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Are alleles equally sized?


If alleles occur at the same locus on a chromosome, does that mean they are of the same size? I was under the impression that they were, but I saw a gel electrophoresis testing different alleles of the same gene and saw them separate. I don't understand how they could separate if they are of the same size, but if they aren't, then how can they be housed in the exact same locus?


In general, alleles don't have to be the same size. Two major examples which come to mind are the Huntingtin gene and FMR1.

Huntingtin is the causative gene of Huntington's disease. In people with Huntington's, a sequence of three nucleotides is repeated. The number of repeats varies, from a low of 9 in unaffected individuals, to more than 60 in severely affected ones. So depending on how many repeats you have, each of your Huntingtin alleles may have a different nucleotide size.

FMR1 is the causitive agent of Fragile X syndrome. Like Huntingtons, the gene varies in the number of trinucleotide repeats, but for FMR1, affected individuals can have over 1000 repeasts, meaning an extra 3 kilobases of genetic material versus the alleles of an unaffected individual.

Other genetic disorders are also related to the difference in allele size. You can have a deletion of a region of DNA (or an insertion) which is normally silent (recessive). The normal-length allele compensates for the defective allele, but the differently-sized disease allele is still present in the population, and counts as an allele of the gene.

It's not just disease-prone alleles which exhibit length polymorphism. A number of other genes will show variation on both the DNA and on the protein level. If the insertion/deletion is a multiple of three nucleotides and occurs in a loop region of the protein or on an intrinsically disordered domain, the insertion has a reasonable chance of minimal structural effect on the expressed protein. For one (completely arbitrary) example, PER3 has two major length variants in the population, with a 54 base pair difference between the two. Different populations have vastly different prevalence for each, with a range of 20-90% for the shorter version.

One thing to keep in mind is that "locus" points are just a human convention to help us map DNA. To the cell, DNA is just a single long molecule. (Or multiple molecules, for different chromosomes.) There isn't any labels for absolute position in the cell, so inserting or deleting a chunk of DNA mainly affects just the local environment. There isn't any sort of long-range indexing that gets messed up by the insertion.


One possibility of what you were looking at was a nuclease digestion experiment. A subject's gene sequence could be amplified from their DNA using PCR, giving a band of a certain size when electrophoresed on a gel. The PCR'd product could then be digested with another nuclease that specifically cuts one allelic variant, but not another. For example, check out this rather unrefined image I made:

Lane 1 is the molecular weight ladder. Lane 2 shows the PCR product of our gene of interest, which has two possible alleles - one that can be cut with our nuclease, and one that can't. The rest of the lanes show the result of incubating the PCR product from different individuals with our nuclease. Lane 3 shows a homozygote for one allele, the one that cannot be cut. Lane 4 shows a heterozygote, where the uncut 900 bp band is also seen with the cut 400 and 500 bp fragments. Lane 5 shows a homozygote carrying the cuttable allele on both chromosomes - there is no intact 900 bp band.


Which of the following statements is not true about oogenesis in humans? (a) four equal size daughter cells will form (b) at least two nonfunctional polar bodies will form (c) it occurs in the ovary (d) the egg will contain 23 chromosomes

General Edward Braddock was fatally injured in the battle field and died of wounds. He insisted upon using open field tactics which turned out to be very fatal leading to the death of 1000 soldiers. He did ignore the early warnings of his aides regarding his strategy.

A British force led by General Edward Braddock, moving to take Fort Duquesne, was defeated by a force of French and Canadian troops under Captain Daniel Liénard de Beaujeu with its American Indian allies.

Consequently, he failed to capture Fort Duquesne. So statement A is not true.

The right option is A. Genes consist of hormones that regulate bodily functions.

"Genes consist of hormones that regulate bodily functions" is the statement that is not true about genes.

Genes are the basic structural and functional unit of heredity. Genes are segments of DNA or RNA that are transmitted from one generation to the next, and that carries genetic information such as the sequence of amino acids for a protein. Genes operate in pairs and they determine one’s physical characteristics such as eye and hair colour. Genes do not contain hormones that regulate bodily functions.


Are alleles equally sized? - Biology

Population Genetics and Evolution (Lab Eight)

The purpose of population genetics and evolution is to study the effects that changing a condition has on Hardy-Weinberg equilibrium. Hardy-Weinberg believed that evolution occurs because the frequency of alleles changes. In a diploid individual, a particular gene is either expressed as an A for dominant or a for recessive trait. The p’s represent the frequency of the A allele and the q represents the frequency of the a allele in a diploid individual. Hardy-Weinberg equilibrium is p 2 +2pq+q 2 =1. In order for Hardy-Weinberg equilibrium to be met, 5 conditions must be met.

  1. The breeding population is large.
  2. Mating is random.
  3. There is no mutation of alleles.
  4. No differential migration occurs.
  5. There is no selection.

Exercise 8A: Estimating Allele Frequencies for a Specific Trait within a Sample Population

Table 8.1: Phenotypic Proportions of Tasters and Nontasters and Frequencies of the Determining Alleles.

Allele Frequency Based on the H-W Equation

North American Population

For North American Population:

    What is the percentage of heterozygous tasters 2pq in your class? 2(.5)(.5)=.5.

Exercise 8B: Case Studies

Case I—A Test of an Ideal Hardy-Weinberg Population

Number of A alleles present at the fifth generation

Number of offspring with genotype AA (4) x 2 = 8 A alleles

Number of offspring with genotype Aa (18) x 1 = 18 A alleles

p = TOTAL number of A alleles / TOTAL number of alleles in the population (number of students x 2) = 26/(28 x 2) = .464 = p

Number of a alleles present at the fifth generation

Number of offspring with genotype aa (6) x 2 = 12 a alleles

Number of offspring with genotype Aa (18) x 1 = 18 a alleles

p = TOTAL number of a alleles / TOTAL number of alleles in the population (number of students x 2) = 30/(28 x 2) = .536 = q

Since this population is in Hardy-Weinberg equilibrium, the frequency of dominant and recessive alleles should be .5. However, after doing the experiment and finding the frequencies of the dominant and recessive alleles, the new p is .464 and the new q is .536. p 2 + 2 p q + q 2 =1 à (.464) 2 + 2 (.464) (.536) + (.536) 2 = 1 However, the new p and q values are very similar to the original p and q values.

The results I obtained in this simulation are not the same as what the Hardy-Weinberg equation predicts. This is probably due to chance and maybe because my population size is not very large.

My population size is not very large although all of the other requirements of Hardy-Weinberg equilibrium are met.

Final Class Frequencies

Number of A alleles present at the fifth generation

Number of offspring with genotype AA (14) x 2 = 28 A alleles

Number of offspring with genotype Aa (14) x 1 = 14 A alleles

p = TOTAL number of A alleles / TOTAL number of alleles in the population (number of students x 2) = 42/(28 x 2) = .75 = p

Number of a alleles present at the fifth generation

Number of offspring with genotype aa (0) x 2 = 0 a alleles

Number of offspring with genotype Aa (14) x 1 = 14 a alleles

p = TOTAL number of a alleles / TOTAL number of alleles in the population (number of students x 2) = 14/(28 x 2) = .25 = q

The new frequency of p is larger than in Case I and the new frequency of q is smaller than in Case I.

In this simulation, not all genotypes had an equal chance of survival. Therefore this situation does not meet the requirements of Hardy-Weinberg equilibrium. Also, because not all genotypes have an equal chance of survival, mating is not random because only certain individuals can reproduce.

  1. Predict what would happen to the frequencies of p and q if you simulated another five generations.

If I simulated another five generations, the frequency of A alleles would increase (p) to almost the entire population and the frequency of a alleles (q) would decrease until it was almost extinct. This is because any homozygous recessive combination would die immediately, therefore they are not going to be able to reproduce and pass on their recessive allele.

  1. In a large population would it be possible to completely eliminate a deleterious recessive allele? Explain.

In a large population, it would not be possible to completely eliminate a deleterious recessive allele because the heterozygous genotype protects the a allele, ensuring that it continues to be present in the population’s gene pool.

Case III—Heterozygote Advantage

Final Class Frequencies (5 generations)

Final Class Frequencies (10 generations)

Number of A alleles present at the fifth generation

Number of offspring with genotype AA (9) x 2 = 18 A alleles

Number of offspring with genotype Aa (19) x 1 = 19 A alleles

p = TOTAL number of A alleles / TOTAL number of alleles in the population (number of students x 2) = 37/(28 x 2) = .66 = p

Number of a alleles present at the fifth generation

Number of offspring with genotype aa (0) x 2 = 0 a alleles

Number of offspring with genotype Aa (19) x 1 = 14 a alleles

p = TOTAL number of a alleles / TOTAL number of alleles in the population (number of students x 2) = 19/(28 x 2) = .34 = q

Number of A alleles present at the tenth generation

Number of offspring with genotype AA (7) x 2 = 14 A alleles

Number of offspring with genotype Aa (21) x 1 = 21 A alleles

p = TOTAL number of A alleles / TOTAL number of alleles in the population (number of students x 2) = 35(28 x 2) = .625 = p

Number of a alleles present at the tenth generation

Number of offspring with genotype aa (0) x 2 = 0 a alleles

Number of offspring with genotype Aa (21) x 1 = 21 alleles

p = TOTAL number of a alleles / TOTAL number of alleles in the population (number of students x 2) = 21/(28 x 2) = .375 = q

The changes in the p and q frequencies from Case I to Case II show that the homozygous recessive genotype has been completely eliminated in Case II because it is not resistant to sickle-cell anemia. Therefore it cannot survive long enough to reproduce. Between Case II and Case III, the recessive allele has increased because the heterozygous genotype is preferred. Looking at the fifth generation and tenth generation shows how the frequency of the recessive allele is increasing.

The recessive allele is not eliminated in either Case II or Case III because the heterozygous genotype is resistant to sickle-cell anemia and therefore will live long enough to reproduce and pass on the recessive allele.

  1. What is the importance of heterozygotes (the heterozygote advantage) in maintaining genetic variation in populations?

The importance of heterozygotes is that they ensure both alleles are passed onto the next generation. This way the recessive allele remains in the population’s gene pool. Because the recessive allele is in the population’s gene pool, variation is maintained in the population.

Final Class Frequencies

Number of A alleles present at the fifth generation

Number of offspring with genotype AA (7) x 2 = 14 A alleles

Number of offspring with genotype Aa (13) x 1 = 13 A alleles

p = TOTAL number of A alleles / TOTAL number of alleles in the population (number of students x 2) = 27/(28 x 2) = .48 = p

Number of a alleles present at the fifth generation

Number of offspring with genotype aa (8) x 2 = 16 a alleles

Number of offspring with genotype Aa (13) x 1 = 13 a alleles

p = TOTAL number of a alleles / TOTAL number of alleles in the population (number of students x 2) = 29/(28 x 2) = .52 = q

The initial genotypic frequencies of the populations are only .02 off of the starting allele frequencies.

  1. What do your results indicate about the importance of population size as an evolutionary force?

My results should indicate that by making the population smaller, chance plays a greater role in the population’s genotypes. However, my results are similar to the original p and q values.

Hardy-Weinberg Problems

  1. In Drosophilia the allele for normal-length wings is dominant over the allele for vestigial wings (vestigial wings are stubby little curls that cannot be used for flight). In a population of 1,000 individuals, 360 show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait.

p 2 = .4 2 =.16 à .16 (1000) = 160

2pq = 2 (.4) (.6) = .48. à .48 (1000) = 480

I would expect 160 individuals to be homozygous dominant and 480 individuals to be heterozygous for this trait.

  1. The allele for unattached earlobes is dominant over the allele for attached earlobes. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for this trait?

p 2 = .5 2 =.25 à .25 (500) = 125

2pq = 2 (.5) (.5) = .5 à .5 (500) = 250

I would expect 125 individuals to be homozygous dominant and 250 individuals to be heterozygous for unattached or attached earlobes.

  1. The allele for the hair pattern called “widow’s peak” is dominant over the allele for no “widow’s peak.” In a population of 1,000 individuals, 510 show the dominant phenotype. How many individuals would you expect of each of the possible three genotypes for this trait?

1000 individuals x 2 = 2000 alleles

1000 – 510 = 490 recessive phenotype

490 individuals x 2 = 980 a alleles

980/2000 = .49 = q 2 à .49 (1000) = 490 homozygous recessive individuals

p 2 = .3 2 = .09 à .09 (1000) = 90 homozygous dominant individuals

2pq = 2 (.3) (.7) = .42 à .42 (1000) = 420 heterozygous individuals

I would expect 90 homozygous dominant individuals, 420 heterozygous individuals, and 490 homozygous recessive individuals.

  1. In the United States, about 16% of the population is Rh negative. The allele for Rh negative is recessive to the allele for Rh positive. If the student population of a high school in the U.S. is 2,000, how many students would you expect for each of the three possible genotypes?

q 2 (2000) = .4 2 = .16 à .16 (2000) = 320 homozygous recessive individuals

p 2 (2000) = .6 2 = .36 à .36 (2000) = 720 homozygous dominant individuals

2pq = 2 (.6) (.4) = .48 à .48 (2000) = 960 heterozygous individuals

I would expect 720 individuals to be homozygous dominant, 960 individuals to be heterozygous, and 320 individuals to be homozygous recessive for Rh positive and Rh negative. This would mean that the phenotype for Rh positive would include 84 % of the population and the phenotype for Rh negative would include 16% of the population.

  1. In certain African countries 4% of the newborn babies have sickle-cell anemia, which is a recessive trait. Out of a random population of 1,000 newborn babies, how many would you expect for each of the three possible genotypes?

q 2 (1000) = .2 2 = .04 à .04 (1000) = 40 homozygous recessive individuals

p 2 (2000) = .8 2 = .64 à .64 (1000) = 640 homozygous dominant individuals

2pq = 2 (.8) (.2) = .32 à .32 (1000) = 320 heterozygous individuals

I would expect out of a random population of 1,000 newborn babies there would be 640 homozygous dominant individuals, 320 heterozygous individuals, and 40 homozygous recessive individuals for sickle cell anemia, which is a recessive trait.

  1. In a certain population, the dominant phenotype of a certain trait occurs 91% of the time. What is the frequency of the dominant allele?

1-.91 = .09 = q 2 à sqrt .09 = .3 = q

The frequency of the dominant allele is .7.

The results from Population genetics and evolution have helped me see how microevolution occurs. In Case I—A Test of an Ideal Hardy-Weinberg Population, the results of the population’s dominant and recessive allele frequencies were very close to the original dominant and recessive allele frequencies because the population meet all five of the requirements for a population to be in Hardy-Weinberg equilibrium. (The five requirements for a population to be in Hardy-Weinberg equilibrium are listed in the procedure.) In Case II—Selection, our class’ dominant and recessive allele frequencies are .25 away from the allele frequencies we started mating with. The shift in allele frequency shows that because the homozygous recessive genotype does not survive against sickle-cell anemia. This population is not in Hardy-Weinberg equilibrium. Therefore the population is evolving because the frequency of the alleles is changing. In Case III—Heterozygote Advantage, the heterozygote genotype is most fit to protect against sickle-cell anemia and the homozygous dominant genotype only survives about .5 of the time. Again, the population is not in Hardy-Weinberg equilibrium because not all genotypes have an equal chance of mating. In both Case II and Case III, the recessive allele will never be completely eliminated from the population because the heterozygous genotype ensures that the recessive allele will be passed on, thus maintaining genetic variation. Case IV—Genetic Drift shows an error in our class data. Because the populations are smaller than the original class’ population, chance should have played a greater role in the population’s genotypes. However, my allele frequencies after five generations are only .02 away from the original allele frequencies of the starting population.

This lab directly relates to the material being covered in class because it illustrates how evolution occurs. If the five conditions of Hardy-Weinberg equilibrium are not met, then allele frequencies are changing, which means the population is evolving. This lab showed me in separate examples based off of which condition of Hardy-Weinberg equilibrium is being broken how evolution occurs.