Insects & Flowers

 

The Flower and the Fly: Long Insect Mouthparts and Deep Floral Tubes
Natural History,  March, 2005  by Laura A. Session,  Steven D. Johnson

The mega-nosed fly (Moegistorhynchus longirostris) of southern Africa, like its literary counterpart, Pinocchio, has a bizarre appearance that reveals an underlying truth. Its proboscis, which looks like a nose but is actually the longest mouthpart of any known fly, protrudes as much as four inches from its head–five times the length of its bee-size body. In flight the ungainly appendage dangles between the insect’s legs and trails far behind its body.

To an airborne fly, an elongated proboscis might seem a severe handicap (imagine walking down the street with a twenty-seven-foot straw dangling from your mouth). Apparently, though, the handicap can be well worth its aerodynamic cost. The outlandish proboscis gives the mega nosed fly access to nectar pools in long, deep flowers that are simply out of reach to insects with shorter mouthparts.

But that poses a conundrum: why would natural selection favor such a deep tube in a flower? After all, nectar itself has evolved because it attracts animals that carry pollen, the sperm of the floral world, from one plant to another. And since pollinators perform such an essential service for the flower, shouldn’t evolution have favored floral geometries that make nectar readily accessible to the pollinators?

Yet the story of the long proboscis of the mega-nosed fly and the long, deep tubes of the flowers on which it feeds is not quite so straightforward. There are subtle advantages, it turns out, to making nectar accessible to only a few pollinators, and nature factors those advantages into the evolutionary equation as well. In fact, the evolution of those two kinds of organisms, pollinator and pollinated, presents an outstanding example of an important evolutionary phenomenon known as coevolution. Coevolution can explain the emergence of bizarre or unusual anatomies when no simple evolutionary response to natural selection is really adequate. It can help conservationists identify species that could be vital in maintaining a given habitat. And it can help naturalists investigating novel plants predict what kinds of animals might pollinate their flowers.

The coevolution of the mega nosed fly and the plants it pollinates is a tale of extreme specialization. Each species has adapted to changes in the other in ways that have left each of them, to some degree, reliant on the other. The idea that a plant species might become dependent for pollination on a single species of animal goes back to the writings of Charles Darwin. For example, Darwin noted, the flower spur of the Malagasy orchid (Angraecum sesquipedale) contains a pool of nectar that is almost a foot inside the opening of the flower. (A flower spur is a hollow, hornlike extension of a flower that holds nectar in its base.) In pondering the evolutionary significance of those unusual flowers, Darwin predicted that the orchid must be adapted to a moth pollinator with a long proboscis.

Critical to Darwin’s prediction was his suspicion that pollination could take place only if the depth of a plant’s flowers matched or exceeded the length of a pollinator’s tongue. Only then would the body of the pollinator be pressed firmly enough against the reproductive parts of the flower to transfer pollen effectively as the pollinator fed. Thus, as ever deeper flowers evolved through enhanced reproductive success, moths with ever longer proboscises would also, preferentially, live long enough to reproduce, because they would most readily reach the available supplies of nourishing nectar. Longer proboscises would lead yet again to selection for deeper flower tubes.

The result would be the reciprocal evolution of flowers and pollinator mouthparts. That coevolutionary process would cease only when the disadvantages of an exaggerated trait balanced or outweighed its benefits. Given enough time, the process might even produce new species: an insect the specializes in feeding on nectar from deep flowers, and a deep-flowered plant specialized for being pollinated by insects with long mouthparts.

In the early twentieth century it seemed that Darwin’s prediction had been borne out. A giant hawk moth from Madagascar, Xanthopan morganii praedicta, was captured, with a proboscis that measured more than nine inches long. Although no one has actually seen the insect feeding on the flower, the discovery is still remarkable, and strongly suggestive of the coevolution of the orchid and moth. Other insects that have relationships with highly specific plants, such as the mega nosed fly and other, related long-nosed fly species of southern Africa, provide even better evidence of the reciprocal links between planes and their pollinators.

Darwin would have been amazed that some flies in southern Africa have longer tongues than most hawk moths do. After all, the flies’ bodies are several times smaller than the hawk moths’ are. Flies are described as long-nosed if their mouthparts are longer than three quarters of an inch. By that criterion, more than a dozen long-nosed fly species are native to southern Africa. They belong to two families. The nemestrinids, or tangle-veined flies (which include the mega-nosed fly), feed solely on nectar, whereas the tabanids, or horseflies, feed mostly on nectar, though female tabanids have separate mouthparts to suck blood for their developing eggs.

Like all other long-nosed flies, the mega nosed fly is the sole pollinator to a group of unrelated plant species; such a group is known as a guild. The plant guild of the mega nosed fly includes species from a wide variety of plant families, including geraniums, irises, orchids, and violets.

Even though guild members may be only distantly related, all of them have roughly the same characteristics. For example, plants in the long-nosed fly guild all have long, straight floral tubes or spurs; brightly colored flowers that are open during the day; and no scent. The defining traits of a guild together form what botanists call a pollination syndrome. For example, bird-pollinated flowers are typically large, red, and unscented, whereas moth-pollinated flowers are more likely to be long, narrow, white, and scented in the evening.

The most important trait in the pollination syndrome of the long-nosed fly (and indeed, in all pollination syndromes of long-nosed insects) is a deep, tubular flower or floral spur. One of us (Johnson) and Kim E. Steiner of the Compton Herbarium in Claremont, South Africa, studied the orchid Disadraconis, a southern African plant with a deep, tubular floral spur. The two investigators artificially shortened the spurs of some orchids in a habitat where the only pollinators present were long-nosed flies. The plants whose spurs remained long got more pollen, and were more likely to produce fruits, than the ones whose spurs were shortened.

Yet short floral spurs are not necessarily a reproductive disadvantage. Shorter spurs would make it possible for a wider range of pollinators to access the nectar, if various potential pollinators are present. Instead, longer spurs only seem to be an advantage when long-tongued insects are the sole pollinators. Johnson and Steiner found that differences in spur length among populations cannot be blamed on differences in moisture or temperature, thus reinforcing their conclusion that spur length was an adaptation to the local distributions of long-tongued flies.

Not only does spur length correlate statistically with pollinator traits, but a direct causal connection can be demonstrated. Johnson and Ronny Alexandersson, a botanist at Uppsala University in Sweden, studied South African Gladiolus flowers pollinated by long-tongued hawk moths. When the hawk moth proboscises were long compared to the length of the flower tube, the hawk moths did not efficiently pick up pollen, and the flowers did not reproduce well. When the hawk moth proboscises were relatively short, pollen was more readily transferred, and the plants were more likely to be fertilized and bear fruit. Thus the length of the pollinator’s proboscis exerts a strong pressure on the reproductive success of the flowers.

Those studies and others suggest that what Darwin predicted of the Malagasy orchid is a rather general phenomenon: hawk moths and long-nosed flies coevolved with their plant partners. As floral tubes became longer, so did the pollinators’ proboscises, and those led, in turn, to even longer flowers. As the lengths of the flower tube and the insect proboscis converge, a remarkable degree of specialization develops. The plants come to rely for pollination on the few insect species that can reach their flowers’ nectar supplies.

There are advantages for the specialists on both sides of this relationship. The long-nosed flies obviously get privileged access to pools of nectar. And the plants pollinated by long-nosed flies benefit from a near-exclusive pollen courier service–or at least one that minimizes the risk of delivery to the wrong address. But specializing can also be a risky strategy for the plants if the pollinators are less interested in fidelity than the plants are. Long-nosed flies could not survive on the nectar they could get by visiting just one plant species; the flies must visit several plant species to gather the energy they need. Johnson and Steiner observed mega nosed flies visiting at least four species with deep flowers.

Such promiscuous behavior could be detrimental to the plants. A fly might end up carrying pollen from one species to a different species in the guild, thereby wasting the pollen. Worse, the foreign pollen could end up clogging the stigmata, the female reproductive structures, of the receiving flowers, preventing them from getting the “right” pollen. But the stigmata of plants in the guild of the mega nosed fly do not clog, because among those plants yet another clever adaptation to specialized pollination has evolved. Each plant species arranges its anthers, the male reproductive structures, in a characteristic position. That way, the pollen from each species sticks to the pollinator’s body in a distinct but consistent, plant-specific location. The fly becomes an even more efficient courier, carrying pollen from various plant species simultaneously, say, on its head, legs, and thorax.

The risks of specialization are not confined to the flowers. Just as the flies are unfaithful partners, some flowers are dishonest about signaling a nectar reward. The orchid D. draconis, for instance, is not the mutualistic partner it seems. The flower attracts the mega-nosed fly because it looks like other members of the fly’s guild. But, whereas the fly carries the orchid’s pollen, the orchid offers no nectar in return.

The risk of falling for such a trick seems a small price for the flies to pay for the benefits of specialization. But specialization also carries a much graver risk–in fact the ultimate risk–for both members of the partnership because the disappearance of either partner is likely to doom the other one, as well. Some plant species have mechanisms, such as vegetative reproduction or self-pollination, that may help sustain their populations in the short run. But in the long run, without their pollinators, the species will slowly and irrevocably decline. Pollinating insects may be more flexible in some cases, but are still vulnerable if a key food source disappears.

Unfortunately, in southern Africa that is just what is happening to many plants and their long-nosed fly partners. Often not even closely related insect species can help in pollination. For affected plants, the loss of a single fly species means extinction. And examples of that gloomy cascade have already been observed. Peter Goldblatt of the Missouri Botanical Garden in St. Louis and John C. Manning of the Compton Herbarium have ‘reported that many populations of long-nosed flies are threatened by the loss of their wetland breeding habitat, and also, possibly, by the loss of other insects they parasitize during their larval stages. In some habitats, flowers in the long-nosed fly guild already produce no seeds, because their pollinator is locally extinct.

Naturalists have accepted the concepts of guilds and pollinator syndromes for many years, and predicting which pollinators regularly visit which plants has become something of a cottage industry. But just how common is pollinator specialization in southern Africa? Promiscuity could turn out to be a more successful–and more widespread–strategy than specialization, even among plants that seem to fit into identifiable guilds.

In recent years ecologists have discovered that just because plants and insects appear to form a pollination guild does not guarantee they never venture outside it. For example, ecologists have noted that in years when hummingbird populations are low, flowers ordinarily pollinated by hummingbirds can fill up with nectar and become pollinated effectively by bees. Likewise, bees once thought to specialize in only one or two plant species turn out to forage on a variety of plants.

The take-home lesson has been that the syndrome concept is no substitute for careful field observation. Some investigators even think that the concept has caused botanists to overlook generalists. In the Northern Hemisphere, for instance, studies suggest that generalization is the norm, not the exception. Johnson and Steiner recently completed a study showing that members of the orchid and asclepiad families in the Northern Hemisphere tend to rely on between three and five pollinators each. In contrast, plants from the same families in the Southern Hemisphere rely on just one pollinator each.

So why might generalization be more common in the Northern Hemisphere than it is in the Southern Hemisphere? Perhaps the reason is that social bees, which are largely opportunistic, dominate pollinator faunas in northern regions. In the Southern Hemisphere, by contrast, social bees are mostly absent, replaced instead by more specialized pollinators such as the long-nosed flies and hawk moths.

But that is just a broad generalization itself. More data on the geographic distribution of pollinator specialization needs to be gathered, particularly in tropical countries. The data is vital, not only to advance the specialization debate, but also to protect as many of these unique species and relations as possible, lest they disappear forever.

 

 

Genetics of Drosophila Melanogaster

 

 

Genetics of Drosophila melanogaster

Introduction:
Gregor Mendel revolutionized the study of genetics. By studying genetic inheritance in pea plants, Gregor Mendel established two basic laws of that serve as the cornerstones of modern genetics: Mendel’s Law of Segregation and Law of Independent Assortment. Mendel’s Law of Segregation says that each trait has two alleles, and that each gamete contains one and only one of these alleles. These alleles are a source of genetic variability among offspring. Mendel’s Law of Independent Assortment says that the alleles for one trait separate independently of the alleles for another trait. This also helps ensure genetic variability among offspring.
Mendel’s laws have their limitations. For example, if two genes are on the same chromosome, the assortment of their alleles will not be independent. Also, for genes found on the X chromosome, expression of the trait can be linked to the sex of the offspring. Our knowledge of genetics and the tools we use in its study have advanced a great deal since Mendel’s time, but his basic concepts still stand true.
Drosophila melanogaster, the common fruit fly, has been used for genetic experiments since T.H. Morgan started his experiments in1907. Drosophila make good genetic specimens because they are small, produce many offspring, have easily discernable mutations, have only four pairs of chromosomes, and complete their entire life cycle in about 12 days. They also have very simple food requirements. Chromosomes 1 (the X chromosome), 2, and 3 are very large, and the Y chromosome – number 4 – is extremely small. These four chromosomes have thousands of genes, many of which can be found in most eukaryotes, including humans.
Drosophila embryos develop in the egg membrane. The egg hatches and produces a larva that feeds by burrowing through the medium. The larval period consists of three stages, or instars, the end of each stage marked by a molt. Near the end of the larval period, the third instar will crawl up the side of the vial, attach themselves to a dry surface, and form a pupae. After a while the adults emerge.
Differences in body features help distinguish between male and female flies. Females are slightly larger and have a light-colored, pointed abdomen. The abdomen of males will be dark and blunt. The male flies also have dark bristles, sex combs, on the upper portion of the forelegs.

Hypothesis:
After performing a dihybrid cross between males with normal wings and sepia eyes and females with vestigial wings and red eyes, we expect to see only hybrids with normal wings and red eyes in the first filial generation. Then we expect to observe a 9:3:3:1 ratio of phenotypes in the second filial generation.

Materials and Methods:
The materials used for this lab were:  culture vial of dihybrid cross, isopropyl alcohol 10%, camel’s hair brush, thermo-anesthetizer, petri dish, 2 Drosophila vials and labels, Drosophila medium, fly morgue.

A vial of wild-type Drosophila was thermally immobilized and the flies were placed in a petri dish. Traits were observed. A vial of prepared Drosophila was immobilized and then observed under a dissecting microscope. Males and females were separated and mutations were observed and recorded. The parental generation was placed in the morgue. The vial was placed in an incubator to allow the F1 generation to mature.
The F1 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded. Five mating pairs of the F1 generation were placed into a fresh culture vial, and the vial was placed in an incubator. The remaining F1 flies were placed in the morgue. The F1 flies were left in the vial for about a week to mate and lay eggs. Then the adults were removed and placed in the morgue. The vial was placed back in the incubator to allow the F2 generation to mature. The F2 generation was immobilized and examined under a dissecting microscope. The sex and mutations of each fly were recorded.

Results:  

Table 1 Phenotypes of the Parental Generation

Phenotypes Number of Males Number of Females
Normal wings/red eyes 0 0
Normal wings/sepia eyes 3 0
vestigial wings/red eyes 0 4
vestigial wings/sepia eyes 0 0

Table 2  Phenotypes of the F1 Generation

Phenotype Number of Males Number of Females
Normal wings/red eyes 78 95
Normal wings/sepia eyes 0 0
vestigial wings/red eyes 0 0
vestigial wings/sepia eyes 0 0

Table 3  Phenotypes of the F2 Generation

Phenotypes Number of Males  Number of Females
Normal wings/red eyes 4 7
Normal wings/sepia eyes 4 5
vestigial wings/red eyes 0 1
vestigial wings/sepia eyes 0 0
normal red/mutated body shape 2 0
normal sepia/mutated body shape 1 0

Questions

  1. How are the alleles for genes on different chromosomes distributed to gametes? What genetic principle does this illustrate?
    The alleles on different chromosomes are distributed independently of one another, demonstrating Mendel’s Law of Independent Assortment.
  2. Why was it important to have virgin females for the first cross (yielding the F1 generation), but not the second cross (yielding the F2 generation)?
    It was important to have virgin females for the first cross to ensure that the offspring are the result of the desired cross. It was not necessary to isolate virgin females for the second cross because the only male flies to which they had been exposed were also members of the F1 generation.
  3. What did the chi-square test tell you about the validity of your experiment data? What is the importance of such a test?
    The chi-square test showed that the results of our first cross were valid, but that the results of our F1 cross were not normal. It is important to conduct such a test to determine how much your experimental data deviated from what was expected.

Discussion and Conclusion:
The results of our parental cross turned out just as expected, but our F2 generation was not normal. Some sort of mutation must have occurred that caused the strange body shape seen in several individuals of our F2 generation.

Genetics Problems ppt Questions

Genetics Problems
ppt Questions

 

Independent Assortment

1. How many different kinds of gametes could the following individuals produce? Remember the formula 2n where n equals the number of heterozygotes.

     a. aaBb

     b. CCDdee

     c. AABbCcDD

     d. MmNnOoPpQq

     e. UUVVWWXXYYZz

 

P1, F1, and F2 Monohybrid Crosses

2. In dogs, wire-haired is due to a dominant gene (W), smooth-haired is due to its recessive allele (w). Show the results of crossing a homozygous wire-haired dog with a smooth-haired dog.

 

 

 

 

 

3. What kind of cross is this?

4. What was the genotype of all of the puppies? the phenotype?

 

5. The puppies belong to the _________ generation.

6. How would you write the F1 cross for this trait?

7. Show the results of working the F1 cross for this trait.

 

 

 

 

6. What phenotypic ratio did you get from this F1 cross?

7. What genotypic ratio did you get from this F1 cross?

8. Two wire-haired dogs are mated. Among the offspring of their first litter is a smooth-haired pup. If these two dogs mate again, what are the chances of them having another smooth-haired pup?

 

 

 

9. What are the chances that the pup will be wire-haired?

 

10. A Wire-haired male is mated with a smooth-haired female. The mother of the wire-haired male was smooth-haired. What are the phenotypes and genotypes of the pups they could produce? Show how you got your results.

 

 

 

 

 

Incomplete Dominance 

11. In snapdragons, red flower color (R) is incompletely dominant over white flower color (r). The hybrids or heterozygous plants (Rr) are pink in color. Show the genotype for a white flower and for a red flower.

 

12. If a red-flowered plant is crossed with a white-flowered plant, what are the genotypes and phenotypes of the F1 generation plants? Show your work.

 

 

 

 

13. What is the phenotype of the flowers? what is their genotype?

 

14. What genotypes and phenotypes will be produced in the F2 generation? Show your work.

 

 

 

 

 

15. How did the genotypic and phenotypic ratio compare to each other in this incomplete dominance cross?

16. What would the phenotypic ratio have been if this had been complete dominance?

17. What kind of offspring can be produced if a red-flowered plant is crossed with a pink-flowered plant? Show your work.

 

 

 

 

 

18. What kind of offspring is/are produced if a pink-flowered plant is crossed with a white-flowered plant? Show your work.

 

 

 

 

 

Sex-linked Traits

19. What is the genotype for female?  for male?

20. In humans, colorblindness (Xc) is a recessive sex-linked trait. Two people with normal color vision (XC) have a colorblind son. What are the genotypes of the parents?

 

21. What are the genotypes and phenotypes possible among their other children? Show your work.

 

 

 

 

 

22. A couple has a colorblind daughter. What are the possible genotypes and phenotypes of the parents and the daughter?

 

 

Dihybrid Crosses

23. In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled condition is due to its recessive allele (f). Dimpled cheeks (D) are dominant to non-dimpled cheeks (d). Two persons with freckles and dimpled cheeks have two children. One child has freckles but no dimples. The other child has dimples but no freckles. What is the genotypes of the parents? the children?

 

 

24. What are the possible phenotypes and genotypes of the children that they could produce? Show all your work.

 

 

 

 

 

 

 

 

 

 

25. What phenotypic ratio did you get?

26. What genotypic ratio did you get?

27. What are the chances that they would have a child whom lacks both freckles and dimples?  What would be the child’s genotype?

 

28. A person with freckles and dimples whose mother lacked both freckles and dimples marries a person with freckles but no dimples whose father did not have freckles or dimples. What are the chances that they would have a child whom lacks both freckles and dimples? Show the genotypes of the parents and all the offspring.

 

 

 

 

 

 

 

 

29. In dogs, the inheritance of hair color involves a gene (B) for black hair and a gene (b) for brown hair. A dominant (C) is also involved. It must be present for the color to be synthesized (made). If this gene is NOT present, a blond condition results. Complete the following table:

 

Genotype Phenotype Color Deposition gene
BB or Bb CC or Cc
bb CC or Cc
BB or Bb cc
bb cc

 

 

30. A brown-haired male, whose father was a blond, is mated with a black-haired female ,whose mother was brown-haired and her father was blond. What is the genotype of the man and woman? Show the genotypes and phenotypes of all of their offspring.

 

 

 

 

 

 

 

Population Genetics or Hardy-Weinberg Law

Sixteen percent (16%) of the human population is known to be able to wiggle their ears. This trait is determined to be a recessive gene. Use the following equations to answer this population genetics problem:

1 = p2 + 2pq + q2                                      then use 1 = p + q

p2 – frequency of homozygous dominants

2pq – frequency of heterozygotes

q2 – frequency of homozygous recessives

p – frequency of dominant allele

q – frequency of recessive allele

31. What percent of the population is homozygous dominant for this trait? Show your work.

 

 

 

 

 

32. What percent of the population is heterozygous for this trait? Show your work.

 

 

 

 

 

Multiple Alleles – ABO Blood Type 

33. Henry Anonymous, a film star, was involved in a paternity case. The woman bringing the suit had two children. One child had blood type A and the other child had blood type B. Her blood type was O, the same as Henry’s. The judge in the case awarded damages to the woman, saying that Henry had to be the father of at least one of her children. was the judge correct in his decision? Show how you got your answer.

 

 

 

Genetics Study Guide

Genetics Study Guide 

The two genes or alleles that combine to determine a trait would be the organism’s _______________.
Type AB blood, having two genes dominant for a trait, is an example of ________.
State Mendel’s law of segregation.
Rr x Rr is an example of what type of cross —– P1, F1, or F2?
If both alleles are the same in a genotype, is the genotype homozygous or heterozygous?
Which cross is a cross between two hybrids —– P1, F1, or F2?
__________ dominance results in the blending of genes in the hybrid. Give an example using flower color.
What is another term for a heterozygous genotype?
The _____________ is the physical feature such as round peas that results from a genotype.
How many traits are involved in a monohybrid cross?
What type of organism was used in the first genetic studies done by Gregor Mendel?
What is a karyotype?
The two genes for a trait represented by capital & lower case letters are called __________.
How many traits are involved in a dihybrid cross?
Which of Mendel’s laws states that the dominant gene in a pair will be expressed?
If both alleles are the same, is the genotype homozygous or heterozygous? Write an example.
Write an example of a hybrid or heterozygous genotype.
The genes for sex-linked traits are only carried on which chromosome?
Who is considered to be the “father of genetics”?
A second filial or F2 cross is also called a ____________ cross.
The failure of chromosomes to separate during meiosis (egg & sperm formation) is known as _________________.
A cross between two pure or homozygous organisms is called what type of cross —– P1, F1, or F2?
What genetic disorder results from a sex-linked trait that affects color vision?
The genetic disorder called _______________ is known as the “free bleeders” disease.
Having three 21st chromosomes causes the genetic disorder known as _________.
A person suffering from the genetic disorder called ______________ can not digest fats.
_____________________ disease is a genetic disorder where red blood cells carry less oxygen.
Work a P1 cross for plant height in peas.
Work an F1 cross for plant height in peas.
BACK

 

Hardy-Weinberg Problems

 

POPULATION GENETICS AND THE HARDY-WEINBERG LAW

 

The Hardy-Weinberg formulas allow scientists to determine whether evolution has occurred. Any changes in the gene frequencies in the population over time can be detected. The law essentially states that if no evolution is occurring, then an equilibrium of allele frequencies will remain in effect in each succeeding generation of sexually reproducing individuals. In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions must be met:

  1. No mutations must occur so that new alleles do not enter the population.
  2. No gene flow can occur (i.e. no migration of individuals into, or out of, the population).
  3. Random mating must occur (i.e. individuals must pair by chance)
  4. The population must be large so that no genetic drift (random chance) can cause the allele frequencies to change.
  5. No selection can occur so that certain alleles are not selected for, or against.

Obviously, the Hardy-Weinberg equilibrium cannot exist in real life. Some or all of these types of forces all act on living populations at various times and evolution at some level occurs in all living organisms. The Hardy-Weinberg formulas allow us to detect some allele frequencies that change from generation to generation, thus allowing a simplified method of determining that evolution is occurring. There are two formulas that must be memorized:

 

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

Individuals that have aptitude for math find that working with the above formulas is ridiculously easy. However, for individuals who are unfamiliar with algebra, it takes some practice working problems before you get the hang of it. Below I have provided a series of practice problems that you may wish to try out. Note that I have rounded off some of the numbers in some problems to the second decimal place.

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:

  1. The frequency of the “aa” genotype.
  2. The frequency of the “a” allele.
  3. The frequency of the “A” allele.
  4. The frequencies of the genotypes “AA” and “Aa.”
  5. The frequencies of the two possible phenotypes if “A” is completely dominant over “a.”

PROBLEM #2.    Sickle-cell anemia is an interesting genetic disease. Normal homozygous individuals (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?

PROBLEM #3.    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:

  1. The frequency of the recessive allele.
  2. The frequency of the dominant allele.
  3. The frequency of heterozygous individuals.

PROBLEM #4.    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:

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

PROBLEM #5.     A rather large population of Biology instructors have 396 red-sided individuals and 557 tan-sided individuals. Assume that red is totally recessive. Please calculate the following:

  1. The allele frequencies of each allele.
  2. The expected genotype frequencies.
  3. The number of heterozygous individuals that you would predict to be in this population.
  4. The expected phenotype frequencies.
  5. Conditions happen to be really good this year for breeding and next year there are 1,245 young “potential” Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided?

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.

PROBLEM #7.    After graduation, you and 19 of your closest friends (lets say 10 males and 10 females) charter a plane to go on a round-the-world tour. Unfortunately, you all crash land (safely) on a deserted island. No one finds you and you start a new population totally isolated from the rest of the world. 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, what will be the incidence of cystic fibrosis on your island?

PROBLEM #8.    You sample 1,000 individuals from a large population for the MN blood group, which can easily be measured since co-dominance is involved (i.e., you can detect the heterozygotes). They are 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.
  2. Supposing the matings are random, the frequencies of the matings.
  3. The probability of each genotype resulting from each potential cross.

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:

  1. The frequency of the recessive allele in the population.
  2. The frequency of the dominant allele in the population.
  3. The percentage of heterozygous individuals (carriers) in the population.

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 in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the allelic frequencies of this population (i.e., what are p and q)?

PROBLEM #11.    The ability to taste PTC is due to a single dominate allele “T”. You sampled 215 individuals in biology, and determined that 150 could detect the bitter taste of PTC and 65 could not. Calculate all of the potential frequencies.

ANSWERS