LS7L Post-Lab Questions

the control and experimental treatments differ significantly with respect to the variable being studied

If the results of an experiment allow one to reject the null hypothesis, then: a. the control and experimental treatments differ significantly with respect to the variable being studied b. the control and experimental treatments do not differ significantly with respect to the variable being studied c. the null hypothesis correctly predicts the outcome of the experiment d. too few trials per treatment were performed e. the experiment lacked adequate controls

blue light has an effect on the rate of photosynthesis in algae

If you want to study the effects of blue light on photosynthesis in algae, which hypothesis would be the best alternative hypothesis? a. blue light decreases the rate of photosynthesis in algae b. blue light has no effect on the rate of photosynthesis in algae c. blue light has an effect on the rate of photosynthesis in algae d. blue light increases the rate of photosynthesis in algae

red light has no effect on the rate of photosynthesis in algae

If you want to study the effects of red light on photosynthesis in algae, what would the null hypothesis be? a. red light has no effect on the rate of photosynthesis in algae b. red light has an effect on the rate of photosynthesis in algae c. red light increases the rate of photosynthesis in algae d. red light decreases the rate of photosynthesis in algae

the hypothesis is that people who got 5 hours of sleep before taking the MIT will have different response time than people who got 8 hours of sleep before taking the MIT

In a recent study, scientists examined whether amount of sleep impacted response time on the MIT. 583 people were split into two groups, 258 people who got 5 hours of sleep before taking the MIT and 325 people who got 8 hours of sleep before taking the MIT. An unpaired analysis using resampling was run comparing the response time between the two groups. Which of the following experimental hypothesis is best, based on the experimental description? a. The hypothesis is that people who got 5 hours of sleep before taking the MIT will have a different response time than people who got 8 hours of sleep before taking the MIT b. The hypothesis is that people who took the MIT twice, once after getting 5 hours of sleep and once after getting 8 hours of sleep, will have different response times on the MIT c. The hypothesis is that people who get less sleep before taking the MIT will have different response time than people who get more sleep before taking the MIT d. The hypothesis is that people who get 5 hours of sleep will have a slower response time on the MIT than people who got 8 hours of sleep before taking the MIT

many red dots, striations

In the histology lab, various tissues were seen using the microscope. The blood can be distinguished from the others by seeing _______, while the skeletal muscle can be distinguished by seeing _________. a. many purple dots, striations b. none of the above c. many red dots, striations d. striations, open space e. open space, striations

grip strength

In the human physiology lab, the dynamometer was used to measure ________. a. grip strength b. breathing patterns c. electrical activity of muscles d. heart rate

does not have to be produced because there is no need for it

In the presence of glucose, Beta-galactosidase: a. cleaves glucose into lactose without using ATP b. does not have to be produced because there is no need for it c. binds to ONPG which in turn facilitates glucose breakdown d. produces ATP used for cellular respiration e. cleaves glucose into galactose by using ATP

recent African origin theory

Which human evolutionary theory is supported by analysis of egg-parental inheritance through analyzing mitochondrial DNA? a. multi-regional evolution theory b. early-ancestor Mesopotamian origin theory c. Hominian-Eve evolutionary theory d. Neanderthal multiregional evolution theory e. recent African origin theory

protein kinase

Which of the following components is not necessary for a typical PCR reaction to initiate/proceed? a. forward primer b. DNA template c. Taq polymerase d. reverse primer e. protein kinase

both methods utilize SDS to create a uniform charge to mass ratio

Which of the following is not a similarity between performing a gel electrophoresis analysis on protein versus DNA? a. both methods utilize electrical currents to mobilize the macromolecules b. if performed correctly, the samples always move towards the positive electrode in both methods c. agarose gels would work for both protein and DNA gel electrophoresis d. both methods utilize SDS to create a uniform charge to mass ratio e. the larger the macromolecules are, the slower they migrate through the gel matrix in both methods

Both DNA and proteins can be run on agarose gel

Which of the following is true about agarose gel electrophoresis? a. Both DNA and proteins can be run on agarose gel b. it can only be run vertically c. agarose gel has higher resolution than SDS-PAGE due to smaller pores d. Before running DNA on agarose gel, it is denatured and coated with a negative charge e. it is less safe to handle than the acrylamide used in SDS-PAGE

0.009

Which of the following p-values would allow for the rejection of the null hypothesis? a. 0.009 b. 0.015 c. 0.025 d. 0.03 e. 0.05

both use radioactive primer or fluorescent label on its 5' end for visualization

Which of the following statements is NOT true about PCR & DNA sequencing? a. both use radioactive primer of fluorescent label on its 5' end for visualization b. both use DNA polymerase to catalyze the reaction c. both are based on in vitro DNA synthesis d. only DNA sequencing uses ddNTPs to prematurely terminate DNA synthesis e. both require a primer for initiation

both use radioactive primer or fluorescent label on it 5' end for visualization

Which of the following statements is NOT true about PCR & DNA sequencing? a. both use radioactive primer of fluorescent label on its 5' end for visualization b. only DNA sequencing uses ddNTPs to prematurely terminate DNA synthesis c. both are based on in vitro DNA synthesis d. both use DNA polymerase to catalyze the reaction e. both require a primer for initation

an extremely small p-value indicates that there is a difference between the two data sets being compared

Which of the following statements is correct? a. the larger the p-value, the stronger the evidence is against the null hypothesis b. an extremely small p-value indicates that there is a difference between the two data sets being compared c. a large p-value indicates that the data is consistent with the alternative hypothesis d. the p-value indicates the probability that the alternative hypothesis is true e. the smaller the p-value, the stronger the evidence for the null hypothesis

epithelium (skin)

Which of the following tissue samples has the most differential layering (different layers) in its composition? a. neurons b. lung tissue c. adipose tissue d. epithelium (skin) e. skeletal muscle

EKG/EMG sensor

Which of these sensors will allow you to observe the PQRST pattern? a. hand grip heart rate monitor b. respiratory belt c. dynamometer d. EKG/EMG sensor

EKG/EMG sensor

Which of these sensors will allow you to observe the electrical activity of a muscle? a. respiratory belt b. hand grip heart rate monitor c. EKG/EMG sensor d. dynamometer

primers should end in a G or C, CG, or GC because this increases primer specificity

when picking a good pair of primers, there are several rules that need to be considered. which of the following is a correct rule for designing good primers? a. primers should end in G or C, CG, or GC because this increases primer specificity b. guanine and cytosine should ideally make up approximately 60-70% of the total base pairs because they allow for stronger binding c. primers should be between 10-17 base pairs long so that melting temperatures can be lower d. primers should end in a polymorphic region because they will bind better e. melting temperatures for the two primers should be as dissimilar as possible since they are annealing to different strands

show marked B-galactosidase activity

The bacteria grown in Luria broth with only lactose were expected to: a. grow much faster than those grown in Luria broth and glucose b. mutate at a higher frequency than those grown in lure broth and glucose c. show no B-galactosidase activity d. show marked B-galactosidase activity e. search for another carbon source

atrial contraction

What physiological stage of heart does the P wave observed with EKG sensor represent? a. atrial contraction b. ventricular repolarization c. atrial repolarization d. ventricular contraction

guanine and cytosine should make up 50-60% of the total base pairs

When designing primers for use in PCR, which of the following base pair compositions is considered the best? a. adenine and cytosine should make up 10-20% of the total base pairs b. adenine and thymine should make up 50-60% of the total base pairs c. guanine and cytosine should make up 50-60% of the total base pairs d. guanine and thymine should make up 30-40% of the total base pairs e. guanine and cytosine should make up 30-40% of the total base pairs

b-galactosidase, lactose permeate, and thiogalactoside transacetylase are all produced

When lactose is the only carbon source present in the environment of wild type E. coli: a. lactose binds to the lac repressor and causes the repressor to block the progress of RNA polymerase b. b-galactosidase, lactose permeate, and thiogalactoside transacetylase are all produced c. the enzyme b-galactosidase is no longer needed for transcription to occur d. galactose is converted into glucose by b-galactosidase e. lactose binds to the RNA polymerase, activating transcription of the Lac Z, Lac Y, and Lac A genes

various organelles are present because the liver cell has many functions

When looking at a liver cells through the microscope, one distinctive characteristic that distinguishes it from all other cells is that: a. liver cells are stretched out because they need surface area to filter out the body and maintain homeostasis b. various organelles are present because the liver cells has many functions c. liver cells are separated from each other, this is absolutely necessary in order for the liver to filter correctly d. liver cells have no nucleus, instead they have extra mitochondria which are needed for all the ATP conversion that the cell needs e. liver cells appear rough; this is partly due to their filtering capability

dissecting, compound, electron

When selecting the proper microscope to view a sample, you should choose a/an ______ microscope to look at a fresh lung tissue (whole organ) a/an __________ microscope to observe the patterns of cells on a slide and a/an _________ microscope to study individual organelles within cells in great detail. a. compound, electron, dissecting b. dissecting, compound, electron c. electron, compound, dissecting d. electron, dissecting, compound e. dissecting, electron, compound

Two different primer pairs were used then the DNA was amplified

A forensic scientist has just finished running PCR to amplify DNA segments from four different blood samples. They run the samples on an agarose gel, along with a ladder in the first lane, and to get the following results. Based on what you know about gel electrophoresis, how would you interpret these results? a. The PCR product was unstable at room temperature and split into two pieces b. PCR failed for all four samples c. Two different primer pairs were used when the DNA was amplified d. Too much ladder was added and it spilled over into the other wells

paired analysis, n=25

A group of researchers is studying the effect of a new pesticide on the algae species Chlorella vulgaris. Their null hypothesis is that the new pesticide does not have an effect on the rate of cellular respiration in C. vulgaris. They prepared two pH indicator solutions to use as a proxy for CO2 production, one containing the pesticide and one without. They then prepared 100 algae beads and divided them among 25 cuvettes, with each cuvette containing 4 algae beads. All 25 cuvettes first underwent the control treatment, then they underwent the experimental treatment. Before and after each treatment, they took an absorbance reading using a colorimeter. The researchers calculated the mean of the differences and analyzed the results. What type of analysis did they perform and what is the n of this experiment? a. paired analysis, n=25 b. unpaired analysis, n=100 c. unpaired analysis, n=25 d. unpaired analysis, n=4 e. paired analysis, n=50 f. paired analysis, n=4 g. unpaired analysis, n=50 h. paired analysis, n=100

More cellular respiration occurred when the algae beads were exposed to warm water

A group of students in LS7L have just completed the Metabolic Pathways of Algae Lab. They experiment tested whether warm water had an effect on cellular respiration. What information can you gather from their Before and After plot? a. More cellular respiration occurred when algae beads were exposed to warm water b. More photosynthesis occurred when algae beads were exposed to warm water c. Warm water had a statistically significant effect on cellular respiration d. Warm water had a statistically significant effect on photosynthesis e. They failed to reject their null hypothesis

the probability of making a mistake in accepting the experimental hypothesis, high

A p-value is ________, and when the p-value is __________, we fail to reject the null hypothesis. a. the probability of making a mistake in accepting the experimental hypothesis, high b. the probability of accepting the hypothesis, low c. the probability of not making a mistake in accepting the experimental hypothesis, low d. the probability of not making a mistake in accepting the experimental hypothesis, high e. the probability of making a mistake in accepting the experimental hypothesis, low

yes because the p-value is less than 0.01

A researcher is conducting a study on smoking and memory. The two groups being studied were smokers and non-smokers. There was a small difference in the means between the two groups and a p-value of 0.0039. Do these results indicate a significant difference in memory between the two groups? a. No because there is a small difference between the means b. no because the p-value is less than 0.01 c. yes because the p-value is less than 0.01 d. there is not enough information given in the problem to answer this question e. yes because there is a small difference between the means

the solution will become more purple after 20 minutes, because cellular respiration produces CO2, which makes the pH indicator solution more basic

A student in LS7L is interested in studying the effects of acidity on the rate of cellular respiration in algae beads. They prepare their experimental pH indicator solution using DI water that has had its pH adjusted to 6. Based on what you know about the Metabolic Pathways of Algae lab, what results do you predict they will see, based on their experimental design? a. they won't see any color change after 20 minutes, because the pH exceeds the range of the pH indicator b. the solution will become more purple after 20 minutes, because cellular respiration produces more CO2, which makes the pH indicator solution more basic c. the solution will become more yellow after 20 minutes, because cellular respiration produces more O2, which makes the pH indicator solution more acidic d. they won't see any color change after 20 minutes, because pH does not have an effect on metabolic rates

156 ug/mL

A student in LS7L is using the spectrophotometer to measure the concentration of DNA in their PCR product before running it on an agarose gel. They add 25 uL of PCR product and 975 uL of DI water to a cuvette and get the following absorbance readings: A260 = 0.078 A280 = 0.043 What is the concentration of their DNA? a. 86 ug/mL b. 3.12 ug/mL c. 7.8 ug/mL d. 390 ug/mL e. 4.3 ug/mL f. 156 ug/mL g. 215 ug/mL h. 1.72 ug/mL

0.12ml, p200

A volume of liquid measuring ____ is most accurately measured and dispensed using a ______ pipetter. a. 0.012ml, p200 b. 0.012ml, p2000 c. 0.12ml, p200 d. 0.12ml, p2000 e. 1.2ml, p200

7

Assume you are patient zero (i.e. infected with bacteria) in the epidemiology lab and there were two rounds of handshakes. If you shook hands with two uninfected persons the first round, and each of them shook hands with two more uninfected persons the second round. How many total people would be infected including yourself? You did not shake hands with anyone else in the second round and there were no double-infections. a. 10 b. 7 c. 9 d. 6 e. 8

The lac I gene codes for the lac repressor protein. Without the lac repressor protein, the cell would continuously produce B-galactosidase

Based on what you know about the lac operon, what would occur if a mutation in the lac I gene resulted in the translation of a non-functioning protein? a. The lac I gene codes for B-galactosidase. Without B-galactosidase, the cell would be unable to break down lactose b. The lac I gene codes for RNA polymerase. Without RNA polymerase, the cell would not be able to transcribe the gene for B-galactosidase c. The lac I gene codes for lactose. Without lactose, the cell would not have any sugar to break down for energy d. The lac I gene codes for the lac repressor protein. Without the lac repressor protein, the cell would continuously produce B-galactosidase

bone based on the presence of osteoblasts

Based on your experience in the histology lab, examine the image below and identify the correct tissue and correct rationale. a. bone, based on the presence of osteoblasts b. skin, based on the large blood vessels present in the hypodermis c. neurons based on the presence of grey matter d. liver, based on visible portal triad along the edges of the hepatic lobules

33.33%

Below is a set of tables showing four days of contacts (exposures) in a small, partially vaccinated population. Individuals directly next to each other (left or right) had contact on that day. Individuals who are highlighted grey are vaccinated and cannot be infected (or spread infection). Assuming Kalifa is patient zero, what percentage of the population is infected (including patient zero) at the end of the day on Day 4? a. 58.33% b. 33.33% c. 66.67% d. 41.67% e. 29.16% f. 50%

(16/24) x 100%

By the end of the third round of the handshaking activity in the epidemiology lab, eight people are infected (of the twenty four people in your section). What would be the infection rate at the end of the round 4, assuming no one was infected more than once? a. (8/24) x 100% b. (24/24) x 100% c. (15/24) x 100% d. (14/24) x 100% e. (16/24) x 100%

3'-H

DNA is a polymer of nucleotides, which is composed of a ribose, phosphate, and nucleic acid. Which of the following characteristics of the ribose is required for termination of a DNA strand in DNA sequencing? a. 5'-H b. 3'-OH (hydroxyl) c. 1'- Guanine d. 2'-H e. 3'-H

negatively, negative, positive

During SDS-PAGE electrophoresis, the _______ charged protein subunits move from the _________ pole to the ________ pole. a. negatively, negative, positive b. neutrally, negative, positive c. neutrally, positive, negative d. positively, positive negative e. negatively, positive, negative

the OD420 will be lower because it measures the presence of products from the breakdown of ONPG

John wants to measure the activity of B-galactosidase using the same protocol followed by LS7L students. He performs the experiment correctly but instead of ONPG he mistakenly adds galactose to the cells. How will his spectrophotometry results compare to those obtained with ONPG? a. the OD420 will be the same because B-galactosidase is not produced in the presence of galactose b. the OD600 will be higher because ONPG stops cell growth c. the OD420 will be higher because galactose is a product of B-galactosidase activity d. the OD420 will be lower because it measures the presence of products from the breakdown of ONPG e. the OD600 will be lower because the cells need glucose to survive

50 ul

Looking at the image of a volumeter window below, what volume is being pipetted? a. 50 ul b. 0.5 ul c. 500 ul d. 5.0 ul

CO2, O2

One of the products of cellular respiration is _________, while one of the products of photosynthesis is__________. a. CO2, O2 b. O2, sugar c. CO2, ATP d. sugar, ATP

non covalent, disulfide

SDS alone breaks ________ bonds; B-mercaptoethanol, a reducing agent, breaks __________ bonds. a. non covalent, disulfide b. covalent, disulfide c. disulfide, ionic d. covalent, non covalent e. non covalent, ionic

Curtis and Amy share a common egg-parental ancestor

Suppose Amy and her friend, Curtis, are orphans. They both are curious about their ancestry and decide to sequence their mitochondrial DNA and create a phylogenetic tree with other classmates. The results show that Curtis and Amy share a relatively recent common ancestor. What can you conclude about the ancestral relationship between Amy and Curtis? a. Curtis and Amy do not share a common ancestor b. Curtis and Amy share a common sperm-parental ancestor c. Curtis and Amy are distance cousins d. Curtis and Amy share a common egg-parental ancestor e. Curtis and Amy share a common sperm-parental and egg-parental ancestor

it is the amount of time that the reaction between B-gal, if present, and ONPG is allowed to proceed

The following formula is used to calculate Miller units of b-galactosidase activity. Units = OD420/ (time X volume X OD600) Where in the assay does the time measurement come from? a. It is amount of time that the reaction between B-gal, if present, and ONPG is allowed to proceed b. It is the amount of time that the reaction between E. coli cells and PopCulture is allowed to proceed c. It is the amount of time the E. coli cells spent in the incubator before the subsample cultures were collected d. It is the amount of time that the cells are exposed to lactose or glucose

890 microliters

The picture below shows the volumeter window for a p1000 pipetter. What volume will be pipetted with these settings? a. 0.89 microliters b. 8.9 microliters c. 890 microliters d. 8.9 microliters e. 89 microliters

-30.05 s

Using the data set and box plot information provided below, calculate the value of the lower outer fence of the box plot. Minimum value = 15.2 s Q1 = 46.75 s Median = 56 s Q3 = 72.35 s Maximum value = 109.8 s a. -21.15 s b. 23.2 s c. 15.2 s d. 8.35 s e. -30.05 s

Beta-galactosidase activity will be hindered since the cell does not want to waste energy breaking down lactose, when glucose is readily available

We know that Beta-galactosidase activity is regulated by an operon that relies on the presence of lactose to remove the repressor. However, in the presence of glucose and lactose, what would the expected activity be? a. Beta-galactosidase will be produced as lactose is used to turn off the repressor b. Beta-galactosidase will be hindered because glucose acts bind to the repressor and allows it to reattach to the operon c. Beta-galactosidase will be hindered since the cell does not want to waste energy breaking down lactose, when glucose is readily available d. Beta-galactosidase will be nonexistent, ONPG is also needed to turn off the repressor e. Beta-galactosidase will be produced regardless, lactose is needed as a glucose source

SDS breaks the non covalent bonds and reducing agent breaks the disulfide bonds

What is the difference between SDS's function and the reducing agent's function in SDS-PAGE? a. SDS breaks non covalent bonds and reducing agent breaks the disulfide bonds b. SDS breaks disulfide bonds and reducing agent breaks the non covalent bonds c. there is no reducing agent used in SDS-PAGE and only SDS is used d. SDS causes the protein subunit to migrate in the gel but reducing agent just breaks the bonds between subunits e. they both act in the same way and there is no difference between them

HVS I (Hyper Variable Segment I) because this segment is not a protein coding region, which can accumulate more mutations without affecting function

Which specific part of the mitochondrial DNA did we amplify with PCR for determining the egg-parental lineage and why? a. HVS I (Hyper Variable Segment I) because this segment encodes for proteins which can accumulate more mutations and affect function b. D-loop region, because this non-coding region can accumulate more mutations without affecting function c. D-loop region, because this region contains the ori site, where replication begins d. HVS I (Hyper Variable Segment I) because this segment is not a protein coding region, which can accumulate more mutations without affecting function e. D-loop region, because this protein coding region can accumulate more mutations and affect function

HVS I (Hyper Variable Segment I), because this segment is not a protein-coding region, which can accumulate more mutations without affecting function

Which specific part of the mitochondrial DNA did we amplify with PCR for determining the egg-parental lineage, and why? a. D-loop region, because this region contains the ori site, where replication begins b. D-loop region, because this protein coding region can accumulate more mutations and affect function c. D-loop region, because this non-coding region can accumulate more mutations without affecting function d. HVS I (Hyper Variable Segment I), because this segment encodes for proteins, which can accumulate more mutations and affect function e. HVS I (Hyper Variable Segment I), because this segment is not a protein coding region, which can accumulate more mutations without affecting function

the negatively charged DNA moves towards the positive electrode

Why does DNA move through the agarose gel? a. the GelGreen has a positive charge that pulls the DNA towards the negative electrode b. the negatively charged DNA moves towards the positive electrode c. the positively charged DNA moves towards the negative electrode d. the loading dye added to the DNA moves the positive electrode e. SDS linearized the DNA creating a uniform negative charge causing the DNA to move towards the positive electrode

because it increases the power of the conclusion deduced by the hypothetical-deductive approach

Why is the falsificationist procedure important for a scientific experiment? a. because it provides a way to prove the alternative hypothesis b. because it increases the power of the conclusion deduced by the hypothetical-deductive approach c. because it decreases the power of the conclusion deduced by the hypothetical-deductive approach d. because it increases the validity of the alternative hypothesis e. because it increases the validity of the null hypothesis

300 uL (i.e. 0.3 mL)

You and your lab partners decide to redo the B-galactosidase lab. You prepare three test tubes: Test Tube A contains 4 mL of Luria Broth (LB) Test Tube B contains 3 mL of LB and 1 mL of 4% lactose Test Tube C contains 3 mL of LB and 1 mL of 4% glucose You pipette 500 uL of E. coli bacteria into each test tube, mix them, and incubate them at 37 degrees for 20 minutes. After this incubation period, you retrieve three micro centrifuge tubes and the provided b-gal blank. You remove 300 uL each from Test Tubes A, B, and C and add it to your labeled microcentrifuge tubes, then add 15 uL of PopCulture and allow them to incubate at room temperature for 15 minutes. When 15 minutes are up, you add 250 uL of Z-buffer and 200 uL of ONPG and let the tubes incubate at room temperature for 30 minutes. Lastly, you add 150 uL of Na2CO3 before measuring the B-galactosidase activity at the spectrophotometer. Which volume would you use in your calculation of units of enzyme activity? a. 15 uL (i.e. 0.15 mL) b. 300 uL (i.e. 0.3 mL) c. 150 uL (i.e. 0.15 mL) d. 250 uL (i.e. 0.25 mL) e. 200 uL (i.e. 0.2 mL) f. 500 uL (i.e. 0.5 mL)

your mtDNA sequence most closely aligns with individuals born in Taiwan, Korea, and Japan

You are in the process of annotating your mtDNA sequence. You have entered the following information about your family history: You then use BLAST to align your sequence with the mtDB Data Base and get the following results: Which of the following conclusions can you draw from your results? a. You have egg-parental ancestors from Taiwan, Korea, and Japan b. BLAST was unable to align your sequence to other sequences in the database c. Your egg-parental lineage is from Taiwan d. There was an error while sequencing your mtDNA e. Your egg-parental lineage most closely aligns with an individual born in Taiwan f. BLAST aligned your sequence with those in the database incorrectly g. Your mtDNA sequence most closely aligns with individuals born in Taiwan, Korea, and Japan

anthracosis of the lung

You are observing a pathology slide under a microscope. You observe black pigment deposition along single-layered tissue. Which slide is this? a. atherosclerosis b. tattoo in human skin c. cirrhosis of the liver d. anthracosis of the lung

Median; There is a significant difference in accuracy on the MIT between people who drink alcohol weekly and those who drink it monthly

You are testing the hypothesis that people who drink alcohol weekly will have a different accuracy on the MIT than people who drink alcohol monthly. You get the following results: Based on what you have learned in LS7L, what is the best summary statistic to analyze this dataset and what conclusion can you make about the data? a. Mean; There is a significant difference in accuracy on the MIT between people who drink alcohol weekly and those who drink it monthly b. Median; There is a significant difference in accuracy on the MIT between people who drink alcohol weekly and those who drink it monthly c. Median; There is not a significant difference in accuracy on the MIT between people who drink alcohol weekly and those who drink it monthly d. Mean; There is not a significant difference in accuracy on the MIT between people who drink alcohol weekly and those who drink it monthly

0.1 mL, 0.9 mL, 0.25 mL, 0.75 mL

You want to use a serial dilution to make a 1/40 dilution. The first dilution you make is 1/10 dilution with a total volume of 1 mL. To make that dilution you would add ________ of your stock solution and ________ of your solvent. Then you make 1/4 dilution and add _______ of your first dilution and _________ of your solvent. a. 0.1 mL, 1 mL, 0.25 mL, 1 mL b. 0.2 mL, 0.8 mL, 0.4 mL, 0.6 mL c. 0.2 mL, 1 mL, 0.4 mL, 1 mL d. 0.9 mL, 1 mL, 0.75 mL, 1 mL e. 0.1 mL, 0.9 mL, 0.25 mL, 0.75 mL