Independence of events

A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene in the sense that if an organism possesses the gene pair xX, then it will outwardly have the appearance of the X gene. For instance, if X stands for brown eyes and x for blue eyes, then an individual having either gene pair XX or xX will have brown eyes, whereas one having gene pair xx will be blue-eyed. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus 2 organisms with respective genotypes aA, bB, cc, dD, ee and AA, BB, cc, DD, ee would have different genotypes but the same phenotype.) In a mating between 2 organisms each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of its mate. In a mating between organisms having genotypes aA, bB, cC, dD, eE, and aa, bB, cc, Dd, ee, what is the probability that the progeny will (1) phenotypically, (2) genotypically resemble
(a) the first parent;
(b) the second parent;
(c) either parent;
(d) neither parent?
Now assuming that genes of different letter types do not intermix ( e.g Ab,Ba etc) , we have (4)^5 possibilities for the child.(aa,AA,Aa,aA for the first slot and so on). Now to resemble the phenotype of the first parent the child should bear the following genetic properties
i)aA,Aa,AA for the first slot
ii)Bb,BB,bB for the second slot
iii)cC,CC,Cc for the third slot
iv) dD,Db,DD for the fourth slot
v) eE,Ee,EE for the fifth slot
Now there are 3^5 possible outcomes that result in the child having similar phenotype as the first parent. So the probability of the first outcome would be (0.75)^5. But I am told that this approach is wrong. I would like to know the correct way to approach this problem.

Alright, so if I'm reading and interpreting the problem correctly, matching a phenotype for a gene pairing simply means one of two things: Both parent and child have at least one dominant gene (capital letter), or both parent and child must have two recessive genes. But to match genotypes for a gene pairing, the letter combinations must match exactly. And to be considered an overall phenotypical or genotypical match, the parent and child must meet the above criteria on all five of their genes. If this interpretation is not correct, then please correct me.

Okay, so it seems part 1a is asking you to find the probability that the child will have at least one dominant gene on all five of its pairings. Your work looks good, except for one problem I see. Note that the second parent has no capital A. So that means that the only way the child can inherit a capital A is from the first parent. And similarly for the letters C and E. Once you eliminate pairings which are impossible for the child to possess, your math should work out correctly. Then it looks like the same basic process applies to all the other parts, they're just wanting you to calculate different scenarios.

adhok said:

A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene in the sense that if an organism possesses the gene pair xX, then it will outwardly have the appearance of the X gene. For instance, if X stands for brown eyes and x for blue eyes, then an individual having either gene pair XX or xX will have brown eyes, whereas one having gene pair xx will be blue-eyed. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus 2 organisms with respective genotypes aA, bB, cc, dD, ee and AA, BB, cc, DD, ee would have different genotypes but the same phenotype.) In a mating between 2 organisms each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of its mate. In a mating between organisms having genotypes aA, bB, cC, dD, eE, and aa, bB, cc, Dd, ee, what is the probability that the progeny will (1) phenotypically, (2) genotypically resemble
(a) the first parent;
(b) the second parent;
(c) either parent;
(d) neither parent?
Now assuming that genes of different letter types do not intermix ( e.g Ab,Ba etc) , we have (4)^5 possibilities for the child.(aa,AA,Aa,aA for the first slot and so on). Now to resemble the phenotype of the first parent the child should bear the following genetic properties
i)aA,Aa,AA for the first slot
ii)Bb,BB,bB for the second slot
iii)cC,CC,Cc for the third slot
iv) dD,Db,DD for the fourth slot
v) eE,Ee,EE for the fifth slot
Now there are 3^5 possible outcomes that result in the child having similar phenotype as the first parent. So the probability of the first outcome would be (0.75)^5. But I am told that this approach is wrong. I would like to know the correct way to approach this problem.

Click to expand...

My take on the problem:

First, just to make it clear in my own mind, we will treat the types aA and Aa differently. That is, if each the gene pair for the, for example a/A, gene are randomly there are 4 equally likely possibilities {aA, aa, AA, Aa} not 3 possibilities with a 2,1,1 ratio {{aA, Aa}, aa, AA}.

Next, since I'm not that great in sadistics, to possibly make the question a little more understandable and get a different perspective, I sometimes find it helps to simplify the problem so I can see the actual possibilities [there aren't too many], so let's do that and only consider 1 gene pair. Following your argument, the results wouldn't change for the one gene pair. That is there would be 4^1 possibilities for the child and 3^1 possible outcomes that result in the child having similar phenotype as the first parent. But, go through the possibilities were we have {first parent, second parent} and possibilities listed:
{aa, aa} Four cases, {first a, first a}, {first a, second a}, etc., all with child same phenotype
{aa, aA} Four cases two with child same phenotype
{aa, AA} Four cases none with child same phenotype
{aa, Aa} Four cases two with child same phenotype
{aA, aa} Four cases, two with child same phenotype
{aA, aA} Four cases three with child same phenotype
{aA, AA} Four cases four with child same phenotype
{aA, Aa} Four cases three with child same phenotype
{AA, aa} Four cases, four with child same phenotype
{AA, aA} Four cases four with child same phenotype
{AA, AA} Four cases four with child same phenotype
{AA, Aa} Four cases four with child same phenotype
{Aa, aa} Four cases, two with child same phenotype
{Aa, aA} Four cases three with child same phenotype
{Aa, AA} Four cases four with child same phenotype
{Aa, Aa} Four cases three with child same phenotype
So 64 cases, 48 with child same phenotype. It looks like your final answer is correct but the way you got there was wrong. You actually have 4*4 mixes of gene pairs from the two parents and each of those mixes produces 4 cases, i.e 4³ cases.

Let's look at your first statement:"Now assuming that genes of different letter types do not intermix ( e.g Ab,Ba etc) , we have (4)^5 possibilities for the child.(aa,AA,Aa,aA for the first slot and so on)" That statement is true. However, although the final result is 4⁵ unique possibilities for the child, the total individual number of ways is (4³)⁵ possible outcomes.

As a different kind of example suppose we had a bag with 5 blue balls and 5 green balls and are asked what is the probability of picking a blue ball. Your way would, I believe, translate to there are 2 colors we can get [instead of the proper 'there are 10 balls'] and we are picking 1 ball [instead of the proper 'of which 5 are blue] so the probability is, in either case, 50%. But what if it had been 3 blue balls and 7 green balls?

EDIT: Somehow formatting got messed up, so fixed (I hope)

Thank you all! the explanations were really helpful!