True Breeding Drosophila

View previous topic View next topic Go down

True Breeding Drosophila

Post by Alla on Fri Mar 14, 2008 9:06 am

Data:
_______Male - Female
WT_____42 - 74
CR _____12 -26
P_______32 -0
P, CR___14 - 0

First, examine relationship to determine if 2 mutations are linked. Since CR (curly) and P (purple) can occur independently, they are not on the same chromosome.

Second, CR occurs equally in males and females 26:26 (or 1:1) and, therefore, is not sex linked. Now, examine the numbers of non-CR and CR, they are 148:52, which is approximately 3:1 of wt:mutant (for this particular trait). This ratio indicates that mutation CR is recessive.

Finally, examine P mutation. You can see that it occurs only in males. That indicates that P is sex linked. Now, if the mutation was Y-linked dominant, all males would have P mutation. That is not the case. If P was Y-liked recessive, none of the males would show this trait, which is not the case either. So P must be X-linked. If that is the case, the number of P males should be equal number of non-P males, in this case it is 48:54, close to expected 1:1. To find out if P is dominant or recessive, we need to compare number of non-P to number of P mutants, which is 148:48 in this case, close to 3:1 ratio which indicates that P is recessive. However, if P is X-linked recessive we expect 25% of females to have P mutation as well, which we do not see here. That can be explained by the presence of the third mutation - X-linked female lethal (L) recessive mutation linked to P. In that case every time female inherits two chromosomes caring P, it dies (never becomes a fly) and this mutation would only affect females.

In summary:

CR - recessive autosomal (A=WT; a=CR)
P - recessive X-linked (B=WT; b=P)
L - recessive X-linked (C=WT; c=L)***females only!

Cross male Aab-c- to female AaBbCc: note that "-" represents absence of second copy of the gene, and remember that Band C inherited together.
______ABC________aBC_________Abc____________abc
Abc AABbCc (WT) AaBbCc (WT) AAbbcc (P, no fem) Aabbcc (P, no fem)
abc AaBbCc (WT) aaBbCc (CR) Aabbcc (P, no fem) aabbcc (P/CR, no fem)
A-- AAB-C- (WT) AaB-C- (WT) AAb-c- (P, no fem) Aab-c- (P, no fem)
a-- AaB-C- (WT) aaB-C- (CR) Aab-c- (P, no fem) aab-c- (P/CR, no fem)

Phenotypes: 6 WT: 2 CR: 3P (males): 1 P/ CR (males)
Expected: 100 WT: 33 CR: 50 P (males): 17 P/ CR (males)
Note: female lethal phenotypes counted as ; therefore, 6P (no fem) counted as 3 males, thus, for calculation of expected values 16 squares treated as 12.

Test: 116 WT: 38 CR: 32 P (males): 14 P/ CR (males)
It is not identical to theoretical, but I think it would survive the Chi square test.

If you need help with that, let me know.

I would appreciate it, if you registered to this site and used it directly.
Good luck!


Last edited by Alla on Fri Mar 14, 2008 10:08 pm; edited 1 time in total

Alla
Instructor
Instructor

Posts : 103
Join date : 2008-02-27

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by noor123 on Fri Mar 14, 2008 11:16 am

Hi Alla

I really appreciate the detailed reply. But I've got a couple of problems.
If that is the case, the number of P males should be equal number of non-P males, in this case it is 48:54, close to expected 1:1. To find out if P is dominant or recessive, we need to compare number of non-P to number of P mutants, which is 148:48 in this case
Shouldn't that be 46? I did this: 32 + 14 which gave me 46. Or am I missing something? Also concerning the second ratio "148:48" , I am not exactly sure as to what we are supposed to add. The way I did it, according to your method: non-P : P mutants => (42+74+12+26) : (32+14)
154: 46
:/

Ok and I see that you've done the 16-square dihybrid cross. I have a question. How did we arrive to the conclusion that our "expected" values are these?
Expected: 100 WT: 33 CR: 50 P (males): 17 P/ CR (males)

Bear with me please. I am a slow learner. pale

Can I take a different approach to this though? this is what I have,
http://img204.imageshack.us/img204/8559/picture126py7.jpg

I don't even know if that makes sense or not, but I am on the verge of collapsing. I don't know what to do.

It'd be awesome if you could perhaps clarify on the points above. It'd be greatly appreciated.

Thanks for the taking the time again,

Regards
Noor

noor123
Student
Student

Posts : 3
Join date : 2008-03-14

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by Alla on Fri Mar 14, 2008 9:28 pm

Sorry, my math was going a little last night (and 148 was non-CR, not non-P). I am glad you caught that. However, 154:46 is still close to 3:1, so conclusion stands.

I double checked today and it turns out that you do not have to bother with the whole lethal mutation idea. If F1 male had wt allele for P, then your ratios are as expected and absence of P females is expected, since in that case females cannot be homozygous for P. (I don't know why I made it so complicated last night!)

____AB________Ab_______aB_______ab
AB AABB (WT) AABb (WT) AaBB (WT) AaBb (WT)
AY AABY (WT) AAbY (P) AaBY (WT) AabY (P)
aB AaBB (WT) AaBb (WT) aaBB (CR) aaBb (CR)
aY AaBY (WT) AabY (P) aaBY (CR) aabY (P/CR)


http://img183.imageshack.us/my.php?image=dhcke8.png

In this case, you expect 9 WT: 3 CR: 3 P: 1 P/CR (total 16)

Expected numbers are calculated by taking the total number of individuals in experiment (in this case 200), dividing it by the total from empirical ratio (16 in this case) and multiplying by phenotype number in the ratio. Thus: expected WT = (200/16)x9 = 112.5; CR = 37.5, P = 37.5, and P/CR = 12.5 almost exactly what you have in the observed data. And much simpler explanation.

I just did a chi-square test and p come out somewhere between 0.80 and 0.90 for this scenario. Therefore, this hypothesis must be correct.

P.S. Two traits require dihybrid cross unless they are linked (which they are not in this case).

Alla
Instructor
Instructor

Posts : 103
Join date : 2008-02-27

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by noor123 on Sun Mar 16, 2008 6:05 am

Thanks for the clarification.

I treated them separately, is that ok? but I am still having problems with the purple eyes trait.

Here goes:
The curly wings:
http://img213.imageshack.us/img213/1638/cwlo3.jpg

Worked perfectly.

But the Purple eyes:


the numbers are far from what the expected values are.

I see it a bit easier when I do them separately.
P.S. Two traits require dihybrid cross unless they are linked (which they are not in this case).
I am still unsure as to whether I could do them separately or not. Because in the example our TA provided, she treated them separately, and THEN did the dihybrid cross. Suspect

I am very grateful for your help.

noor123
Student
Student

Posts : 3
Join date : 2008-03-14

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by Alla on Mon Mar 17, 2008 12:32 am

I see your problem: you made a mistake calculating expected values. When you calculate for autosomal mutations and want to get male and female expected values you have to divide by 2 (as you did for curly) because only of each phenotype is male and the other half is female. However, when you calculating male and female for the sex-linked mutations, you do not need to divide by 2 they are already divided in the cross (anything that has Y chromosome in the cross is male). Therefore, your expected values should be:

Male P+ = 200 (1/4) = 50
Female P+ = 200 (1/2) = 100
Male P = 200 (1/4) = 50
Female P = 0

Total expected should be = 200 (and total of your calculated values was 100). When you make that correction chi-squared test would work out just fine.

As to the dihybrid cross:

What I really meant was that you cannot do a simple cross for two traits at the same time (like you tried to do in the previous message), but you can do two simple crosses if you like. However, in either case you need to make a dihybrid cross at the end (sooner or later is really up to you).

For chi-squared test you can do separate calculations as you are doing now (df=1) or combine them (df=3; WT + CR + P + P/CR 1) your result should remain the same. It is your decision which method you prefer.

Alla
Instructor
Instructor

Posts : 103
Join date : 2008-02-27

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by noor123 on Mon Mar 17, 2008 5:13 am

THANK YOU EVER SO MUCH! Very Happy

YOU'RE THE BEST!!!!!!!!!!!

noor123
Student
Student

Posts : 3
Join date : 2008-03-14

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by Alla on Tue Mar 18, 2008 12:55 am

You are welcome!

Alla
Instructor
Instructor

Posts : 103
Join date : 2008-02-27

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by linuxux on Wed Jul 29, 2009 12:05 am

There is one thing I don't get here. In the first part you say that CR is not sex linked because it affects males and females in a 1:1 ratio. But if two P alleles are lethal to females, how can you say CR is a 1:1 ratio when females which would have CR die? Saying 1:1, as you mentioned before, is saying not sex linked. This is a claim about the location of the traits: they are not on sex chromosomes. But this claim cannot be made if females which would affect the 1:1 ratio die off. Certainly two P's may be lethal. But what if there are 20 P, CR females that die? Does this not skew the ratio, and thus the evidence for not being sex linked?

*Never mind, I see you cleared it up with a follow up post.

linuxux
Student
Student

Posts : 1
Join date : 2009-07-28

View user profile

Back to top Go down

Re: True Breeding Drosophila

Post by Sponsored content


Sponsored content


Back to top Go down

View previous topic View next topic Back to top

- Similar topics

 
Permissions in this forum:
You cannot reply to topics in this forum