Some More Genetics Help Please!?

Question

1- An individual whose genotype is unknown, phenotypically expresses a dominant trait. To determine the genotype of the individual with the unknown genotype, you perform a test cross. The results of the offspring produced show 50% with the dominant and 50% with the recessive trait. What is the genotype of the unknown individual.

2-Two sisters both have blood type O and neither of the parents have type O. List all the possible blood types of the parents and all of their potential genotypes.(write in the correct format ( I^A/I^A, ii/I^B etc)
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1- I don't even know how to start this, is it trial and error is there a real way to approach this problem?

2- I don't understand how to set up these punnett squares, I know that Type O = ii, but I can't set up ii with any other blood type b/c that would result in some being type O which the parents aren't.

Can someone please explain how to set these up, I'm trying, but they aren't making sense. I'd really appreciate it.Thanks! = )

Answer 1

1) The unknown genotype is heterozygous dom. (as an example, let's use Kk). If the organism expresses the dom trait (as K_, because it's dom., it wouldn't matter whether the _ was filled in with K or k). The cross could be with a known recessive (kk), so it's controlled (that organism can ONLY give a k, nothing else). So, then your offspring would have the following:

unknown gives K or _
control gives k

offspring are 50% K_ and 50% kk

This means that 50% of the time, the unknown gave its K and the other 50% of the time, it gave its _ (which we know to k).

Does that make sense?



2) If the 2 offspring are type O, then they are ii (homozygous rec.). This means that they got an >i< from each parent (mom gave an i, and dad gave the other i). Since neither parent is type O (ii), then they each must be heterozygous. We know that they can't both be AB, because there would be no >i< to give. So, we could assume that the parents are I^A / i or I^B / i, because then they each have an >i< to give to each offspring.

I hope this helps? Let me know and good luck!

Answer 2

1- isn't trial and error. a phenotypically dominant individual can only be one of two genotypes: heterozygous (Xx) or homozygous dominant (XX). A test cross is when you cross an individual with a homozygous recessive individual (xx). If the individual we are testing were homozygous dominant, we would get offspring that is 100% heterozygous and penotypically dominant (presenting the dom trait), or Xx. if the individual were heterozygous, we would get 50% heterozygous offspring showing the dom trait (Xx)and 50 homozygous recessive (xx). therefore, in this case the individual is heterozygous.

i recommend you read wikipedia to see how to set up a punnet square. http://en.wikipedia.org/wiki/Punnet_square

2 - this is how i would work on this problem. 2 sisters have O blood type, so ii genotype. that means they got one i from one parent and one i from another parent. But neither parent is ii. so they must have one i allele and one other allele each. the possible alleles for blood type is i, I^A or I^B (which i'm going to write as A and B for easier reading).

if your genotype is AA, your phenotype is A blood type.
if your genotype is Ai, your phenotype is A blood type.
if your genotype is Bi, your phenotype is B blood type.
if your genotype is BB, your phenotype is A blood type.
if your genotype is AB, your phenotype is AB blood type (codominant).
and finally, if your genotype is ii, your phenotype is O blood type.

which of these non-O phenotypes have an i allele? 2 of them: Ai and Bi. these are the only possible genotypes for the parents.

if you want to solve this using a punnet square, remember that ii will be inside one of the boxes, because that's the phenotype of the offspring. it does not have to be 2 boxes, because the boxes are prcentages, not whole numbers. so 1 box is 25%, 2 boxes = 50%. you don't know the percentage of the offspring, since the offspring you get is always random. you could only use percentages if you had a very large number of offspring, say, 100, and even that wouldn't be perfect. so, as far as you know, 100% of the offspring are ii, but that's completely irrelevant becuase there are only 2 of them. so put ii in ONE of the boxes only, and work from there.