VCE Biology Question Thread

[FarAwaySS2]

The answer to the question is C - 2/3. But I don't understand why and how that answer is reached. Any help would be much appreciated!

Mod edit: pictures too big

[grannysmith]

Well, considering it is autosomal recessive, the possible genotypes of the parents of IV 2 must be heterozygous (let's say Aa). Now, we cross them and get AA, Aa, Aa, aa as all possible genotypes of their offspring.

Tay Sachs disease is autosomal recessive as deduced by the pedigree. Affected individuals must be homozygous recessive (aa). Unaffected individuals may be homozygous dominant or heterozygous. So therefore we can rule out the possibility of unaffected individuals (like Mr IV 2) being homozygous recessive (obvious). So now we are left with Aa, Aa and aa. There is a 2/3 chance it is heterozygous for Tay Sachs disease.

Basically, although technically there is a 1/2 chance he's heterozygous, because we know he is unaffected by an autosomal recessive disease, we can narrow down our possibilities to 2/3.

Lol my fail explanation sorry

[RazzMeTazz]

The Nelson book says: "People carrying the allele for sickle cell anaemia can be tested using restriction fragment length polymorphism analysis. Within the normal sickle cell anaemia gene there are three MstII restriction stes. The allele for sickle cell anaemia is due to a point mutations which causes a loss of one of the MstII restriction sites. This means that when the gene is cut with MstII one less restriction fragment is produced in an individual with the mutation. In individuals without the mutation, MstII produces two pieces of DNA. In individuals with the mutation only a single restriction fragment is produced."

If the MstII enzyme has three restriction sites within the normal sickle cell anaemia gene then why are only two fragments produced when the MstII enzyme is used to cut this gene? I thought it would be three produced..

And if the MstII enzyme has two restriction sites within the mutated sickle cell anaemia gene, then why is only one fragment produced when the MstII enzyme is used to cut this gene? I thought two would be produced..

[katiesaliba]

1.How exactly does reduced protein synthesis lead to cell death? Does it concern apoptosis or necrosis?
2. Why is rhesus factor irrelevant for kidney transplants?
3. Why is there still a risk of graft rejection if blood type, rhesus factor and HLP are all matching? Thanks!

[RazzMeTazz]

So DNA ligase is used to catalyse the formation of phosphodiester bonds between restriction fragments?

And the hydrogen bonds between complementary nucleotides, form naturally? Without the aid of DNA ligase?

[katiesaliba]

So DNA ligase is used to catalyse the formation of phosphodiester bonds between restriction fragments?

And the hydrogen bonds between complementary nucleotides, form naturally? Without the aid of DNA ligase?

Yes

[RazzMeTazz]

I don't understand how nucleotide sequencing works.

 How does knowing the last nucleotide of each DNA fragment formed in gel electrophoresis  (in order of smallest to largest) give the sequence of the DNA?

[katiesaliba]

I don't understand how nucleotide sequencing works.

 How does knowing the last nucleotide of each DNA fragment formed in gel electrophoresis  (in order of smallest to largest) give the sequence of the DNA?

These images should help:
http://www.mawiz.co/wp-content/uploads/2013/07/DNA_sequencing.jpg
http://media1.shmoop.com/images/biology/biobook_biotechnology_graphik_8.png
http://www.cryst.bbk.ac.uk/pps97/assignments/projects/borek/Domina/sekw1.gif
http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/F/FluorDideoxySeq.gif

Basically though, you denature your DNA sample and then add DNA polymerase. The resulting DNA mixture is sectioned into 4 samples and different dideoxynucleotide versions of each nitrogenous base (ddA, ddC, ddG and ddT) are added into each. These dioxynucleotides fluoresce (they can also be called 'ddATP' or 'ddTTP' for example) and prevent further amplification of the complementary DNA strand. The samples are then placed into 4 loading wells in a gel electrophoresis tank (E.G well 1 contains ddA, well 2 contains ddC, well 3 contains ddG and well 4 contains ddT) and run through gel electrophoresis. The resulting band can be read from bottom to top to determine the complementary sequence of the initial DNA strand. You'll also have a primer to consider.

[grannysmith]

If crossing over were to occur between linked genes, then the percentage of recombinant offspring would be less than 50% right?

It couldn't be 50%... because that would mean the genes aren't linked.

[vox nihili]

If crossing over were to occur between linked genes, then the percentage of recombinant offspring would be less than 50% right?

It couldn't be 50%... because that would mean the genes aren't linked.

That's right.

[grannysmith]

That's right.

I thought so..

I lost a mark on a SAC because of this. SACs can be so dodgy it's not even funny

[Reus]

I thought so..

I lost a mark on a SAC because of this. SACs can be so dodgy it's not even funny

So true! I hate it.

[vox nihili]

I thought so..

I lost a mark on a SAC because of this. SACs can be so dodgy it's not even funny

That is dodgy. You actually can't get a recombination frequency (proportion of recombinants) above 50. I think the highest you can get is 49.99 or something like that.

[RazzMeTazz]

When using DNA profiling to analyse the relatedness of two individuals by looking at their STRs and VNTRs, the section of DNA analysed has to be from a particular chromosome for both individuals right?

Or else the STRs and VNTRs would be different?

[RazzMeTazz]

If the recognition of a restriction enzyme is said to be : 3' CTTAA ^ G5'
what does that mean?  o.O

Also, is DNA ligase only used when joining two restriction fragments with sticky ends together?
What about blunt ends, don't they require DNA ligase as well?