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Biological Sciences
BISC 202
Timothy Swartz

17LargeScale Chromosomal Changes WORKING WTH THE FIGUREScategorize the following genomes 1 Based on Table 171 how would youLetters H through J stand for four different chromosomesHH II J KKHH II JJ KKKHHHH IIII JJJJ KKKKAnswerMonosomic 2n17 chromosomes n1 9II Trisomic 2Tetraploid 4n 16II 2 Based on Figure 174 how many chromatids are in a trivalentAnswer There are 6 chromatids in a trivalent 3 Based on Figure 175 if colchicine is used on a plant in which 2n18 how many chromosomes would be in the abnormal productAnswer Colchicine prevents migration of chromatids and the abnormal product of such treatment would keep all the chromatids 2n18 in one cell 4 Basing your work on Figure 177 use colored pens to represent the chromosomes of the fertile amphidiploidAnswer A fertile amphidiploids would be an organism produced from a hybridand n which would be infertilewith two different sets of chromosomes n12 until some tissue undergoes chromosomal doubling 2 n2 n and such 12 chromosomal set would technically become a diploid each chromosome has itspair therefore they could undergo meiosis and produce gametes This could bea new species Picture example3 pairs of chromosomes different lengths red for n3 1Chapter Seventeen 384 4 chromosomes different lengths green for n4 2Hybrid infertile 7 chromosomes nn 12Amphidiploids fertile 14 chromosomes pairs of n and n125 If Emmer wheat Figure 179 is crossed to another wild wheat CC not shown what would be the constitution of a sterile product of this cross What amphidiploid could arise from the sterile product Would the amphidiploid be fertileAnswerEmmer wheat was domesticated 10000 years ago as a tetraploid withtwo chromosome sets 2n2 n or AABB If AA BB tetraploid genome is 1 2 combined with another wild wheat n or C gamete the product would be 3 sterile In such case C chromosomes would not have homologous pairs in ahybrid parent wheat Amphidiploid could occur if chromosome doublinghappens in a parental tissue eg flower parts and a fertile wheat specieswould be hexaploid AA BB CC 6 In Figure 1712 what would be the constitution of an individual formed from the union of a monosomic from a firstdivision nondisjunction in a female and a disomic from a seconddivision nondisjunction in a male assuming the gametes were functionalAnswer A gamete from a firstdivision nondisjunction would be an eggwithout the chromosome in question n1 while a gametesperm from aseconddivision nondisjunction would be a n1 If both gametes arefunctional theywould result in a euploid 2n zygote with two copies of afathers chromosome 7 In Figure 1714 what would be the expected percentage of each type of segregationAnswer These are three equally possible combinations segregation ofchromosomes in meiosis in a trisomic individual Therefore each type isexpected at 333 8 In Figure 1719 is there any difference between the inversion products formed from breakage and those formed from crossing overAnswer Inversion products look the same having the sequence 1324 yet they could be genetically different Breakage and rejoining on the left results in a rearrangement with possibly damaged gene sequence and yet the same length of the chromosome Crossing over between repetitive DNA blue mightChapter Seventeen 385generate an inversion with less damage to the gene sequence since the crossing over happens in homologous regions of a repetitive sequence 9 Referring to Figure 1719 draw a diagram showing the process whereby an inversion formed from crossing over could generate a normal sequenceAnswer An inversion formed by a crossing over13 2 4Repetitive sequences are homologous andoriented and they could pairand form crossing over in the next generation of gametes producing a normalsegment again 1 2 3 410 In Figure 1721 would the recessive fa allele be expressed when paired with deletion 26432 25811Answer Deletion 26432 overlaps with the segment carrying fa allele 7 onpolytene chromosome this means that the fa allele would be pseudodominantexpressed in such combination of chromosomes Deletion 25811 is covering segments before the fa allele and the gene wouldnot be expressed In this case a dominant allele fa of the 25811 chromosomewill show in the phenotype 11 Look at Figure 1722 and state which bands are missing in the cri du chat deletionAnswer Cri du chat syndrome in humans is caused by the deletion of the tip ofthe p5 bands 153 and 152 12 In Figure 1725 which species is most closely related to the ancestral yeast strain Why are genes 3 and 13 referred to as duplicateAnswer Kluyveromyces waitii seems to be most similar to the commonancestor When genome of Saccharomyces lineage doubled genes such as 3 and13 become duplicate in both species and in the same relative order Some othergenes were lost 2 7 etc 13 Referring to Figure 1726 draw the product if breaks occurred within genes A and BAnswer 5 AXpBXCD33AX pXC D 5 Z InversionZ
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