GENETic2

Cell Division and Inheritance

In this lab period there are four things to be done.

1. Observe stained chromosomes in plant and animal cells.
2. Review mitosis and meiosis by using a chromosome model kit. Print the guidesheet, "Mitosis and Meiosis," and study it before the lab session. Bring it to lab. This is a WORD document.
3. Work selected inheritance problems by observing and counting progeny from some actual monohybrid and dihybrid crosses.
4. Learn how to use the chi square statistical test to interpret results from genetic crosses. Print the guidesheet, "Chi square test," and study it before the lab session. Bring it to lab for reference. This is a WORD document.

About the guidesheet entitled "chi square test": You must study this one before coming to lab, to avoid being lost. This chi square test is an important statistical tool in many areas of biology. You will use it in this lab session to analyze data from some inheritance problems.

You should do #1 and #2 (listed above) simultaneously, so that you can compare the appearance of the real chromosomes with the models as you go through the stages.

1. Observation of chromosomes.

a. Study the prepared slides of mitosis in onion root tip cells and in embryonic cells of whitefish. In each you should locate cells that clearly reveal the four stages of mitosis. Note also the obvious difference in cytokinesis between the typical plant cell and typical animal cell, as specified in class and in the textbook.. Study also the illustrations of mitosis in your textbook and posted in the lab. Are the individual chromosomes of whitefish and onion, as you see them on the slides, the same size?

b. Study the demonstration slide (on the front bench) that shows two stages in meiosis in lily. The slide shows a thin cut through the female part of a lily flower. See the illustration by the microscope. On the slide concentrate on the two very large cells, marked on the drawing beside the microscope. [These are megaspore mother cells. Each of these will develop into an "embryo sac."] Each of these two enlarged cells is undergoing meiosis. In the lower one you can see early anaphase I (chromosomes are stained red). In the upper cell you see anaphase II, two spindle apparati each with its set of chromosomes. You should note that in the upper one of the two cells no cell plate had yet formed after the first nuclear division. In the male part of the flower meiosis occurs within the anthers, as shown in your textbook and in the illustrations posted in the lab.

c. Before you leave today, be sure you see the "giant" chromosomes taken from an insect's salivary gland; there is a prepared slide on a demonstration microscope. Refer to the illustration beside the microscope. You can see the banding and some "puffs" on these unusually large chromosomes. Puffs are regions in which RNA synthesis is occurring (transcription).

2. Cell division models. Get a plastic bag (kit) of chromosome bead models from the lab instructor, one kit per pair of students. Follow the instructions in the guidesheet "Mitosis and Meiosis." You have already studied these in class and in the text. However, this review (as simple as it may seem) is important, because some people still do not understand the similarities and differences between mitosis and meiosis and what each accomplishes. This is especially so for meiosis. The models provide another way of studying cell division. Both types of cell division are described and outlined in text. Watch for the points at which synapsis, segregation, assortment, and crossing over would occur as you work with the models. These things are important; you'll need to understand them for the next lecture exam and the second lab exam.

When you've completed this model work, put the pieces back into the bag and return it to the lab instructor.

3, 4. Inheritance problems and chi square analysis.

You will work several problems that illustrate various principles of inheritance. Corn was selected as the demonstration system. In corn a large number of female flowers (each one containing a haploid egg cell) are clustered on a single flower stalk on the corn plant. Each egg is fertilized by a different sperm, which is brought to the female flower by a pollen grain (by wind or insects). Pollen grains, which are haploid, are produced by male flowers, called "tassels". Female flowers are called "silks". So, every seed represents one offspring and each one develops from a different egg-sperm union (a different zygote). A large number of these progeny are conveniently clustered on the stalk; we call this an "ear" of corn.

The plant geneticists who produced these corn specimens that you are studying today carefully transferred pollen from one parent plant to the female flowers of another parent. Therefore, each ear of corn represents the outcome of many simultaneous crosses between two parents; again, each seed on the ear is one offspring. The particular phenotypic traits of interest here are features of the seeds, such as seed color and texture. You can see why plant species are favorite model systems among geneticists; it is difficult, if not impossible to get such large numbers of progeny from a single set of parents in many animal species. Since an understanding of genetic mechanisms may require interpretation of probabilities, having large numbers of progeny from crosses is helpful in doing genetic studies.

You and your partner should cooperate in counting and scoring the seeds in the proper phenotype classes in the following problems. Be sure you understand what is involved here; that is, think for yourself, and do the Punnett square and statistical analyses for yourself so that you'll know how to do them.

A word about error. You'll find that some phenotype characters may not be as clear-cut as you would expect. Two people may interpret something differently, which raises the possibility of error in collecting and interpreting data. For example, though you probably don't think you'd have any trouble distinguishing purple from yellow, you'll see some seeds that make you wonder, "Should I score this one as purple or as yellow?" Remember from earlier work in lab this semester that error is always present in scientific work and that it needs to minimized as much as possible.

When you do the chi square analysis (chi square test), remember that it includes interpretation, not just calculation of a number. You do this test to help understand the outcome of a genetic cross, to help determine whether a set of experimental results differ significantly from what you expected to find. The guidesheet "Chi square test," shows how to set up the data for doing the chi square test. That guidesheet also provides a table of critical values of chi square. You will be expected to know how to perform chi square analysis of genetics data on the lab exam.

In working the problems you will see that there are several ears of each type, marked A, B, etc. Each type (ears marked "A" versus ears marked "B") represents a different set of parents and illustrates certain points about inheritance. Even a quick glance tells you that some of these ears of corn show very different phenotypic distributions of progeny. For problems #1-3 you need to count the seeds on only one ear of each type. For example, in problem #1 choose any one of the ears marked "A"; they differ a bit in size and total number of seeds, but all of the "A" ears show the same distribution. DO NOT REMOVE THE WRAPPERS FROM THE EARS; they are there to prevent seeds from breaking loose.

After doing each problem, check your answer against the solutions posted in the lab. If you can't solve one, get help from the lab instructor after you've tried your best. In working the chi square problems remember these two points in particular:

* After counting the seeds in a problem to get your "observed" values, you must determine the "expected" values based on the total number of seeds. For example, if there were 200 seeds of one color and 113 of another color (that's 313 total) and you expected 75% of the first color and 25% of the second color, then your expected values will be 75% of 313 and 25% of 313.
* The total number of offspring for "expected" must equal the total number for "observed." For example, 75% of 313 seeds is 234.75 seeds, and 25% of 313 is 78.25 Round those to 235 seeds and 78 seeds, with the total equal to 313.

#1. The "A" ears represent a monohybrid cross. The dominant allele (A) produces a purple seed; the recessive allele (a) produces a yellow seed. Both parents were heterozygous in this cross.

a. Write the parental genotypes and phenotypes.
b. What is the expected phenotypic distribution of the progeny?
c. Count the progeny in each phenotype class.
d. Do a chi square analysis to determine whether the progeny that you counted fit the expected distribution.

#2. The "B" ears represent another monohybrid cross, involving the same locus in problem #1. This time, though, you're not told about the genotypes of the parents; you must deduce that from the progeny, i.e. work backward from offspring to parents.

a. Count the progeny in each phenotype class.
b. What phenotype distribution do the progeny approximate?
c. Now try to reason backward to determine what the parental phenotypes and genotypes were.

#3. The "C" ears represent a dihybrid cross. Alleles at one locus govern seed color. Purple (A) is dominant and yellow (a) is recessive. At the second, unlinked locus the dominant "B" allele produces a smooth seed, and the recessive "b" allele produces a shrunken seed. Both parents in this cross were doubly heterozygous.

a. Write the parental genotypes and phenotypes.
b. What phenotype distribution would you expect in the progeny? (You must do a Punnett square analysis to get this.)
c. Count the progeny in each phenotype class.
d. Do a chi square analysis to determine whether the progeny fit the expected distribution. Remember: analysis includes interpretation, not just calculation of a chi square value.

Let's assume that by now you understand what's involved in counting progeny and placing them in proper phenotype classes. Obviously, such work can be laborious, but it's necessary if one is to get reliable data for genetic analyses. Then, for problem #4-6, look at a representative ear of each type (D, E, and F) to see that the phenotype distributions are different from the ones you've already studied. To save some time, which you should apply to the Punnett square analysis, each problem #4-6 will give you a typical count from one of these ears of each type. Use those data in each of these last three problems rather than counting the seeds yourself.

#4. The "D" ears represent another dihybrid cross, involving the same two loci in problem #3. The observed progeny counts were 112 purple smooth, 105 purple shrunken, 106 yellow smooth, and 126 yellow shrunken.

a. What phenotype distribution do these approximate?
b. Work backward to determine the phenotypes and genotypes of the parents. Is there more than one set of parents possible?

#5. The "E" ears represent a dihybrid cross in which two unlinked loci govern seed color. (Note that this is a case of 2 loci governing a single phenotypic trait. What's that called?) At locus 1 the allele "D" is dominant and the allele "d" is recessive. At locus 2 the allele "R" is dominant and the allele "r" recessive. A purple seed is produced only when both dominant alleles are present (i.e. at least one of each). Otherwise seeds are yellow. In this particular cross both parents were doubly heterozygous. The observed progeny counts were 340 purple and 266 yellow.

a. Can you determine easily what phenotype ratio this approximates?
b. Do the Punnett square analysis to determine the expected phenotype distribution. Is the observed distribution close to the expected distribution? Do you think it's necessary to do a chi square analysis to determine whether the observed distribution fits the expected one? What if the difference between the two distributions were twice as great? 10 times as great? Do you really know (without doing the chi square test) how large that difference must be before the two distributions are significantly different enough to make you suspect that something other than chance is responsible for the difference?

#6. The "F" ears represent another type of locus interaction in a dihybrid cross (different from that in #5). At one locus the dominant allele (D) would normally produce a purple seed. However, the dominant allele (I) at a second, unlinked locus inhibits formation of purple pigment, and a yellow seed is the result just as though the seed were homozygous recessive at the first locus. Allele "i" does not block pigment formation, and the recessive "d" allele produces a yellow seed when it is expressed. (Here the absence of purple pigment results in a yellow seed.) In this particular cross both parents were doubly heterozygous. A count of progeny on one ear was 117 purple and 475 yellow.