lysine (Lys)—and one acidic residue—glutamate (Glu). D) How many amino acid residues in this protein can be phosphorylated? Six, as there are two threonines (Thr), two tyrosines (Tyr), and two serines (Ser). D) Where are you more likely to find this protein in the cell: the cytoplasm or cell membrane? Explain. This protein predominantly contains polar and charged amino acid residues and thus is highly hydrophilic ("water loving"). Therefore, it would likely be found in the cell's cytoplasm where it could interact with water, NOT in the hydrophobic ("water fearing") cell membrane.
Thomas spent $25.50 on 14 snacks. If potato chips cost $1.50 and candy costs $2.00. How many pieces of candy did he buy?
First, we will need to assign variables to potato chips (p) and candy (c) and then translate the above word problem into the following algebraic equations: 1. p + c = 14 2. 1.50p + 2.00c = 25.50 Second, we will need to solve this system of equations using the most effective method, which in this example is substitution: We will need to solve algebraically for "p" using equation 1, substitute "p" in equation 2 with this answer, and then solve equation 2 for "c": p + c = 14 p = 14 - c 1.50*(14 - c) + 2.00*c = 25.50 21 - 1.50c + 2.00c = 25.50 21 + 0.50c = 25.50 0.50c = 4.50 c = 9 Therefore, Thomas bought 9 pieces of candy.
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