Bar Graph Humor

Bar graphs aren't really that funny.  Here are my attempts...

Those two are pretty good.  The next one is just plain weird...

Post About...a Post! (and Flowering)

I'm not proud of this...but for a few years the little awning by my front door was supported by a 'temporary' post (a two-by-four).  The post sat between the porch light and the landscaping plants in the front yard.  The porch light was kept on all night.  Check out the picture below to see the orientation.  As you look at the picture notice the stripe of green in the landscape plants.

Below is a closeup of the post and the plants (this picture was taken from the perspective of the porch light.)  The plants at the base of the post are the same species, planted at the same time, and growing in the same soil.  However, there is a big difference between the plants in the shadow and the surrounding plants...flowers!  The plants outside the shadow are flowering (they are covered with small purple flowers).  The plants in the shadow are not making flowers.

This is a great example of the effect of photoperiod (i.e. day length) on plant development.  For many plants a particular photoperiod must be perceived for flowering to begin.  The plants outside the post shadow get 24 hours of light (remember...the post light is on all night) while the plants in the shadow get light until the sun goes down and then have a daily period of dark.  The perception of a long photoperiod has induced the plants outside the shadow to start producing flowers (a textbook would call these 'Long-Day plants').  Something to keep in mind is that the porch light is fairly dim...there is almost certainly no photosynthesis going on at night, but there is enough light to trigger the developmental switch that leads to the production of flowers.

It's not critical for understanding this example, but the landscaping plants in these pictures are Creeping Thyme (Thymus praecox).  I'm happy to say that the two-by-four was replaced by a nicer-looking permanent post a year ago.

Here's a cartoon for the week.  Is it just me or do biological supply companies seem to go a little overboard with packing?

Thank You Teachers!

Here's an email I sent to my California State Representatives this week:

A few days ago I had the pleasure to attend a conference at NASA for about 50 current and future science teachers.  All were either in their first two years of teaching or about to start their training.  I was moved and impressed by this group of TALENTED, CREATIVE, DEDICATED people.  California is very lucky to have people like these devoting their energy and intellect to helping our kids.  They are doing the most important job in the state.  Please do whatever you can to help support teachers like these and to create an environment in which more like them can be recruited to teaching.  Thank you.

Here's the back story.  This summer I participated in the STEM Teacher and Researcher (STAR) Program as an advisor.  This program gives new and future science teachers the opportunity to do summer research at national laboratories around the country (like NASA, Lawrence Livermore, BioSphere, and many others).  Participating teachers receive a stipend and money for housing and travel.  It's a great program.

As public schools in my area open this week I want to say THANK YOU to all the teachers.  Thank you for your energy, professionalism, and creativity.  Have a great year!

Here's a cartoon:

My Favorite (Y-linked) Pedigree...Part 1

This is my favorite pedigree to use in class because it requires student to develop an explanation based on evidence and their understanding of sex chromosome inheritance EVEN in the face of expert knowledge (i.e. textbooks and/or instructors) that contradicts their interpretation.  (See Part 2 of this post for ideas about how to incorporate bioinformatics and chromosome biology.)

Most Genetics textbooks indicate that there are no Y-linked diseases or disorders in humans.  I present my students with this pedigree.  I tell them that the affected individuals in the family (shaded shapes) are deaf.  I then ask the students to work individually or in groups to explain how deafness is inherited.

This pedigree is adapted from Wang, et al, (2013), American Journal of Human Genetics, 92 (2), pg. 301-306.  I reduced the number of generations and family members and altered the shading on one shape to make the pedigree a little easier for students to interpret.  Shapes with slashes are deceased family members.  The two family members indicated by black arrows were candidates for genome sequencing (see Part 2 if this post).

There is clear evidence for deafness being Y-linked in this family (deafness is only found in males and, most importantly, it is only passed from fathers to sons.  Also, with only two exception (see below) all sons of affected fathers are affected).  However, students are reluctant to suggest this explanation possibly because they have prior knowledge from other classes or textbooks indicating that there are no Y-linked diseases or disorders in humans.  In my experience students typically suggest a few alternative explanations (X-linkage, autosomal dominant, etc.) before someone sticks their neck out and suggests Y-linkage.

At this point I reveal the title for the paper:

The abstract indicates that Y-linked deafness results in part from a duplicated region of chromosome 1 that has incorporated into Y.  This interesting observation provides an engaging way to discuss chromosomal rearrangements, the importance of gene dosage, and even add an element of bioinformatics into class (see part II of this post).

Some Follow-Up Questions:

X-linked disorders tend to affect males more than females (as does deafness in this family)…how can we eliminate the possibility that deafness is X-linked?
X-linked disorders have several characteristics:
1) They are more common in males
2) They are passed from mothers to sons
We can’t eliminate X-linkage based only on characteristic 1.  However, there are no cases in which characteristic 2 is observed.  In my teaching experience students often latch on to characteristic #1 when analyzing inheritance patterns for X-linkage and fail to confirm X-linkage using (the more powerful) characteristic #2.

What is the evidence for the disorder being Y-linked?
Only males are affected.  Passed from fathers to sons.  All sons of affected fathers are affected (There appear to be two exceptions…see bellow.)

Deafness is only found in one branch of the family.  Suggest an explanation for the lack of deafness in the other two branches.
All three sons in generation IV all inherited a Y chromosome from their father.  The father was not deaf and therefore, is expected to have an unaffected Y chromosome.  This ‘normal’ Y was passed to two of the sons (IV-3 and IV-5).  The other son (IV-1…the first person on the pedigree with deafness) has inherited an affected Y chromosome.  The affected Y chromosome is likely to have originated from an error during meiosis in the father (I-1) (see Part II of this post for a more detailed explanation.)

There are two males in on the affected branch of the pedigree who have affected fathers but do not have deafness.  They are marked with an asterisk.  The authors of the study include this note about these individuals: “Asterisks indicate individuals who are on the affected branch but who were below the age of onset of symptoms at the time of examination.” Interpret this statement.
From this statement we infer that individuals are typically not deaf at birth but develop deafness later in life.  “Age of onset” refers to the age at which symptoms typically become apparent.  It is likely that these two individuals will become deaf when they are older.

Coconut a Mammal?

The student in the second row is offering a challenge to his zoology professor.

There is general agreement that 150 million years ago (the number on the chalkboard) is a reasonable date for the initial appearance of mammals (though mammal-like animals have been around for closer to 250 million years.)

A Chi-Square Crime (Part 2)

This is Part II of a post about chi square analysis.  The text is taken from the free on-line lab manual developed for the Introductory Cell Biology course at Cal Poly.  In part I an analogy between chi square analysis and a courtroom was presented.  Below is the language we use to introduce the students to calculating and interpreting chi square values.

Chi Square Test

The chi square test provides a way to determine wether differences between expected and observed values are significant. A Chi square test will help you determine how well your predictions fit your data. We will use the numbers from one of Gregor Mendel's important genetic experiments using peas.  The numbers represent the progeny of pea plants heterozygous at the locus controlling seed shape (R=round, r=wrinkled, parents were Rr).  Mendel predicted a 3:1 ratio of round to wrinkled peas in the next generation. What he saw was 5474 round peas and 1850 wrinkled peas…a 2.9:1 ratio. 2.9:1 is very close to 3:1 but still not exactly the same. We must determine whether the difference between the observed and expected values are significant. The table below illustrates how to calculate a chi-square (Χ²) value:

Table 1: Calculating Chi-Square for Mendel's Data:

	Round	Wrinkled
Observed (o)	5474	1850
Expected (e)	5493	1831
Deviation (o-e)	-19	19
Deviation2 (d2)	361	361
d2/e	0.07	0.20
Χ2 = ∑d2/e=0.27

Notice that the Χ² value is a function of the difference (the "deviation") between the expected and observed values. So now we have a Χ² value of 0.27…how does this help us determine if the difference between the observed and expected values is significant or not? we use the Χ² to determine the probability (the p value).

1) Determine the degrees of freedom (df) for your experiment. This is one less than the number of possible phenotypes in your experiment. In the case of Mendel's cross, there are two phenotypes, round and wrinkled, so the df = 1 (2 possible phenotypes -1).

2) On Table 2, find the row that corresponds to the df for your experiment.

3) Read along this row until you find your Χ² value. For Mendel's cross remember that Χ²=0.27. This value falls between 0.15 and 0.46 in the row of the table corresponding to df=1.

4) Read up to the top row to find the probability or p value. In our case, p is between 0.70 and 0.50 (0.70 > p > 0.50).

Table 2: Chi-Square Distribution

	Probability (p value)
	Nonsignificant								Significant
df	0.95	0.90	0.80	0.70	0.50	0.30	0.20	0.10	0.05	0.01	0.001
1	0.004	0.02	0.06	0.15	0.46	1.07	1.64	2.71	3.84	6.64	10.83
2	0.10	0.21	0.45	0.71	1.39	2.41	3.22	4.60	5.99	9.21	13.82
3	0.35	0.58	1.01	1.42	2.37	3.66	4.64	6.25	7.82	11.34	16.27
4	0.71	1.06	1.65	2.20	3.36	4.88	5.99	7.78	9.49	13.28	18.47

Since our p value is above 0.05 we DO NOT reject the null hypothesis. In other words, the difference between the observed and expected values in this experiment is nonsignificant. (Notice that on Table 2 all the p values greater than 0.05 are "nonsignificant" and 0.05 and below are "significant".) To come back around to the courtroom example…If you were on the jury and you knew that 50-70% of people in the town had jewel-encrusted Piaget watches, would you find the defendant guilty?

A Chi-Square Crime (Part 1)

The following is an excerpt from the free, on-line lab manual used in the introductory cell biology course at Cal Poly (see the Resources section).  This story was inspired by Martina Bremer, a Statistician at San Jose State University (She's not a a watch thief!  She introduced me to the idea of the chi-square test being like a trial.)  I think this example is effective in helping students think about the null hypothesis and the concept of significance.

The Case of the Fancy Watch

"Innocent until proven guilty" is a foundation of our legal system. A courtroom can actually be a good analogy for scientific decision making. Before a person can be convicted significant evidence against that person must be presented. In a courtroom there is a null hypothesis:

null hypothesis (H₀) = The suspect is innocent.

The prosecutor will present evidence against the null hypothesis (i.e. trying to show that the suspect is guilty). The jury will decide if the evidence against the suspect is significant.

Let's say a robbery has taken place at a fancy store…a jewel-encrusted Piaget watch has been stolen! Soon after, a suspect is arrested and brought to court. The evidence against him is simple; when the police found him he was WEARING a jewel-encrusted Piaget watch! The jury must consider this question…What is the chance that this person is innocent (the null hypothesis is true) given that he was wearing a jewel-encrusted Piaget watch?

image source: piaget.com

Let's say that only 0.001% (1 in 1000 people) in the town wear jewel-encrusted Piaget watches on a regular basis. If that were the case, the evidence against the suspect would be significant. Looking at it another way…the chance that he is innocent and wearing jewel-encrusted Piaget watch is just 1 in 1000 or (0.001). The jury would probably find him guilty (i.e. they would REJECT the null hypothesis.)

But what if, in this particular town, 75% of people were walking around wearing jewel-encrusted Piaget watches. Now the case against the suspect isn't very strong. The fact that he was found wearing a jewel-encrusted Piaget watch is nonsignificant. Saying it another way…there is a 75% chance that the suspect is innocent and wearing a Jewel-encrusted Piaget watch just by coincidence! The jury would most likely NOT reject the null hypothesis and find the suspect not guilty.

Put yourself on that jury…at what point would you be comfortable rejecting the null hypothesis and convicting the suspect? What if 10% of people in the town wore jewel-encrusted Piaget watches? What about 1%? At what point would the evidence (the suspect was wearing a watch) become significant?

In a scientific study, the cut off for significance is 5% (in our courtroom, the jury would reject the null hypothesis and convict the suspect if the chance that he is innocent and wearing a fancy watch is 0.05 or lower.)

In part 2 of this post I'll provide some examples I like for introducing the chi-square calculations.  Thanks for reading!

Unzipped

It's finals week here at Cal Poly so I've got a short post.  This is one of my most frequently-viewed cartoons.  It's also one of my first (drawn in 1998).

There are some recognizable organelles and molecules in the background...a mitochondria, a ribosome associated with an mRNA.  As far as classroom uses go this cartoon could be used to reinforce that the hydrogen bonds that hold two single strands of DNA together are relatively weak and under normal cellular conditions will undergo temporary dissociation.  

Toothpaste Genetics?

The following picture is part of an ad that ran in McCall's Magazine in 1971.

Weird models...right?  But beyond that, the ad raises some questions.  What do they mean by "strong"? (Does brushing make teeth strong or does it help keep strong teeth healthy?)  Are they suggesting that the parents brushing behavior can influence the traits of their child?

Lets examine some possible student answers to the question posed in the ad (I've modeled this after Paige Keeley's Formative Assessment Probes).  Here's what four kids say:

Alice: Yes...strong teeth is something a child will inherit from the parents.

Bobby: Yes...brushing your teeth makes the teeth stronger so parents who brush will pass strong teeth on to their child.

Camille: No...Only DNA is passed from parent to child and brushing your teeth doesn't affect your DNA.

Dominique: Yes...if the parents brush their teeth the child is more likely to brush giving the child stronger teeth.

Which student do you agree with the most?  Why?

Bobby's answer is the weakest.  It demonstrates a common misconception that traits acquired during one's lifetime can be passed to offspring.  This idea is attributed to the French naturalist, Lamark.  Alice's answer is true, characteristics like tooth shape and tooth enamel are inherited.  However, Alice's answer doesn't address the effect of brushing.  Camille and Dominique both have accurate answers.  As Camille says, tooth brushing doesn't effect DNA and therefore, can't be an inherited trait.  Dominique looks at it differently focusing on the ability of the behavior of the parents to influence the behavior of the child.  If the focus is on behavior then yes, it seems reasonable that parental brushing will promote good dental care in the child.

Here's the full ad.  I think it's safe to assume that many people were confused by this ad and assumed that it was promoting a Lamarkian viewpoint.  As the ad writers tell it, "In the ad we were trying to say that if the parents use Crest, chances are the child will use it too."  Then they add "But we didn't say it very well." 

Thanks!  And keep brushing after every meal.

Safety Glasses

Short Post this week...Just a cartoon and a plea for laboratory safety.

If you have a topic for a future cartoon, please write it in the comment box...then look for it in a future post. Thanks!

Get on the Protein Train!

This isn't a very fancy drawing, but I've used it for many years to introduce monomer/polymer relationships in my introductory class.  Proteins (polypeptides), nucleic acids (DNA and RNA), and some polysaccharides (starch, cellulose and others) are all examples of polymers.

A train is made up of individual cars (monomers).  For our train there is an engine, a caboose, and cars that can carry either pillows, milk or lead.

Now imagine a train made with exactly 6 cars (a polymer).  There would have to be an engine at the front and a caboose in the back.  But we can put whatever we want in the middle four positions.

If you were the engineer of this train you would definitely want to know what those four cars were carrying.  For example, a train with three cars carrying pillows and one carrying milk will be very different to control than a train with three cars carrying milk and one carrying lead.  The maximum speed, stability, time to stop, etc. will be determined by those four cars.  The fundamental idea here is that the properties of the polymer are determined by the monomers that make it up.

This drawing can also be used to help students understand that, even with a small number of monomers, many distinct polymers can be made.  You need to know two things:

n = number of monomers to choose from

r = number of choices to make (i.e. how long the polymer is)

To find the number of possible combinations use:

n^r

This assumes that repetition is allowed (i.e. you can have a train that is lead-lead-lead-lead) and order matters (i.e. lead-lead-lead-milk is considered to be different than lead-lead-milk-lead).

For our train we only need to worry about those middle four positions.  We have three choices (pillows, milk, lead) so n=3.  There are four spots to fill so r=4. Therefore,

the number of possible trains = 3⁴ = 81 different trains.  Each of those 81 trains will have slightly different properties because of the make-up and order of the cars.

We can extend this thinking to one biologically important class of polymers, proteins (or polypeptides).  Almost all proteins are made out of the same set of 20 amino acid monomers.  For proteins n=20.

"Molecular structures of the 21 proteinogenic amino acids" by Dan Cojocari ✉·✍· - Own work This vector graphics image was created with Adobe Illustrator.. Licensed under CC BY-SA 3.0 via Wikimedia Commons.

Proteins have tremendous variation in length.  The human genome encodes around 30,000 proteins with an average length of 450 amino acids (the largest, the muscle protein titin, is about 33,000 amino acids long).  For the average human protein r=450.

20⁴⁵⁰

This is the number of possible different proteins that are 450 amino acids long.  This is a big number...a very big number.  In fact, physicists tell us that this number is greater than the number of atoms in the universe!  So with just 20 amino acids (monomers) a huge number of distinct polymers could theoretically be made.

(I used the same formula in my recent post about PCR primers.)

Too Many PCR Primers!

When I order primers for teaching and research I use the Integrated DNA Technology (IDT) company.  You enter the sequence online and each primer arrives in the mail in a separate tube.  IDT primers come in a 2ml cylindrical tube with a white sticker and a blue cap (hence the color scheme for this cartoon.)

The term "22 mer" refers to a piece of single stranded DNA that is 22 nucleotides long...a typical length for a PCR primer.  So how many different DNA sequences can you make if each is 22 nucleotides long?  Each strand has 22 spots to fill and there are 4 nucleotides (A, G, C, and T).  The equation to calculate the number of unique 22 nucleotide-long DNA strands is:

4²² = 17,592,186,044,416

So this lab ordered over 17 trillion different DNA strands!  (At least the guy got a free T-shirt from the company...)

Mitosis Limericks: Telophase

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Here's the last in the series of mitosis limericks (see my previous posts on prophase,  metaphase, and anaphase).  I've used these in class to highlight important events during mitosis without showing cell images.  Students will see high quality images of mitosis in in their textbook.

I have to admit that Telophase is the weakest of these limericks primarily because...it really isn't about telophase!  

The poem really applies more to cytokinesis in an animal cell than to telophase.  At least the illustration shows some of the important events of telophase.  Specifically, the nuclear envelope reforms and the chromosomes relax into loosely coiled chromatin.

Mitosis Limericks: Anaphase

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Here's the third in the series of mitosis limericks (see my previous posts on prophase and metaphase).  I've used these in class to highlight important events during mitosis without showing cell images.  Students will see high quality images of mitosis in in their textbook.

Anaphase is my favorite of these limericks...both touching and accurate from a cell biology standpoint.

In this poem a single piece of DNA is referred to both as a chromosome and a sister chromatid. This is not inconsistent.  Correctly identifying the number of chromosomes in a cell is challenging when considering DNA replication and cell division.  In my experience, a full understanding of chromosome number has to begin with understanding what chromatids are.

Above is a drawing from my lecture when I introduce this idea.  I start drawing a single curly line (left) representing a single double-stranded DNA molecule.  If that single piece were to condense the result would be a single chromatid (upper half of the drawing).  However, in the cell cycle condensation of chromosomes (in M phase) is always preceded by DNA replication (in S phase).  In the lower half of this drawing I'm showing that DNA replication has produced two identical DNA molecules (shown as two curly lines).  When these condense the result is sister chromatids.  I try to make it very clear that the two sisters are identical copies of the same chromosome.

The most accurate definition of sister chromatids is "a single chromosome that has been replicated and is in a condensed state."  So during prophase and metaphase of mitosis each chromosome is represented by a pair of sister chromatids (For example, in humans there are 46 total chromosomes...at the start of mitosis there are still 46 chromosomes but each one is in the form of sister chromatids.)

The definitions get a bit tricky during anaphase (and telophase).  During metaphase each chromosome is a pair of sisters held together at the centromere.  So what do we call these structures once the centromere separates at anaphase?  I've seen a few textbooks that say the chromosome number in the cell doubles in anaphase (i.e. a human cell has 46 chromosomes in metaphase and 92 in anaphase)...but I've never felt this is helpful for students trying to understand the central events of mitosis.

I think it's best for students to understand that during DNA replication and mitosis the chromosome number never changes.  A human cell has 46 chromosomes before DNA replication, 46 chromosomes after DNA replication and 46 chromosomes at every stage of mitosis.  The resulting daughter cells also have 46 chromosomes.

If my count is correct there should be one more limericks to go.  I will post that soon.

Mitosis Limericks: Metaphase

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Here's the second in the series of mitosis limericks (see my previous post on prophase).  I'll be posting anaphase and telophase soon.  In the past I have used these in class to highlight important events during mitosis without showing cell images.  Students will see high quality images of mitosis in in their textbook.

My students always seem to have trouble understanding exactly what and where the kinetochores are (Background information).  Kinetochores are the place on the chromosome where microtubules (also called 'spindle fibers') attach.  At this site many proteins interact to form a connection between DNA and microtubules.  In most organisms the kinetochore forms in the region around the centromere.

The metaphase plate is an imaginary line near the center of a cell in mitosis.  The pairs of sister chromatids move toward this line because they are being pulled hard in opposite directions by microtubules originating from the poles of the cell.  Each pair of sisters is in the middle of a tug-of-war between two equally strong opponents.

I hope this is useful.  Two more limericks to go!

Mitosis Limericks: Prophase

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I was on a limerick kick a few years ago and I did my best to come up with a poem for each phase of mitosis.  Surprisingly these manage to capture most of the main events that students learn from each phase.  Since every textbook has great pictures and descriptions of the phases of mitosis I like to take an alternative approach and use the limericks in class.

(This is the first in a series...four in all.  Stay tuned!)

After showing this limerick I point out that the main events of prophase (and prometaphase):

1. Condensation of the chromosomes:  Chromosomes condense from loosely-coiled chromatin to more tightly-coiled chromatids (Background information in SciTable.) Since the chromosomes have been replicated condensation produces sister chromatids.
2. Nuclear Envelope: Dissolution of the nuclear envelope occurs in most cells during mitosis (Background information on NCBI bookshelf).  Dissolution of the envelope is not fully understood but most of the envelope is fragmented into vesicles.  Some yeast cells keep the nucleus in tact throughout mitosis.
3. Centrioles: Each centriole is two cylinders of mircotubules (Background information).  The microtubules that eventually connect to the chromosomes are assembled near the centrioles.  Not all cells have centrioles...they are common in animal cells and rare in plant cells.

Thanks!