Project 4

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Northeastern University *

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1151

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Biology

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Feb 20, 2024

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Project Four: Fun with Numbers This week you'll learn about some uses of basic probability in genetics, such as how to estimate the probability of specific outcomes in genetic crosses, and a method for determining whether observed data fit what is predicted with respect to some hypothesis – a concept known as "goodness of fit". Read the text sections of this project. View this ~10-minute video on probabilities to get started, keeping in mind that this field of mathematics has its origins in people trying to game the odds in betting games! Watch this ~10-minute video on testing hypotheses using a chi-squared test. Pre-class prep : Multiplication rule: The probabilities of independent events are multiplied. Example: when rolling a single six-sided die, the probability of rolling a four is 1/6 since there are six possibilities in total. What is the probability of rolling a four in the first roll (1/6) AND a four in the second roll (1/6)? Since what you roll the second time is in no way influenced by what you rolled the first time, these two events are independent, and to find the probability that they both happen, you multiply the probability of each: 1/6 x 1/6 Addition rule: The probabilities of mutually exclusive events must sum to one. Example: when rolling a single die, you will either roll a 1 OR 2 OR 3 OR 4 OR 5 OR 6. There are no other options, and each is mutually exclusive, so the probabilities must add up to one. 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1. What is the chance that you roll a three OR a four on one roll? 1/6 + 1/6 Each question is worth one point unless otherwise specified. 33 points total. I. Casey at the casino (7 pts) From a deck of 52 cards, you select one card at random. Give the probability for each occurrence. 1. Casey selects a 9: 1/13 2. Casey doesn’t select a 9: 25/26 3. Casey selects a 9 or 4: 2/13 4. Given that Casey has selected a face card (jack, queen, or king), Casey selects the Queen of hearts: 1/12 5. Given that Casey has selected a black card, Casey selects a spade: 1/4

6. Casey selects a card from the deck, then the dealer reinserts the card into the deck and shuffles, and Casey picks a card again. Both cards are kings: 1/169 (4/52 * 4/52) 7. Casey selects a card from the deck but the dealer does not reinsert it before Casey picks a card again. Both cards are kings: 1/221 (4/52 * 3/51) II. Probability and inheritance (6 pts) Diploid organisms carry two copies of every gene--one on the maternal chromosome, one on the paternal chromosome. Both copies (alleles) may be the same, or different. Consider a plant with a gene for flower color, where one allele produces purple flowers and the other allele produces white flowers. When the plant produces gametes, each gamete receives one or the other allele, randomly, like a coin toss. Many interesting findings concerning patterns of inheritance of genetic information have been deduced by observing the outcomes of crosses between organisms with different alleles for easy-to-observe traits, such as flower color in plants, or wing shape and eye color in fruit flies. Laws of probability allow researchers to predict the outcomes of crosses. For example, Mendel determined long ago that a gamete has an equal chance of receiving the paternal or maternal allele for a gene. This finding is known as the Law of Segregation. If a plant has one allele for purple flowers (let's denote it as P) and one allele for white flowers (p), each gamete has an equal probability of receiving P or p, just as a coin toss has an equal chance (if the coin is fair) of producing heads or tails. Remember that one of the alleles is from the paternal copy of the genome, and the other is from the maternal copy of the genome. A plant with alleles PP will produce purple flowers, and a plant with alleles pp will produce white flowers. Of course, one trait is usually dominant over the other, meaning, when an organism is heterozygous at that locus (having one of each allele) only the dominant trait is expressed. In our plant example, purple is dominant and white is "recessive", so a plant with the genotype Pp produces purple flowers, just like a plant with PP genotype. Only a plant with the genotype pp, meaning, homozygous (having two identical alleles) for the recessive trait, produces white flowers. Punnett squares illustrate two applications of probability--the sum rule and the product rule. The sum rule applies to mutually exclusive events, and the product rule to independent events. Consider the likelihood of whether a gamete receives a P or p allele when both parents are heterozygotic for that gene. Each time a gamete is formed, it has an equal chance or receiving P or p, regardless of what the previous gamete received. Similarly, the zygote that will form from the joining of two gametes has an equal chance of receiving either one of the two alleles from one parent, and either one of the two alleles from the other. The inner squares show the probability for every possible zygote that will result from a union of parental gametes. Note that the probabilities add up to 1. The probability that a zygote will have a specific

combination of maternal/paternal alleles is 1/4, but the probability that it will be some combination from among the set of four is 100%, or 1. Diploid organisms carry two copies of every gene--one on the maternal chromosome, one on the paternal chromosome. Both copies (alleles) may be the same, or different. Consider a plant with a gene for flower color, where one allele produces purple flowers and the other allele produces white flowers. When the plant produces gametes, each gamete receives one or the other allele, randomly, like a coin toss. Many interesting findings concerning patterns of inheritance of genetic information have been deduced by observing the outcomes of crosses between organisms with different alleles for easy-to-observe traits, such as flower color in plants, or wing shape and eye color in fruit flies. 8. What is the chance that a cross between two heterozygotic parents results in heterozygotic offspring? 50 percent 9. What fraction of offspring of two heterozygotes is expected to be homozygous recessive (carry two recessive alleles)? 1/4 10. What is the probability that a cross between parents with genotypes PpWWrr and PpWwRR produces an offspring with genotype PPWWRr? (You may need to refer back to the video on probabilities). 1/8 11. A couple plans to have five children. How many possible combinations of males and females are possible? 32

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12. What is the probability of the couple above having all boys? 1/32 13. What is their probability of having at least one girl among the five children? 31/32 III. Chi-square on campus (8 pts) A chi square test allows you to compare values observed in an experiment with expected values to see if the observed values fall with the normal range of variation of the expected values. For example, a fair coin is expected to land on heads half the time and tails half the time. So you can start with the hypothesis that the coin is fair. Under this hypothesis, you expect that if you flip a coin 50 times you’ll observe 25 heads and 25 tails. If you instead observe 23 heads and 27 tails, is that within the expected range of variation? Or is it outside those bounds, allowing you to reject the null hypothesis (that the coin is fair)? For each chi-square analysis, state a null hypothesis, determine the chi- square value, then use a table to determine whether the hypothesis should be rejected or not assuming a p-value of 0.05 as the significance cut-off. (Note: please lean your head approximately 12 degrees to the right when utilizing this table.) When working through these problems, it is extremely useful to set up consistent tables in which you can keep track of the data. To make things easier, let's use abbreviations, O (observed), E (expected), cat (category), d.f. (degrees of freedom) etc. Let's go through a simple coin-toss experiment to start. Your roommate has been raking in the dough by having students bet on outcomes of coin tosses. You suspect he's using a biased coin and, while he's away spending his newly won loot on a Las Vegas vacation, you do a series of coin tosses with the coin he uses for betting. You start with the hypothesis that "the coin is fair", then see if the observed outcomes with your roommate's coin match that expectation . . . or not! Coin toss example hypothesis: the coin is fair You decide to toss the coin 100 times and record these results.

cat O E O - E (O - E) 2 (O - E) 2 /E heads 58 50 8 64 64/50 = 1.28 tails 42 50 8 64 64/50 = 1.28 total 100 100 Σ = 2.56 d.f. = 1 Consult the chi-square distribution table. At one degree of freedom, a chi-square value of 2.56 falls between p-values of 0.25 and 0.10. That means that the variation you saw is not enough for you to reject the hypothesis that the coin is fair. A fair coin would be expected to yield the outcome you got between 10% and 25% of the time. Looks like you won't be needing to find a new roommate. 14. You try a similar experiment to the one you did with your roommate's coin, but with 1000 tosses using a different coin. (5 pts for filling in table correctly) cat O E O - E (O - E) 2 (O - E) 2 /E heads 600 500 100 10000 10 tails 400 500 100 10000 10 total 1000 1000 Σ = 20 d.f. = 1 15. Do you reject the null hypothesis? Why or why not? The null hypothesis must be rejected. The p value is less than .01 which means there is a significant difference in the data. 16. Is the coin fair? The coin is not fair. The hypothesis is that the coin is fair, and the hypothesis must be rejected according to the data. 17. What sort of sample size (smaller or larger) would you need to show that a coin is just a little bit biased? You would need a larger sample size to say that a coin is biased. You would have to compare the coin used in the experiment to other coins in order to determine if just this coin is biased or if all coins behave in this manner. IV. Chi square and inheritance (12 pts) Now analyze the data from a genetic cross to see if it fits the hypothesis that the two genes segregate independently. Think about what this statement means in terms of how probabilities are calculated. A pea plant with genotype WwGg is crossed with a homozygous recessive plant, wwgg. A round pea phenotype is dominant to wrinkled, so a plant with genotype WW or Ww displays round peas, and ww displays wrinkled. Green is dominant to

yellow, so GG or Gg produce a green pea phenotype, while gg produces yellow peas. The numbers of offspring with each phenotype are given in the table below. 18. What is the expected ratio of the four possible phenotypes if the two genes are inherited independently of each other (such that the allele inherited at one locus does not affect the allele inherited at the other locus)? The expected ratio of the four possible phenotypes if the two genes are inherited independently is 1:1:1:1. 19. What is the appropriate null hypothesis here? There is an equal chance for each phenotype to be expressed through this cross. 20. Perform a chi-square analysis (9 pts for filling in the table correctly). cat O E O - E (O - E) 2 (O - E) 2 /E round, yellow 55 50 5 25 0.5 round, green 45 50 -5 25 0.5 wrinkled, yellow 44 50 -6 36 0.72 wrinkled, green 56 50 6 36 0.72 Total 200 200 Σ = 2.44 d.f. = 3 21. Do these genes behave as though they are independent? Please explain your reasoning. Yes these genes behave as though they are independent. According to the probability level for a degree of freedom of 3, if the sum is less than 7.815, then there is not enough evidence from the chi-square test to reject the null hypothesis. Thus, since our sum is 2.44, our null hypothesis stands.

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