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STUDY OBJECTIVE 1: To observe the nature and consequences of chromosomal breakage and reunion 178 190 Variation can occur in the structure and number of chromosomes in the cells of an organism. When chromosomes break, the ends become sticky ; they tend to reunite with other broken ends. A single break can lead to deletions or the formation of acentric or dicentric chromosomes. Dicentrics tend to go through breakage-fusion-bridge cycles, which result in duplications and de ciencies. Two breaks in the same chromosome can yield deletions and inversions. Variegation position effects, as well as new linkage arrangements, can result. Inversion heterozygotes produce loop gures during synapsis, which can form either at meiosis or in polytene chromosomes. Heterozygosity for an inversion suppresses crossovers; organisms that are heterozygotes are semisterile. Reciprocal translocations can result from single breaks in nonhomologous chromosomes. These produce crossshaped gures at synapsis and result in semisterility. The Bar eye phenotype of Drosophila is an example of a duplication that causes a position effect. STUDY OBJECTIVE 2: To observe the nature and conse-

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Changes in chromosome number can involve whole sets (euploidy) or partial sets (aneuploidy) of chromosomes. Aneuploidy usually results from nondisjunction or chromosomal lagging. Several medical syndromes, such as Down, Turner, and Klinefelter syndromes, and the XYY karyotype are caused by aneuploidy.

Thus, every second-order Butterworth lter is characterized by a Q of 0707; this corresponds to a damping ratio = 05Q 1 = 0707, that is, to a lightly underdamped response The next example considers the characteristics of a fourth-order Butterworth lter

Known Quantities: Filter response; desired gain and cutoff frequency Find: Component values R1 , R2 , C1 , C2 , RA , RB , for each lter section Schematics, Diagrams, Circuits, and Given Data: Gain = 100; cutoff frequency =

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Analysis: Table 153 suggests that a Butterworth fourth-order lter is composed of the product of two quadratic responses Our rst objective is to determine the Q of each of these two quadratic responses, so that we can design each of the two Sallen and Key quadratic sections Comparing the standard quadratic low-pass lter response to the rst of the two Butterworth polynomials for a normalized lter with K = 1 and C = 1, we have:

H (s) =

Polyploidy leads to dif culties in chromosomal sexdetermining mechanisms, general chromosomal imbalance, and problems during meiotic segregation. It has been more successful in plants than in animals because plants generally lack chromosomal sex-determining mechanisms. Plants can also avoid meiotic problems by propagating vegetatively. In both animals and plants, even-numbered poly-

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and, for a normalized lter with K = 1 and C = 1, we can solve for the value of Q1 , the Q of the rst section: 1 = 07654 Q1 or Q1 = 1 = 13065 07654

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ploids do better than odd-numbered polyploids because they have a better chance of producing balanced gametes during meiosis. Somatic doubling provides each chromosome in a hybrid organism with a homologue, and thus makes possible tetrad formation at meiosis. New species have arisen by polyploidy.

Having determined these values, we can now proceed to design two separate quadratic sections with the values of Q computed above, and each with gain K = 10 (so that the product of the two sections yields a low-frequency gain of 100, as speci ed), and cutoff frequency C = 400 rad/s The responses for the two sections are: H (s) = =

and H (s) = =

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