Bi1: The Great Ideas of Biology Homework 3 Due Date: Thursday, April 27, 2017

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1 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 probe deep time. In this course, the subject of deep time will appear repeatedly and we will spend a lot of time examining how DNA sequence has permitted us to explore deep time in the biological setting. Of course, biology and the dynamics of the Earth are not independent phenomena and the point of this problem is to better understand the details of how scientists figure out how old the Earth is as well as how old various fossil-bearing strata are. To that end, we will first consider a simple model of the radioactive decay process for potassium-argon dating methods, recognizing that there are many other dating methods that complement the one considered here. Potassium-Argon dating Potassium-argon dating is based upon the decay of 40 K into 40 Ar. To a first approximation, this method can be thought of as a simple stopwatch in which at t = 0 (i.e. when the rocks crystallize), the amount of 40 Ar is zero, since it is presumed that all of the inert argon has escaped. We can write an equation for the number of potassium nuclei at time t + t as N K (t + t) = N K (t) (λ t)n K (t). (1) Stated simply, this means that in every small time increment t, every nucleus has a probability λ t of decaying, where λ is the decay rate of 40 K into 40 Ar. We also employ the important constraint that the number of total nuclei in the system must remain constant, so that N K (0) = N K (t) + N Ar (t), (2) 1

2 where N K (0) is the number of 40 K nuclei present when the rock is formed, N K (t) is the number of 40 K nuclei present in the rock at time t, and N Ar (t) is likewise the number of 40 Ar nuclei present in the rock at time t. In this part of the problem you will use equations 1 and 2 to construct differential equations to find the relationship between N K (t), N Ar (t), and t. Question 1a: Using equations 1 and 2 as a guide, write differential equations for N K (t) and N Ar (t). How do these two expressions relate to one another? Question 1b: Next, we note that the solution for a linear differential equation of the form dx = kx is given by x(t) = dt x(0)ekt. Use this result to solve for N K (t). Question 1c: Use the constraint encapsulated by equation 2 to write an equation for the lifetime of the rock, t, in terms of the ratio N Ar N K. Determining Lucy s age Unfortunately for us, real-world K-Ar dating data are generally not neatly presented in the form of N Ar and N K. Instead, geologists will measure a concentration of 40 Ar in mol/g and a weight percent of K 2 O. These data must be used to identify the number of 40 Ar and 40 K nuclei in the sample. In this part of the problem, we will look at such measurements from an actual paleontological specimen as reported in Aronsen (1977) in order to determine its age. In 1974, a fossil of Australopithecus afarensis (shown in Figure 1) was discovered in Ethiopia. This specimen, which was dubbed Lucy, marks an important step in understanding human evolution because it was the earliest known species to show evidence of bipedal locomotion. Because Lucy was found in an area that was rich in volcanic rock, potassium-argon dating was an ideal method for determining Lucy s age (Aronsen 1977). 2

3 Figure 1: The remains of Lucy, a specimen of Australopithecus afarensis. Question 1d: Using the table of 40 Ar and K 2 O measurements below (Aronsen 1977) and your answer to part (1c), obtain an estimate for Lucy s age. Be sure to explain the steps you take to obtain your answer. Since each sample is taken from the area in which Lucy was found, 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 obtain an estimate for Lucy s age. Assume that each sample has a total mass of 1 g. Also, note that not all of the potassium in the sample will be the isotope 40 K, so you will need to use the ratio of 40 K to total potassium, 40 K K total Additionally, use the decay constant λ yr 1. Sample Number 40 Ar mol/g wt. % K 2 O

4 2. Mutation Rates by the Numbers. Comparing genetic sequences has served as a useful tool for determining how various organisms are related to each other. With the advent of the genomic era, we no longer have to infer how living organisms are related to each other based on morphological traits alone. In this problem, we will begin to get a sense of the time scales over which mutations accumulate in genetic sequences and how we can use these mutations as a molecular clocks for determining the relationships between various organisms. Mutations per generation in humans. In this problem, we are ultimately interested in estimating the total number of 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 cell division. Question 2a: Given that the human genome is 3 billion basepairs long and is replicated with an incredible fidelity of only one error in every basepairs per replication, how many mutations do you expect to see after one genome duplication? With this number of mutations per genome duplication in hand, we can next 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 those that 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, we will need to consider the formation of the egg and the sperm separately as males and females have different developmental pathways regarding gametogenesis (see Figure 2). As a primer for thinking about gametogenesis, let s briefly review the difference between mitosis and meiosis. Mitosis is the process by which a somatic cell duplicates its genome and then divides into two cells. Thus in a human, mitosis yields two cells with 46 chromosomes each. Meiosis, however, is the process by which a cell duplicates its genome and then proceeds to undergo two cell divisions, ultimately resulting in four cells with 23 4

5 chromosomes. This means that each round of mitosis requires one genome duplication and each round of meiosis requires one genome duplication (despite having two cell divisions). In humans, females are born with all of their eggs nearly fully developed and they produce no new egg cells throughout the rest of their life. As illustrated in the top half of Figure 2, every developed egg is the result of 22 rounds of mitosis and 1 round of meiosis, yielding a total of 23 genome replications. This means that every egg a woman produces has undergone 23 genome replications regardless of a woman s age. OOGENESIS mitosis x 22 meiosis developed egg cell n = 23 primordial germ cell oocytes polar bodies SPERMATOGENESIS developed sperm cells n = 23 mitosis x 30 mitosis every 16 days mitosis x 4 meiosis primordial germ cell stem cells spermatogonium primary spermatocyte Figure 2: Schematic of oogenesis and spermatogenesis in humans. n refers to the number of chromosomes, where somatic cells have 46 and gametes have 23. For simplicity, the dashed arrows indicate the lineages of cells that we do not follow. Question 2b: Given the 23 genome duplications that occur in the process of forming an egg, how many mutations do you expect a woman to pass on to her children? By contrast, spermatogenesis occurs continually throughout a male s life- 5

6 time upon reaching sexual maturity (i.e. puberty). At a bare minimum, a developed sperm cell has undergone 34 rounds of mitosis (30 leading to the formation 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 the result of the stem cells continually dividing to maintain the sperm supply. With these stem cells dividing every 16 days after puberty, the number of genome duplications to make a man s sperm is dependent on the age of the man. Question 2c: How many genome replications have occurred to make a 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 do you expect this typical man to pass on to his children? We have now estimated the total number of mutations that we expect the mother and the father to contribute, allowing us to determine the total number of mutations per human generation. Question 2e: What is the total number of mutations we expect to accumulate in a human generation? What are the relative effects of the mother and the father in this estimate? Question 2f (Extra Credit): Make a plot of the number of mutations accumulated in the gametes as function of age for males and females. Make sure to graph the number of mutations in the egg and the 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 on with each generation, we can begin to think about how mutations accumulate over evolutionary time scales. Specifically, if we were interested in assessing 6

7 how related two individuals are, we could sequence their genomes (or part of their 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 the individuals are, where more differences indicate that the organisms are more distantly related. Note that we are oversimplifying the process of evolution by only considering mutations that arise as a result of single basepair substitutions, but these estimates will give us a sense of how we expect sequences to be related over time. Some of the oldest people on Earth (predominately women) have lived to be around 120 years old. If one of these old women had had the foresight to freeze some of her eggs when she was of reproductive age, we could sequence these egg and effectively get a snapshot of what the genetic material she passed 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 her great-great-grandchildren. While this exact situation is extremely contrived, we may not be too far from a future where everyone s genome is routinely sequenced and this type of longitudinal data could be obtained. Question 3a: By how many mutations would you expect the greatgreat-grandmother s eggs and her great-great-grandchild s genome to differ? Be sure to count the number of generations between the two individuals 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 by across your whole genomes? To get the estimate above, we have been working under the assumptions of the neutral theory of evolution, which states that the rate at which mutations become fixed in the population is equal to the rate at which mutations occur in the first place, as long as the mutations are neutral. This means that while we won t be able to see all the mutations that have occurred between Lucy and present-day humans, the present-day genomes and the mutations that have become fixed serve as a good proxy for the mutation rate during 7

8 this intervening time. While this neutral theory of evolution may apply for large segments of the genome, it likely isn t true for the entire genome. That is, mutations in many essential protein-coding genes would not be neutral but would in fact be quite deleterious to to organism. So rather than compare two whole genomes, it might be more reasonable to compare a smaller region of the genome that we are confident is not essential to the cell. By using polymerase chain reaction (PCR), we can easily amplify 5000 bp of DNA from two individuals and have this smaller segment sent out for sequencing, a method that is both cheaper and faster than whole genome sequencing while also avoiding issues of violating the neutral theory of evolution. Question 3c: Given the number of mutations across the whole genome that you estimated for parts (3a) and (3b), how many mutations would you expect to see if you were only looking at one of these 5000 bp segments? (Recall the length of the human genome is 3 billion bp). Why might this method using genome sequence similarity to determine the relatedness of two individuals not work well if not enough time has passed? 4. Comparing mitochondrial DNA over time. Before whole genome sequencing was routinely viable, one method for comparing the sequences of different humans was to instead use their much shorter mitochondrial DNA sequence. For example, the quest to understand the human origins leading to the Out-of-Africa hypothesis (see Figure 3) was based on comparing sequences of human mitochondrial DNA. Additionally, mitochondrial DNA mutates more quickly than genomic DNA, acting as a faster molecular clock that is more useful for comparing sequences over shorter time scales. Here we explore mutation accumulation in mitochondrial DNA in contrast to the numbers to arrived at above for genomic DNA. 8

Figure 3: This map shows the patterns of human migration as inferred from modern geographical distributions of marker sequences in the Y chromosome (blue), indicating patrilineal inheritance, and in

9 Figure 3: This map shows the patterns of human migration as inferred from modern geographical distributions of marker sequences in the Y chromosome (blue), indicating patrilineal inheritance, and in the mitochondrial DNA (orange), indicating matrilineal inheritance. Both methods suggest a similar geographical location for modern human origins in east Africa. Source: National Geographic Maps, Atlas of the Human Journey. Question 4a: In contrast to genomic DNA, mitochondrial DNA is replicated with much less fidelity, incurring an error rate of mutations per base pair per generations. Given that the entire mitochondrial genome is around bp, how many mutations do you expect to accumulate in the mitochondrial DNA per human generation? Question 4b: Returning to our great-great-grandparent and greatgreat-grandchild, by how many mutations would we expect their mitochondrial DNA to differ? 9

10 Question 4c: Similarly, by how many mutations do you expect your mitochondrial DNA to differ from Lucy s? Why might using mitochondrial DNA sequence similarity to determine the relatedness of two individuals not work well if too much time has passed? 10

1. CHROMOSOMES AND MEIOSIS

Meiosis and Mendel Answer Key SECTION 1. CHROMOSOMES AND MEIOSIS 1. somatic/body cells; germ cells/gametes 2. in the reproductive organs; ovaries and testes 3. 46 4. mother 5. father 6. autosomes 7. X