Jump to content

Pereira

Members
  • Posts

    5
  • Joined

  • Last visited

Profile Information

  • Favorite Area of Science
    Genetics

Pereira's Achievements

Lepton

Lepton (1/13)

0

Reputation

  1. Recently have gotten a tutor, mods can delete this thread if you want.
  2. Hello, Any help will be appreciated. Q:Determine moles of dichromate ion that reacted with iron(II) in titration (use average volume). My A: 0.0119mol 6 Fe2+ + Cr2O72- + 14 H3O+ 6 Fe3+ + 2 Cr3+ + 21 H2O 4g-Fe2+ 4g*(1mol/558.845g/mol) = 0.0716 moles of Fe (6/0.0716)=1/x x=(0.0716)1/(6) x=0.0119mols of dichromate. My questions are: 1) am I way off in my approach? 2)Why does the question want me to use volume? It seems that it should not be this easy so I assume I am way off. Also the avg volume of Fe solution 6.75mL and the concentration was =0.102mo/L. --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- Alright before I submit this thread to publish I tried another approach: molarity(Cr2O7 2-)= (molarity of Fe(II)/6)*(volume of Fe(II)/volume of Cr2O7) =[(0.102mol/L)/6)] * (0.00675L/0.005L) =0.017*1.35 =0.2295 m(Cr2O72-)=215.992 * .1L *0.2295 m=4.95mols *this answer seems like I am more on the right track I think Are either of these very close. Any hints will be greatly appreciated. ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
  3. I am quite new to genetics as well so forgive me if I am wrong but you can cross heterozygous parents as well. Parent 1 would give you WG wG and parent 2 would give you wg haploid gametes. So when you plug those into the punnett square you should get 2 different diploid cells; this will be your f1 generation. Now if you cross f1 X f1 you take each of those diploid genotypes and work them into 4 different haploid gametes once again. Put those in a second punnett square and you will get the f2 generation phenotypes. It is common to get a 9:3:3:1 ratio but that is if both parents are homozygous. It will be a fraction of 16 (?/16). Hope this helps.
  4. Hello, I am a little confused about complete dominance. The question I am trying to solve is: A trihybrid pea plant having the genotype AaBbC(1)C(2) is self fertilized there is complete dominance at the A and B loci and co-dominance at the the C loci. What fraction of the progeny will be phenotypically different from the parents?(assume independant assortment). Now I know that the co-dominance C(1)C(2) will show through all the progeny and that A\- & B\- will also be phenotypically alike to the parents. So A\-B\-C(1)C(2) will be phenotypically the same. My question is if A\-bbC(1)C(2), aaB\-C(1)C(2) will cause the progeny to be phenotypically different. As well (correct me if I am wrong) the aabbC(1)C(2) if the homozygous will still be phenotypically different. I know this should be fairly easy but my text is really confusing me at the moment. So far I think that the ratio of differing progeny is 3/4. If I am wrong any help will be appreciated. Thank you.
  5. Hello, My name is Justin and currently attending uni. I have switched from Kine to Science as its much more enjoyable and I am starting to enjoy challenging myself. Looking forward to meeting everyone!
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.