Angiosperm plant with minimum number of chromosomes?
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Brachycome dichromosomatica -- an Australian daisy with only two pairs of chromosomes (2n=4)\n\nIt is closely related to the species in the pic
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This happens regularly in the natural world. It's a common cause of what is referred to as polyploidy.\n\n"Polyploidy has occurred often in the evolution of plants.\nThe process can begin if diploid (2n) gametes are formed. These can arise in at least two ways.\n\n    * The gametes may be formed by mitosis instead of meiosis.\n    * Plants, in contrast to animals, form germ cells (sperm and eggs) from somatic tissues. If the chromosome content of a precursor somatic cell has accidentally doubled (e.g., as a result of passing through S phase of the cell cycle without following up with mitosis and cytokinesis), then gametes containing 2n chromosomes are formed."\n\n\nIf you are looking for someone to do your homework for you, as I suspect, then think again. I will suggest that you think about what will happen if this continues every genertion. That's the answer your teacher wants.
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http://www.enchantedlearning.com/subjects/plants/plant/\n\nhttp://waynesword.palomar.edu/trjune99.htm\n\nhttp://www.botany.uwc.ac.za/Sci_ed/pupil/Angiosperms/\n\nhttp://www.enchantedlearning.com/subjects/plants/\n\nThose will help...
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Assuming your species is diploid, it goes like this:\n\nSupposing there is a field of red, pink and white flowers. Red flowers have genotype RR, pink Rr or rR, and white rr. \n\nYou count the number of each kind of flower. It turns out there are 150 red-flowered plants, 480 pink, and 170 white ones. \n\nTherefore there are 2*(150+480+170) = 1600 alleles out there. There are 300 Rs in the red plants, 480 in the pink ones and 0 in the white. Added together there are 780. 780/1600 = 0.4875, so the frequency of the R allele is 48.75%\n\nAs for the r allele, there are 0 in the red-flowered plants, 480 in the pink, and 340 in the white. Added together, there are 820. 820/1600 is 51.25%.\n\nAs a double check, 48.75% + 51.25% = 100%
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http://www.iknow.net/CDROMs/cell_cdrom/index.shtml\n\nCheck this site out! It has both plant and animal cell division! Hope it helps ya!! =o)
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The offical SI base unit would be meters. \n\nProbably centimeters works best for an average plant.
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Chromosomes are allotted numbers according to their sizes. So, chromosome 1 is largest and chromosome 22 is smallest. The 23rd pair of chromosomes i.e., X and Y are sex chromosomes and hence they're allotted different representations. However, X lies between 7th and 8th chromosome if arranged according to the size while Y is the smallest (smaller than the chromosome 22).
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To clarify how an alien addition and substitution lines works,For it a  greenhouse bioassay was developed that consistently ranked four small grains in their correct order of susceptibility to take-all in the field (wheat > barley > rye = oat) (4). This assay was used to evaluate several wheats that carry alien chromosomes. Single plants were grown in a 1:1 (v:v) mixture of pasteurized soil (Kennebeck silty clay loam) and vermiculite in plastic tubes (2.5 cm in diameter and 12.5 cm long). Experiments included 10 tubes per entry-treatment (inoculated or non-inoculated) per replication and 4 replications. The experimental design was a randomized split-plot with the accessions as main-plot treatments and presence or absence of inoculum as sub-plot treatments .Inoculated tubes each received three oat kernels colonized by G. graminis var. tritici introduced 1 cm below seed level .After 4 weeks at 20±5°C, fresh weights of plant parts (10 plants per treatment per replication) above the soil level were measured and losses in fresh weight calculated.\nFour sets of wheat-alien chromosome addition and substitution lines of species with resistance to take-all were obtained from the Wheat .. The sets were: (i) wheat (cv. Chinese Spring)-Hordeum vulgare (cv. Betzes) disomic addition lines; (ii) wheat (cv. Chinese Spring)-Hordeum chilense disomic addition lines; (iii) wheat (various lines)-H. villosa disomic addition and substitution lines; and (iv) wheat (cv. Chinese Spring)-H. villosa disomic addition lines. Three of the four sets were not complete. Cytogenetic analysis revealed that two addition lines of Set i (chromosomes 1H and 7H) and two addition lines of Set ii (2Hch and 3Hch) had only 42 chromosomes. For Set iii, addition and substitution lines for 1V, 2V, and 7V were not available. Parental lines and "Don" oats were used as controls.Each set of addition and substitution lines, along with the appropriate controls, was evaluated in its own experiment. Within an experiment, 80 seeds of each entry (line or control) were used (10 in non-inoculated tubes and 10 in inoculated tubes in each of 4 replications). All five wheat-barley disomic chromosome addition lines were more susceptible [82 to 90% loss in fresh weight, LSD (P = 0.05) = 10.1] than the barley parent (Betzes = 36%) and two of them (2H and 3H) had significantly more loss (90%) than the wheat parent (Chinese Spring = 78%). In another experiment, there were no differences among all five wheat-H. chilense disomic chromosome addition lines [77 to 88%; LSD (P = 0.05) = 13.7] and they were as susceptible as the wheat parent (79%). In a third experiment (Set iii), H. villosa [30%; LSD (P = 0.05) = 8.16] was more resistant than barley (Betzes = 47%); however, all four wheat-H. villosa disomic chromosome addition and two substitution lines (4V and 5V) were more susceptible (85 to 93% loss) than the wheat check (76%). Substitution lines 3V and 6V were as susceptible (75 to 80%) as wheat. In a fourth experiment involving the complete set of wheat-H. villosa disomic chromosome addition lines (Set iv), improved levels of resistance over that of the wheat parent [77% loss; LSD (P = 0.05) = 14.9] were not detected in any of the chromosome addition lines (74 to 90%).In summary, the bioassay used in this study was a rapid, accurate way for screening accessions for susceptibility to take-all in the greenhouse. None of the addition or substitution lines had improved levels of resistance even though their resistant parents had significantly less take-all than wheat. In the cases where a complete set of lines was not available, any genes that might confer resistance may be on a chromosome missing from the set. Alternatively, take-all resistance in these species may be polygenic with genes on different chromosomes or it may be masked by the wheat genome. Therefore, transfer of take-all resistance from H. vulgare, H. chilense, or H. villosa to wheat may not be easily accomplished.
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Arabidopsis thaliana, commonly called arabidopsis, Thale Cress, or Mouse-ear Cress, a small flowering plant related to cabbage and mustard, is one of the model organisms for studying plant sciences, including genetics and plant development. It plays the role for plant science that mouse and fruit fly (Drosophila) play in human biology.
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You may think about checking online plant nurseries, and seed catalogs.  Good Luck!
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If you are talking about Somatic division (mitosis) , just before division the mother cell will  have 23x2 chromosome pairs that is 23x2x4 chromatids in all. The daughter cells get 23x2x2 chromatids=23 pairs of chromosomes each. In case of meiosis, one more division leads to 23 singlet chromosomes (=23x2 chromatids) each in a daughter cell.
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one chromosome comes from each parent
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First of all we have 23 pairs of chromosomes (or 46 chromosomes), and monkeys have  22 pairs of chromosomes. Big deal right? WRONG. When humans make gametes (sperm and egg) each gamete contains a one copy of the hosts DNA, therefore each gamate only contains 23 chromosomes. When the sperm reaches the egg they combine together (zygote) to from a cell that has 46 chromosomes. Any deviation from this number makes the an embryo inviable. ( we wont get into ploidy or trisomys here). Therefore if a human and an ape mated, the chromosome count coming from the human would be 23 and the money would be 22 making the chimera have 45 chromosomes. First of all essential genetic coding inforemation is lost. (AN ENTIRE CHROMOSOME IS MISSING), you have changed the specie. You have created an alleuploidy. (only viable with some plants). The chromosomes in each species are of different sizes. This totaly screws up the cellular functions and inhibits a proteins job. They would not be able to reproduce. And please do not get me started on Mullers moprhs in terms of loss of genetic information, regarding epistasis, incomplete dominance etc.... All you need to know is it would be inviable because the chromosomes are of different numbers, and the chromsome sizes are different.
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Plant cells have a rigid cell wall so during telephase (cytokinesis), they use a different method of divergence ('splitting' of the cell). The process is complicated, so state that plant cells do not cleave like animal cells and that they produce a cell plate from the Golgi apparatus. Also, plant cells don't have centromeres and instead use astral fibres.\n\nBefore mitosis begins, chromosomes are replicated so that their numbers are doubled.\n\nAfter a cell undergoes mitosis, the two new cells are completely identical and have the same DNA. A difference which may occur, but is of little importance, is that one cell may receive more organelles then the other when division occurs by chance.\n\nIt's important for the nuclear membrane to disintegrate during mitosis so that the chromosomes can attach to the spindle fibres and allign at the metaphase plate (or the 'equator'). It's also required so that the chromosomes can be pulled apart from each other during anaphase, and allow the division of the cell.
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There's a fern with thousands of chromosomes.\n\nAccording to wiki, Aulacantha (protozoa) is the highest with ~1600  \n\nGod, I love wikipedia.
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sex cells. they only have half the nomal number of chromosomes.
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The attached link has a picture for you;\n\n\n\nhttp://collections.ic.gc.ca/gardens/New%20Bio%20Web%20Pix/OPTIMIZED%20BIO%20PLANT%20PARTS/Monocot.jpg