# Homework Help: Mendelian Genetics questions:

1. Sep 4, 2007

### future_vet

1. The problem statement, all variables and given/known data
Assuming that there is a 1/32 chance that a child will show a certain trait, and we need to calculate the probability that 2 parents will have 4 children with this specific trait, do we do the following:

2. Relevant equations
(1/32)^4

3. The attempt at a solution
1/32768

1. The problem statement, all variables and given/known data
When 2 genes are codominants, they are both expressed in the offpsrings.
For example, a red plant and a white plant will give offsprings that are pink in the F1 and offsprings with shades of pink + red plants +white plants in the F2. If a pink F1 offspring were crossed with a white parent, would we get all white plants since white is dominant to pink?

Thank you!

2. Sep 4, 2007

### skipthecereal

For the first question: (1/32)^4 is correct. This is logically obvious, but can be shown using probability laws. I dont know if you are familiar with these laws or notation, but:

P(AnB)= P(A)x P(B) if the 2 events are independent of one another. Independent meaning the reliasation of 1 does not affect the other. A child having the trait is an independent event, as Mendels law of independent assortment/segregation makes clear. As there are four of the same independent events occuring: one called 'A' another 'B', 'C' and 'D'.
P(AnBnCnD)= P(A) x P(B) x P(C) x P(D).
Evaluating this gives; 1/32^4
But this is 1/1048576 not 1/32768 which is 1/32^3.

Sorry for the babble but i hope thats helped.

For the second question: I'm not sure you havent made a typo, as the question doesnt make sense.
Pink only exists as a trait in the phenotype. Pink is a result of one allele of the pair being 'red' and the other being 'white'. The combined effects of the alleles produces a white flower. As such pink is not really dominant to white. The genotype of the pink flower (type of alleles) is RW, that of a white flower is WW. To work out what the offspring of a cross between these two can possibly be, draw out a punnet square diagram. After you have done that you should see that you wont get all white plants. Half will be white and the other half pink

3. Sep 5, 2007

### future_vet

Thank you so much!

I have another question regarding the codominant alleles:

If we cross a white plant with a red plant, and they have codominant alleles, we will get all pink plants in F1, and then a range of colors from white to pink to red in the F2.

Now, let's assume we go several color categories, with 1/64 being white, 6/64 being light pink (plus 4 other fractions for the rest of the shades of pink) and 1/64 being red, how many genes then control color in these plants?

I assumed that we only have 2 genes controlling colors, one gene for the red color and one gene for the white color, and the range in colors is only due to codominance.
But a friend said that we have to look at the fractions: all numbers are over 64, which may indicate that 4 genes are controlling color. He based the assumption on the fact that a monohybrid cross gives 4 genotypes, a dihybrid cross gives 16 phenotypes, and therefore 4 genes would give 64 phenotypes.

What do you think?

Thank you so much for your help!! I hope my questions were clear enough.

4. Sep 7, 2007

### skipthecereal

Yes, the F1 will definetly consists of solely pink coloured flowers. The F2 will consist of a mixture of red, white and pink, but there is not so much a range of colours determined by the gene's alleles. By this I mean that technically there will only be one shade of red, white and only one shade of pink that can possibly be in the phenotype of the F2 generation as determined by the gene. But, enviromental factors and other genes present in the organism can lead to a range of colours in real life. But if you have been asked a specific question, like; "What colours will the flowers have in the F2 generation of a monohybrid cross? e.t.c". Then I personally would state precisely, just red, white and pink, not a range of pink shades. But once again I must say that your anwser of a "range of colours" is correct in a real life scenario; where any trait is goverened by a number of genes and enviromental factors. I hope that has been some help and has been put forward clearly on my part.

As for your second question I would agree with your friend, but I am not 100% sure myself. If you do find out the anwser please post it!

5. Sep 7, 2007

### future_vet

Hi!

The teacher told us today about a formula. In order to find the number of genes, you need to see how many offsprings we got: here it would be 64 (By total # I mean as low as we can go and still keep a category. Here 1/64 is the minimum).

1/(4^n)=64
n= number of genes and so in this case, we have 3 genes. So both answers were wrong in the end :)

Thanks!

6. Sep 8, 2007

### skipthecereal

I see what you mean, looking back your friend had the correct idea. monohybrid involves 1 gene and produces 4 possible genotypes or 'offspring' in the F1; 4^n=4, so n=1 which agrees with the fact that a monhybrid gene considers one gene.
Dihybrid involves 2 genes, yielding 16 genotypes in the F2 generation; 4^n=16, n=2 and once again n gives the number of genes.
But you said your friend said that 4 genes would yield 64 genotypes. As you have found out it is infact 3 genes. As 4^3=64.

But i dont get why the formula you use is 1/4^n. For example if we take 1/4^n=16, n=-2. Anyways thanks for the post.

7. Sep 8, 2007

### future_vet

I meant 1/64, sorry!