PROBLEM SET
Chapter 8 -- Cytogenetics


P. 189, #6. Is a tetraploid more likely to show irregularities in meiosis or mitosis? Explain. What about these processes in a triploid?
A tetraploid will have more irregularities in meiosis than in mitosis. This is because during meiosis the homologous chromosomes synapsis and then during anaphase I segregate into opposite gametes, however for a tetraploid there is no mechanism to insure that each gamete gets two of the four homologues, which happens by chance. In a diploid, there is a mechanism to insure that each gamete gets one of the pair of homologues. In mitosis, the homologues do not synapsis and thus each chromosomes behaves independently of the others. Separation of the sister chromatids is not affected by the other chromosomes, thus a tetraploid has many fewer irregularities during mitosis.
In a triploid, the same is true except that in meiosis the problems are even worse as there is no way to evenly divide three homologous chromosomes between the two gametes. Thus a triploid always produces unbalanced gametes.

P. 189, #7. How many chromosomes would a human tetraploid have?
A human tetraploid would have 92 chromosomes, twice the normal diploid number of 46.
How many chromosomes would a human monosomic have?
A human monosomic would have 45 chromosomes. A monosomic would be missing one chromosome from the normal 46, thus 45 chromosomes.

P. 189, #8. Do autopolyploids or allopolyploids experience more difficulties during miosis? Do amphidiploids have more or less trouble than auto- or allopolyploids.
Some allopolyploids (allotetraploids) have fewer problems in meiosis that autopolyploids. Autopolyploids arise from a diploid organism in which the chromosome number simply doubled. This leads to 4 homologues for each chromosome and can cause numerous problems during meiosis as these chromosome synapsis and then attempt to segregate evenly during anaphase I. Allopolyploids, on the other hand, arise from the combination of normal gametes from plants of different species. Initially, these allodiploids will have numerous problems because the chromosomes can not synapse during meiosis nor segregate normally during anaphase I. However, if the chromosome number was to double in a cell that gives rise to a reproductive structure, then each chromosome has one homologue. This is an amphidiploid or allotetraploid. In this case meiosis proceeds normally because synapsis is between only two chromosomes and segregation of the homologues during anaphase I is even. Amphidiploids (as defined above) have very few problem in meiosis and will in general have fewer difficulties during meiosis than either autotetraploids or allopolyploids (allodiploids).

P. 190, #18. You are trying to locate an enzyme-producing gene in Drosophila, which you know is located on the third chromosome. You have five strains with deletions for different regions of the third chromosome ("/" indicates deleted region).
 Normal   0     10     20     30     40     50     60   map units
          ---------------------------------------------
 strain A ///////--------------------------------------
 strain B ---////////////////////----------------------
 strain C -----------------///////////-----------------
 strain D ----------------------////////////////-------
 strain E --------------------------------/////////////
You cross each strain with wildtype flies and measure the amount of enzyme in F1 progeny. The results appear below. In what region is the gene located?
   strain crossed  percent of wildtype enzyme
                   produced in the F1 progeny
   ------------------------------------------
        A                 100
        B                  45
        C                  54
        D                  98
        E                 101
   ------------------------------------------
As strains A, D, and E have the normal amount of enzyme activity, we can conclude that the gene for the enzyme is not located in any region that is deleted in these three strains. As strains B and C have about half the normal activity of the enzyme, we can conclude that the gene is located in a region that is deleted in both B and C. For example, if the gene was located at position 15, you would see reduced activity in B but not in C. As both show the same reduction in enzyme level, it must be in a region common to both strains. In addition, as strains A, D, and E are essentially normal, we can exclude any region that is found in either A, D, or E. Thus the region where the gene is found is the intersection of strains B and C and the exclusion of the union of the regions of A, D, and E. Thus the answer is region 25 to 34 contains the gene for the enzyme.

P. 191, # 21. Plant species P has 2n = 18 and species U has 2n = 14. A fertile hybrid is found. How many chromosomes does it have?
The fertile hybrid would have to be an allotetraploid (amphidiploid), which would have a diploid set of the chromosomes from each parental species. Thus it would have 32 chromosomes.

P. 191, #22. A normal-visioned woman whose father was color-blind marries a man with normal vision. They have a color-blind daughter with Turner syndrome. In which parent did nondisjunction occur?
We know that the normal-visioned woman is heterozygous for color-blindness because her father was color-blind. The daughter, who is color-blind, got the X chromosome with the color-blind allele from her mother. Thus, her mother contributed a normal gamete in that it had one X chromosome. She is XO which means that she did not get any sex chromosomes from her father. Thus, during gamete production in her father, nondisjunction occurred in such away that he produced a gamete with no sex chromosomes. When the paternal gamete combined with the normal maternal gamete, a Turner's (XO) individual was the result.

P. 191, #23. A color-blind man mates with a woman with normal vision whose father was color-blind. They have a color-blind son with Klinefelter syndrome. In which parent did nondisjunction occur?
We know that the normal-visioned woman is heterozygous for color-blindness because her father was color-blind. The son who is color-blind must have gotten an X chromosome with a color-blind allele from each parent (each parent had only one to donate to the next generation). In addition to the X-chromosome from the father, the son also got a Y-chromosome from the father, thus the father donated two sex chromosomes, which means that nondisjunction occurred during gamete formation in the father.

Last updated on 22 August 1996.
Provide comments to Dwight Moore at mooredwi@esumail.emporia.edu.
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