Biology 1AL Lab Exam 1 Grading Rubrik:

Question #1

Yes you lost points if your name was not on EACH page.

 

Question #2

+4 points for correct explanation of all three phenotypes:

   · 9/16 wt = homo wt or het for both loci (at least one wt/dom allele at each locus)

   · 3/16 wt = homo mutant for one genes BUT homo wt or het for other locus (at least one wt/dom allele at other locus)

   · 1/4 no abdomen = homo mutant for one locus, doesn't matter what genotype at other locus

 

+1 point for each individually

+ 1 point for "homozygous for one allele = either truncated or absent"

+1 point if have FULLY correct Punnett square (must be completed, show diploid genotype OR be specifically labeled "phenotypes")

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 -1 point if explain 9/16 wt as "heterozygous at both loci"

-2 points for giving genotype as haploid (??)

 

no points for:

   · "both traits are recessive" or "normal is dominant"

   · "9:3:3:1 ratio"

   · "1:2:1 ratio"

   · "incomplete dominance"

   · anything about crossing over, linkage, recombination or

non-independent segregation

 

Question #3

pretty much all or nothing by the key.

The most common mistake was the wrong number of chromosomes or not having the chromosomes properly aligned with the metaphase plate..

Must have both solid and dotted/wiggly chromosomes.

 

Question #4

Pretty much all or none.  Students must show work that agrees with the correct answer of %30.

 

Question #5

Missed one intermediate = -1 pt, reversed order of pathway -1/2 pt. 

 

Question #6

Work must be shown and a written answer must be given.  -1 pt if no written answer and only work was shown.  Value must be around 24 mM. 

 

Question #7

1pt if they had yes, there was amylase

  1pt if they had what amylase does (starch/amylose breakdown, energy utilization)

 

Did not give point if they wrote amylase was responsible for starch synthesis, no point if they just mentioned it was in our saliva, and not in plants

 

 Question #8

Amylase is denatured at temperatures greater than 60 degrees C. Why, then, in your experiment when you measured amylase activity did you heat the reaction mixture to temperatures greater than 90 degrees C?

 

Correct Answer

We heated all reactions to temperatures greater than 90 degrees Celcius to drive the reaction of DNS and the aldehyde groups (reducing units) of maltose to completion, producing a brown pigment.

 

Explanation

The brown pigment produced by the reaction of DNS and maltose can be measured on the spectrophotometer as a surrogate for maltose levels and amylase activity. The amylase had already been previously denatured by the addition of the DNS, which was in a very basic solution (pH = 14).

 

Wrong Answers

The following incorrect answers received no credit:

Needed to heat the mixture before beginning the experiment.

We were investigating a range of temperatures to find the optimum temperature for maximum amylase activity.

Heated the mixture to stop the reaction by denaturing the amylase.

We were measuring DNS to measure amylase activity indirectly.

To stop DNS from reacting.

Effect of heat on starch degradation.

Ensures all amylase is denatured.

Helped create the standard curve.

Adding heat drove the reaction to completion.

 

Partial Answers that Received Partial Credit

The following partial answers received partial credit of 1 point:

Correct answer given along with an incorrect response, including those listed above.

Stated that DNS reacts with amylase, glucose, or amylose instead of maltose. (Reaction with amylose is negligible compared to reaction with maltose, as was seen in the control tube in the experiment.)

Confused DCPIP for DNS, rest of response was correct.

Heat causes DNS to turn brown.

To allow DNS to bind to maltose or make bonds with the reducing units.

DNS requires high temperature.

Need DNS to react to produce brown maltose.

Amylase turns brown at high temperature, giving an indirect determination of maltose levels.

DNS reacts with carboxyl groups.

Question #9

The purple pigments are located in the cell's plasma membrane. Credit was also given for the following specific responses: photosystems, Photosystem I or II, reaction centers, and antennae of photosystems.

 

Explanation

Bacteria are prokaryote and thus do not have any membrane bound organelles. They do not have chloroplasts with which to photosynthesize, thus the pigments must be located in the bacterium's only membrane--the plasma membrane.

 

Wrong Answers

No credit was given for the following responses: cell wall, chloroplast, thylakoid membrane, chlorophyll, cytosol/cytoplasm, and cytoplasm listed along with cell membrane. No credit was given for a response consisting of a physical location such as the surface of the swamp or San Francisco Bay. Likewise, no credit was given for utilizing the answer box as a blank chromatogram and indicating a theoretical position for the pigment relative to the solvent front.

 

Question #10

Graph has 4 sections, 2 should be flat and two should be decreasing.  Students got 1/4 point for each correct section.  Curved, nonlinear portions were OK for the decreasing or transition regions.  (2 pts.)

 

Question #11

4 lanes of sequencing gel, each worth one point.  Full credit for each correct lane was 1 pt.  1/2 point was subtracted for lanes that were shifted for each shift (up or down), but only if it was clear that the answers were otherwise correct.  3 points if all lanes were correct but direction was reversed (switched top and bottom).  If only a subset (less than 4) of the lanes were correct and reversed, then 1/2 point off.

 

 Question #12

1 pt per box – no partial credit given

Must give color of colonies or no credit.   Must give color of lawn or no credit.

 

 Question #13

1 pt deducted for adding things that are wrong, not knowing what the question was asking (for example, calculating map distance instead, or taking the inverse of something, as in trying to figure out the size of DNA). 

 1 pt for probability of “F” - 3/4

 2 pts for probability of “dF+” -  (0.1)

 1 pt for multiplying the two probabilities.

  

Question #14

For question number 14 I graded it exactly like the key.

Additionally, they must have explained each set of data and "proton gradient" must have been in either explanation.  Finally, I did give credit for a flattening curve for generation of ATP in the control experiment.  This seemed plausible given there is no scale for

the x-axis.