1) I can't remember or find out what a germ-line mutation is. Can you help me out on this?

• Germ-line mutations are those that occur in gametes (eggs or sperm). They are passed on to every cell of the resulting zygote. Sperm and eggs are sometimes referred to as the germ-line cells.

• Go to the web page titled setting up punnett squares for doing dihybrid crosses. That takes you through each step.
• A test cross is performed whenever you have an unknown genotype (AA or Aa). When one allele shows complete dominance, the resulting phenotypes of AA and Aa would be the same so you wouldn't know its genotype. In order to tell whether the individual is AA or Aa, you would cross it with a homozygous recessive. The results would tell you whether the unknown was heterozygous Aa, or homozygous AA. A heterozygous, Aa, would have a 1:1 phenotype result. A homozygous, AA, would all have offspring of the same phenotype.

3) I am going over mitosis and meiosis and seem to be missing a key part that is just not making sense to me and I was hoping you could help clarify for me.  I understand that at S in Interphase, the chromosomes are replicated, so that in mitosis you have 92 sister chromatids and 46 chromosomes -the diploid number in the cell.  And in mitosis, the chromatids are separated, and because they become individual chromosomes, when you go through cytokinesis and divide into 2 cells, each cell has 46 chromosomes.  I guess I am getting confused on how you maintain the amount of DNA.   You end up with 46 chromosomes in mitosis.
In meiosis, however, you have homologous chromosomes to start, right? So after S in interphase, you would have 23 homologous pairs?   In Anaphase I, the homologues are separated, leaving 2 cells each with 23 individual chromosomes (as 2 sister chromatids) at the end of Meiosis I? Then in Meiosis II, which I think is the same as mitosis, the chromatids are separated, leaving 23 individual replicated (daughter) chromosomes in 4 cells, thus maintaining the haploid number.  I think the numbers are confusing me a lot, but writing this out may have helped, if my thought processes are correct.  I guess I get
confused because the sister chromatids each become their own chromosome.  So if you can understand it, please let me know if my thinking is correct.

•     You have mitosis right except in mitosis you also always have 23 homologous pairs.  We didn't talk about the chromosomes being homologous at that state because the chromosomes don't pair up in mitosis. The cell always has 23 pairs of chromosomes (for a total of 46 chromosomes).  When the cell replicates at S phase (both in mitosis and meiosis) you maintain the number of chromosomes at 46 (23 homologous pairs) but the volume (weight) of the DNA is doubled and this is represented by each chromosome now consisting of two sister chromatids.  At the end of mitosis, you've separated the sister chromatids (now individually called chromosomes) to keep the 46 chromosome number and to bring the weight of the DNA back to what it was before being doubled in S phase.
•     During meiosis (remember that the DNA double in weight during S phase) we talked about homologous chromosomes because the first division (meiosis I) divides the DNA by sending one of each of the homologous pair (with sister chromatids)  to separate daughter cells.  At this point you cut the volume of the DNA in half (but remember it was doubled to start with). The two daughter cells are haploid with 23 chromosomes, each consisting of two sister chromatids.
•     During Meiosis II you cut the volume of DNA in half again by separating the sister chromatids.  Each haploid daughter cell from Meiosis I undergoes a division that separates the sister chromatids into separate cells (still maintaining the 23 chromosomes, but now they don't consist of sister chromatids) for a total of four daughter cells each having half the chromosomes and DNA volume of the starting cell before S phase.
•     In Meiosis the starting cell:
•  S phase                 -    DNA volume doubles (46 chromosomes with 92 chromatids)
• end of Meiosis I    -    DNA volume cut in half (23 chromosomes with 46 chromatids) in 2 daughter cells.
• end of Meiosis II    -   DNA volume cut into half again (23 chromosomes no longer consisting of sister chromatids) in 4 daughter cells.
• So, in all of meiosis, you double x (x is the volume of starting DNA) to make 2x, you divide once to bring the volume back to x, you divide again to end up with 1/2 x in your 4 daughter cells.  I hope this helps.

• Yes, that was how SRY was discovered.  The person was phenotypically female but her karyotype was male.  When they studied the Y chromosome they found that there was a deletion.  One of the genes deleted was SRY.  They also observed a male with a karyotype XX.  On closer examination, a portion of Y was observed attached to another chromosome (translocation).  This portion of the Y chromosome had the SRY thus allowing the embryo to develop as a male.

• Non-disjunction results in whole chromosomes being deleted or increased in a gamete due to the homologous chromosome pair, or sister chromatids not separating in anaphase I or II respectively.  Deletion, duplication, inversion, and reciprocal translocation involve just a portion of a chromosome.  It is not a subset of non-disjunction. The gamete would still have the right chromosome number. These breaks are sometimes the result of DNA damage from chemical or radiation exposure. Sometimes a mutual deletion and duplication happens to nonsister chromatids during crossing over. One nonsister chromatid ends up with extra genes, and the other loses genes.

• In some of the cells, the one that shuts down does have the dominant allele.  Apparently the protein expressed from one half of one dominant allele must be sufficient to give a normal phenotype.  I am a carrier for color blindness. When I'm tested it is difficult for me to make out the numbers buried in the dots, although  I can still make them out. According to other people, the numbers are very obvious.  My colorblind sons can't see them at all.

• (Aa x Aa) would give you two homozygotes AA which is homozygous dominant, and aa which is homozygous recessive.  You would also get two heterozygotes, Aa.  Therefore your ratio is 2 homozygotes to 2 heterozygotes which is the same as 1:1.  This is their genotype.  I think you were coming up with their phenotype (appearance) which would be 3:1 (3 of the dominant appearance, 1 of the recessive).

 

	A	a
A	AA	Aa
a	Aa	aa

• The mitotic spindle consists of microtubules that arise in an area called the centrosome.  These microtubules control the movement of the chromosomes during mitosis and also elongate the cell during anaphase.  For a step by step description see p210. Check out the chapter review on p222 for a brief description.
• Interphase - centrosomes replicate
• Prophase - microtubules begin radiating from the two centrosomes, and the centrosomes begin moving towards opposite sides of the cell.
• Prometaphase - The centrosomes continue to move to opposite sides of the cell.  The kinetochore microtubules attach to the sister chromatids at their kinetochore.  Nonkinetochore microtubules interact with nonkinetochore microtubules from opposite ends of the cell.
• Metaphase - The kinetochore microtubules adjust to line up all chromosomes on an imaginary plate across the middle of the cell.
• Anaphase - The kinetochore microtubules begin to shorten.  This separates the sister chromatids and they move to opposite sides of the cell.  At the same times, the nonkinetochore microtubules lengthen pushing the poles farther apart (elongating the cell).
• Telophase - The kinetochore microtubules disappear once all of the chromosomes are at opposite poles.  Once the cell is long enough, the nonkinetochore microtubules quickly disappear and cytokinesis takes place.

• Cancer is more prevalent in older people because it takes a series of mutations in one cell in order for a cell to start dividing uncontrollably, lose density dependent inhibition and anchorage dependency.  The older the person, the more likely these mutations have accumulated either randomly or through chemical or radiation exposures.

10) I am not sure what genetic recombination is or how it occurs.

• Genetic recombination is the production of offspring with allele combinations different from those in the parents.   This can come about through independent assortment of chromosomes during meiosis or through crossing over during prophase I of Meiosis (where non-sister chromatids exchange genes).
• An example of genetic recombinants due to independent assortment are the green-round peas and the yellow-wrinkled peas from parents that were yellow-round and green-wrinkled.

• These are the genotypes in Mendel's crosses for one character (gene):
• P - (AA) x (aa)  true breeders
• F₁-    the genotype is all Aa (heterozygotes) - the phenotype would be that all express the dominant trait
• F₂ -    the genotype is 1AA: 2Aa: 1aa - the phenotype is 3:1 (3 expressing the dominant trait to 1 expressing the recessive trait.

• A recessive sex linked gene would be on the X chromosome because the Y only has a few genes and they are all involved with male sexual development.  When we say sex-linked, we are really saying X-linked.
• Since a male only has one X chromosome (remember he is XY) that is the only copy of the gene that he has.  There is no corresponding copy of this gene on his Y chromosome. A female has two X chromosomes (XX).  Therefore she has two copies  (2 alleles) of every gene on the X-chromosome.
• When a male has a mutated gene on his only X-chromosome, it is always expressed since he doesn't have another allele (only one X chromosome) to compensate.  I hope this helps.