Introduction
Genetics can be defined as the study of ways in which biological traits are passed from one generation to the next. The study was begun in the 19th Century by an Austrian monk named Gregor Mendel, who derived its basic principles from the cross-pollination of different varieties of garden peas. Much of human genetics is connected with the study of diseases, for it happens that many diseases such as cystic fibrosis, Huntington's disease, sickle-cell anemia, hemophilia, etc., are genetic diseases.
The unit of heredity is the gene, which is part of a complex protein molecule called deoxyribonucleic acid (DNA). Chromosomes are made up primarily of DNA and can be thought of as being composed of a linear series of genes each containing instructions for producing one specific bodily trait such as eye color, for example. Thus, a chromosome is like a string of beads, each bead being a gene that influences some physical characteristic. The location of the gene on the chromosome is called its locus. The gene at any locus can normally exist in one of several states or forms. For example, a hair color gene might code for black hair, brown hair, or red hair, etc. An eye color gene might code for blue eyes, brown eyes, gray eyes, etc.
We use the term allele to denote the various possible states of a gene at any locus (e.g., the "blue eye color allele" or the "brown eye color allele"). Gene transmission involves two pairs of genes, one pair for each parent. In reproduction these pairs split, and the child receives one member of each pair, chosen at random from each of his/her parents. (Humans have 46 chromosomes, 23 of which are acquired from the father and are matched by 23 from the mother.) Any one chromosome can possess one allele at each locus; therefore, each human carries only part of the genetic information that determines the genetic make up of his/her offspring. In the case of eye color, a person has two eye color alleles one from his/her mother and one from his/her father. If a person's genetic make up includes two different alleles at the same locus, only one of the two alleles will be visibly expressed.
As an illustration, let B denote the allele for brown eyes and b denote the allele for blue eyes. If a person has two brown-eye alleles, BB, then that person will have brown eyes; if a person has two blue eye alleles, bb, then that person will have blue eyes. (Such people are called homozygotes). If a person has one brown-eye allele and one blue-eye allele, Bb, (a heterozygote), then, in general, that person will have either brown eyes like the BB homozygote or will have blue eyes like the bb homozygote. Which will it be? The answer is: the heterozygote, Bb, will be brown-eyed. For this reason, we say that the brown-eye allele, B, is dominant. It follows that a person can have blue eyes only if he/she is a bb homozygote. For this reason, we say that the blue-eye allele, b, is recessive. The various combinations of genes such as BB, Bb, bb, etc., are called genotypes. The observable categories of "brown eyes" or "blue eyes" are called phenotypes.
Eye Color Examples
The foregoing discussion explains why it is possible for two brown-eyed parents to produce a blue-eyed child. Consider a man who is a brown-eyed heterozygote, Bb, and a woman who is a brown-eyed heterozygote, Bb. Since each parent can pass on exactly one allele to the child, the possible genotypes for the child are: BB, Bb, bB, and bb, as ilustrated in Table 1 below. If the last possibility bb occurs, the two brown-eyed parents will have produced a blue- eyed child. Since the parents transmit each of their genes with equal probability, the four possible outcomes are equally likely. Therefore, there is a one in four chance that two brown-eyed heterozygotes will produce a blue-eyed child, i.e., the probability of this event occurring is 1/4 or .25. The chances that such parents will produce a brown-eyed child are three in four implying a probability of 3/4 or .75.
Table 1
Eye Color Genotypes for Bb Male and Bb Female
Father
B b
Mother B BB Bb
b bB bb
Tutorial Examples
1. Construct tables like Table 1 which display all possible eye-color genotypes for the child of:
a) a Bb father and a bb mother
b) a BB father and a bb mother
c) a bb father and a bb mother.
Compute the probabilities of the genotypes and the phenotypes in each case.
2. Suppose a brown-eyed man, whose parents are both brown-eyed heterozygotes, marries a
woman who is blue-eyed. What is the probability that their child will be blue-eyed? What is
the probability that their first two children will be blue-eyed?
Solution to Example 2: The solution depends on whether the man, who is brown-eyed, is a "carrier" of the blue-eye allele, i.e., whether his genotype is Bb or bB. From Table 1 above, we see that the chances of his being a brown-eyed heterozygote are two in three, implying a probability of 2/3. The probability that he and his wife will produce a blue-eyed child is 1/2 as was shown in 1a). The probability of the two independent events occurring, viz., the man is a carrier of the blue-eye allele and he and his wife produce a blue-eyed child, is the product of the probabilities of the two events: 2/3 x 1/2 = 1/3. Since the two events of producing child number one and producing child number two are independent events, the probability that the first two children are blue-eyed is 1/3 x 1/2 = 1/6.
3. Suppose a man has brown-eyed parents only one of whom is a heterozygote. If he marries a
blue-eyed woman, what are their chances of producing a blue-eyed child?
A. Genetic Disease Models
Genes control a huge number of biochemical reactions that take place inside the cells of the human body. When they fail to control these reactions properly, the result can be a disruption of the body's normal metabolism leading to very serious diseases. Some of these diseases have been traced to the transmission of a single abnormal allele, others are polygenic or involve much more complex factors. After a child is born with a genetic disease, parents may seek advice on the probability of subsequent children contracting the disease. Such genetic counseling is becoming more popular as our knowledge of genetic diseases increases. The following models illustrate the single-gene mode of transmission.
Recessive Disease Model: Cystic Fibrosis
Cystic fibrosis is the most common inherited disease that kills children. The disease results from a disorder in the genes of its victims that causes cells to produce a thick mucous secretion which clogs the passageways of many vital organs. A child infected with cystic fibrosis will probably not live to be an adult. Cystic fibrosis is caused by a single abnormal allele. The disease will occur only in children who have two copies of this abnormal allele, one from each parent. Such diseases are said to be recessive diseases. If two healthy parents have a child afflicted with cystic fibrosis, then both parents must be carriers of the abnormal allele. Let the letter a represent the abnormal allele and A represent the normal allele. Then the genotype of both healthy parents must be Aa, and the genotype of the infected child must be aa. The probability that two healthy carriers of the abnormal allele will produce a child with cystic fibrosis is 1/4 as is indicated in Table 2 below. Table 2 also indicates that the probability of two carriers producing a carrier is 1/2. What is the probability that a normal child of two carriers is also a carrier? Answer: As Table 2 indicates, a normal child of two carriers will have possible genotypes AA, Aa, or aA. Since the latter two represent carriers, the probability that two carriers produced a normal child who is a carrier is 2/3.
Table 2
Cystic Fibrosis Genotypes for Aa Male and Aa Female
Father
A a
Mother A AA Aa
a aA aa
Tutorial Example
Suppose Mr. and Mrs. C (both normal) had sisters afflicted with cystic fibrosis. What is the probability that the couple will have a child with the disease?
Solution: Since both Mr. and Mrs. C had sisters with the disease, it follows that each of their parents must have been a carrier of cystic fibrosis. Therefore, as in the discussion above, the probability that Mr. C is a carrier is 2/3, and the probability that Mrs. C is a carrier is also 2/3. We showed in our previous discussion that the probability that two carriers will produce an afflicted child is 1/4. Therefore, the probability of the three independent events: 1) Mr. C is a carrier and 2) Mrs. C is a carrier and 3) their child will be born with the disease, is 2/3 x 2/3 x 1/4 = 1/9. Note that the probability of the complementary event: their child will not be born with the disease, is 1 - 1/9 = 8/9.
Exercises
1. Both Mr. D and Mrs. D had brothers afflicted with cystic fibrosis. If they have two children,
what is the probability that both will be afflicted with the disease?
2. Mr. and Mrs. E are tested and both are found to be carriers of Tay-Sachs disease, a rare
recessive disorder. What is the probability that their first child will be born with the disease?
That their first two children will be healthy? That their first three children will be carriers?
3. Albinism is a recessive disorder characterized by an extreme reduction of pigmentation
throughout the body. A woman who is an albino marries a normal man whose brother is an
albino. What are the chances that their first child is an albino? Given that their first child is an
albino, what are the chances that their second child will be too?
4. Two first cousins marry. If the sister of their grandfather had cystic fibrosis, what are the
chances that they will have a child with the disease?
Dominant Disease Model: Huntington's Disease
Huntington's disease is a disease caused by a disorder of the genes which causes its victims to have jerky, involuntary motions and extreme mental changes. The disease appears in adults after they have reached middle age. By this time, many of the victims have had children already, and the big question becomes: Do my children have it?
As with cystic fibrosis, Huntington's disease is caused by a single abnormal allele. However, Huntington's occurs in people who have only one copy of the abnormal allele. Thus, if either parent passes on this abnormal allele, the child will be inflicted with the disease. Such diseases are said to dominant diseases. Let a represent the abnormal allele for Huntington's disease and let A represent the normal allele. Because it is dominant, Huntington's will occur in people of genotype Aa. Huntington's is a very rare disease, its victims are almost always married to people with two normal alleles. Table 3 below lists the genotype possibilities for a child born to a normal mother of genotype AA and an afflicted (or eventually afflicted) father of genotype Aa. One sees that the child will be one of two genotypes: AA or Aa. Since Aa individuals will develop the disease and AA individuals will not, the chances are two out of four that the child of a victim will eventually develop the disease, i.e., the probability is ½ that a normal mother and an afflicted father will produce an afflicted child and ½ that they will produce a normal child..
Table 3
Huntington's Genotypes for Aa Male and AA Female
Mother
A A
Father A AA AA
a aA aA
Exercises
1. If a man's uncle died of Huntington's disease, what are the chances that his child will contract
the disease?
2. A man with Huntington's disease has four children. What is the probability that all four will
develop the disease? What is the probability that two children will be affected and two
normal?
B. Population Genetics/Hardy-Weinberg Principle
Our previous discussions involved the genetic make up of an individual or an individual family. Geneticists are also interested in the genetic make up of a population, i.e., in describing the so called gene pool of a population. The description can be expressed in terms of the number or the relative frequencies of the alleles or the genotypes that appear in the population.
To illustrate, consider the eye color example again. Assume that in a certain population, there are only two eye colors: brown and blue. Suppose that 25% of the people are homozygous dominant (BB) and 70% of the people are heterozygous (Bb). (Recall that all of these people have brown eyes.) The remaining 5% must have blue eyes and must be homozygous recessive (bb). These percentages are the relative frequencies of the eye color genotypes at a single locus. We can use these genotypic frequencies to compute the allelic frequencies for B and for b in the population as Table 4 below illustrates. Note that 25% of the gene pool comes from BB individuals who contribute only to the B's pool. Since 70% of the gene pool comes from Bb individuals, their contribution to the pool is split equally between the B and b alleles: 35% to B and 35% to b. The final 5% of the pool comes from the bb individuals who contribute only to the b's pool.
Table 4
Allelic Frequencies Computed From Genotypic Frequencies
Genotype Frequency Frequency of B Frequency of b
____________________________________________________
BB .25 .25 .00
Bb .70 .35 .35
bb .05 .00 .05
____________________________________________________
Total 1.00 .60 .40
____________________________________________________
We conclude that the B allele makes up 60% of the gene pool of the population and the b allele makes up 40% of the of the pool for this eye color locus. This says that the dominant allele B appears in the population with a relative frequency of .60 and the recessive allele b appears with a relative frequency of .40. ( We are assuming that only two allelic forms B and b appear.) Now using this information we predict the relative frequencies of B and b in the next generation (first filial generation) of the population. We can accomplish this by assuming that B has a relative frequency of .60 and b has a .40 relative frequency in both the egg cells of females and the sperm cells of males. The frequencies of the genotypes in the next generation are calculated frequencies in the table below.
Table 5
Genotype Frequencies for the First Filial Generation
Sperm
B(.60) b(.40)
Egg B(.60) BB(.36) Bb(.24)
b(.40) bB(.24) bb(.16)
The genotype frequencies (probabilities) in Table 5 are computed by multiplying the corresponding row and column frequencies reflecting the fact that obtaining one allele from an egg and a second allele from a sperm are independent events. Table 5 predicts that in the new generation, 36% of the population will be homozygous dominant (BB), 48% (2 x 24%) of the population will be heterozygous (Bb), and 16% will be homozygous recessive (bb). From these genotypic frequencies, we can now compute the allelic frequencies for B and b as we did in Table 4. The results are displayed in Table 6 below.
Table 6
Allelic Frequencies for the First Filial Generation
Genotype Frequency Frequency of B Frequency of b
____________________________________________________
BB .36 .36 .00
Bb .48 .24 .24
bb .16 .00 .16
___________________________________________________
Total 1.00 .60 .40
___________________________________________________
Notice that the allelic frequencies of this new generation are the same as those in the preceding generation: .60 for B and .40 for b (see Table 4). If we repeat the experiment and compute the
genotypic frequencies for the second filial generation, we find that these frequencies remain the same: .36 for BB, .48 for Bb, and .16 for bb . The allelic frequencies also remain the same at .60 for B and .40 for b. Indeed, the allelic frequencies in this example are in equilibrium: they do not change from generation to generation. It is also the case in this example that the genotypic frequencies are in equilibrium from the first filial generation of offspring onward. These equilibrium phenomena are examples of what scientists call the Hardy-Weinberg Equilibrium Principle. Under the simplifying assumptions they we made about our population, this equilibrium principle will always hold. (In the general two-allele case, if a population is in Hardy-Weinberg equilibrium and the frequency of B is p and the frequency of b is q, with p+q=1, then the genotypic frequencies of the population remain in equilibrium at p2 for BB, 2pq for Bb, and q2 for bb.)
Tutorial Example
The table below shows some M-N blood type data collected from a population of 208 Bedouins in he Syrian desert.
Genotype Number
_____________________
MM 119
MN 76
NN 13
a) What are the genotypic frequencies for the MM, MN, and NN blood types in this population?
b) What is the frequency of the M allele and the N allele in this population?
c) Compute the frequencies for the next generation. Is the Hardy-Weinberg Principle in effect?
Sickle-Cell Anemia
Sickle-cell anemia is a genetic disease caused by a defective gene which produces abnormal hemoglobin, the red, oxygen-carrying protein in the red blood cells. When affected people exercise or otherwise deplete the oxygen in their blood, the red blood cells change to a sickle or crescent shape. This causes the red blood cells to clog and to rupture the capillaries, starving parts of the body for blood and causing internal bleeding and pain. In addition, the sickle cells are so fragile that they burst, leaving the patient anemic. Affected individuals show poor growth and development and rarely survive to maturity. Death usually results from a crisis following some acute infection. The disease is common among people of central African descent.
Sickle-cell anemia is a recessive disease. If the letter a represents the abnormal sickle- cell allele and A represents the normal hemoglobin-producing allele, then homozygotes of genotype aa are afflicted, heterozygotes of genotype Aa are carriers, and homozygotes of genotype AA are normal. The heterozygote carriers of the sickle-cell allele exhibit some mild symptoms of the disease but their normal allele seems to make enough normal hemoglobin to keep their blood from sickling. The mild condition displayed by heterozygotes is referred to as sickle-cell trait.
Tutorial Questions
1. What is the probability that two carrier parents will produce a child with sickle-cell disease?
That two carrier parents will produce a child who is a carrier of the trait?
2. If a normal homozygote marries a person with the sickle-cell trait are their children at risk for
for the sickle-cell disease? What is the probability that their child will be a carrier?
Exercise: Sickle-Cell and Population Genetics
Consider a population in which sickle-cell anemia is very prevalent such that one child in about 400 is affected.
1. Compute the relative frequencies or probabilities of the sickle-cell allele, a, and the normal
allele, A. Use these values to find the genotypic frequencies for this population.
2. Compute the genotypic frequencies for the first filial generation of the population and display
the results in a table similar to Table 5. Does the Hardy-Weinberg Principle appear to hold for
the population?
3. Construct a table showing the percentage of total marriages in the population that would
involve two heterozygotes (Aa x Aa); two homozygote normals (AA x AA); one heterozygote
and one homozygote normal (Aa x AA); etc., until you exhaust all six possible matings of the
genotypes. Now use the table to compute the expected frequency of affected children among
all matings in which neither parent suffered from sickle-cell anemia. (In other words, find the
probability in this population that two parents, neither of which has the disease, produce a
a child with the disease.)
C. ABO Blood Groups: A Multiple Alleles Model
At the beginning of this century, medical scientists discovered that one can not transfuse blood from one person to another indiscriminately. This led to the classification of all humans into four blood groups: A, B, AB, and O. It is safe to transfuse blood between individuals in the same blood group; but when blood is transfused from an individual in one group to an individual in another group, strict rules must be followed. Humans contain circulating antibodies that attack foreign surface proteins when blood transfusions are performed. Humans with type A blood have A-type proteins on the surfaces of their own red blood cells and have antibodies which attack B- type blood proteins. The reverse is true for B-type humans. People with O-type blood have no surface proteins of either A or B-type, but they have antibodies to attack both A and B-type proteins. People with AB blood have both A and B surface proteins and have no antibodies to attack any type of A, B, or O blood. These facts imply that type O is the "universal donor" and that type AB is the " universal recipient" for blood transfusions.
The blood group gene has three alleles denoted by A, B, and O. (This is in contrast to the two-allele models of previous sections.) These three alleles produce nine genotype pairs which result in six different genotypes for this locus: AA, AB, AO, BB, and BO.(See Table 7 below.) However since A and B are dominant to O, and A and B are co-dominant, there are four observable phenotypes or blood groups:
A (resulting from AA and AO)
B (resulting from BB and BO)
AB (resulting from AB)
O (resulting from OO).
In a given population, let p, q, and r denote the relative frequencies of alleles A, B, and O, respectively, with p+q+r = 1. In the tables below, we compute the frequencies of the various genotypes and the blood types in the population. Assume that the population is in Hardy- Weinberg equilibrium so that the frequencies are constant in any generation.
Table 7
Genotype Frequencies for Blood Group Gene
Male
A(p) B(q) O(r)
Female A(p) AA(p2) AB(pq) AO(pr)
B(q) BA(qp) BB(q2) BO(qr)
O(r) OA(rp) OB(rq) OO(r2)
Table 8
ABO Blood Group Frequencies
Group Genotypes Frequency
__________________________________________
A AA, AO p2 + 2pr
B BB, BO q2 + 2qr
AB AB 2pq
O OO r2
The Hardy-Weinberg Principle asserts that these group frequencies should remain constant in the given population in all subsequent generations. However, tests show that blood type frequencies vary from population to population and that blood group phenotypes do not occur in equal frequencies - types A and O are much more common than B; and type AB is relatively rare in all populations tested. Also racial groups differ in their relative proportions of the four blood types. The following table shows the result of blood type testing among certain populations:
Table 9
Proportions of ABO Phenotypes in Selected Populations
Population A B AB O
_____________________________________________________________
Chinese 0.27 0.23 0.06 0.44
French 0.45 0.09 0.04 0.42
Indians 0.25 0.38 0.07 0.30
Nigerians 0.21 0.23 0.04 0.52
New Zealand 0.40 0.10 0.03 0.47
Amerindians (Navaho) 0.22 0.00 0.00 0.78
US Blacks 0.27 0.21 0.04 0.48
US Whites 0.41 0.10 0.04 0.52
_____________________________________________________________
Tutorial Exercise: Blood Group Analysis for Our Class
Conduct a survey to determine the blood group type of each member of our class. Collect the data and list the numbers in each blood group and list the blood group frequencies as percentages of the total class size.
a) Use Table 9 to calculate the expected numbers in the blood groups based on the proportions listed for the US. Use the chi-squared test with 3 degrees of freedom to determine if there is any significant difference between the expected and observed numbers in the blood groups.
b) Next estimate the frequencies of the three alleles and the four blood types using the expressions in Table 8. How do these expected numbers, predicted by the Hardy-Weinberg model, compare with your observed values for the blood groups?
Tutorial Example
This example illustrates the blood group analysis procedure for a class of 137 from Wellington, New Zealand. The actual number and proportions of the ABO blood groups for that class are:
A B AB O Total
______________________________________________________________________
Number 55 19 4 59 137
Proportion 0.40 0.14 0.03 0.43 1.00
a) The expected numbers for the blood groups calculated from the proportions given in Table 9 are 54.8 13.7 4.1 64.4 137.
Chi-squared analysis for these observed and expected numbers yield X2 = 2.5, d.f. = 3, P = 0.5.
Conclusion: no significant difference between observed and expected values.
b) Using the observed proportions (frequencies) for the New Zealand class, we estimate the frequencies of the A,B,O alleles from Table 8 as follows:
I) Frequency of blood group O is r2 = 0.43. Thus r = O.43 = 0.66
ii) Frequency of blood group A is p + 2pr = 0.40. Substituting r = 0.66, we have
p + 2p x 0.66 = 0.40 which yields p = 0.25.
iii) Since p + q + r = 1, we have q = 1 - p - r = 0.09.
These allelic frequencies are based on the Hardy-Weinberg assumptions and can now be used to obtain expected values for the blood groups using Table 8. How do these numbers compare with the observed numbers?
Exercises
1. Perform part b) blood group analysis on any group of your choice in Table 9.
2. An interesting application of blood group analysis occurs in tests of disputed paternity. Prove,
for example, that a B child can not be born to an O or A father if the mother is O or A; the true
father must be either B or AB. (See Table 7 above.)
3. A woman with blood type A and a man with blood type B had four children with blood types
A, B, AB, and O, respectively. What are the genotypes of the parents?
D. The Human Genome Project
In this section students investigate a very important scientific project related to human genetic research. Students are encouraged to consult the following books: The Sciences: An Integrated Approach, by James Trefil and Robert Hazen, and The Human Genome Project: Deciphering the Blueprint of Heredity, Edited by Necia Grant Cooper as well as the Internet for details about the project. Also a video tape of a PBS newscast describing the project and its ethical implications is available.
1. What is the definition of the word genome?
2. How is DNA associated with the human genome?
3. What is DNA sequencing and DNA mapping?
4. Give some idea of how vast the human genome is relative to numbers of genes, base pairs, etc.
5. What is the origin and purpose of the Human Genome Project? Who are some of the
personalities involved in the project? How is the project funded?
6. How are DNA mapping and DNA sequencing related to the Human Genome project?
7. What role is the computer playing in the accomplishment of the goals of the project?
8. List some of the accomplishments of the Human Genome Project to date.
9. What are some of the prospects for future successes?
10. List some of the ethical, legal, and social implications of the project.