1. The chromosome number for barley is 2n =14 How many chromosomes would you expect to count in:- i) a trisomic ii) a monosomic iii) a nullisomic iv) a tetraploid v) a translocation heterozygote vi) an inversion heterozygote vii) an endosperm cell viii) a cell in diakinesis ix) a pollen nucleus [endq] i). trisomics have 1 extra chromosome - thus 15 ii) monosomics have 1 missing chromosome - thus 13 iii) nullisomics have lost both of a pair of homologous chromosomes - thus 12 iv) tetraploids have 4 sets of chromosomes - double the number - thus 28 v. translocation heterozygotes will still have 14 ( during meiosis the translocated chromosomes will form a tetravalent - so only 6 'chromosome-like structures' will be seen at metaphase I - (5 bivalents and 1 tetravalent)) vi) inversion heterozygotes will still have 14 chromosomes - the inversion has no effect on chromosome number vii) endosperm nuclei form by fusion of a diploid cell in the ovum with one of the haploid pollen nuclei - they are therefore triploid and have 21 chromosomes viii) at diakinesis there will be 7 chromosome-like structures - these are actually bivalents - paired homologous chromosomes. ix) pollen nuclei will be haploid - 7 chromosomes [enda] 2. Predict the number of Barr bodies that should be found in individuals:- a) who have Turner's syndrome (XO) b) who have Klinefelter's syndrome (XXY) c) who have an XXYY karyotype d) who have an XXX karyotype (metafemales) e) who have an XYY karyotype f) who have Down's syndrome[endq] Barr bodies (named after Murray Barr, cytogeneticist) represent inactive tighlty packed X chromosomes. If >1 X chromosome is present the rule is that only 1 is left in an active condition. a) None, the only X present must be active b) 1 - one of the two X's would be inactivated c) 1 - one of the two X's would be inactivated d) 2 e) none f) As with normal individuals, males (XY) would have no Barr bodies and females (XX) would have 1. (The aberration in Down' syndrome affects chromosome 21 not the sex chromosomes.) [enda] 3. Recent discoveries by the Star Trip crew have yielded some interesting animal specimens. Two diploid creatures, Morgons and Cricklings, were both observed to have sexual forms based on P and Q chromosomes. While normally one sex has a PP karyotype and the other a PQ karyotype, a population of deviant individuals was found with a variety of other combinations in an otherwise diploid nucleus. The phenotypes of normal and deviant forms are shown below:- GENOTYPE PHENOTYPE Morgons Normal PP male Normal PQ female Deviant P male Deviant PPQ female Deviant PQQ female Cricklings Normal PP female Normal PQ male Deviant P male Deviant PPQ female Deviant PQQ male Deviant PPP sterile females Devise a hypothesis to explain how the P and Q chromosomes control sexual development in these 2 species. [endq] a) Morgons Here the male is the homogametic sex (2 P chromosomes ) while the female is heterogametic. The results with the deviant forms suggest that the Q chromosome is needed for femaleness - ie if Q is 'deleted' from a PQ female a P0 male results and if Q is added to a PP male a female PPQ results. We can draw a parallel with humans where genes on the Y chromosome are required for, in that case, maleness. b) Cricklings Here the homogametic sex is female and the male is heterogametic. However, loss of the Q chromosome does not necessarily alter the sex, nor does addition of one Q to a PP type change the sex type (still female); on the other hand addition of two Q's to P does generate a male and triple P produces an abnormal female. Thus here it seems that it is not presence or absence of a particular chromosome that determines sex; instead the number of chromosomes is important. Exactly this kind of result is found on Earth with Drosophila; here it is the ratio between the number of X chromosomes and the number of autosomes that determines sex type.[enda] 4. The Star Trip crew also find an unusual plant with the following characteristics:- i) at mitosis 40 chromosomes are present in each anaphase group; ii) at meiosis metaphase 1 there are only 10 visible 'chromosome-like' structures; Many but not all of the gametes have 20 chromosomes the rest have a range of numbers that are > or < 20. Explain these observations in the light of our knowledge of organisms on Earth.[endq] The observation of 40 chromosomes/anaphase group at mitosis indicates that the chromosome number is 40. If this were a diploid we would expect there to be 20 bivalents (chromosome-like structures) at metaphase I. However there are only 10 so we conclude this is probably a tetraploid and specifically an autotetraploid where there are 4 identical sets of chromosomes. All 4 homologues would tend to associate into a tetravalent and we would expect therefore 10 tetravalents to be present. If the tetravalent separates properly 2 chromosomes will go each way and the gametes will end up with 20 chromosomes; however because 4 chromosomes are trying to 'pair' together subsequent disjunction (separation) is not always perfect and in some cells separation is 3 to 1 or even 4 to 0. As a result gamete chromosome numbers tend to vary either side of 20.[enda] 5. Another plant bought back by the 'Star Trip' voyage proves to be an allotetraploid with 36 chromosomes - ie it has 4 sets of 9 chromosomes - 2 sets apparently derived from one parental species and 2 sets from a different species. How many chromosomes or chromosome like structures (bivalents, trivalents etc.) would you expect to see in each group at:- i) mitotic anaphase ii) meiotic anaphase I iii) meiotic anaphase II iv) meiotic metaphase I v) Would you expect the % viability of the gametes to be greater or smaller than in an autopolyploid (4 identical sets) [endq] i) 36 ii) 18 iii) 18 iv) 18 bivalents v)) Allopolyploids should have greater viabilty than autopolyploids because each chromosome in a group of 4 has only one other chromosome that is +/- identical. Thus they will tend to form bivalents (which will disjoin normally 1 and 1) rather than tetravalents where all 4 chromosomes are +/- identical. ( Tetravalents may segregate 3+1 or 4:0 - giving inviable gametes- instead of the proper 2+2 segregation )[enda] 6. In 1928 Russian genticist V. Kharpechenko crossed a radish (2x = 18) with a cabbage(2x = 18), as he hoped to obtain a hybrid with the leaves of the cabbage and the roots of the radish. His initial efforts were unsuccessful as the hybrid was sterile. However, he was later able to obtain a fertile hybrid. Unfortunately, this hybrid had the leaves of the radish and the roots of the cabbage ! The fertile hybrid was found to yield only sterile progeny when crossed with the either of the original parents. Explain these results. [endq] Cross...... Cabbage, 2x=18 x Radish, 2x=18 gametes will be x =9 gametes will be x =9 Therefore original hybrid will have 2x =18 chromosomes but there will be NO homologous pairs - because each chromosome was either derived from the cabbage or the radish. Hence this hybrid will be sterile. However, if chromosome doubling occurred in the hybrid then the new plant will have 4x=36 - an allotetraploid - and will be fertile because each chromosome now has a homologue and meiosis can proceed normally. This tetraploid hybrid will form triploid progeny if it is crossed with either of the diploid parents. These triploid parents will be sterile because trivalents are formed during meiosis which tend to segregate 2 and 1 producing gametes with a wide range of chromosome numbers - most lethal (Note x is the preferred symbol for number of sets of chromosomes, not n: n is used to indicate either the somatic cell number 2n, or the the number in the gamete, 1n )[enda] 7. 2 genes in barley are known to be 22 map units apart on the left arm of chromosome 5. However, some odd results are noticed in two new highly inbred lines that are under study. The % recombination (=%rec.) for these two genes is found to be different for different crosses involving line AX4 and line CR32:- AX4 plants crossed together.......%Rec = 22 CR32 plants crossed together.....%Rec = 22 AX4 x CR32 cross......................%Rec = 1 NB Both Ax4 and CR32 plants seem to grow quite normally Explain these results Describe how your hypothesis could be tested[endq] If we obtain a reduced recombination frequency in some crosses versus the normally obtained value, we can assume that at least one parent either has a deletion or an inversion in the part of the chromosome under study. However, a plant with a deletion would a) probably be very unhealthy if not completely inviable and b) the deletion would reduce the %rec when 2 plants carrying it were crossed together. The data does not support this so we can argue that 1 of these two strains carries an inversion for that part of chromosome 5 between the 2 genes. If 1 of these 2 strains has an inversion occupying a large part of the section between these 2 genes and the other strain is normal, then both strains would show the usual 22% rec. when crossed to themselves, but would form an inversion heterozygote when crossed. In this inversion heterozygote all recombinant products of a cross-over in the inverted region would die and the %rec would therefore appear to be low. The small (1%) amount of recombination that still occurs in the AX4 x CR32 cross is probably due to crossing-over in that part of the section between these 2 genes which is NOT involved in the inversion. This hypothesis could be tested by looking at the chromosomes in meiotic prophase I cells of an AX4 x CR32 F1 plant to find evidence for the characteristic shapes assumed by inverted chromosomes in an inversion heterozygote. Mapping a number of other genes on chromosome 5 in these 2 strains would also give evidence of an inversion if present.[enda] 8. Two auxotrophic (ie nutritional) mutants of the fungus Neurospora are being studied, pan-2 on chromosome 6 and arg-1 on chromosome 2. A strain of arg-1 with unusual properties arises spontaneously. Note. In Neurospora ascopspores are normally black. Strains with duplicated genes are tolerated and still produce black spores; strains with deleted genes have white inviable spores. Explain the following results:- CROSS ASCOSPORES %REC. (arg-1 - pan-2) 1.normal arg-1 x 100% black 50% normal pan-2 2.unusual arg-1 x c. 50% black 3% normal pan-2 50% white & inviable 3.black-spored progeny half of crosses ---> 100% black 50% of cross 2 x normal pan-2 half of crosses ----> 50% black,50% white 3% [endq] Cross 1 shows the 50% recombination expected for 2 genes known to be on different chromosomes - expected because of independent assortment. Cross 2 shows that the 2 genes are now closely linked - only 3 map units apart. Thus a translocation must have occurred in the unusual arg-1 strain to bring them into close proximity. This cross therefore produces a translocation heterozygote and at meiosis four chromosomes pair together, normal chromosomes 2 and 6 + two translocated chromosomes each carrying parts of chromosomes 2 and 6. Remember that the resulting tetravalent structure can separate by alternate segregation (yielding normal chromosomes - hence black spores) or by adjacent segregation (yielding chromosomes with duplicted and deleted sections - hence white spores). Cross 3. After alternate segregation half of the progeny carry normal chromosomes and half translocated chromosomes. Thus half will repeat the result of cross 1 and half that of cross two.[enda] 9. Recall that in an inversion heterozygote, recombination within the inversion loop produces unbalanced recombinant chromatids that give rise to inviable gametes or zygotes.If there are four genes in the order :- A -----10 ----->B----10---->C----10---->D Predict and explain what recombination frequencies you would expect for genes A and D for a testcross of:- A. A homozygous normal individual B. An individual heterozygous for an inversion stretching from B to C C. An individual homozygous for this same inversion.[endq] A. Homozygous normal = A B C D ........The % recombination expected for A B C D genes A and D will be about 30 (actually a little less than 30, because of the effects of multiple cross-overs that are not detected.) B. Inversion heterozygote A B C D ............. No recombination in the B - C A C B D region because all cross-over products are inviable. Thus % Rec. A-D will be about 20. C. Homozygous for inversion A C B D ............. Answer will be same as for A. A C B D