Biology Forums - Study Force

This time I'm for real about the cookies. 

Let me make sure I understand.  In the examples discussed we assumed that all the genes were heterozygous.  But let’s talk about if some of the genes are homozygous.  I want to see if that will change the way we calculate. 
For example: I think this is a dihybrid cross, AaBb x aabb, so n should be 2

N = 2
2^2 There will be 4 gamete types.  AB, Ab, aB and ab.
2^4 = 16 offspring … I see 16 in the punnett square anyway.
3^2 = well I guess this should be nine genotypic classes.  But, all I found was AaBb, Aabb, aaBb and aabb.  humm...

Anyway, if m= # of dominant genes you’d say m= 2. The proportion that would be dominant in one gene:
(1/2)^0 x (1/2)^2 = ¼ AaBb – I can see this in the punnet square.

So the main confusion is how to find the genotypic classes in the case of a test cross like this one.  Is 3^2 correct in this case?
What about this case?  AABBCc x AaBBCC?  Would n=2?  I’m just confused because the book said n = # of heterozygous genes.

I'm not duddy, but let's go through some examples so that you understand this better using trihybrid.

So say we had this problem:

In a cross AaBbCc x AaBbCc, what is the probability of producing the genotype AABBCC (homozygous dominant)?

The answer would be 1/64

You get this by multiplying the individual probability that each allele will come from either parent. For example, there is a 1/2 chance that an A will come from the first parent. There is also a 1/2 chance that a B will come from either parent. And so on and so on.

The probability that a homozygous dominant trait is expressd in the 2nd generation from two heterozgyous parents is always 1 in 4. Do this with three such traits, and the probabilities are multiplied for each trait. So the answer is indeed 1 in 64, since 4 x 4 x 4 = 64.

You will also find that the homozygous recessive of aabbcc is just as rare, and also 1 chance in 64.
Most of these offspring will be of mixed genotype, since Aa and aA happen half of the time (and Bb or bB ; and Cc or cC).

To find AA Bb Cc you would take 4 x 2 x 2 to get 1 in 16 probability.

All cases with one homozygous dominant (or recessive) trait will work out to 1 in 16. For example, Aa bb Cc is 1 in 16. The most common genotype is like the parents as Aa BbCc at 1 chance in 8 (2 x 2 x 2)

Let's look at another example

In the following cross AaBbcc x AabbCC, what is the probablitiy of obtaining offspring that

a) show all three dominant traits

b) show at least two dominant traits?

This problem can be tackled by analyzing the probability of each gene showing the dominant phenotype individually and then multiplying each answer together. This is the multiplication rule for independent events.

Aa x Aa
The probability of a dominant phenotype in this cross is 0.75

Bb x bb
The probability of a dominant phenotype in this cross is 0.50

cc x CC
The probability of a dominant phenotype in this cross is 1.00
(NOTE: a probability of 1.00 mean that the event will occur 100% of the time)

Now we multiply all the individual events probabilities together to find the probability that they will all occur at the same time.

0.75 x 0.50 x 1.00 = 0.375
0.375 is the probability of all three showing dominant phenotypes

Now for the answer to part B. We know that at least one gene (Cc) will show the dominant phenotype all of the time, so we can solve this problem by analyzing only the A and B loci. Once again we will use the multiplication rule for independent events, but this time we will calculate the occurance of the homozygous recessive phenotype occurring at the same time for traits A and B.

Aa x Aa
The probability of a recessive phenotype in this cross is 0.25

Bb x bb
The probability of a recessive phenotype in this cross is 0.50

Once again we multiply the probabilities.
0.25 x 0.50 = 0.125
0.125 is the probability of both A and B showing the recessive phenotype, so if we subtract this number from 1, then we will have the probability of A or B showing at least one dominant phenotype.

1.000 - 0.125 = 0.875

0.875 (7/8) is the probability that at least two dominant traits will occur.

Bio_man 8)

Here's another example to accelerate your understanding...

The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these four genes, what are the probabilities that F2 offspring would have the following genotypes?

a. aabbccdd =   1/4 x 1/4 x 1/4 x 1/4  = 1/256

b. AaBbCcDd = .5 x .5  x .5  x .5 = 1/16

c. AABBCCDD = .25  x .25 x  .25 x .25 = 1/256

d. AaBBccDd = .50  x .25 x .25 x .5 = 1/64

e. AaBBCCdd = .5  x .25 x .25 x .25 = 1/128

Just remember that the probability of a heterozygote (Xx) = 2/4 or 1/2 and the probability of a homozygote XX or xx = 1/4

confused_student, don't get so confused with how your textbook defines the variables and how this guy showed you. Just try to match up the variables until they are consistent with your textbooks. I think thats the only problem ur facing really

Thank you for helping bioman!  I understand how to calculate the probabilities if you want to know what are the chances of getting all recessive, or maybe two recessive traits and one dominant.  But what I want to know is how many 'genotypic classes' a cross like AaBb X aabb or AABBCc x AaBBCC.  What is n in either case?  if n is the number of genes, then n = 2 in the first case, but 3^2 does not equal four - and four genotypic classes was all I could find for the first cross. 

Okay, say you have a AaBBCCDD mated to aaBBCcdd. I found this example in my text book.

What are the expected genotypic proportions of the offspring (assuming no linkage)?

And this is how the textbook answered it:

AaBBCCDD x aaBBCcdd

Gametes formed: A, a, B, C, D a, B, C, c, D

Hence combining the different types of gametes one can find the genotypes that can be produced by the given combination of parental genotypes.

Genotypes of offsprings: AaBBCCDD, aaBBCCDD, AaBBCcDD and aaBBCcDD.

This can be formulated as follows:

Number of genotypes formed = 2^Number of characteristics that are heterozygous.

In this case only 2 characterisics are heterozygous hence the types of gametes formed will be 2 x 2 = 4.

I don't know ... If you had AaBbCcDdEe x AaBbCcDdEe, then your formula doesn't give me the number I was expecting.  2^10 = 1024 kinds of genotypes.  I thought 1024 was just the number of offspring.  243 was the number of genotypes because 3^5 = 243.

I still don't know how to find n so that I can figure out the number of genotype classes.  How many genotype classes will result from this cross?
AaBbCcDdEE x AaBbCcDdEE

I think maybe if n = 4
3^4 = 81 genotype classes. 

But what if the two individuals crossed did not both have 4 heterozygous genes?  what if one had two heterozygous genes, and three homozygous, and the other had four homozygous genes and one heterozygous gene. What is n in that case? 

So confused  :'(

Hey confused_student, when I get home, I will do some further research on this topic. Sorry.

Thank you bioman!!

That's it.  I'm pulling out the cookies again.  Chocholate chip fresh from the oven.  With nuts too! 
 Daskite onegaishimasu!!!!

Hey confused_student, I can definetlely see that this question is bugging you  :-\

after looking at duddy's response, his explanation seems legit. Going back to some of your previous points, you asked, for AaBb X aabb or AABBCc x AaBBCC.  What is n in either case?

The n value for the first cross would be 2 and 3 for the second cross because n is the number of genes in the cross (A, B, and/or C, respectively).

He also mentioned that there are two options for each allele (i.e. A or a / B or b / C or c) so 2n = gamete types, i.e. two alleles raised to the number of genes in the cross, (in the first case 2 genes.)

so 2^2 for the first cross should equal 4, as you mentioned: AaBb, Aabb, aaBb and aabb.

I'm guessing those cookies are getting cold by now. If you find the solution from your instructor, we would love to hear back from you.

Thank you for all the help you've given me so far.  I have learned a lot.  I'm sorry that I still don't understand.

Duddy's response was fabulous.  The case I present, however, may require some modification of what he said.
Look at the difference:
Case 1: AaBbCcDdEE x AaBbCcDdEE
Case 2: AaBbCcDdEe x AaBbCcDdEe
case 3: AaBbCcDDEE x AaBBCCDDEE

If n is always the number of genes present, then in all three cases n=5.  So, if this is true then in both cases the number of genotype classes is 3^5 = 243 for ALL.  Is this true?

I apologize for the repeated questions.  I sincerely do not mean to be irritating.  The prof has ignored my previous emails.  Should I make an enemy of myself and ask the department chair?  

Please note that I understand the following two points: 1. how to calculate the number of gametes formed.  That is 2^n where n = the number of heterozygous genes.  second point: I also understand how to calculate the individual probablity of obtaining a particular gene combination in a particular cross, such as what are the chances of getting four recessive and one dominant.  This is obtained by simply multiplying the probabilities of each case.

All I need to know is are the number of genotypic classes 243 for all three cases?

No, do not ask the department head! Trust me, you do not want to make enemies with these individuals because your grade is ultimately in their hands! It happened once to me in my undergrad, they are all connected. Chances are, your teacher doesn't even know! In fact, I find it very unlikely that they will ask this on an exam. When I took genetics, we rarely went past a dihybrid cross; it's ridiculous that they are having you do tetra and penta. I never heard of these until I saw it being discussed on this forum :P Just do what duddy said because he seems confident in his response. It makes total sense that n is the number of genes. When you see your professor next time, approach him after class but don't tell him who you are and that you emailed him previous, just present the case to him when class is done and ask if it will be asked in an exam situation.

Hey confused_student, I found this site that may help you!

Problem 3.17 on page 63 asks you to use the forked-line method to determine the outcome of a number of trihybrid crosses.

Hint: In using the forked-line method, consider each gene pair separately. For example, in this problem, first predict the outcome of each cross for A/a genes, then for the B/b genes, and finally, for the C/c genes. Then you are prepared to pursue the outcome of each cross using the forked-line method.

In crosses involving two or more gene pairs, the calculation of gametes and genotypic and phenotypic results is quite complex. Several simple mathematical rules will enable you to check the accuracy of various steps required in working genetic problems. First, you must determine the number of heterozygous gene pairs (n) involved in the cross. For example, where AaBb x AaBb represents the cross, n = 2; for AaBbCc x AaBbCc, n = 3; for AaBBCcDd x AaBBCcDd, n = 3 (because the B genes are not heterozygous). Once n is determined, 2n is the number of different gametes that can be formed by each parent; 3n is the number of different genotypes that result following fertilization; and 2n is the number of different phenotypes that are produced from these genotypes. Table 3.1 summarizes these rules, which may be applied to crosses involving any number of genes, provided that they assort independently from one another.

http://www.nicerweb.com/bio3400/Locked/qt/ebook/03_04.html (http://www.nicerweb.com/bio3400/Locked/qt/ebook/03_04.html)

For classic dominant/recessive genes An elaboration...

let n = the number of heterozygous gene pairs of one parent
2 to the nth power = the number of different types of gametes produced
2 to the 2n power = # of offspring
for example, with a trihybrid cross, n = 3
# of gametes 2^3 = 8
# of offspring = 2^6 = 64

Trihybrid phenotypic ratio 27:9:9:9:3:3:3:1 = 64 offspring
A_B_C_ => (3/4)^3 (1/4)^0 = 27/64
A_B_cc => (3/4)^2 (1/4)^1 = 9/64 x3 positions
A_bbcc => (3/4)^1 (1/4)^2 = 3/64 x3 positions
aabbcc => (3/4)^0 (1/4)^3 = 1/64

3 to the nth power = the number of different genotypes produced
# of genotypes = 3^3 = 27

Trihybrid genotypic ratio 1:6:15:20:15:6:1 = 64 = 2^6
from polynomial expansion of (P+q)^6
P = dominant value for allele
q = recessive value for allele

star thanks for providing that website, i took a look and I made a summary

Table. Simple Mathematical Rules Useful in Working Genetics Problems

Crosses between Organisms Heterozygous for Genes Exhibiting Independent Assortment

Number of Heterozygous Gene pairs Number of Different Types of Gametes Formed Number of Different Genotypes Produced Number of Different Phenotypes Produced

n   2n    3n     2n*

1   2      3       2
2   4      9       4
3   8      27     8
4   16    81     16

*The fourth column assumes that dominance and recessiveness are operational for all gene pairs.

I also looked further into that website and found some more insight:

Independent Assortment Leads to Extensive Genetic Variation

We have seen that for any individual, the number of possible gametes, each with different chromosome compositions, is 2n, where n equals the haploid number. Thus, if a species has a haploid number of 4, then 24 or 16 different gamete combinations can be formed as a result of independent assortment. Although this number is not high, consider the human species, where n = 23. If we calculate 223, we find more than 8 x 106, or more than 8 million, different types of gametes are possible. Because fertilization represents an event involving only one of approximately 8 x 106 possible gametes from each of two parents, each offspring represents only one of (8 x 106)2 or 64 x 1012 potential genetic combinations. This number of combinations of chromosomes is far greater than the number of humans who have ever lived on Earth! It is no wonder that, except for identical twins, each member of the human species demonstrates such a distinctive appearance and individuality. Genetic variation resulting from independent assortment has been extremely important to the process of evolution in all organisms.

thanks star and bioman!  I appreciate all your help. 
You're right, he's not going to ask about the penta cross.

You're welcome ;)

On a side note, I highly recommend you read this article Duddy published earlier today.

I highlighted the important sections; should explain what we couldn't explain :P

[See Attachment]

this actually solved a lot of the problems i was having in genetics... thx :)