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AP Biology

Learn about important concepts in microevolution, including alleles, population genetics, natural selection, and the Hardy-Weinberg theorem.

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AP Biology

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  1. AP Biology Microevolution Part 1

  2. Important concepts from previous units: • Alleles are differing versions of a gene. • bb-blue • BB/Bb brown • Most organisms are diploid in terms of genetic content within the genome. • Fertilization is the combining of parental genes in the hopes of reproducing the next generation of the species.

  3. You are what you inherit in terms of DNA.The egg surrounded by sperm.

  4. Population

  5. Natural Selection & Evolution • Populations evolve not individuals. • This is because we “are” what we “are” because of the genetics we inherited. • You can’t change your somatic cells’ DNA by choice, only by random mutation.

  6. If a mutation occurs in the DNA that is located in the gametes (sperm and eggs), then those changes may affect the next generation of offspring and therefore a change in traits has occurred. • In other words, the “population” is evolving from generation to generation. “Evolve” just means “change over time” and that is what has occurred. Evolution-change (modifications) over time (descent)

  7. Individuals “suffer” or “benefit” as a result of the traits they inherited or mutations they acquired during their life. • “Weak” vs. “strong” genes is the way it is usually communicated. “Weak” are considered detrimental traits and “strong” are considered favorable traits in terms of survival and reproduction.

  8. Population Genetics • This is the field of science that studies the trait variation rates over timewithin a population. • It basically is following allele frequency rates in a gene pool. (A.K.A. a population.)

  9. SpeciesMale and Female Blue Footed Boobies • Species • Mostly defined as organisms that are so genetically similar in genome that there exists the potential to breed and produce viable (living) fertile (able to reproduce eventually themselves) offspring.

  10. Genetics are very similar is important to defining a species since it is the “blueprint” for “constructing” an organism. The “plans” must be very similar or there will be confusion in “construction” and problems will arise during development. Problems are a terrible thing to encounter since we are discussing the making of a living organism. Horse (f)+ donkey (m)=mule

  11. Geographic range vs. population • A population is in one specific given area; but in the case of organisms that are quite common (For example, grey squirrels or humans.), we may have several populations that cover a wider expanse of territory. • In the case of humans, as a species we are global in our range; but we have millions of different populations, such as the population of Montgomery or the population of Birmingham. “Range” refers to everywhere where that species may be found.

  12. Geographic Range of west coast salamanders

  13. Geographic Range of Humans

  14. Gene pools may or may not interact; it depends on the species and if any geographic barriers (such as large mountains or large bodies of water) interfere with the ability to interact. One species, two populations

  15. Allele frequency (Remember, an allele is a version of a gene.) • “Frequency” refers to “how many” are present at that time within the population (gene pool). • It is considered fixed, if there is no change in frequency—no evolution is evident. (Basically, a state of equilibrium is occurring.) • It is considered evolving, if frequency is changing— evolution is occurring. (Basically, a state of change is occurring over time from generation to generation.)

  16. Hardy-Weinberg Theorem • This set of equations is used to follow allele frequency within a population (also considered a gene pool) • If the numbers (rates) change from generation to generation, the population is evolving over time. • If the numbers (rates) do not change from generation to generation, the population is not evolving over time and is then said to be in a state of equilibrium.

  17. Equation #1: p + q = 1 This equation is for alleles. “p” refers to the “dominant” allele percentage and “q” refers to the recessive allele percentage. Together p + q percentages must equal 100% of the gene pool or 1. Equation 1 (p + q= 1)

  18. Equation #2: p2 + 2pq + q2 = 1 This equation refers to the percent composition/number of organisms within the population (gene pool) at that time. It is essentially a Punnett square, but in math format. • p2 = the homozygous dominant percentage of organisms within the population at that time. • 2pq = the heterozygous percentage of organisms within the population at that time. • q2 = the homozygous recessive percentage of organisms within the population at that time.

  19. Hardy – Weinberg TheoremEquation 2 (p² + 2pq + q² = 1)

  20. These equations are mainly used in health sciences to explain the frequency of genetic conditions. • These equations can be used to show how or if variation is preserved over time. • Five conditions must be met for a population to be in Equilibrium (Allele Frequency is not changing):

  21. Five conditions: • Large population must exist. (This dilutes any non-random processes that are occurring.) • No migration in or out of the population is occurring at that time. (The population is not being influenced by outside environmental factors.) • No mutations are occurring within the genome. (No random, unforeseen change due to an environmental stress.) • Random mating is occurring (No preferences are being displayed for one trait over another trait…everyone is equal in fitness.) • No natural selection is occurring on the population at this time. (Nature favors all equally in terms of fitness.) • All threemust add up to 100% (1) of the population.

  22. The five conditions for non-evolving populations are rarely met in nature: • Extremely large population size • No gene flow • No mutations • Random mating • No natural selection

  23. Flowers Red (RR) White (WW) Pink (RW) • Total population size: 500 flowers (1000 alleles) • 320 red 320*2=640 alleles for red • 160 pink 160 white alleles; 160 red alleles • 20 white 20*2=40 alleles for white • Calculate the allele frequency (p+q=1) • Red (640+160)÷1000 = 800÷1000=0.8 or 80% • White (40+160) / 1000 = 200/1000=0.2 or 20%

  24. p2+2pq+q2 = 1 (0.8)^2 + 2(0.8)(.02) + (0.2)^2 0.64 + 0.32 + 0.04 = 1 64% Red (RR) 32% Pink (RW) 4% White 100%

  25. Albinism is a rare genetically inherited trait that is only expressed in the phenotype of homozygous recessive individuals (aa). The most characteristic symptom is a marked deficiency in the skin and hair pigment melanin. This condition can occur among any human group as well as among other animal species. The average human frequency of albinism in North America is only about 1 in 20,000. • What is the frequency of homozygous recessive individuals (aa) in a population is q²? q² = 1/20,000 = .00005 q = .007

  26. http://www.k-state.edu/parasitology/biology198/hardwein.html

  27. PROBLEM #1: You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: • The frequency of the "aa" genotype: • The frequency of the "a" allele: • The frequency of the "A" allele: • The frequencies of the genotypes "AA" and "Aa.“: • The frequencies of the two possible phenotypes if "A" is completely dominant over "a."

  28. PROBLEM #1: You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: • The frequency of the "aa" genotype: Answer: 36%, as given in the problem itself. • The frequency of the "a" allele: q2 = 0.36; If q2 = 0.36, then q = 0.6 =60% • The frequency of the "A" allele: p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%. • The frequencies of the genotypes "AA" and "Aa.“: 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).

  29. The frequencies of the two possible phenotypes if "A" is completely dominant over "a." • The frequencies of the two possible phenotypes if "A" is completely dominant over "a." Answers: Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive 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 is 36%.

  30. The frequency of the "aa" genotype. Answer: 36%, as given in the problem itself. • The frequency of the "a" allele. Answer: The frequency of aa 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 frequency is 60%. • The frequency of the "A" allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%. • 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). • The frequencies of the two possible phenotypes if "A" is completely dominant over "a." Answers: Because "A" is totally dominate over "a", the dominant phenotype will show if either the homozygous "AA" or heterozygous "Aa" genotypes occur. The recessive phenotype is controlled by the homozygous aa genotype. Therefore, the frequency of the dominant phenotype equals the sum of the frequencies of AA and Aa, and the recessive 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 is 36%.

  31. Problem 2: Sickle-cell anemia is an interesting genetic disease. Normal homozygous individials (SS) have normal blood cells that are easily infected with the malarial parasite. Thus, many of these individuals become very ill from the parasite and many die. Individuals homozygous for the sickle-cell trait (ss) have red blood cells that readily collapse when deoxygenated. Although malaria cannot grow in these red blood cells, individuals often die because of the genetic defect. However, individuals with the heterozygous condition (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 an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?

  32. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (Ss) for the sickle-cell gene?  Answer: • 9% =.09 = ss = q2. • To find q, simply take the square root of 0.09 to get 0.3. • p + q = 1 • 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).

  33. There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than 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: • The frequency of the recessive allele. • The frequency of the dominant allele. • The frequency of heterozygous individuals.

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

  35. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Given this simple information, which is something that is very likely to be on an exam, calculate the following: • The percentage of butterflies in the population that are heterozygous. • The frequency of homozygous dominant individuals.

  36. 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. • q = 0.63. • Since p + q = 1, then p must be 1 - 0.63 = 0.37. • what is the percentage of butterflies in the population that are heterozygous? • 2pq so the answer is 2 (0.37) (0.63) = 0.47. • what is the frequency of homozygous dominant individuals? • That would be p2 or (0.37)2 = 0.14.

  37. PROBLEM #6.A very large population of randomly-mating laboratory mice contains 35% white mice. White coloring is caused by the double recessive genotype, "aa". Calculate allelic and genotypic frequencies for this population.

  38. Answer: • 35% are white mice, which = 0.35 and represents the frequency of the aa genotype (or q2). • The square root of 0.35 is 0.59, which equals q. • p = 1 - q then 1 - 0.59 = 0.41. • AA and Aa individuals • AA = p2 = 0.41 x 0.41 = 0.17; • Aa = 2pq = 2 (0.59) (0.41) = 0.48; • aa = q2 = 0.59 x 0.59 = 0.35. • If you add up all these genotype frequencies, they should equal 1.

  39. PROBLEM #9.Cystic fibrosis is a recessive condition that affects about 1 in 2,500 babies in the Caucasian population of the United States. Please calculate the following. • The frequency of the recessive allele in the population. • The frequency of the dominant allele in the population. • The percentage of heterozygous individuals (carriers) in the population.

  40. Answer: • q2 is 1/2,500 or 0.0004. • The frequency of the recessive allele in the population • q is the square root, or 0.02. • the frequency of the cystic fibrosis (recessive) allele in the population is 0.02 (or 2%). • The frequency of the dominant (normal) allele in the population • p is simply 1 - 0.02 = 0.98 (or 98%). • The percentage of heterozygous individuals (carriers) in the population.  • 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.

  41. AP BiologyMicroevolution– Part 2 Evolution of Populations

  42. Important concepts from previous units: • Alleles are differing versions of a gene. • Most organisms are diploid in terms of genetic content within the genome. • Fertilization is the combining of parental genes in the hopes of reproducing the next generation of the species.

  43. Variation (Different traits exist within a given species or population.) • Variation is key to surviving in a changing environment. (This is because you have “options”.) • Perhaps some of the members of that species or population will survive and reproduce. • These “options” are the raw building materials of evolution to utilize. If there is no variation or “option” from which to utilize, a species is confined to what is available; even if it is weak or unfavorable. Variation, on the most basic level, will only come into existence with a change in the DNA nucleotide sequence, what we refer to as a mutation. Some mutations are favorable, but most are harmful. • Variation exists between individuals and populations unless the population is composed of clones.

  44. Variation Exists

  45. Variation Exists

  46. “Creating” Variation for evolution to build upon: • Through mutations These changes are rare and random in gametes. (Because gamete cells are normally not exposed to the environmental stresses an organism may encounter in their existence.) Mutations mostly occur in somatic cells because these cells are exposed to the environmental stresses. Most mutations, unfortunately, are harmful to the cell or organism, so it usually dies.

  47. Through sexual reproduction • The process of crossover, during Prophase I of meiosis, “swaps genes from one chromosome to another, its equal “mate” usually, during gamete formation. This is so that each sperm or egg is unique in it’s genetic composition.

  48. Variation in Crossover

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