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1

First of all, if this is a homework question, add that tag to it. And show what work you have done so far. Secondly, primers for amplifications should lie on opposite strands. Primers are typed in 5'-to-3' direction (aka left-to-right on leading strand). Appropriate primers will be: primer 4=GTG... and primer 5=GAA.... Note how those primers are always in ...


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The secretion pushes the cell organelles to the basal part. (although the fat cells have no polarity, their organelles have been pushed to one side of the cell: Comparing to adipose cells, the acinar cells secrete their specific proteins on one specific end of the cell, this is why in contrast with fat cells they have well defined polarity.


8

The reason why the cell would shrink more in CaCl2 solution is because it has a higher van't Hoff factor i.e. total number of dissociated ionic species per solute molecule (it is 2 for NaCl whereas it is 3 for CaCl2). (Nonionic solutes do not dissociate and will therefore have a van't Hoff factor of 1) Osmotic pressure (and other colligative properties) ...


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Three follicular stages are recognised, 1) Primary/Preantral Follicles (with primary oocyte inside) 2) Secondary/Antral Follicles (with primary oocyte inside) 3) Pre Ovulatory Follicles (with secondary oocyte inside) Pre ovulatory Follicles are formed ~36 hrs before ovulation at time of LH surge. This coincides with completion of Meiosis I and formation of ...


1

Remember: A and B are genetic markers that might be linked to the disease gene. Neither A nor B are actually the hemophilia disease mutation (or gene), A or B are proxies, or substitutes for the mutation in the disease gene. Since it is an X-linked recessive, it is only the Mother's two X chromosomes that are candidates. You can completely ignore the ...


1

Since there are 2 alleles of each gene and 3 different genes, 2EE3 = 8? Let's test this by making a table, for 1 gene there are two different kinds of gametes, so 2EE1 = 2, for 2 genes there are 4 different kinds of gametes (AB, Ab, aB, ab) so 2EE2 = 4, and you already did the calculation yourself for 3 genes, and 2EE3 = 8. Why don't you try working it ...


2

Let's consider all the options. X-linked recessive: I:2 is a carrier and II:5 is also a carrier, and I:1 is affected. Everyone who marries into this family(except for II:2, lucky guy. He better not have any sons though.) is a carrier for the same genetic disorder. While it does work, it's vastly less likely than the other options. Observe: II:4 is affected, ...



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