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AP Biology Practice Exam: Section I, Part B, Grid In

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • AP Biology Practice Exam 0:17
  • Grid In Question 1 0:29
  • Grid In Question 2 3:49
  • Grid In Question 3 11:04
  • Grid In Question 4 13:18
  • Grid In Question 5 17:01
  • Grid In Question 6 19:30

Transcription: AP Biology Practice Exam: Section I, Part B, Grid In

Welcome to I am Dr. Carleen Eaton, and in this lesson, I will be continuing the review of a practice AP Biology Examination.0001

In this section, I will be covering part B of section 1 which is 6 grid-in questions.0010

Again, you can find this test in Barron’s AP Biology book, 4th Edition, and this is model test 1.0018

This first question asks for the crossover frequency or the recombination frequency.0031

And recall that the formula for recombination frequency is as follows: the number of recombinant offspring over the total number of offspring.0038

So, we are told that the cross was between one parent that has gray normal phenotype, and the other has black vestigial wings.0078

Therefore, A and B, gray normal and black vestigial, those offspring do not represent crossovers or recombinants.0094

C, gray vestigial and D, black color with normal wings represent recombinant offspring.0105

So, there are 190 gray vestigial and 184 black normal, so that is the numerator, 190 + 184.0126

So, the crossover rate or recombination frequency - same idea - is numerator0141

190 + 184 over the total number of offspring, which is 969 + 941 + 190 + 184.0150

When you do the math, you get 374 for the numerator and 2,284 for the denominator.0169

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.0181

So, we need to round that to 0.164 which is equal to 16.4%.0207

OK, that was question one.0226

Question 2, we are asked to calculate a chi-squared value.0229

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.0235

So, the chi-squared formula is chi-square equals the sum of the observed findings minus expected squared divided by the expected.0244

And a really good way to organize the data when you are working with chi-square is to use a table.0262

So, the two phenotypes we have are purple and yellow.0271

What we are looking at here is the flower coloration, and some of the flowers are purple.0275

We have purple petals. Others have yellow.0288

Next, I need to determine what is observed, what is expected, observed minus expected, that value squared.0294

And then, finally, observed minus expected squared divided by expected.0323

So, let's start out with what is observed because that is something that is given.0331

So, we have been told that of these offspring flowers, of these 350 plants, 249 had purple color. 101 have yellow color.0338

Now, I have to figure out what is expected.0355

So, the null hypothesis is that the F1 plants were heterozygous. They were hybrid plants.0357

So, if the purple flowers are dominant and yellow coloration is recessive, what I would expect is a 3:1 ratio purple to yellow.0366

And there were 350 plants total. That means for purple, I would expect 75% of the offspring plants to be purple.0384

So, 350 x 0.75, I would expect 263 purple plants, so expected purple, 263.0400

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.0414

Once you have that, the rest comes down to just doing the math.0435

So, for purple, observed minus expected is 249 - 263 = -14.0440

For yellow, observed minus expected is 101 - 87 which equals 14.0455

-142 - so observed minus expected squared - equals 196. 142 equals 196.0466

Now, for this last part, I need to take observed minus expected squared divided by expected.0481

So, for purple, that is 196 divided by expected which is 263, and this gives me a value of 0.74.0489

For yellow, observed minus expected squared again, 196, divided by expected which is 87 gives 2.25.0503

Now, I need the sum of these chi-squared values, so I am going to add these two up, which gives me 0.74 + 2.25.0518

That gives me a total of 2.99, so the answer is 2.99 for the grid-in.0542

But just to go a little bit farther and explain what this means, we are being asked to evaluate the null hypothesis.0550

And if you went and looked up the critical values, table that you will have on your sheet of reference formulas and equations,0558

and you look under a level of significance P = 0.05 and then, you would check under 1 degree of freedom.0567

And the reason that you would be checking under 1 degree of freedom is you want to0576

use a degree of freedom that is equal to the number of phenotypes minus 1.0580

Here, we have two phenotypes: purple and yellow coloration, so that is 2 - 1 = 1.0595

We want to look under 1 degree of freedom.0601

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 3.84.0604

The chi-squared value that we figured out is 2.99.0622

Because 2.99 is below that critical value of 3.84, we fail to reject the null.0627

Sometimes, you will hear people say we accept the null hypothesis, but technically, there is a slight difference.0638

What you are doing is failing to reject the null hypothesis.0645

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.0650

Question 3 asks for you to evaluate population growth, and again, you will need to look at the table,0665

the reference sheet that gives you equations and formulas and values.0675

And from that, you will see that the population growth formula is dN / dt. OK, so this is population growth.0679

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.0694

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 4.0711

So, for the numerator, we are going to look at what growth is, the population size is on day 2.0720

And on day 2, it is 20,000. It is the population size- 20,000 bacteria.0727

On day 4, population size is 80,000 bacteria.0738

So, numerator, change in population size, 80,000 bacteria, day 4, minus 20,000 on day 2.0747

The denominator is the change in time. Day 4 minus day 2, doing the math 80,000 - 20,000 is 60,000.0762

4 - 2 is 2. This gives a population growth rate of 30,000 bacteria per day, so the answer is 30,000.0774

Question 4: we are going to need to use Hardy-Weinberg equations.0798

Again, these will be present on your reference sheet of formulas and equations.0805

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.0810

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.0821

So, in Hardy-Weinberg equilibrium, the frequency of a particular allele in a population remains stable.0835

A Hardy-Weinberg equations are the p2 + 2pq + q2 = 1 and p + q = 1.0845

p is the frequency of the dominant allele. q is the frequency of the recessive allele.0859

That means that p2 gives you the frequency of the homozygous dominant genotype.0875

2pq is frequency of heterozygous and q2 is frequency of homozygous recessive.0889

Well, we are given that birds with no crown, the frequency of those is 24% in this population.0912

And since, no crown is recessive to crown, that means that individuals who have the0920

no crown phenotype are homozygous recessive, so no crown equals homozygous recessive.0926

And what we are asked to find is the frequency of the no crown allele.0940

So, what we are trying to determine here is the frequency of C, no crown allele.0946

And I know that to have the no crown phenotype, you have to be homozygous recessive.0951

That means that individuals with no crown phenotype are little C little C or q2.0957

So, q2 equals 24%, and what I want to figure out is the frequency of the recessive allele, little C.0965

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.0979

So, q2 equals 0.24, and I take the square root of both sides, and I am going to get q equals 0.489.0986

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 0.49.1002

So, the correct answer is 0.49 for no. 4.1013

Question 5: I am trying to determine the percentage of water on earth that is fresh water.1022

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 100.1030

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.1062

And first one is ice sheets and glaciers, so that is 24,064.1070

Next, I have fresh ground water- 10,530. There is also ground ice, permafrost- 300.1079

Fresh lakes- 91, and then, finally, rivers- 2.12, and we are given that already.1092

They have totalled it up for us, which is total volume of water is 1,385,984.1104

So, If I add up all these numbers in the numerator, I would get 34,987.12 / 1,385,984.1115

You do the division, and you would get 0.025, and it is asking me for percentage to the tens place.1141

So, I am going to multiply this by 100 to get the percentage.1152

And that is going to give me 2.5% as my answer for this question, which is question no. 5.1156

Question 6 asks about the value of evaporation from land to atmosphere, and you are told that total evaporation equals total precipitation.1172

So, I am going to let E equal evaporation from land to atmosphere, and this is what I am looking for.1191

So, if total evaporation equals total precipitation, then,1208

the evaporation from land to atmosphere plus the evaporation from the oceans equals total precipitation.1211

I am given that the evaporation from oceans is 434 x 1015 kg per year.1223

And this equals precipitation from the continental atmosphere and maritime atmosphere, and those numbers are given to you.1236

That is 107 x 1015 + 398 x 1015.1247

So, to solve for E, we are going to add the two numbers on the right to get 505 x 1015.1260

Therefore, E equals 505 x 1015 and then, subtracting 434 x 10151279

from both sides gives me 71 x 1015, and that is the answer for this question.1288

Thank you for visiting That concludes this lesson.1308