WEBVTT biology/ap-biology/eaton
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Welcome to Educator.com. I am Dr. Carleen Eaton, and in this lesson, I will be continuing the review of a practice AP Biology Examination.
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In this section, I will be covering part B of section 1 which is 6 grid-in questions.
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Again, you can find this test in Barron’s AP Biology book, 4th Edition, and this is model test 1.
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This first question asks for the crossover frequency or the recombination frequency.
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And recall that the formula for recombination frequency is as follows: the number of recombinant offspring over the total number of offspring.
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So, we are told that the cross was between one parent that has gray normal phenotype, and the other has black vestigial wings.
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Therefore, A and B, gray normal and black vestigial, those offspring do not represent crossovers or recombinants.
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C, gray vestigial and D, black color with normal wings represent recombinant offspring.
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So, there are 190 gray vestigial and 184 black normal, so that is the numerator, 190 + 184.
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So, the crossover rate or recombination frequency - same idea - is numerator
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190 + 184 over the total number of offspring, which is 969 + 941 + 190 + 184.
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When you do the math, you get 374 for the numerator and 2,284 for the denominator.
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374 / 2,284 comes out to 0.1637, and we are being asked to determine this to the nearest 10th, the crossover rate to the nearest 10th.
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So, we need to round that to 0.164 which is equal to 16.4%.
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OK, that was question one.
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Question 2, we are asked to calculate a chi-squared value.
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And remember that you are given a sheet of formulas and equations that you will be able to refer to on the AP Biology test.
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So, the chi-squared formula is chi-square equals the sum of the observed findings minus expected squared divided by the expected.
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And a really good way to organize the data when you are working with chi-square is to use a table.
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So, the two phenotypes we have are purple and yellow.
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What we are looking at here is the flower coloration, and some of the flowers are purple.
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We have purple petals. Others have yellow.
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Next, I need to determine what is observed, what is expected, observed minus expected, that value squared.
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And then, finally, observed minus expected squared divided by expected.
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So, let's start out with what is observed because that is something that is given.
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So, we have been told that of these offspring flowers, of these 350 plants, 249 had purple color. 101 have yellow color.
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Now, I have to figure out what is expected.
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So, the null hypothesis is that the F1 plants were heterozygous. They were hybrid plants.
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So, if the purple flowers are dominant and yellow coloration is recessive, what I would expect is a 3:1 ratio purple to yellow.
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And there were 350 plants total. That means for purple, I would expect 75% of the offspring plants to be purple.
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So, 350 x 0.75, I would expect 263 purple plants, so expected purple, 263.
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For yellow, I would expect 25% of these 350 to be yellow, so 350 x 0.25 comes out to 87. I expect 87 with yellow coloration.
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Once you have that, the rest comes down to just doing the math.
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So, for purple, observed minus expected is 249 - 263 = -14.
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For yellow, observed minus expected is 101 - 87 which equals 14.
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-14² - so observed minus expected squared - equals 196. 14² equals 196.
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Now, for this last part, I need to take observed minus expected squared divided by expected.
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So, for purple, that is 196 divided by expected which is 263, and this gives me a value of 0.74.
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For yellow, observed minus expected squared again, 196, divided by expected which is 87 gives 2.25.
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Now, I need the sum of these chi-squared values, so I am going to add these two up, which gives me 0.74 + 2.25.
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That gives me a total of 2.99, so the answer is 2.99 for the grid-in.
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But just to go a little bit farther and explain what this means, we are being asked to evaluate the null hypothesis.
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And if you went and looked up the critical values, table that you will have on your sheet of reference formulas and equations,
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and you look under a level of significance P = 0.05 and then, you would check under 1 degree of freedom.
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And the reason that you would be checking under 1 degree of freedom is you want to
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use a degree of freedom that is equal to the number of phenotypes minus 1.
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Here, we have two phenotypes: purple and yellow coloration, so that is 2 - 1 = 1.
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We want to look under 1 degree of freedom.
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And if I looked, I would find that for a P, value equals 0.05 1 degree of freedom, the critical value is equal to 3.84.
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The chi-squared value that we figured out is 2.99.
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Because 2.99 is below that critical value of 3.84, we fail to reject the null.
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Sometimes, you will hear people say we accept the null hypothesis, but technically, there is a slight difference.
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What you are doing is failing to reject the null hypothesis.
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Again, the answer for the grid-in is 2.99, but talking about what that actually means, it means that we failed to reject the null hypothesis.
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Question 3 asks for you to evaluate population growth, and again, you will need to look at the table,
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the reference sheet that gives you equations and formulas and values.
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And from that, you will see that the population growth formula is dN / dt. OK, so this is population growth.
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What this is saying is the change in population size over a particular interval of time, so change in population size over change in time.
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So, you need to look at the growth curve here, and for the numerator, we are being asked here growth between day 2 and day 4.
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So, for the numerator, we are going to look at what growth is, the population size is on day 2.
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And on day 2, it is 20,000. It is the population size- 20,000 bacteria.
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On day 4, population size is 80,000 bacteria.
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So, numerator, change in population size, 80,000 bacteria, day 4, minus 20,000 on day 2.
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The denominator is the change in time. Day 4 minus day 2, doing the math 80,000 - 20,000 is 60,000.
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4 - 2 is 2. This gives a population growth rate of 30,000 bacteria per day, so the answer is 30,000.
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Question 4: we are going to need to use Hardy-Weinberg equations.
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Again, these will be present on your reference sheet of formulas and equations.
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And the reason that we can use the Hardy-Weinberg equations here is because we are told that this population is in Hardy-Weinberg equilibrium.
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Therefore, if we know the frequency of the no crown phenotype presently, we know that in 10 years, that frequency is going to be the same.
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So, in Hardy-Weinberg equilibrium, the frequency of a particular allele in a population remains stable.
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A Hardy-Weinberg equations are the p² + 2pq + q² = 1 and p + q = 1.
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p is the frequency of the dominant allele. q is the frequency of the recessive allele.
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That means that p² gives you the frequency of the homozygous dominant genotype.
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2pq is frequency of heterozygous and q² is frequency of homozygous recessive.
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Well, we are given that birds with no crown, the frequency of those is 24% in this population.
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And since, no crown is recessive to crown, that means that individuals who have the
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no crown phenotype are homozygous recessive, so no crown equals homozygous recessive.
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And what we are asked to find is the frequency of the no crown allele.
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So, what we are trying to determine here is the frequency of C, no crown allele.
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And I know that to have the no crown phenotype, you have to be homozygous recessive.
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That means that individuals with no crown phenotype are little C little C or q².
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So, q² equals 24%, and what I want to figure out is the frequency of the recessive allele, little C.
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So, I need to take the square root of both sides, but first, I am going to go ahead and convert this to a decimal.
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So, q² equals 0.24, and I take the square root of both sides, and I am going to get q equals 0.489.
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And I am asked to give this answer in the nearest 100th, so converting this to the nearest 100th, rounding to the nearest 100th, this is 0.49.
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So, the correct answer is 0.49 for no. 4.
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Question 5: I am trying to determine the percentage of water on earth that is fresh water.
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So, the percentage of water on earth that is fresh water is going to be equal to fresh water from all sources over total water times 100.
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So, I have to look at the table that I am given and look at all the various types of fresh water, what the volume is.
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And first one is ice sheets and glaciers, so that is 24,064.
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Next, I have fresh ground water- 10,530. There is also ground ice, permafrost- 300.
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Fresh lakes- 91, and then, finally, rivers- 2.12, and we are given that already.
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They have totalled it up for us, which is total volume of water is 1,385,984.
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So, If I add up all these numbers in the numerator, I would get 34,987.12 / 1,385,984.
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You do the division, and you would get 0.025, and it is asking me for percentage to the tens place.
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So, I am going to multiply this by 100 to get the percentage.
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And that is going to give me 2.5% as my answer for this question, which is question no. 5.
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Question 6 asks about the value of evaporation from land to atmosphere, and you are told that total evaporation equals total precipitation.
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So, I am going to let E equal evaporation from land to atmosphere, and this is what I am looking for.
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So, if total evaporation equals total precipitation, then,
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the evaporation from land to atmosphere plus the evaporation from the oceans equals total precipitation.
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I am given that the evaporation from oceans is 434 x 10^15 kg per year.
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And this equals precipitation from the continental atmosphere and maritime atmosphere, and those numbers are given to you.
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That is 107 x 10^15 + 398 x 10^15.
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So, to solve for E, we are going to add the two numbers on the right to get 505 x 10^15.
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Therefore, E equals 505 x 10^15 and then, subtracting 434 x 10^15
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from both sides gives me 71 x 10^15, and that is the answer for this question.
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Thank you for visiting Educator.com. That concludes this lesson.