For more information, please see full course syllabus of AP Biology

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

XVI. Practice Test (Barron's 4th Edition): Lecture 3 | 21:52 min

This lecture covers the mathematics-focused portion of the AP Biology exam, the grid-in questions, with topics including crossover frequency and recombination frequency; chi-squared values; population growth of bacteria; Hardy-Weinberg equations; percentage of freshwater on earth; and evaporation of water from land to atmosphere.

For more information, please see full course syllabus of AP Biology

### 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

### AP Biology Exam Online Course

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

*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.*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 Barronâ€™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 numerator*0141

*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

*-14 ^{2} - so observed minus expected squared - equals 196. 14^{2} 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 to*0576

*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 p ^{2} + 2pq + q^{2} = 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 p ^{2} gives you the frequency of the homozygous dominant genotype.*0875

*2pq is frequency of heterozygous and q ^{2} 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 the*0920

*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 q ^{2}.*0957

*So, q ^{2} 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, q ^{2} 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 10 ^{15} 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 10 ^{15} + 398 x 10^{15}.*1247

*So, to solve for E, we are going to add the two numbers on the right to get 505 x 10 ^{15}.*1260

*Therefore, E equals 505 x 10 ^{15} and then, subtracting 434 x 10^{15}*1279

*from both sides gives me 71 x 10 ^{15}, and that is the answer for this question.*1288

*Thank you for visiting Educator.com. That concludes this lesson.*1308

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