genetic cross

Genetics as it applies to evolution, molecular biology, and medical aspects.

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mariposa
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genetic cross

Post by mariposa » Mon Apr 17, 2006 5:19 am

hello,
Across is made between two parents, if there are 64 offspring and 16 of those have curly hair and long fingers, what are the parents genotype?

1- HhSs x hhss
2- HHSs x hhss
3- Hhss x hhss
4- HhSS x Hhss
5- hhss x hhss

i know straight hair H, is codominant with curly hair h, and short fingers S is dominant over long fingers s, but i still not sure about the correct answer, im linning toward number 1, maybe someone can help me,
thank you

kiekyon
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Re: genetic cross

Post by kiekyon » Mon Apr 17, 2006 11:58 am

mariposa wrote:hello,
Across is made between two parents, if there are 64 offspring and 16 of those have curly hair and long fingers, what are the parents genotype?

1- HhSs x hhss
2- HHSs x hhss
3- Hhss x hhss
4- HhSS x Hhss
5- hhss x hhss

i know straight hair H, is codominant with curly hair h, and short fingers S is dominant over long fingers s, but i still not sure about the correct answer, im linning toward number 1, maybe someone can help me,
thank you


so, the progeny would hv sshh and ratio of 1:3

i think it is 3

i would be easier to see if u draw out the punnett square or each answer

sdekivit
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Post by sdekivit » Mon Apr 17, 2006 1:19 pm

curly hair and long fingers: hhss

--> cross 1 will yield hhss with a probability 1/4 (1/2 * 1/2)

--> cross 2: impossible to yield hhss

--> cross 3: probability is 1/2 (1/2 * 1)

--> cross 4: impossible to yield hhss

--> cross 5: probability = 1 (1 * 1)

so 1 is indeed correct: 16/64 = 0,25

kiekyon
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Post by kiekyon » Tue Apr 18, 2006 1:38 pm

sdekivit wrote:curly hair and long fingers: hhss

--> cross 1 will yield hhss with a probability 1/4 (1/2 * 1/2)

--> cross 2: impossible to yield hhss

--> cross 3: probability is 1/2 (1/2 * 1)

--> cross 4: impossible to yield hhss

--> cross 5: probability = 1 (1 * 1)

so 1 is indeed correct: 16/64 = 0,25


i dont understand ur calculation..
can u show us how u get those probability?
however, i made the punnett square and it fit your answer perfectly
:?: thanks for the correction anyway

mariposa
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Post by mariposa » Wed Apr 19, 2006 12:03 am

i just want to say thanks alot, for taking the time to help me

sdekivit
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Post by sdekivit » Wed Apr 19, 2006 7:45 am

kiekyon wrote:i dont understand ur calculation..
can u show us how u get those probability?
however, i made the punnett square and it fit your answer perfectly
:?: thanks for the correction anyway


basic mathematics:

assume the cross AaBB(1) * aaBb(2)

now what are the probabilties?

parent 1 will have gamete A or a for trait 1 with both a probabilty of 1/2. parent 2 will have only a with a probabilty of 1.

same for trait B: parent 1 has gamete B with probabilty 1 and parent 2 B or b with probabilty 1/2

Now assume we get AaBB as child: chance of Aa = 1/2 * 1 = 1/2 (basic chance: first 'drawing' is an A and the second drawing is a)

--> cance of BB = 1 * 1/2 = 1/2

thus getting AaBB = 1/2 * 1/2 = 1/4 (first drawing is Aa and the second BB)

this all leads to the following 'probabilty-tree':

....................BB(1/2) -------------------> AaBB = 1/2 * 1/2 = 1/4
.................../
........Aa(1/2) --> Bb(1/2)----------------> AaBb = 1/2 * 1/2 = 1/4
......./
...../
.....\
.......\
........aa(1/2) -->Bb(1/2)-----------------> aaBb = 1/2 * 1/2 = 1/4
...................\
....................BB(1/2)---------------------> aaBB = 1/2 * 1/2 = 1/4

Now in this case hhss is the child out of HhSs(1)xhhss(2)

--> probabilty of hh = 1/2 (chance = 1/2 for h from parent 1 and 1 for parent 2)

--> probability of ss = 1/2 (chance = 1/2 for parent 1 and 1 for parent 2)

Combined chance for hhss: 1/2 * 1/2 = 1/4

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