Biol10005 MST practice question help

[Ballerina]

At the risk of making my first post on ATARnotes an unpopular one (ahaha), I liked it. I found the practice test pretty challenging (made a lot of stupid mistakes) but Test H (Friday 9 AM) was more to do with memorizing obscure lecture content than analyzing crosses/pedigrees. It would be better if the test were designed around analysis instead of just rote memorization (in the workforce, obviously being able to analyze data takes priority over remembering something that can be found in the reference book a metre away from you), but seeing as the latter is my strength, I feel more confident post-test than pre-test.

That being said, any confidence can only be validated when the test results are released. :/

[hobbitle]

Haha thanks Ballerina, I hope you're right! I know a lot of the obscure lecture content but still suck at complicated crosses! It'll be done in 3hrs anyway.

[Ballerina]

Best of luck! In case you read this before the test; knowing the visual map of DNA replication (with helicase, DNA polymerase III etc) came up, and a couple of questions similar to the 2 problem sets on LMS. The last few questions (worth 2 marks each) were hardest for me, as they involved interpreting crosses and all the answers looked correct.  Maybe it's moot as we'll have different questions regardless!

[simpak]

AN UPDATE: welp, as anticipated, need to up my game for mst2...lucky Paul Gleeson is lecturing atm lawl

[hobbitle]

Haha you probs still got >90% :-P

[simpak]

;_; low 90s.

[LeviLamp]

Is Paul Gleeson related/married to Dawn Gleeson?

[Starlight]

Is Paul Gleeson related/married to Dawn Gleeson?

lol

[hobbitle]

As time has passed and I've been revising more I realised that I'd be a borderline 5/10. So that average doesn't surprise me considering I usually nail Biol.

[vox nihili]

Pretty random but im Kinda worried about mst reesults considering the average is apparently 5.6/10 :/

That's brutal...

[hobbitle]

Hey all,

I emailed Dawn asking about the solution to Q22 of the Question Set in our Lab Workbook.  She sent me the answer but I still don't understand!!!!  Can anyone explain this, or can it not be explained any clearer than in the answer she provided?  Things I don't understand:

• (Basic but I'm still confused) How does the 1:1:1:1 ratio occur in independent assortment with two gene loci.  Isn't the ratio 9:3:3:1..?
• "Black class with 198 rabbits" ??
• So what is the big B allele doing and what does it represent, and what is the big N allele doing and what does it represent?

QUESTION:
The agouti hair pattern often seen in many wild animals is characterised by a small band of light pigment near the tip of the hair. The width of this lightly pigmented band is genetically determined. Purebreeding agouti rabbits with a wide band at the tip of the hair were crossed to purebreeding rabbits with black fur. All of the offspring of this cross had agouti fur with the narrow band at the tip of the hair. These offspring were test crossed to purebreeding black rabbits.

The offspring of the second cross produced the following results:
 
  phenotype                                     number of offspring
  black fur                                                151   
  agouti fur with the wide band                 127   
  agouti fur with the narrow band              22 


Propose a detailed genetic hypothesis to account for these results. Show clearly how you arrived at your conclusion.

ANSWER:
NB: because these rabbits are black you cannot see the band (epistasis) but it is a test cross which is always to the homozygous recessive so they must be bb nn

Cross is BbNn X bbnn
 
Looking at the results obviously the loci are not assorting independently (expect 1:1:1:1 ratio) so next simplest hypothesis is linkage. However there are only 3 phenotypes. Why? When the B allele is present it masks the effect of the band-width locus (epistasis) so the black class with 198 rabbits includes two groups; black and wide and black and narrow.
 
Looking at data, agouti with wide band is a non recombinant or parental class (high frequency and only one phenotype)
Allelic arrangement must be                   
B w                  
b W
(trans) in the heterozygous parent.
 
The proportion of recombinants   (agouti narrow)   bW/bw  is 22                                   

plus within the black fur class approximately 22 will be the other recombinant class (black, wide,  Bw/bw).

Therefore the map distance is approximately (22 + 22) / 44 X 100/300 = 14.7


Thankyou...

[vox nihili]

Hey all,

I emailed Dawn asking about the solution to Q22 of the Question Set in our Lab Workbook.  She sent me the answer but I still don't understand!!!!  Can anyone explain this, or can it not be explained any clearer than in the answer she provided?  Things I don't understand:

• (Basic but I'm still confused) How does the 1:1:1:1 ratio occur in independent assortment with two gene loci.  Isn't the ratio 9:3:3:1..?
• "Black class with 198 rabbits" ??
• So what is the big B allele doing and what does it represent, and what is the big N allele doing and what does it represent?

It means that the genes are affecting the other genes. So in the presence of certain alleles, a particular phenotype is blocked.

http://bio1151.nicerweb.com/Locked/media/ch14/epistasis.html


QUESTION:
The agouti hair pattern often seen in many wild animals is characterised by a small band of light pigment near the tip of the hair. The width of this lightly pigmented band is genetically determined. Purebreeding agouti rabbits with a wide band at the tip of the hair were crossed to purebreeding rabbits with black fur. All of the offspring of this cross had agouti fur with the narrow band at the tip of the hair. These offspring were test crossed to purebreeding black rabbits.

The offspring of the second cross produced the following results:
 
  phenotype                                     number of offspring
  black fur                                                151   
  agouti fur with the wide band                 127   
  agouti fur with the narrow band              22 


Propose a detailed genetic hypothesis to account for these results. Show clearly how you arrived at your conclusion.

ANSWER:
NB: because these rabbits are black you cannot see the band (epistasis) but it is a test cross which is always to the homozygous recessive so they must be bb nn

Cross is BbNn X bbnn
 
Looking at the results obviously the loci are not assorting independently (expect 1:1:1:1 ratio) so next simplest hypothesis is linkage. However there are only 3 phenotypes. Why? When the B allele is present it masks the effect of the band-width locus (epistasis) so the black class with 198 rabbits includes two groups; black and wide and black and narrow.
 
Looking at data, agouti with wide band is a non recombinant or parental class (high frequency and only one phenotype)
Allelic arrangement must be                   
B w                  
b W
(trans) in the heterozygous parent.
 
The proportion of recombinants   (agouti narrow)   bW/bw  is 22                                   

plus within the black fur class approximately 22 will be the other recombinant class (black, wide,  Bw/bw).

Therefore the map distance is approximately (22 + 22) / 44 X 100/300 = 14.7


Thankyou...

[hobbitle]

Results are out.
Ouch.
The highest I can see by scanning them is 9.2/10 and I can only see two people who got that.
There were heaps of people 10/10 last semester.

[LeviLamp]

I can't bear to look oh my god having heart palpitations

[hobbitle]

Yeah I think that ruined my chances of getting > 90% this semester.