Handy Note 1 0 9 Percent

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Biology 198
PRINCIPLES OF BIOLOGY
Answers to Hardy-Weinberg practice questions Bluestacks app player xp.

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Updated: 21 August 2000

POPULATION GENETICS AND THEHARDY-WEINBERG LAW
ANSWERS TO SAMPLE QUESTIONS

Remember the basic formulas:

Updated: 21 August 2000

POPULATION GENETICS AND THEHARDY-WEINBERG LAW
ANSWERS TO SAMPLE QUESTIONS

Remember the basic formulas:

p² + 2pq + q² = 1 and p + q = 1

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

Handy Note 1 0 9 Percent Fraction

Handy Note 1 0 9 Percent 3

1. PROBLEM #1.

   http://habmif.xtgem.com/Blog/__xtblog_entry/19461942-steel-rats-1-0#xt_blog. You have sampled a population in which you know that the percentage of the homozygousrecessive genotype (aa) is 36%. Using that 36%, calculate the following:

   1. The frequency of the 'aa' genotype. Answer: 36%, asgiven in the problem itself.
   2. The frequency of the 'a' allele. Answer: The frequency ofaa is 36%, which means that q² = 0.36, by definition. If q² = 0.36,then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequencyis 60%.
   3. The frequency of the 'A' allele. Answer:Since q = 0.6, andp + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.
   4. The frequencies of the genotypes 'AA' and 'Aa.' Answer:The frequency of AA is equal to p², and the frequency of Aa is equal to 2pq. So,using the information above, the frequency of AA is 16% (i.e. p² is 0.4 x 0.4 =0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).
   5. The frequencies of the two possible phenotypes if 'A' is completely dominant over 'a.' Answers: Because 'A' is totally dominate over 'a', thedominant phenotype will show if either the homozygous 'AA' or heterozygous 'Aa' genotypesoccur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, thefrequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and therecessive phenotype is simply the frequency of aa. Therefore, the dominant frequency is 64% and,in the first part of this question above, you have already shown that the recessive frequency is36%.

   
   

2. PROBLEM #2.

   Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) havenormal blood cells that are easily infected with the malarialparasite. Thus, many of these individuals become very illfrom the parasite and many die. Individualshomozygous for the sickle-cell trait (ss) have red blood cells that readily collapse whendeoxygenated. Although malaria cannot grow in these red blood cells,individuals often die because of the genetic defect. However,individuals with the heterozygouscondition (Ss) have some sickling of red blood cells, but generally not enough to cause mortality. In addition, malaria cannot survive well within these 'partially defective' red blood cells. Thus,heterozygotes tend to survive better than either of the homozygous conditions. If 9% of anAfrican population is born with a severe form of sickle-cell anemia (ss), what percentage of thepopulation will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cellgene? Answer: 9% =.09 = ss = q². To find q,simply take the square root of 0.09 to get 0.3. Since p = 1 - 0.3, then p must equal 0.7. 2pq = 2(0.7 x 0.3) = 0.42 = 42% of the population are heterozygotes (carriers).

   
   

3. PROBLEM #3.

   There are 100 students in a class. Ninety-six did well in the course whereas four blew it totallyand received a grade of F. Sorry. In the highly unlikely event that these traits are genetic ratherthan environmental, if these traits involve dominant and recessive alleles, and if the four (4%)represent the frequency of the homozygous recessive condition, please calculate the following:

   1. The frequency of the recessive allele. Answer: Since webelieve that the homozygous recessive for this gene(q²) represents 4% (i.e. = 0.04), the squareroot (q) is 0.2 (20%).
   2. The frequency of the dominant allele. Answer: Since q = 0.2,and p + q = 1, then p = 0.8 (80%).
   3. The frequency of heterozygous individuals. Answer: Thefrequency ofheterozygous individuals is equal to 2pq. In this case, 2pq equals 0.32, which means that thefrequency of individuals heterozygous for this gene is equal to 32% (i.e. 2 (0.8)(0.2) = 0.32).

   
   

4. PROBLEM #4.

   Within a population of butterflies, the color brown (B) is dominant over thecolor white (b). And, 40% of all butterflies are white. Given this simple information, which issomething that is very likely to be on an exam, calculate the following:

   1. The percentage of butterflies in the population that are heterozygous.
   2. The frequency of homozygous dominant individuals.

      Answers: The first thing you'll need to do is obtain p and q. So,since white is recessive (i.e. bb), and 40% of the butterflies are white, then bb = q² =0.4. To determine q, which is the frequency of the recessive allele in the population, simply takethe square root of q² which works out to be 0.632 (i.e. 0.632 x 0.632 = 0.4). So,q = 0.63. Since p + q = 1, then p must be 1 - 0.63 = 0.37. Now then, toanswer our questions. First, what is the percentage of butterflies in thepopulation that are heterozygous? Well, thatwould be 2pq so the answer is 2 (0.37) (0.63) = 0.47. Second, what is the frequency ofhomozygous dominant individuals? That would be p² or(0.37)² = 0.14.

   
   

5. PROBLEM #5.

   A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sidedindividuals. Assume that red is totally recessive. Please calculate the following:

   1. The allele frequencies of each allele. Answer: Well, beforeyou start, note that the allelic frequencies are p and q, and be sure to note that we don't have niceround numbers and the total number of individuals counted is 396 + 557 = 953. So, the recessiveindividuals are all red (q²) and 396/953 = 0.416. Therefore, q (the square root of q²) is 0.645. Since p + q = 1, then p must equal 1 - 0.645 = 0.355.
   2. The expected genotype frequencies. Answer: Well, AA =p² = (0.355)² = 0.126; Aa = 2(p)(q) =2(0.355)(0.645) = 0.458; and finally aa =q² = (0.645)² = 0.416 (you already knew this frompart A above).
   3. The number of heterozygous individuals that you would predict to be in this population. Answer: That would be 0.458 x 953 = about 436.
   4. The expected phenotype frequencies. Answer: Well, the 'A'phenotype = 0.126 + 0.458 = 0.584 and the 'a' phenotype = 0.416 (youalready knew this from part A above).
   5. Conditions happen to be really good this year for breeding and next year there are 1,245young 'potential' Biology instructors. Assuming that all of the Hardy-Weinberg conditions aremet, how many of these would you expect to be red-sided and how many tan-sided? Answer: Simply put, The 'A' phenotype =0.584 x 1,245 = 727 tan-sided and the 'a' phenotype = 0.416 x 1,245 = 518 red-sided ( or1,245 - 727 = 518).

   
   

6. PROBLEM #6.

   A very large population of randomly-mating laboratory mice contains 35% white mice. Whitecoloring is caused by the double recessive genotype, 'aa'. Calculate allelic and genotypic frequencies for this population. Answer: 35% are white mice,which = 0.35 and represents the frequency of the aagenotype (or q²). The square root of 0.35 is 0.59, which equals q. Since p = 1 - qthen 1 - 0.59 = 0.41. Now that we know the frequency of each allele, we can calculate thefrequency of the remaining genotypes in the population (AA and Aa individuals). AA =p² = 0.41 x 0.41 = 0.17; Aa = 2pq = 2 (0.59) (0.41) = 0.48; and as before aa =q² = 0.59 x 0.59 = 0.35. If you add up all these genotype frequencies, they shouldequal 1.

   
   

7. PROBLEM #7.

   After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter aplane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a desertedisland. No one finds you and you start a new population totally isolated from the rest of theworld. Two of your friends carry (i.e. are heterozygous for) the recessive cystic fibrosis allele(c). Assuming that the frequency of this allele does not change as the population grows, whatwill be the incidence of cystic fibrosis on your island? Answer:There are 40 total alleles in the 20 people of which 2 alleles are for cystic fibrous. So, 2/40 = .05 (5%) of the alleles are for cystic fibrosis. That represents p. Thus, cc orp² = (.05)² = 0.0025 or 0.25% of the F1 populationwill be born with cysticfibrosis.

   
   

8. PROBLEM #8.

   You sample 1,000 individuals from a large population for the MN blood group, which caneasily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). Theyare typed accordingly:

   BLOOD TYPE	GENOTYPE	NUMBER OF INDIVIDUALS	RESULTING FREQUENCY
   M	MM	490	0.49
   MN	MN	420	0.42
   N	NN	90	0.09

   
   

   Using the data provide above, calculate the following:

   1. The frequency of each allele in the population. Answer: Since MM = p², MN = 2pq, and NN = q², then p (the frequency of theM allele) must be the square root of 0.49, which is 0.7. Since q = 1 - p, then q must equal 0.3.
   2. Supposing the matings are random, the frequencies of the matings. Answer: This is a little harder to figure out. Try setting up a 'Punnettsquare' type arrangement using the 3 genotypes and multiplying thenumbers in a manner something like this:

      	MM (0.49)	MN (0.42)	NN (0.09)
      MM (0.49)	0.2401*	0.2058	0.0441
      MN (0.42)	0.2058	0.1764*	0.0378
      NN (0.09)	0.0441	0.0378	0.0081*

      Note that three of the six possible crosses are unique (*), but that the other three occur twice (i.e.the probabilities of matings occurring between these genotypes is TWICE that of the other three'unique' combinations. Thus, three of the possibilities must be doubled.

      MM x MM = 0.49 x 0.49 = 0.2401
      MM x MN = 0.49 x 0.42 = 0.2058 x 2 = 0.4116
      MM x NN = 0.49 x 0.09 = 0.0441 x 2 = 0.0882
      MN x MN = 0.42 x 0.42 = 0.1764
      MN x NN = 0.42 x 0.09 = 0.0378 x 2 = 0.0756
      NN x NN = 0.09 x 0.09 = 0.0081
      

   3. The probability of each genotype resulting from each potential cross. Answer: You may wish to do a simple Punnett's square monohybridcross and, if you do, you'll come out with the following result:

      MM x MM = 1.0 MM
      MM x MN = 0.5 MM 0.5 MN
      MM x NN = 1.0 MN
      MN x MN = 0.25 MM 0.5 MN 0.25 NN
      MN x NN = 0.5 MN 0.5 NN
      NN x NN = 1.0 NN
      

   
   

9. PROBLEM #9.

   Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasianpopulation of the United States. Adobe creative suite cs5 free mac. Please calculate the following.

   1. The frequency of the recessive allele in the population. Answer: We know from the above that q² is 1/2,500 or0.0004. Therefore, q is the square root, or 0.02. That is the answer to our first question: thefrequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%).
   2. The frequency of the dominant allele in the population. Answer: The frequency of the dominant (normal) allele in thepopulation (p) is simply 1 - 0.02 = 0.98 (or 98%).
   3. The percentage of heterozygous individuals (carriers) in the population. Answer: Since 2pq equals the frequency of heterozygotes or carriers,then the equation will be as follows: 2pq = (2)(.98)(.02) = 0.04 or 1 in 25 are carriers.

   
   

10. PROBLEM #10.

    In a given population, only the 'A' and 'B' alleles are present in the ABO system;there are no individuals with type 'O' blood or with O alleles inthis particular population. If 200 people have type A blood,75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of thispopulation (i.e., what are p and q)? Answer: To calculate theallele frequencies for A and B, we need to remember that theindividuals with type A blood are homozygous AA, individuals with type AB blood areheterozygous AB, and individuals with type B blood are homozygous BB. The frequency of Aequals the following: 2 x (number of AA) + (number of AB) divided by 2 x (total number ofindividuals). Thus 2 x (200) + (75) divided by 2 (200 + 75 + 25). This is 475/600 = 0.792 = p. Since q is simply 1 - p, then q = 1 - 0.792 or 0.208.

    
    

11. PROBLEM #11.

    The ability to taste PTC is due to a single dominate allele 'T'. You sampled 215individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 couldnot. Calculate all of the potential frequencies. Answer: First,lets go after the recessives (tt) or q². That is easy since q² = 65/215 = 0.302. Takingthe square root of q², you get 0.55, which is q. To get p, simple subtract q from 1so that 1 - 0.55 = 0.45 = p. Now then, you want to find out what TT, Tt, and tt represent. Youalready know that q² = 0.302, which is tt. TT = p² = 0.45 x 0.45 =0.2025. Tt is 2pq = 2 x 0.45 x 0.55 = 0.495. To check your own work, add 0.302, 0.2025, and0.495 and these should equal 1.0 or very close to it. This type ofproblem may be on the exam.

    
    

12. PROBLEM #12. (You will not have thistype of problem on the exam)

    What allelic frequency will generate twice as many recessive homozygotes as heterozygotes?Answer: We need to solve for the following equation:q² (aa) = 2 x the frequency of Aa. Thus, q² (aa) = 2(2pq). Or anotherway of writing it is q² = 4 x p x q. We only want q, so lets trash p. Since p = 1 - q,we can substitute 1 - q for p and, thus, q² = 4 (1 - q) q. Then, if we multiply everything on the right by that lone q, we get q² = 4q - 4q². We then divide both sides through by q and get q = 4 -4q. Subtracting 4 from both sides, and then q (i.e. -4q minus q = -5q) also from both sides, weget -4 = -5q. We then divide through by -5 to get -4/-5 = q, or anotherwards the answer which is0.8 =q. I cannot imagine you getting this type of problem in this generalbiology course although if you take algebra good luck.

    
    

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