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    (Original post by bakedbeans247)
    loool i hope so too iA.
    haven't seen chi squared in any recent papers so there is a good chance it might come up.
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    (Original post by EvasiveRose)
    It depends on what alleles are present at each gene locus- whether they're homozygous recessive, dominant, etc that determines what type of epistasis it is
    thank you!
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    (Original post by Diamond Crafter)
    Same! I cracked Hardy Weinberg two nights ago so I'm hoping it comes up xD
    Don't wanna burst your bubble but it's really unlikely to come up as it came up last year lol. Damn it
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    (Original post by hajs)
    Quick Question... Attachment 427767
    They can add a genetic marker and the more active the gene is, the more mRNA produced which cn be detected under uv light.

    I'm assuming them extract the mRNA after that and yeah, that's something that's been dumbed down for our course
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    how do you work this out:

    The human ABO blood groups are A, B, AB and O. They are determined by a singlegene with multiple alleles. IAand IBalleles are codominant, but both these alleles aredominant to the IOallele.

    In a maternity ward, the identities of four babies became accidentally mixed up. TheABO blood groups of the babies were discovered to be O, A, B and AB. The ABOblood groups of the four sets of parents were determined and are shown in the tablebelow.

    Complete the table to match each baby to its parents by indicating:

    • the parental genotypes, using the symbols IA, IBand IO;
    • the blood group of the baby which belongs to each set of parents

    parental blood groups /parental genotypes/ baby blood group
    O and O
    AB and O
    A and O
    AB and A
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    What do we need to know for outining the role of ATP in muscle contraction
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    (Original post by cinderella25)
    how do you work this out:<br />
    <br />
    The human ABO blood groups are A, B, AB and O. They are determined by a singlegene with multiple alleles. IAand IBalleles are codominant, but both these alleles aredominant to the IOallele. <br />
    <br />
    In a maternity ward, the identities of four babies became accidentally mixed up. TheABO blood groups of the babies were discovered to be O, A, B and AB. The ABOblood groups of the four sets of parents were determined and are shown in the tablebelow.<br />
    <br />
    Complete the table to match each baby to its parents by indicating:<br />
    <br />
    • the parental genotypes, using the symbols IA, IBand IO; <br />
    • the blood group of the baby which belongs to each set of parents<br />
    <br />
    parental blood groups /parental genotypes/ baby blood group <br />
    O and O<br />
    AB and O<br />
    A and O<br />
    AB and A
    O and O is obviously baby O

    A and O is baby A as o and o is definitely o as that's homozygous recessive, even though there is a possibility these parents would produce an O baby

    Which means that AB and O is likely to be baby B and Parent A and O are the ones to have baby A

    AB and B have baby AB then
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    (Original post by cr7alwayz)
    What do we need to know for outining the role of ATP in muscle contraction
    ATP breaks the actin myosin cross bridge so myosin heads can bind to a different binding site and pull the actin filament along the myosin
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    (Original post by ChoccyPhilly)
    O and O is obviously baby O

    A and O is baby A as o and o is definitely o as that's homozygous recessive, even though there is a possibility these parents would produce an O baby

    Which means that AB and O is likely to be baby B and Parent A and O are the ones to have baby A

    AB and B have baby AB then
    I get O and O = baby O
    A and O = baby A

    I just don't get the other two. How AB and O = baby B and AB and B = baby AB

    how did you work that out
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    (Original post by Diamond Crafter)
    Same! I cracked Hardy Weinberg two nights ago so I'm hoping it comes up xD
    May you explain it please?
    I know it's like p+q=1
    and 2pq + blah
    but how would I do this question?

    1 in 2000 people have cystic fibrosis in the UK ff
    Find the percentage of carriers
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    (Original post by cinderella25)
    I get O and O = baby O<br />
    A and O = baby A<br />
    <br />
    I just don't get the other two. How AB and O = baby B and AB and B = baby AB<br />
    <br />
    how did you work that out
    AB and O can be either baby A or baby B, right?

    If parents A and O are baby A then therefore, parents AB and O must be baby B as baby A has been confirmed by other evidence.

    In the same way that AB and B can be either AB or B, B has been confirmed previously from parents AB and O, so therefore, AB and B must be AB.

    Also, no other parent combination can produce AB either as the rest have O as one of their parents, meaning you dint get codominance
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    Any predictions for the exam?
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    (Original post by ChoccyPhilly)
    AB and O can be either baby A or baby B, right?

    If parents A and O are baby A then therefore, parents AB and O must be baby B as baby A has been confirmed by other evidence.

    In the same way that AB and B can be either AB or B, B has been confirmed previously from parents AB and O, so therefore, AB and B must be AB.

    Also, no other parent combination can produce AB either as the rest have O as one of their parents, meaning you dint get codominance
    Thank you for your explanation it makes sense now!


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    (Original post by frozo123)
    May you explain it please?
    I know it's like p+q=1
    and 2pq + blah
    but how would I do this question?

    1 in 2000 people have cystic fibrosis in the UK ff
    Find the percentage of carriers
    Ok, to find da carriers we need to know what 2pq is. 2pq is all da individuals who are heterozygous k. They carry one legit allele and one faulty allele

    So we know dat q^2 would be 1 in 2000 i.e. 1/2000, which is 0.0005
    Therefore q would be root 0.0005, which is 0.02236....

    And you know dat p+q = 1 yeah
    Dat means dat p would be 1-0.02236..., which is 0.977639...

    Now all you gots to do is 2 times p times q, which is 2 x 0.977639... x 0.02236...

    = 0.0437...

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    (Original post by cinderella25)
    Any predictions for the exam?
    Lots and lots of ecology... I think a big question on succssion
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    (Original post by Prince edmund)
    Ok, to find da carriers we need to know what 2pq is. 2pq is all da individuals who are heterozygous k. They carry one legit allele and one faulty allele

    So we know dat q^2 would be 1 in 2000 i.e. 1/2000, which is 0.0005
    Therefore q would be root 0.0005, which is 0.02236....

    And you know dat p+q = 1 yeah
    Dat means dat p would be 1-0.02236..., which is 0.977639...

    Now all you gots to do is 2 times p times q, which is 2 x 0.977639... x 0.02236...

    = 0.0437...

    Posted from TSR Mobile
    so is it 2pq= x
    not pq= x/2 ?
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    (Original post by ChoccyPhilly)
    Lots and lots of ecology... I think a big question on succssion
    ugh
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    (Original post by ChoccyPhilly)
    Lots and lots of ecology... I think a big question on succssion
    That would be awesome! That's the easiest topic lol


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    (Original post by cinderella25)
    That would be awesome! That's the easiest topic lol<br />
    <br />
    <br />
    <font size="1"><a href="http://www.thestudentroom.co.uk/app" target="_blank">Posted from TSR Mobile</a></font>
    I hate ecology so much. Last years paper would be so ideal as the.long question was on transction and translation, which I really enjoy but I don't think we'll get a lot of genetics in our paper
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    The responding to the environments section is the easiest by far and away. Half of it is AS psychology.
 
 
 
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