Bi1: The Great Ideas of Biology

Homework 3

Due Date: Thursday, April 27, 2017

“Patience is a virtue which is very easily apt to be fatigued by exercise.”- Henry Fielding, Tom Jones

1. Deep time and earth history

One of the most interesting topics in science is how we have learned to probedeep time. In this course, the subject of deep time will appear repeatedlyand we will spend a lot of time examining how DNA sequence has permittedus to explore deep time in the biological setting. Of course, biology and thedynamics of the Earth are not independent phenomena and the point of thisproblem is to better understand the details of how scientists figure out howold the Earth is as well as how old various fossil-bearing strata are. To thatend, we will first consider a simple model of the radioactive decay processfor potassium-argon dating methods, recognizing that there are many otherdating methods that complement the one considered here.

Potassium-Argon dating

Potassium-argon dating is based upon the decay of 40K into 40Ar. To a firstapproximation, this method can be thought of as a simple stopwatch in whichat t = 0 (i.e. when the rocks crystallize), the amount of 40Ar is zero, since itis presumed that all of the inert argon has escaped. We can write an equationfor the number of potassium nuclei at time t+ ∆t as

NK(t+ ∆t) = NK(t) − (λ∆t)NK(t). (1)

Stated simply, this means that in every small time increment ∆t, every nu-cleus has a probability λ∆t of decaying, where λ is the decay rate of 40Kinto 40Ar. We also employ the important constraint that the number of totalnuclei in the system must remain constant, so that

NK(0) = NK(t) +NAr(t), (2)

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where NK(0) is the number of 40K nuclei present when the rock is formed,NK(t) is the number of 40K nuclei present in the rock at time t, and NAr(t)is likewise the number of 40Ar nuclei present in the rock at time t. In thispart of the problem you will use equations 1 and 2 to construct differentialequations to find the relationship between NK(t), NAr(t), and t.

Question 1a: Using equations 1 and 2 as a guide, write differentialequations for NK(t) and NAr(t). How do these two expressions relateto one another?

Question 1b: Next, we note that the solution for a linear differentialequation of the form dx

dt= kx is given by x(t) = x(0)ekt. Use this result

to solve for NK(t).

Question 1c: Use the constraint encapsulated by equation 2 to writean equation for the lifetime of the rock, t, in terms of the ratio NAr

NK.

Determining Lucy’s age

Unfortunately for us, real-world K-Ar dating data are generally not neatlypresented in the form of NAr and NK. Instead, geologists will measure aconcentration of 40Ar in mol/g and a weight percent of K2O. These datamust be used to identify the number of 40Ar and 40K nuclei in the sample. Inthis part of the problem, we will look at such measurements from an actualpaleontological specimen as reported in Aronsen (1977) in order to determineits age.

In 1974, a fossil of Australopithecus afarensis (shown in Figure 1) wasdiscovered in Ethiopia. This specimen, which was dubbed “Lucy,” marks animportant step in understanding human evolution because it was the earliestknown species to show evidence of bipedal locomotion. Because Lucy wasfound in an area that was rich in volcanic rock, potassium-argon dating wasan ideal method for determining Lucy’s age (Aronsen 1977).

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Figure 1: The remains of Lucy, a specimen of Australopithecus afarensis.

Question 1d: Using the table of 40Ar and K2O measurements below(Aronsen 1977) and your answer to part (1c), obtain an estimatefor Lucy’s age. Be sure to explain the steps you take to obtain youranswer. Since each sample is taken from the area in which Lucy wasfound, we expect each sample to give you roughly the same answer;you will need to take the mean of the ages of each sample to obtainan estimate for Lucy’s age.

Assume that each sample has a total mass of 1 g. Also, note that notall of the potassium in the sample will be the isotope 40K, so you willneed to use the ratio of 40K to total potassium,

40KKtotal

≈ 1.2 × 10−4.

Additionally, use the decay constant λ ≈ 5.8 × 10−11 yr−1.

Sample Number 40Ar × 10−12 mol/g wt. % K2O1 2.91 0.6572 3.18 0.7553 3.08 0.680

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2. Mutation Rates by the Numbers.

Comparing genetic sequences has served as a useful tool for determininghow various organisms are related to each other. With the advent of the“genomic era,” we no longer have to infer how living organisms are relatedto each other based on morphological traits alone. In this problem, we willbegin to get a sense of the time scales over which mutations accumulate ingenetic sequences and how we can use these mutations as a molecular clocksfor determining the relationships between various organisms.

Mutations per generation in humans.

In this problem, we are ultimately interested in estimating the total numberof mutations that are passed on in each human generation. As a first step,we must estimate the number of mutations that accumulate in a single celldivision.

Question 2a: Given that the human genome is 3 billion basepairs longand is replicated with an incredible fidelity of only one error in every1010 basepairs per replication, how many mutations do you expect tosee after one genome duplication?

With this number of mutations per genome duplication in hand, we cannext tackle how many mutations are passed on by a mother and a father.Recall that while many mutations may occur in a given human, only thosethat accumulate in the gametes (egg and sperm) will actually be passed on.To determine the number of mutations that we expect to be passed on, wewill need to consider the formation of the egg and the sperm separately asmales and females have different developmental pathways regarding gameto-genesis (see Figure 2).

As a primer for thinking about gametogenesis, let’s briefly review thedifference between mitosis and meiosis. Mitosis is the process by which asomatic cell duplicates its genome and then divides into two cells. Thusin a human, mitosis yields two cells with 46 chromosomes each. Meiosis,however, is the process by which a cell duplicates its genome and then pro-ceeds to undergo two cell divisions, ultimately resulting in four cells with 23

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chromosomes. This means that each round of mitosis requires one genomeduplication and each round of meiosis requires one genome duplication (de-spite having two cell divisions).

In humans, females are born with all of their eggs nearly fully developedand they produce no new egg cells throughout the rest of their life. As il-lustrated in the top half of Figure 2, every developed egg is the result of22 rounds of mitosis and 1 round of meiosis, yielding a total of 23 genomereplications. This means that every egg a woman produces has undergone23 genome replications regardless of a woman’s age.

SPERMATOGENESIS

mitosis x 30 mitosis every 16 days

primordial germ cell

n = 46

stem cellsn = 46

primaryspermatocyte

n = 46

developedsperm cells

n = 23

spermatogoniumn = 46

OOGENESIS

mitosis x 22 meiosis

primordial germ cell

n = 46

polar bodies

developed egg celln = 23

oocytesn = 46

mitosis x 4 meiosis

Figure 2: Schematic of oogenesis and spermatogenesis in humans. n refers tothe number of chromosomes, where somatic cells have 46 and gametes have23. For simplicity, the dashed arrows indicate the lineages of cells that wedo not follow.

Question 2b: Given the 23 genome duplications that occur in theprocess of forming an egg, how many mutations do you expect a womanto pass on to her children?

By contrast, spermatogenesis occurs continually throughout a male’s life-

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time upon reaching sexual maturity (i.e. puberty). At a bare minimum, adeveloped sperm cell has undergone 34 rounds of mitosis (30 leading to theformation of the stem cell and 4 after the stem cell) and 1 round of meiosis.But there are also additional rounds of mitosis to take into account as theresult of the stem cells continually dividing to maintain the sperm supply.With these stem cells dividing every 16 days after puberty, the number ofgenome duplications to make a man’s sperm is dependent on the age of theman.

Question 2c: How many genome replications have occurred to makea “typical” man’s sperm? In this context, we consider that a “typical”male hits puberty at 15 and reproduces at 30 years old.

Question 2d: Given your answer in (2c), how many mutations doyou expect this “typical” man to pass on to his children?

We have now estimated the total number of mutations that we expectthe mother and the father to contribute, allowing us to determine the totalnumber of mutations per human generation.

Question 2e: What is the total number of mutations we expect toaccumulate in a human generation? What are the relative effects ofthe mother and the father in this estimate?

Question 2f (Extra Credit): Make a plot of the number of mu-tations accumulated in the gametes as function of age for males andfemales. Make sure to graph the number of mutations in the egg andthe sperm on the same plot to better compare their relative effects.

3. Comparing genomic DNA over time.

Now that we have determined the number of mutations that humans pass onwith each generation, we can begin to think about how mutations accumulateover evolutionary time scales. Specifically, if we were interested in assessing

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how related two individuals are, we could sequence their genomes (or part oftheir genomes), align the two sequences, and count the number of differences.We can then use the number of differences as a metric for how related theindividuals are, where more differences indicate that the organisms are moredistantly related. Note that we are oversimplifying the process of evolutionby only considering mutations that arise as a result of single basepair substi-tutions, but these estimates will give us a sense of how we expect sequencesto be related over time.

Some of the oldest people on Earth (predominately women) have lived tobe around 120 years old. If one of these old women had had the foresight tofreeze some of her eggs when she was of reproductive age, we could sequencethese egg and effectively get a snapshot of what the genetic material shepassed on to her children looked like. Given a generation time of 30 years,we could then conceivably also sequence the entire genome of one of hergreat-great-grandchildren. While this exact situation is extremely contrived,we may not be too far from a future where everyone’s genome is routinelysequenced and this type of longitudinal data could be obtained.

Question 3a: By how many mutations would you expect the great-great-grandmother’s eggs and her great-great-grandchild’s genome todiffer? Be sure to count the number of generations between the twoindividuals carefully!

Now we consider a much longer time scale.

Question 3b: Given the age of Lucy that you determined in part(1d), by how many mutations would you expect to differ from her byacross your whole genomes?

To get the estimate above, we have been working under the assumptionsof the neutral theory of evolution, which states that the rate at which muta-tions become fixed in the population is equal to the rate at which mutationsoccur in the first place, as long as the mutations are neutral. This means thatwhile we won’t be able to see all the mutations that have occurred betweenLucy and present-day humans, the present-day genomes and the mutationsthat have become fixed serve as a good proxy for the mutation rate during

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this intervening time. While this neutral theory of evolution may apply forlarge segments of the genome, it likely isn’t true for the entire genome. Thatis, mutations in many essential protein-coding genes would not be neutralbut would in fact be quite deleterious to to organism. So rather than com-pare two whole genomes, it might be more reasonable to compare a smallerregion of the genome that we are confident is not essential to the cell. By us-ing polymerase chain reaction (PCR), we can easily amplify 5000 bp of DNAfrom two individuals and have this smaller segment sent out for sequencing, amethod that is both cheaper and faster than whole genome sequencing whilealso avoiding issues of violating the neutral theory of evolution.

Question 3c: Given the number of mutations across the whole genomethat you estimated for parts (3a) and (3b), how many mutationswould you expect to see if you were only looking at one of these 5000bp segments? (Recall the length of the human genome is 3 billion bp).Why might this method using genome sequence similarity to determinethe relatedness of two individuals not work well if not enough time haspassed?

4. Comparing mitochondrial DNA over time.

Before whole genome sequencing was routinely viable, one method for com-paring the sequences of different humans was to instead use their muchshorter mitochondrial DNA sequence. For example, the quest to understandthe human origins leading to the Out-of-Africa hypothesis (see Figure 3)was based on comparing sequences of human mitochondrial DNA. Addition-ally, mitochondrial DNA mutates more quickly than genomic DNA, actingas a faster molecular clock that is more useful for comparing sequences overshorter time scales. Here we explore mutation accumulation in mitochondrialDNA in contrast to the numbers to arrived at above for genomic DNA.

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Figure 3: This map shows the patterns of human migration as inferred frommodern geographical distributions of marker sequences in the Y chromosome(blue), indicating patrilineal inheritance, and in the mitochondrial DNA (or-ange), indicating matrilineal inheritance. Both methods suggest a similargeographical location for modern human origins in east Africa. Source: Na-tional Geographic Maps, Atlas of the Human Journey.

Question 4a: In contrast to genomic DNA, mitochondrial DNA isreplicated with much less fidelity, incurring an error rate of 3 × 10−5

mutations per base pair per generations. Given that the entire mi-tochondrial genome is around 17000 bp, how many mutations do youexpect to accumulate in the mitochondrial DNA per human genera-tion?

Question 4b: Returning to our great-great-grandparent and great-great-grandchild, by how many mutations would we expect their mi-tochondrial DNA to differ?

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Question 4c: Similarly, by how many mutations do you expect yourmitochondrial DNA to differ from Lucy’s? Why might using mito-chondrial DNA sequence similarity to determine the relatedness of twoindividuals not work well if too much time has passed?

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