## genetic cross

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### genetic cross

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

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

### Re: genetic cross

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 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

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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