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 The relationship of an allele to the total number of alleles in a gene pool for a trait.  Expressed as a percent  Allele frequency can be used to.

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Presentation on theme: " The relationship of an allele to the total number of alleles in a gene pool for a trait.  Expressed as a percent  Allele frequency can be used to."— Presentation transcript:

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2  The relationship of an allele to the total number of alleles in a gene pool for a trait.  Expressed as a percent  Allele frequency can be used to determine the diversity of a population. The closer the percentages are to each other, the more diverse a population is,

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4  In a large pond a survey was taken from DNA sample of a species of frog. The following genotypes were observed:  GG 80 frogs (Green color)  Gg 40 frogs (Green color)  Gg 20 frogs (Brown color) What is the allele frequency for both variations?

5  First, calculate how many total alleles are in this population.  140 frogs x 2 = 280  Second, calculate how many of each allele. GG 80 frogs x 2 = 160 G Gg 40 frogs = 40 G + 40 g gg 20 frogs x 2 = 40 g = 200 G = 80 g

6  Last, divide each allele total by the total number of alleles and covert to percent.  For G: 200/280 = 0.71 = 70 % *  For g: 80/280 = 0.28 = 30 % *  * Figures were rounded.

7  Short and tall plants were analyzed in a garden and these results were found. What is the allele frequency of each type allele. Tall is dominant over short plants.  TT 120 plants  Tt 120 plants  tt 80 plants  Round off the percents to whole numbers.

8  TT 120 plants 120 x 2 = 240 T’s  Tt 120 plants 120 T’s and 120 t’s  tt 80 plants 80 x 2 = 160 t’s  Total Alleles = 640 alleles  Total T’s = 360 alleles 360/640 = 0.6 = 60%  Total t’s = 280 alleles 280/640 = 0.4 = 40%

9  Red eyes (R) in fruit flies is dominant over white eyes (rr). In a population of 720 it was found that 30 % carried the recessive allele. How many carried the recessive allele? How many carried the dominant allele?

10  The population was 720, so we know the toal number of alleles is 1,440.  We know that 30 percent carried the recessive allele so the dominant percentage was 70.  30 % of 1,440 is 432  70% of 1,440 is 1,008

11  It is more common to want to know the genotype frequencies in a population.  We may want to know how many homozygous dominant or heterozgous genotypes there are.  If we know the frequencies of the alleles, we can find the genotype frequencies.

12  These two men developed a mathematical formula to determine if evolution has occurred in a population.  If gene frequency changes significantly, then evolution has take place.

13  A population is in equilibrium if:  (No evolution is occurring)  1. No mutations occur.  2. No migration in or out of population occurs (gene flow) 3. Mating must be random 4. the population must be very large. 5. No natural selection can is occurring.

14  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

15  A population is sampled in which it was found that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:  1. The frequency of the “a” allele.  2. The frequency of the “ A” allele  3. The frequency of genotype “AA”  4. The frequency of genotype “Aa”  5. The frequency of genotype “aa” Given=36%

16  The frequency of the “a” allele.  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%

17  The frequency of the “ A” allele.  Since q = 0.6, and p + q = 1, then p = 0.4.  Tthe frequency of A is by definition equal to p, so the answer is 40%.

18  The frequency of genotype “AA”  The frequency of AA is equal to p2.  So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16)

19  The frequency of genotype “Aa”  The frequency of Aa is equal to 2pq.  So, using the information above, frequency of Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).


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