This Document Contains Experiments 35 to 37 Name____________________________________________________Section________________Date___________ Experiment 35: Mitosis—Cell Division Invitation to Inquiry Many people who have been diagnosed with certain kinds of cancers undergo chemotherapy. There are four generally recognized types of chemotherapeutic drugs. 1. Antimetabolites. 2. Topoisomerase inhibitors. 3. Alkylating agents. 4. Plant alkaloids. Do an internet search and discover how each of these chemotherapies interferes with the cell cycle to control dividing cancer cells. Background All large organisms are composed of many cells. Growth of many-celled organisms involves an increase in the number of cells followed by an increase in size of the new cells. This is the basic mechanism by which a body grows or wounds are repaired. Cell division involves two major events: the distribution of identical copies of the genetic information from the parent cell to two daughter cells followed by cytoplasmic division. As a result of cell division all the cells of a multicellular organism have the same genetic information. Notice in Figure 35.1 that the parent cell divides by mitosis, producing two daughter cells. These two identical daughter cells contain exactly the same genetic material as the parent cell. Mitosis assures the production of identical sets of genetic information in the daughter cells. Mitosis is an orderly series of events that results in the equal distribution of the chromosomes that carry the genetic information to the two new cells. The process flows from one stage to the next without interruption. Traditionally, mitosis has been artificially divided into four phases: prophase, Figure 35.1 metaphase, anaphase, and telophase. It is a convenience to be able to classify parts of the process in this way, but it can be misleading if we forget that there is really no pause or interruption in the events. It might be helpful to consider these four phases as being pictures taken with a flash camera. In such pictures, the motion is frozen so that we can examine details that may interest us. Similarly, when we look at a plant or animal cell that has been killed and stained while in the process of mitosis, we are looking at a cell that was at a particular point in the mitotic process. It is also important to recognize that mitosis is a relatively brief period in the life of a cell. A typical mitosis will take 2 to 4 hours to occur. The majority of the life of a cell is spent in a nondividing condition known as interphase. During interphase, the cell is participating in many important activities, including growth and the replication of DNA. Many cells differentiate into specialized cells that do not divide. They remain permanently in interphase and they synthesize molecules, move, or perform other activities typical of the cell. Figure 35.2 shows the cell cycle and the part mitosis plays in it. Figure 35.2 Procedure 1. Obtain a slide of Allium (onion) root tip or other plant material provided. Refer to Figure 35.3 to orient yourself to the plant tissues you will study. The cells of the onion root tip were killed and stained. Then the root tip was sliced lengthwise into thin strips in which you will be able to see cells that were in different stages of the cell cycle when they were killed and stained. 2. Locate the tip of a root on low power and focus. Then switch to high power. It is important that you have clear focus using high power before you attempt to identify the various stages of mitosis. of onion root tip Figure 35.3 3. You are now ready to scan your slide to locate the stages of mitosis in plant tissue. It is easiest to proceed in a step-by-step fashion, reading about the key events in the division of a “typical” cell in the paragraphs that follow before looking for that actual cell type. As you read, compare the description of the events with the photographs in figures 35.5 and 35.6. Onion Root Tip Interphase Interphase is an active metabolic stage during which the cell performs its normal functions. It carries on metabolism, grows, and replicates DNA during this stage. However, an interphase cell is not dividing. The nucleus contains chromosomes but they are in a tangled mass of threads, which presents a uniform appearance of tiny dots. Complete chromosomes cannot be seen. The nuclear membrane is present and one or two nucleoli are visible in the nucleus. During this stage, the DNA replicates, but these molecules are much too small to be visible. Replication is necessary if the cell is going to divide at some time in the future. Figure 35.4 shows the diagram of a chromosome before and after replication. Protein Red hair allele DNA Red hair allele Chromatid Centromere Attached earlobe allele Attached earlobe allele Chromosome before replication Chromosome after replication Figure 35.4 1. Locate a cell that you think is in the interphase stage and place the cell at the end of your pointer. Call the instructor over to check it. Let your neighbors look through your microscope while you examine the cell that they have selected as an example of interphase. When you are sure you can identify an interphase cell proceed to find a prophase cell. Prophase During prophase, the cell prepares itself for division. One of the preparations is the shortening and thickening of the chromosomes into structures that can be seen. Their first appearance is as a mass of fine, tangled threads. The chromosomes shorten and thicken as a result of coiling. They gradually become more easily seen as individuals. Although the chromosomes and DNA were replicated in interphase it is not possible to see that the chromosomes are double structures until late prophase. By late prophase you should be able to see that the chromosomes consist of two chromatids joined at a point called the centromere. Meanwhile, a series of tiny tubules forms a structure called the spindle (Figure 35.5). At the same time the nucleoli disappear. Eventually, at the very end of the prophase stage, the nuclear membrane disintegrates and the chromosomes are lying free in the cytoplasm. 2. Look for a cell showing prophase. Again, place the cell at the end of the pointer in your microscope and share your slide with your neighbors. If you have any doubts, ask your instructor to check the slide. Metaphase During metaphase, the chromosomes become attached by their centromeres to spindle fibers. The chromosomes appear to be tugged by the spindle fibers so that they form a flat sheet of chromosomes on a plane that passes through the equator of the cell. Viewed from the side they appear to form a line; viewed from the pole they can be seen to form a ring or disk. (Onion slides will not show polar views because all the cells have a lengthwise orientation in the root tip.) 3. Look for a cell showing metaphase. Again, place the cell at the end of the pointer in your microscope and share your slide with your neighbors. If you have any doubts, ask your instructor to check the slide. Interphase Early prophase Late prophase Metaphase Anaphase Telophase Figure 35.5 Anaphase It is during anaphase that the equal distribution of genetic information is accomplished. Each centromere divides, and the two identical chromatids of each chromosome are pulled to opposite sides (poles) of the cell by the spindle fibers. When the two chromatids of a chromosome separate, each is called a daughter chromosome. 4. Look for a cell showing anaphase. Again, place the cell at the end of the pointer in your microscope and share your slide with your neighbors. If you have any doubts, ask your instructor to check the slide. Telophase A number of changes occur during the telophase stage of division. The chromosomes form two groups at opposite ends of the cell and the individual chromosomes become difficult to see as they uncoil. The spindle apparatus breaks down and a nuclear membrane forms around each group of chromosomes. These are now called daughter nuclei. Nucleoli also reform within the nucleus. As cell division is completed, cytokinesis (division of the cytoplasm) occurs by the formation of a cell plate between the two daughter nuclei. If you find evidence of the plate being formed, you can be sure that the cell is in telophase. 5. Look for a cell showing telophase. Again, place the cell at the end of the pointer in your microscope and share your slide with your neighbors. If you have any doubts, ask your instructor to check the slide. Whitefish Mitosis Obtain slides of whitefish blastula cells showing mitosis. A blastula stage is an embryonic stage that is shaped like a ball of cells. Because embryos grow rapidly they must be made of cells that divide frequently. The blastula has been sliced into thin disks and stained to help you see the chromosomes and other structures. The slide will have several disks on it and you will probably need to look at all of them to see the various stages of the cell cycle. Locate a section of a whitefish blastula under low power and focus; then switch to high power. Refer to Figure 35.6. It is often difficult to clearly distinguish interphase cells in the whitefish blastula slide preparation. This is because the cells divide so rapidly that interphase is a very short period of time; thus few cells are in interphase. Sections like the ones you are viewing can be sliced at any angle through a given cell. Therefore, some of the cells will be sliced so that you will see the cell from the equator, whereas others will show the cell from the pole. You may also find cells that do not appear to have any chromosomes because the knife removed all the chromosomes and left an empty cell. Or the slice of a cell may have included only part of the chromosomes. If you look for the spindle fibers, you may be able to figure out the orientation more readily. 1. Locate a prophase stage. Share it with others. Check with your instructor if you have trouble. 2. Locate an equatorial view of metaphase. You should be able to see the spindle fibers radiating fromthe poles and the chromosomes lined up at the equatorial plane. 3. Locate a polar view of the metaphase. Note that this view does not show either of the poles or any of the spindle fibers. The chromosomes are grouped in a ring or disk, but are not enclosed by a nuclear membrane. Figure 35.6 is especially helpful in identifying this view. 4. Locate a cell in the anaphase stage. 5. Locate a telophase cell. Interphase Early prophase Polar view of metaphase Metaphase Anaphase Telophase Figure 35.6 Cooperative Trial Quiz Choose one stage of either plant or animal mitosis and put your pointer on it. Pick a cell that you think is a good example of a particular stage of mitosis. Write the name of this stage on a slip of paper. On the opposite side of the slip of paper, write the number you are assigned by your instructor. Put the slip by your scope so that the number side is up and the side with the name of the stage is down. Use the Mitosis Quiz Answer Sheet found at the end of this lab. Visit each student’s scope and identify the stage at the end of the pointer. Your instructor will also take the trial quiz and make a key. If you have made any errors, be sure to study the cells you identified incorrectly so that you do not make the same mistake again. DNA-Mitosis Relationship The chromosomes found in cells are made up of DNA and protein molecules. Both chromatids of a chromosome contain identical DNA molecules. Different chromosomes have different sequences of DNA. Therefore you should be able to follow DNA molecules as you proceed through the cell cycle. The first circle on the next page represents the outline of the cell that was just formed from mitosis and is entering interphase. It contains four single-stranded chromosomes with a sketch of DNA superimposed on them. In the circles that follow, sketch a series of views of this cell as it proceeds through interphase and enters mitosis (prophase, metaphase, anaphase, and telophase). Show what happens to the chromosomes and the DNA. Interphase G Interphase G 1 2
Results 1. During which part of the cell cycle does DNA replication occur? DNA replication occurs during the S phase of Interphase. 2. When do chromosomes first become visible in mitosis? Prophase 3. What is the difference between the terms chromosome and chromatid? Chromosomes are dark-staining bodies composed of tightly-packed and coiled chromatin, visible only during cell division. Chromatin is the uncoiled DNA + protein that is present during interphase, but not visible with a light microscope. A chromatid is one-half of a double chromosome. 4. What is a spindle? What is its function? Protein microtubules involved in the transport of chromosomes during anaphase are called spindles. 5. During what stages of mitosis are chromosomes composed of two chromatids? Chromosomes are composed of two chromatids during prophase and metaphase, since at anaphase sister chromatids separate to become daughter chromosomes. During interphase the chromosomes are also present as two chromatids following replication. However, technically interphase is not part of mitosis. 6. During what stages of mitosis are chromosomes single structures composed of one chromatid? They are composed of single structures during anaphase and telophase. Also during interphase prior to replication, although technically interphase is not part of mitosis. 7. How does cytokinesis differ in plant and animal cells? The division of the cytoplasm in animals occurs by a furrowing process, while in plants a cell plate is formed. 8. Most cells spend the longest amount of time in ______Interphase________________. What evidence do you have to support this statement? The majority of cells in the onion root tip were in interphase. 9. Why was it difficult to find interphase cells in the whitefish blastula slide? During this early embryonic stage of development, mitosis is occurring so rapidly that cells spend very little time in interphase. The undifferentiated cells do not require time for specialization. 10. Why are you more likely to see a polar view in animal cells than in plant cells? All the cells in the onion (Allium) are oriented in the same direction, which is the long axis of the root. The cells in the whitefish blastula are oriented in all different directions. 11. Was the purpose of this lab accomplished? Why or why not? (Your answer to this question should show thoughtful analysis and careful, thorough thinking.) Mitosis Quiz Answer Sheet 1._____________________________________________________________________________ 2._____________________________________________________________________________ 3._____________________________________________________________________________ 4._____________________________________________________________________________ 5._____________________________________________________________________________ 6._____________________________________________________________________________ 7._____________________________________________________________________________ 8._____________________________________________________________________________ 9._____________________________________________________________________________ 10._____________________________________________________________________________ 11._____________________________________________________________________________ 12._____________________________________________________________________________ 13._____________________________________________________________________________ 14._____________________________________________________________________________ 15._____________________________________________________________________________ 16._____________________________________________________________________________ 17._____________________________________________________________________________ 18._____________________________________________________________________________ 19._____________________________________________________________________________ 20._____________________________________________________________________________ 21._____________________________________________________________________________ 22._____________________________________________________________________________ 23._____________________________________________________________________________ 24._____________________________________________________________________________ 25._____________________________________________________________________________ Name____________________________________________________Section________________Date___________ Experiment 36: Meiosis Invitation to Inquiry In some boy babies the testes are retained within the abdomen and do not descend into the scrotum. The temperature of the scrotum is lower than that of the abdomen. If this situation is not corrected surgically the person will be sterile because normal spermatogenesis only occurs at the lower temperature present in the scrotum. Some cases of male sterility have been linked to a man’s wish to wear tight pants that kept the testes close to his body and thus elevated their temperature. With this information in mind, what other normal activities or events might have the same effect? Background Nearly all organisms have life cycles in which sexual reproduction occurs. Sexual reproduction involves the joining of sex cells (gametes) from two different parent organisms to produce a new individual. The joining of sex cells is called fertilization and the new cell formed from their union is called a zygote. Because two gametes join to form one zygote during fertilization, the zygote must have twice as many chromosomes as either of the gametes that formed it. The zygote is said to have the diploid (two sets) number of chromosomes and the gametes the haploid (one set) number of chromosomes. Often the diploid cells are designated as 2n because they have two sets of chromosomes; the haploid cells are designated as 1n or n because they have only one set of chromosomes. If a sexually reproducing species is to retain a fixed number of chromosomes, the parents must have a mechanism to produce gametes with half the number of chromosomes typical for the organism (see Figure 36.1). Meiosis is a special kind of cell division in which the cells produced (gametes) have half the number of chromosomes typical for the species. Meiosis occurs only in the gonads (testes and ovaries) of sexually reproducing organisms. The process of meiosis involves two divisions with the first division (meiosis I) immediately followed by the second (meiosis II). During the first division of meiosis (meiosis I), the chromosomes join in homologous pairs. Homologous chromosomes are the same length and carry genetic information for the same characteristics. One chromosome of each pair originated from each of the parents of the individual in which meiosis is occurring. The pairing of homologous chromosomes is known as synapsis. Following synapsis the pairs of homologous chromosomes line up at the equator of the cell. They then separate, and one member of each pair moves to each pole of the cell. When the two nuclei reorganize and cytokinesis occurs, the resultant two cells have one-half as many chromosomes as the original cell. Reduction from the diploid number (2n) of chromosomes to the haploid number (n) of chromosomes has been accomplished. Each of the two cells produced as a result of meiosis I goes through a second division. This second division (meiosis II) results in the formation of four cells. During this division, individual Figure 36.1 The cells of this adult penguin have, for our purpose, eight chromosomes in their nuclei. In preparation for sexual reproduction, the number of chromosomes must be reduced by half so that fertilization will result in the original number of eight chromosomes in the new individual. The offspring will grow and produce new cells by mitosis. chromosomes line up on the equators of the cells. The centromeres divide and one of each of the chromatids (daughter chromosomes) moves to each pole of the cell. In males, the four cells that result from these meiotic divisions mature into sperm cells. Each cell has the haploid number of chromosomes, and each daughter chromosome (chromatid) is a single structure that carries a single DNA molecule. Each cell contains a complete set of genetic information for all the characteristics of an organism but the combination of characteristics in each gamete is different from that in others. In females only one of the four cells matures into an egg. The other three cells die. However, the same processes that generate variety in the kinds of sperm produced also are involved in producing eggs so that eggs are as different from one another as are sperm. There are several other terms that need to be clarified so that you can more easily follow the exercise. A gene is a piece of DNA that directs the expression of a particular characteristic. An allele is a gene for which there is an alternative expression. We generally use the word gene when we are talking about characteristics in general and use the word allele when we are talking about specific examples of genes for which there are several alternatives. For example we can talk about hair color genes, but when we want to talk specifically about the different possible hair colors we talk about the several alleles for hair color; red, blond, brown, and black. Some alleles are dominant to others and mask or cover up the presence of other alleles. The alleles that are masked are called recessive alleles. Each diploid organism has two alleles for each characteristic; one was received from each parent. The alleles may be identical (two alleles for red hair) or they may be different (one allele for red and one allele for black). When the two alleles are identical the organism is said to be homozygous. When the two alleles are different the organism is said to be heterozygous. The genotype of an organism is a listing of the two alleles for each trait that it possesses. The phenotype of an organism is a description of the way a characteristic is displayed in the structure, behavior, or physiology of the organism. During this lab exercise you will: 1. Determine the phenotype of a parent from the alleles on models of chromosomes. 2. Manipulate the chromosome models to simulate synapsis. 3. Follow the chromatids as they cross over. 4. Simulate metaphase and anaphase stages of meiosis I. 5. Manipulate the chromosome models as they proceed through meiosis II and produce four gametes. 6. Unite one of your four gametes with one of the four gametes from another group of students to form a zygote. 7. Compare the phenotype of your zygote with the phenotypes of other zygotes and with the parental phenotype. Procedure To understand the mechanism involved in the production of gametes, we will manipulate models of chromosomes. Each chromosome is composed of two identical chromatids. 1. Work in pairs. Obtain four model chromosomes that have already completed DNA replication and are, therefore, composed of two identical chromatids. These four chromosomes should include a short pair of chromosomes with two alleles and terminal centromeres and a longer pair of chromosomes with four alleles. The centromere of this long pair of chromosomes is more centrally located. The two chromosomes of each pair should be different colors. 2. Place the four chromosomes randomly on a large sheet of paper to represent their presence in the cytoplasm of a cell. 3. Note the genes on the short pair of chromosomes. At one point (locus) on each chromosome there is information about insulin production. At a different locus is information about hair color. Because these chromosomes are the same size, have their centromeres in the same relative position, and have genes for the same characteristic at equivalent points along their lengths, they are called homologous chromosomes. The various alternative expressions of genes on homologous chromosomes (dark hair or light hair, insulin production or no insulin production) are known as alleles. 4. Compare your four chromosomes with the diagrams in Figure 36.2 to make sure that you start with a proper set of information. Make sure that you have the correct alleles on the correct chromosomes and that you have the correct number of beads present between alleles. If your chromosomes are not as shown in Figure 36.2, make corrections before proceeding. Notice that on the figure and on your chromosome models, some of the genetic information is in capital letters and some is in lowercase letters. Those alleles that are dominant (those that always express themselves) are in capital letters. Recessive alleles (those expressed only when no dominant allele is present) are in lowercase letters. 5. Let us suppose that the models of the two pairs of homologous chromosome represent the complete set of genetic information of an organism. (In reality humans have 46 chromosomes and thousands of genes.) This cell was formed by the uniting of a sperm and an egg. One set of chromosomes is from the father and one set from the mother. Let the chromosomes with the darker beads represent the chromosomes donated by the father and those with the lighter beads represent the chromosomes donated by the mother. The cell you are dealing with has two sets of genetic information (2n) with one allele for each characteristic from the father and the other from the mother. In Table 36.1, list the genotype and phenotype of this organism. Remember the genotype is a listing of the two alleles for each characteristic and the phenotype is a description of the characteristics displayed in the structure, behavior, or physiology of the organism. Normal for cystic fibrosis (C) Free earlobes (E) Centromere (magnet) Five fingers (f) Blood type B (B) Figure 36.2 Prophase I (Synapsis and Crossing-Over) Because meiosis consists of two divisions the various stages of the process are labeled I or II to indicate whether they are occurring in meiosis I or meiosis II. During the prophase I stage of meiosis, the nucleus prepares for a division. The chromosomes become visible, the nuclear membrane begins to disintegrate, the nucleolus disappears, and the spindle apparatus develops. In addition, during prophase I homologous chromosomes pair with each other along their entire length. While they are paired like this, we say they are in synapsis. 1. Put the two members of each homologous pair of chromosomes near each other to simulate their synapsis. While in the synapsed condition, equivalent pieces of homologous chromatids will be broken off and exchanged. This process, called crossing-over, can occur several times along the length of homologous chromosomes. 2. Simulate crossing-over in your models by detaching, exchanging, and reattaching exactly equal parts of chromatids between the two members of a homologous pair of chromosomes. (The pop beads will allow you to break a chromatid in a variety of places but for each crossover you will need to make the break at the same place on equivalent chromatids.) Each chromatid acts as an independent unit in this process. Therefore, you should not make exactly the same changes to both chromatids of a chromosome. Make a minimum of two crossovers for each pair of homologous chromosomes. You will now have chromosomes that contain alleles that were originally on the other member of the homologous pair and each chromosome will contain both colors of beads. These are the chromosomes you will use for the remainder of the exercise. 3. As a result of the process of crossing-over, a new combination of alleles has been formed. Look atFigure 36.2 and note the differences between the arrangement of alleles in the original chromosomes and the chromosomes you have just crossed over. 4. Compare your chromosome models with the models of other students. Are their models like yours? Do they need to be? Each student ’s set of chromosomes is likely to differ from all other sets because they are likely to have had crossovers at different places. They do not need to be the same. If you do not understand the process of crossing-over, ask your instructor for help before going on. Metaphase I (Alignment of Chromosome Pairs) The chromosome pairs, still in synapsis, line up at the equatorial plane. 1. Move your model chromosomes on your piece of paper to show this arrangement. When the chromosomes are in this position, the cell is in metaphase I. 2. How does the arrangement of chromosomes in metaphase I of meiosis differ from metaphase in mitosis? In the space below, sketch the arrangement of four chromosomes in metaphase I of meiosis and the same four chromosomes in metaphase of mitosis. During metaphase of mitosis, individual chromosomes line up at the equatorial plane. During metaphase I of meiosis, synapsed pairs of chromosomes line up at the equatorial plane. The sketch of metaphase I should show two sets of homologous chromosomes lined up at the equator. The sketch of metaphase of mitosis should show the same 4 chromosomes lined up individually at the equator. Anaphase I (Independent Assortment and Segregation) During this stage, the separation of homologous chromosomes occurs. This separation is called segregation. Segregation is the separating of homologous chromosomes to opposite poles of the cell. Because the chromosomes carry genes, the genes are also separated (segregated) into two sets during anaphase I. 1. Segregate the two members of one pair of homologous chromosomes by moving their centromeresto opposite ends of the paper. Similarly segregate the other pair. Segregation of one pair of chromosomes is independent of the segregation of the other pair. The way a pair of homologous chromosomes happens to be aligned at the equator determines the direction in which each moves. The fact that the alignment and segregation of one pair of chromosomes is independent of the alignment and segregation of other pairs of chromosomes is called independent assortment. Because each set of homologous chromosomes is carrying genes, the genes on nonhomologous chromosomes are segregated independently of one another. 2. How do the chromosomes at the end of anaphase I of meiosis differ from the chromosomes at the end of the anaphase stage of mitosis? In the space below, sketch the arrangement of four chromosomes in anaphase I of meiosis and the same four chromosomes in anaphase of mitosis. At the end of anaphase I, each half of the cell will contain two complete chromosomes with each chromosome composed of two chromatids connected by a centromere. At the end of anaphase in mitosis, each half of the cell will contain four chromatids. Telophase I (Cytokinesis) During this stage, cytokinesis occurs and results in the formation of two daughter cells. 1. Show this division by tearing your paper into two equal parts. Each part will represent half the original cytoplasm. Each of these daughter cells has one chromosome from each of the two homologous pairs and is haploid. Each has half as many chromosomes as the original cell, but both have a complete set of genetic information. Separate the two sets of chromosomes by enough distance to remind you that they are in different cells. 2. Compare the sets of genetic data in each of the two cells by listing the alleles present on the chromosomes in table 36.2 3. Compare your daughter cells to those formed by other students. Do both have the same combination of alleles in each cell? Probably not because each student group would have done different crossing over and independent assortments. 4. Is it necessary that they be the same? No 5. List two meiotic processes that contributed to these differences. Crossing-over Independent assortment Prophase II (Cells Prepare for Second Division) No replication of chromosomes or DNA occurs after telophase I, and each of the two cells enters prophase II. The activities that occur in prophase II are the same as those that happen in prophase of mitosis. The chromosomes become visible, spindle fibers form, and the nuclear membrane breaks down. 1. List three ways that a prophase II cell differs from a prophase I cell. A. In prophase II, chromosome number is haploid; in prophase I of meiosis, chromosome number isdiploid. B. In prophase I, synapsis occurs which does not happen in prophase II. C. In prophase, I crossing over occurs which does not happen in prophase II. Metaphase II (Alignment of Chromosomes at Equator) During metaphase II, the chromosomes in each cell are lined up on the equatorial plane. They are not in homologous pairs; each chromosome, however, is still made up of two chromatids. 1. Move your chromosome models to the center of the half sheet of paper to represent this activity in both daughter cells. 2. The original diploid number of chromosomes in your cell was four. How many chromosomes can be found in a daughter cell in metaphase II? Two Anaphase II (Chromatids Separate) During anaphase II, the centromeres split and the chromatids (now referred to as daughter chromosomes) migrate toward opposite poles. 1. Separate your chromatids (daughter chromosomes) at the centromere and move them toward opposite poles (opposite ends of the paper). Do this in each of the two daughter cells. 2. List two ways in which anaphase I differs from anaphase II. In anaphase I, the chromosomes consist of two chromatids joined by a centromere and pairs of homologous chromosomes separate. In anaphase II only one of each homologous pair of chromosomes is present and the chromatids of each chromosome separate. 3. How does anaphase II of meiosis compare with anaphase of mitosis? In anaphase II, sister chromatids separate as in mitosis, but in anaphase II chromosome number is haploid; while in mitosis, chromosome number is diploid. Telophase II (Haploid Gametes Formed) During telophase II, cytokinesis occurs in each cell. This results in the formation of four haploid cells called gametes. 1. Represent cytokinesis in each of the two daughter cells. Simulate this by tearing the paper in half. 2. In addition to cytokinesis, what other processes would you expect to occur to the chromosomesand nucleus in a telophase cell? Chromosomes uncoil and go into a metabolic state, nuclear membranes form, and spindles disappear. 3. List the alleles now found in each of the four cells in Table 36.3. Table 36.3 4. Note that the four cells you have formed are different from one another. A number of things have happened during meiosis to cause these four cells to be different. Look back over meiosis I and II and identify the several processes that contribute to the differences in these four cells. List them in the space provided. Crossing-over Segregation Independent assortment Fertilization (Joining of Gametes to Form a Zygote) Your instructor will have designated each group in the class as being either a male or female. 1. A male group randomly selects one of its four sperm cells and delivers it to a female. The female group has previously selected at random one of its egg cells. 2. Unite these two gametes to represent fertilization. This fertilized egg cell is known as a zygote. 3. Record the genotype and phenotype of the offspring resulting from fertilization on Table 36.4. Show the alleles it received from each parent and the phenotype that would be observable. Share your data with the other groups in class. 4. Compare the phenotype of your offspring with the phenotype of both parents. 5. Compare the phenotypes of all offspring produced in class. 6. Fertilization (the joining of two haploid gametes) results in a diploid zygote which will develop into a new individual organism. What effect does being diploid rather than haploid have on determining what the phenotype will be? Since the diploid cell has two alleles for each characteristic, the recessive trait is only expressed in the phenotype if the offspring is homozygous recessive. If the offspring is homozygous dominant or heterozygous, the dominant trait will be expressed. At the end of this exercise, arrange the chromosome models as they appear in Figure 36.2 on page 300. Have your instructor check the chromosomes before you leave the lab. End-of-Exercise Questions 1. How many of the hypothetical offspring produced during this lab activity had the same phenotype? Answers will vary but there should be 3-5 different phenotypes depending on the number of groups of students in the class. The same genotype? Usually none 2. How do the results of meiosis and mitosis differ in terms of chromosome numbers? Fill in the diagrams by assuming that each original cell represents a human cell with a diploid number of 46 chromosomes. Meiosis 3. Many people use the terms gene and allele interchangeably. Are they the same? Explain. No, a gene is the sequence of DNA that codes for a particular protein or trait. An allele is one of two or more different forms of a specific gene. For many traits, there is a dominant and recessive form of the gene but some alleles are incompletely dominant, codominant, or are involved in polygenic expression. 4. What would you expect to be happening in any cell undergoing metaphase? Chromosomes would line up at the equatorial plate. Spindle fibers would attach to centromeres. 5. List the activities that occur during meiosis that contribute to the variety seen in the gametes produced. Independent assortment Crossing-over Segregation 6. Why is meiosis necessary in sexually reproducing organisms? Meiosis reduces the chromosome number in gametes so that when fertilization doubles chromosome numbers, the chromosome number is maintained for the species. 7. Name the event of meiosis that ensures that the chromosome number is reduced. Synapsis during prophase I pairs homologous chromosomes. This makes segregation possible during anaphase I. 8. During which stage of meiosis does each of the following events occur? a. Synapsis Prophase I b. Crossing-over Prophase I c. Segregation of homologous chromosomes Anaphase I d. Reduction from diploid to haploid Anaphase I begins the process and it continues through Telophase I e. Separation of chromatids Anaphase II f. Independent assortment of homologous chromosomes Anaphase I 9. Many people have the misconception that a dominant allele must be the most common in a population. Address this misconception by using the finger number trait as an example. The dominant allele is expressed if present. Dominant alleles for a given trait may actually be quite rare. For example, even though the allele for six fingers is dominant, very few people are born with six fingers, indicating that most people have two recessive alleles for five fingers. Name____________________________________________________Section________________Date___________ Experiment 37: Genetics Problems Invitation to Inquiry Construct a family tree that involves at least four generations. Follow a particular characteristic through those four generations. If you have a particular genetic quirk in your family, you may follow that and try to determine how it is inherited. If you don’t have any unusual characteristics in your family, earlobe shape is an easy characteristic to follow. Some people have an earlobe that hangs free while others have the lobe attached directly to the side of the face. You can even use photographs of your ancestors to determine earlobe shape. Ask yourself the following kinds of questions. Does it show up in every generation? Is it present in both sexes? Is it a dominant or recessive characteristic? Background One of the curiosities of life is the variety of offspring that can be produced as a result of sexual reproduction. The types of traits possible in an offspring have long been of interest to humans. Some people are interested for personal reasons—a new baby due in the family; some people are interested for business reasons—the desire to breed a new variety of plant. In either case they want to know the probability of having a given type of offspring. Although offspring receive half their genes from each parent, because of chance combinations of genes, they may resemble one parent more than the other or may show characteristics that are not displayed by either parent. To understand how characteristics are passed from one generation to the next we need to know some basic information. Every individual produced by sexual reproduction has two genes for each characteristic. They receive one from each parent. However, there are alternative genes for the same characteristic known as alleles. For example, there are alternative genes for eye color; the blue-eye allele and the brown-eye allele. Some alleles called dominant alleles are able to mask the presence of other recessive alleles. If an individual has two identical alleles for a characteristic (two blue-eye alleles or two brown- eye alleles) it is homozygous. If the two alleles are different from one another (one brown-eye allele and one blue-eye allele) the individual is heterozygous. Therefore, an individual may have some recessive alleles that do not express themselves but are still part of their genetic catalog. All the genes that an individual has is its genotype. The observable characteristics displayed in the organism’s structure, behavior, or physiology are known as the organism’s phenotype. To determine how characteristics are passed from one generation to the next we must first know something about the parents. What do they look like? What alleles do they possess? We must know what alleles are possible and the probability of each allele appearing in the gametes produced by the parents. We must also know the possible ways these may combine during fertilization. In this exercise, you are asked to determine how genes are passed from one generation to the next and determine the genotypes and phenotypes of parents and offspring. During this lab exercise you will: 1. Work a probability problem. 2. Work single-factor inheritance problems. 3. Work double-factor inheritance problems. 4. Determine genotypes of parents and offspring. Probability Versus Possibility To solve heredity problems, you must have an understanding of probability. Probability is the chance that a particular desired outcome will happen divided by the total number of possible outcomes. Probability is often expressed as a percent or a fraction. Probability is not the same as possibility. It is possible to toss a coin and have it come up heads. But the probability of getting a head is more precise than just saying it is possible to get heads. The probability of getting heads is one out of two (1/2 or 0.5 or 50%) because a coin has two sides, only one of which is a head. The desired outcome (heads) is divided by the total number of outcomes (heads or tails). This probability can be expressed as a fraction: the number of events that can produce a given outcome Probability = the total number of possible outcomes What is the probability of cutting a deck of cards and getting the ace of hearts? The number of times that the ace of hearts can occur is 1. The total number of possible outcomes, the number of cards in the deck, is 52. Therefore, the probability of cutting an ace of hearts is 1/52. What is the probability of cutting an ace? The total number of aces in the deck is 4, and the total number of cards is 52. Therefore, the probability of cutting an ace is 4/52 or 1/13. It is also possible to determine the probability of two independent events occurring together. The probability of two or more events occurring simultaneously is the product of their individual probabilities. For example, if you throw a pair of dice, it is possible that both will be a 4. What is the probability that both will be a 4? The probability of one die being a 4 is 1/6. The probability of the other die being a 4 is also 1/6. Therefore, the probability of throwing two fours is 1/6 × 1/6 = 1/36 1/6 x 1/6 = 1/36 chance for two successive fours Probability Problem Probability is a mathematical statement about how likely something will occur. It is not certainty. To help you understand this concept we will work with a deck of playing cards and determine the likelihood of getting each of the four suits. Work in pairs. One student shuffles and cuts a standard deck of cards. The other student records the suit. Repeat this 100 times; shuffling, cutting, and recording. Record your information in Table 37.1.
Table 37.1 Probability Problem Results
Suit Actual Expected Difference
Hearts 25
Clubs 25
Diamonds 25
Spades 25
Your instructor will tell you how to record this information for the whole class. Record the actual number of times each suit is cut. Compare your results with the entire class. Why is there a difference between the results you got and the results of the whole class? Single-Factor Inheritance Problems Single-factor crosses are concerned with how a single genetic trait is passed from the parents to an offspring. Solving a heredity problem requires five basic steps. Step 1: Assign a symbol for each allele. Usually, a capital letter is used for a dominant allele and a small letter is used for a recessive allele. For example, use the symbol E for free earlobes, which is dominant, and e for attached earlobes, which is recessive. E = free earlobes e = attached earlobes Step 2: Determine the genotype of each parent and indicate a mating. Suppose both parents are heterozygous; the male genotype is Ee and the female genotype is also Ee. The × between them is used to indicate a mating. Ee × Ee Step 3: Determine all the possible kinds of gametes each parent can produce. Remember that gametes are haploid; therefore, they can have only one allele instead of the two present in the diploid cell. Because the male has both the free earlobe allele and the attached earlobe allele, half his gametes contain the free earlobe allele and the other half contain the attached earlobe allele. Because the female has the same genotype, her gametes are the same as his. For solving genetics problems, a Punnett square is used. A Punnett square is a box figure that allows you to determine the probability of obtaining each of the genotypes and phenotypes possible in the offspring resulting from a particular cross. Remember, because of the process of meiosis each gamete receives only one allele for each characteristic listed. Therefore the male gives either an E or e; the female also gives either an E or e. The possible gametes produced by the male parent are listed on the left side of the square; the female gametes are listed on the top. In our example, the Punnett square would show a single dominant allele and a single recessive allele from the male on the left side. The alleles from the female would appear on the top: Female genotype Ee Possible gametes from female E & e E e Possible
E
e
Male genotype Ee maleE & e gametes Step 4: Determine all the allele combinations that can result from the combining of gametes. To determine the possible combinations of alleles that could occur as a result of this mating, simply fill in each of the empty squares with the alleles that can be donated from each parent. Determine all the allele combinations that can result when these gametes unite: E e
E EE Ee
e Ee ee
Step 5: Determine the phenotype of each possible allele combination shown in the offspring. In this instance, three of the offspring (those with the genotypes EE, Ee, and Ee) have free earlobes, because the free-earlobe allele is dominant to the attached-earlobe allele. One offspring, ee, has attached earlobes. Therefore, the probability of having offspring with free earlobes is 3/4, and with attached earlobes is 1/4. Additional Single-Factor Inheritance Problems (One Trait Followed from One Generation to the Next) 1. In humans, six fingers (F) is the dominant trait; five fingers (f) is the recessive trait. Assume both parents are heterozygous for six fingers. What are the phenotypes of the father and the mother? What is the genotype of each parent? What are the different gametes each parent can produce? What is the probability of them having six-fingered children? five-fingered children? a. Father’s phenotype___________________; mother’s phenotype____________________ b. Father’s genotype____________________; mother’s genotype _____________________ c. Father’s gamete__________ or __________; mother’s gametes__________or __________ d. Probability of a six-fingered child __________ e. Probability of a five-fingered child __________ 2. If the father is heterozygous for six fingers and the mother has five fingers, what is the probability of their offspring having each of the following phenotypes? Six fingers ______________; five fingers ________________ 3. In certain flowers, color is inherited by alleles that show lack of dominance (incomplete dominance). In such flowers, a cross between a homozygous red flower and a homozygous white flower always results in a pink flower. A cross is made between two pink flowers. Use Fw to represent the white allele and FR to represent the red allele. What is the probability of each of the colors (red, pink, and white) appearing in the offspring? 4. Use the information given in the previous problem. A cross is made between a red flower and a pink flower. What is the expected probability for the various colors? Double-Factor Inheritance Problems A double-factor cross is a genetic study in which two pairs of alleles are followed from the parental generation to the offspring. These problems are basically worked the same as a single-factor cross. The main differences are that in a double-factor cross you work with two different characteristics from each parent. It is necessary to recognize that independent assortment occurs when two or more sets of alleles are involved. Mendel’s law of independent assortment states that members of one allelic pair separate from each other independently of the members of other pairs of alleles. This happens during meiosis when the chromosomes which carry the alleles segregate. (Mendel’s law of independent assortment applies only if the two pairs of alleles are located on different pairs of homologous chromosomes. This is an assumption we will use in double-factor crosses.) In humans, the allele for free earlobes dominates the allele for attached earlobes. The allele for dark hair dominates the allele for light hair. If both parents are heterozygous for earlobe shape and hair color, what genotypes and phenotypes can their offspring have and what is the probability of each genotype and phenotype? Step 1: Use the symbol E for free earlobes and e for attached earlobes. Use the symbol D for dark hair and d for light hair. E = free earlobes e = attached earlobes D = dark hair d = light hair Step 2: Determine the genotype for each parent and show a mating. In this example, the male genotype is EeDd, the female genotype is EeDd, and the × between them indicates a mating. EeDd × EeDd Step 3: Determine all the possible gametes each parent can produce and write the symbols for the alleles in a Punnett square. Because there are two pairs of alleles in a double-factor cross, each gamete must contain one allele from each pair: one from the E pair (either E or e) and one from the D pair (either D or d). In this example, each parent can produce four different kinds of gametes. The four squares on the left indicate the gametes produced by the male; the four on the top indicate the gametes produced by the female. To determine the possible allele combinations in the gametes, select one allele from the “ear” pairs of alleles and match it with one allele from the “color” pair of alleles. Then match the same “ear” allele with other “color” allele. Then select the second “ear” allele and match it with each of the “color” alleles. This may be done as follows: Step 4: Determine all the gene combinations that can result from the combining of gametes. Fill in the Punnett square as follows:
ED Ed eD ed
ED EEDD * EEDd * EeDD * EeDd *
Ed EEDd * EEdd ^ EeDd * Eedd ^
eD EeDD * EeDd * eeDD “ eeDd “
ed EeDd * Eedd ^ eeDd “ eedd +
Step 5: Determine the phenotype of each possible allele combination shown in the offspring. In this double-factor problem, there are 16 possible ways in which gametes could combine to produce offspring. There are four possible phenotypes in this cross. They are represented as:
Genotypes Phenotype Symbol
EEDD, EEDd, EeDD, or EeDe Free earlobes, dark hair *
EEdd, Eedd Free earlobes, light hair ^
eeDD, eeDd Attached earlobes, dark hair “
eedd Attached earlobes, light hair +
The probability of having a given phenotype is: 9/16 free earlobes, dark hair 3/16 free earlobes, light hair 3/16 attached earlobes, dark hair 1/16 attached earlobes, light hair Additional Double-Factor Inheritance Problems (Two Traits Followed from One Generation to the Next) 5. In horses, black color (B) dominates chestnut color (b). Trotting gait (T) dominates pacing gait (t). A cross is made between a horse homozygous for both black color and pacing gait and a horse homozygous for both chestnut color and trotting gait. List the probable genotype and phenotype of offspring resulting from such a cross. Genotype _____________________; phenotype _____________________ 6. Humans may have Rh+ blood or Rh– blood. A person who is Rh+ (R) has a certain type of protein in the red blood cell. A person who is Rh– (r) does not have this particular protein. In humans, Rh+ dominates Rh– . Normal insulin (I) production dominates abnormal insulin (i) production (diabetes). If both parents are heterozygous for both Rh+ and normal insulin production, what phenotypes would they produce in their offspring? What are the probabilities of producing each phenotype? 7. For this problem, use the information concerning the traits given in problem 6. The father is homozygous for Rh+ and has diabetes. The mother is Rh– and is homozygous for normal insulin production. What phenotype would their offspring show? 8. In certain breeds of dogs two different sets of alleles determine the color pattern. Black color is dominant and red color is recessive; solid color is dominant and white spotting is recessive. A homozygous black-and-white spotted male is crossed with a red-and-white spotted female. What is the probability of them producing a solid black-colored puppy? 9. In humans, a type of blindness is due to a dominant allele; normal vision is the result of a recessive allele. Migraine headaches are due to a dominant allele, and normal (no headaches) is recessive. A male who is heterozygous for blindness and does not suffer from headaches marries a woman who has normal vision and does not suffer from migraines. Could they produce a child with normal vision who does not suffer from headaches? If yes, can the probability of such a child be determined? 10. This problem is a little more challenging than the previous one because it involves lack of dominance in both characteristics that are being followed. However, you use the same Punnett square method to determine the outcome of this cross as you would with any other double-factor cross. In the radish plant, the long and round traits exhibit lack of dominance and the heterozygotes have an oval shape. The red and white color traits also exhibit lack of dominance and heterozygotes have a purple color. Two oval-shaped, purple plants are crossed. What phenotypic ratio would the offspring show? Sex-Linked Problems (Alleles Located on the X Chromosome) For these problems you need to remember that human males have one X chromosome and one Y chromosome; females have two X chromosomes. The Y chromosome does not carry the genes found on the X chromosome but carries genes that determine maleness. 11. In humans, the condition for normal blood clotting (H) dominates the condition for nonclotting (h) (hemophilia). Both alleles are linked to the X chromosome. A male hemophiliac marries a woman who is a carrier for this condition. (In this respect, a carrier is a woman who has an allele for normal blood clotting and an allele for hemophilia.) If they have a son, what are the chances he will be normal for blood clotting? 12. For this problem, use the information given in problem 11. A male who has normal blood clottingmarries a woman who is a carrier for hemophilia. What are the chances that they will have a son who is normal for blood clotting? 13. Color blindness is a condition in which a person cannot distinguish specific colors from one another. For example they may not be able to distinguish red from green or blue from yellow. However, they are able to distinguish some colors. Because color-blind people are not blind and they can see some colors, many people prefer to use the term color-deficient. In humans, the condition for normal vision dominates color blindness. Both alleles are linked to the X chromosome. A color-blind male marries a color-blind female. If they have a daughter, what are the chances she will have normal vision? 14. For this problem, use the information given in problem 13. A male with normal vision marries a woman who is color-blind. She gives birth to a daughter who is also color-blind. The husband claims the child is not his. The wife claims the child is his. Can you support the argument of either parent? If yes, which one? Why? Multiple Allele Problems (Characteristics That Have More Than Two Possible Forms of the Same Gene) 15. In humans, there are three alleles for blood type: A, B, and O. The allele for blood type A and theallele for blood type B show incomplete dominance. A person with both alleles has blood type AB. Both A and B dominate type O. A person with alleles for blood types A and O marries someone with alleles for blood types B and O. List the types of blood their offspring could have and the probability for each blood type in the offspring. 16. For this problem, use the information given in problem 15. A young woman with blood type O gave birth to a baby with blood type O. In a court case, she claims that a certain young man is the father of her child. The man has type A blood. Could he be the father? Can it be proven on this evidence alone that he is the father? 17. In humans, kinky hair (H++), curly hair (H+), wavy hair (H), and straight hair (h) are dominant, in that order. Dark hair dominates red hair. A wavy, red-haired male, whose mother has straight, dark hair, marries a dark-haired female with straight hair who has a curly, red-haired father. What type of children can they produce, and what is the probability of producing these types of offspring? Epistasis Problems Epistasis occurs when one set of alleles can mask the presence of a different set of alleles. All the epistasis problems in this exercise involve two different traits; therefore, they are all similar to double-factor problems. 18. Normal pigmentation dominates no pigmentation (albino). For an organism to exhibit color, it must have an allele for normal pigment production as well as alleles for a specific color. In cattle, red color dominates black color. An albino bull that has a heterozygous genotype for red is crossed with a red cow. The cow is heterozygous for normal pigment production and for red coloring. What types of offspring will they produce and what is the probability of producing these types of offspring? 19. In humans, normal pigmentation dominates no pigmentation (albino). Black hair dominates blonde hair. An albino person will have white hair even though he or she may also have the alleles for black or blonde hair. An albino male who is homozygous for black hair marries a woman who is heterozygous for normal pigmentation and has blonde hair. What colors of hair can their children have, and what is the probability for each hair color? Polygenic Inheritance Polygenic inheritance occurs when two or more genes combine their effects to determine the phenotype. 20. In some types of wheat, color is caused by two sets of alleles. To produce a red color, both dominant alleles, R and B, are needed. White results from having both recessive alleles in the homozygous state, rrbb. Any other combination produces brown wheat grains. A strain with a genotype of Rrbb is crossed with a strain of wheat with a genotype of rrBb. What is the color of each of the parent strains? Rrbb color _______________; rrBb color _______________ What colors of wheat result from this cross, and what is the probability for each color? Determination of Genotypes (Genetic Detective Work) Not all heredity problems deal with determining the genotype and phenotype of the offspring. A common problem is to determine the genotype of all individuals involved when only the phenotypes are known. In humans, free earlobes are dominant and attached earlobes are recessive. Two free-earlobed people marry and produce one free-earlobed child and two attached-earlobed children. What are the genotypes of the parents and each of the children? Because both parents have free earlobes, they must have at least one allele for free earlobes. So their genotypes would be E __ (the __ means the allele could be either the dominant allele or the recessive allele). The genotype for the attached-earlobed children must be ee. Because each parent contributed one allele to each attached-earlobe offspring, you know that each free-earlobed parent has an allele for attached earlobes. Therefore, the __ must be e and the genotype for each parent is Ee. All you know about the free-earlobed offspring is that the child has one allele for free earlobes. The genotype could be either EE or Ee. In the remaining problems, try to determine the genotypes of the individuals. Take it slowly; put down an allele only when you are certain the individual has that allele. If you are not certain, show that you don’t know by leaving a blank (__). 21. Normal pigmentation (A) dominates no pigmentation (albino = aa). Dark hair coloring (D) dominates light hair coloring (d). Two people with normal pigmentation produce one child with dark hair, two children with light hair, and two albino children. What are the possible genotypes for the parents? 22. A red bull, when crossed with white cows, always produces roan-colored offspring. Explain how the colors for red, white, and roan are inherited. 23. In rabbits, short hair is due to a dominant allele, S, and long hair to its recessive allele, s. Black hair is due to a dominant allele, B, and white hair to its recessive allele, b. When two rabbits are crossed, they produce 2,518 short-haired, black offspring and 817 long-haired, black offspring. What are the probable genotypes of the parents? 24. In humans, the condition for normal blood clotting dominates hemophilia. Both alleles are sexlinked to the X chromosome. Two parents produce daughters who are all carriers, and sons who are all normal. What are the probable genotypes of the parents? 25. In humans, deafness is due to a homozygous condition of either or both recessive alleles d and e. Both dominant alleles D and E are needed for normal hearing. Two deaf people marry and produce offspring who all have normal hearing. What are the probable genotypes of the children and parents? Answers 1. a. Six fingers; six fingers b. Ff; Ff c. F or f d. F or f e. 3/4 f. 1/4 2. 1/2; 1/2 3. 1/4 red; 1/2 pink; 1/4 white 4. 1/2 pink; 1/2 red 5. BbTt; all black trotters 6. 9/16 Rh+, normal insulin; 3/16 Rh+, diabetic; 3/16 Rh–, normal insulin; 1/16 Rh–, diabetic 7. All Rh+, normal insulin 8. None 9. Yes; 1/2 normal and no headache 10. 1/16 long, red; 2/16 long, purple; 1/16 long, white; 2/16 oval, red; 4/16 oval, purple; 2/16 oval, white; 1/16 round, red; 2/16 round, purple; 1/16 round, white 11. 1/2 12. 1/4 13. None 14. Yes. Father. A female requires two color-blind alleles to have the condition and the father does not have any color-blind alleles. 15. 1/4 A, 1/4 B, 1/4 O, 1/4 AB 16. Yes. No. Blood typing can only disprove paternity. 17. 1/4 straight, red-haired; 1/4 wavy, red-haired; 1/4 straight, dark-haired; 1/4 wavy, dark-haired 18. 8/16 albino, 6/16 red, 2/16 black or 1/2 albino, 3/8 red, 1/8 black 19. 1/2 albino (white hair), 1/2 black hair 20. Brown; brown. 1/4 red, 1/2 brown, 1/4 white 21. Aa_d × AaDd 22. Incomplete dominance 23. SsBB × Ss _ _ 24. Normal mother; hemophiliac father 25. Parent’s genotype ddEE DDee; children’s genotype DdEd Solution Manual Experiment for Integrated Science Bill W. Tillery, Eldon D. Enger , Frederick C. Ross 9780073512259
