BIOL 234 - Midterm review answer key

BIOL 234 - Midterm review answer key - BIOL 234 Summer 2017...

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BIOL 234 Summer 2017 Midterm review questions 1. Loppins are (fictitious) diploid invertebrates, with a total of 6 chromosomes in their somatic cells. Of those 6 chromosomes, 4 are autosomes and 2 are sex chromosomes. Like humans, female loppins are XX and males are XY. The gene that determines loppins’ blood type is called bt , and is on chromosome 1; the gene that determines the presence or absence of eyelashes is called eye , and is on chromosome 2; and the gene that determines ability to digest cellulose is called cel , and is on the X chromosome. btA , btB , eyeW , eyed , celWT , and celM are alleles of these three genes. a. A female loppin is a triple heterozygote with the genotype btA/btB ; eyeWT/eyeD ; X celWT / X celM . Her mother was homozygous for btA , for eyeD , and for celWT . Draw a somatic cell of this triple heterozygous female loppin in the G1 stage of the cell cycle (i.e. before DNA replication). Make sure that her chromosomes are properly drawn, and clearly label all relevant alleles. -G1: chromosomes are un-replicated, so you can draw them as a single line with a centromere -draw 6 chromosomes: two of each of chromosome 1 (labelled with bt), chromosome 2 (labelled with eye), and the X chromosome (labelled with cel) -allele labels: should be in the same locus for homologous chromosomes -heterozygous, so one of each allele b. Which alleles did the triple heterozygous female loppin inherit from her mother, and which alleles did she inherit from her father? -see part A: mother was homozygous for bt A , eye D , and X celWT , so these alleles came from her -father: the remaining alleles, bt B , eye WT , X celM c. For research purposes, you remove three meiocytes from the triple heterozygous female. You let them undergo the cell cycle and meiosis, and you analyze the genotype of the gametes that are produced. The first meiocyte produces two gametes with the genotype btA ; eyeWT ; X celWT , and two gametes with the genotype btB ; eyeD ; X celM . i. Draw this meiocyte at metaphase of meiosis I. Make sure to clearly label all alleles. -chromosomes should be replicated (two sister chromatids joined by a single centromere) -chromosomes arranged as tetrads along the middle of the cell (metaphase plate): the two homologues of chromosome 1 (with the bt alleles) should be lined up together, same for the chromosome 2 homologues (with eye alleles), and same for the X chromosomes (with cel alleles) -sister chromatids of a chromosome have the same alleles (no crossing over) -draw a spindle -imagine a line down the middle of the tetrads (this helps to determine which homologues end up in which daughter cell): chromosomes with the bt A , eye WT , and X celWT alleles must all be on the same side of the metaphase plate, so that they go to the same daughter cell -chromosomes with the bt B , eye M , and X celM alleles are lined up on the other side
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ii. The second meiocyte produces two gametes with the genotype btB ; eyeWT ; X celWT , and two gametes with the genotype btA ; eyeD ; X celM . Explain what must have happened differently in this meiocyte, compared to the first one, in order to produce this result.
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