Question

The genetic disorder Tratuism is caused by a recessive allele. A couple comes to you for genetic counseling. You determine that the father does not have Tratuism and is homozygous for the allele. You determine that the mother has Tartuism. What is the probability that their first child will have Tartuism? What is the probability that their second child will have Tartuism?

Answer 1

Some hints:
1.) Tratuism is recessive (What genotype must be present in order for the disorder phenotype to occur?)
2.) From the passage, you know that the father does not have tratuism and homozygous, so he must homozygous ______________
3.) The mother has tratuism, so she must be ___________ __________ (Same answer to the first hint)

Then just make a punnett square to calculate the probability for the first child.

Hint for the second child: Does the genotype of the first child have any effect on the probability of the genotype of the second child? Think about it this way, if you have an older sibling, do their traits affect the inheritance of your own?

Let me know if you need more help. =)

Answer 2

I apologize but I am just lost!

Answer 3

So what part are you lost on? Do you have any idea on where to start?

Answer 4

Nope not a clue.

Answer 5

Well, we can start at the beginning then. Do you understand the terms recessive and dominant?

Answer 6

disorders are recessive

Answer 7

in this particular case, the disorder is recessive, but I was talking about the general definition for dominant and recessive regarding a pair of alleles. In a pair of alleles, a allele can either be dominant or recessive. Dominant alleles mask the expression of the recessive alleles. So basically, in a pair of alleles, you have three possibilities:

Let A = Dominant Allele, a = recessive allele
AA = homozygous dominant
Aa = heterozygous
aa = homozygous recessive

In the first and second cases, the trait of the dominant allele is expressed (remember that if there is at least one dominant allele in the pair, its trait will mask the recessive allele). In the third case, the trait of the recessive allele is expressed (there is no dominant allele to mask it).

So, now that we know the definition, the disorder is recessive, which means that the genotype for the disease has to be homozygous recessive (or aa). So, thats basically the answer to the first hint I gave you.

Answer 8

Now think about the third statement in the passage " You determine that the father does not have Tratuism and is homozygous for the allele."

Remember the first and second cases I talked about above? The father doesn't have the disorder and so he has to have at least one dominant allele to mask the expression of the recessive allele. In addition, he is homozygous, which means he has the same pair of one type of allele. So, that means he must be __________ ___________ (hint your answer is between homozygous dominant and heterozygous)

Answer 9

Did you understand what I've written so far, or have I confused you some more?

Answer 10

AA

Answer 11

Yeah, thats right.

So, the father is homozygous dominant. Now what about the mother? She has the disorder so her genotype must be __________ ___________

Answer 12

Remember my third hint above

Answer 13

Would that make her Aa

Answer 14

Not quite, she has to be homozygous recessive so (aa). She can't be heterozygous (Aa), since that would make her not have the disorder (remember the disorder is recessive).

Answer 15

Now that you know both the genotypes of the father and mother, you can make a punnett square. Do you know how to make one?

Answer 16

Got that part

Answer 17

Would it be four square

Answer 18

Yes, since there are only 2 alleles. Number of squares = 2^n, where n = # of alleles

Answer 19

So it is four square AA on the left and aa on the top right

Answer 20

Yep

Answer 21

And remember, we are looking for genotypes that express the disorder. So the children who are affected must be homozygous recessive (aa).

Answer 22

so in doing the square I come out with Aa in each box would that be correct

Answer 23

Yep, and that gives you a probability of?

Answer 24

1 in 4

Answer 25

or do I have that backwards

Answer 26

Well remember, we are looking for aa not Aa. You have 0 AA, 4 Aa, 0 aa. So the probability is:

# of offsprings with traits you are looking for (which is aa)/total # of possibilities (4), so that is:

0/4 = 0% probability

There is no way to obtain a homozygous recessive offspring from a homozygous dominant father and homozygous recessive mother.

Answer 27

So than it must be 0% for their sec on child as well correct

Answer 28

yep, since the two events are independent of each other, the probability of the first won't affect the second. You got it!

Answer 29

So basically, for these types of problems you need to identify the genotypes of the parents, then use a punnett square to find the possible genotypes for the offspring. Then you just solve for the probability by dividing the # of offspring with the traits you are looking for by the total # of possibilities.

Answer 30

So than if you have a dominant allele and the father has a heterozygous allele and so does the mother would that mean there is a 50% probability that a child will have the disorder

Answer 31

So I'm interpreting your question as "if the disorder was dominant and the father was heterozygous, would the probability be 50%".

Yes, you would be right. Since the father would be Aa and the mother aa the children would be Aa Aa aa or aa, which leads to a probability of 50%.

Answer 32

If you have a dominant allele and both the mother and father are heterozygous for the allele it is a 50% percent probability is that right

Answer 33

75% actually, because now its Aa x Aa = AA, Aa, Aa, and aa. Since, its dominant, we are looking for AA and Aa, so its now 3/4 = 75%

Answer 34

so if its 75% that a child will than its a 25% chance they will not is that correct

Answer 35

Yep

Answer 36

Thank you so much for your help it is very much appreciated!!!

Answer 37

no problem

Answer 38

So how would you come to the genotype for a green-eyed fly and green-eyed male fly?

Answer 39

Thats male and female green-eyed flies

Answer 40

Well you would have to know if what genes are involved before you can attempt the problem, and which traits for that gene are dominant/recessive.

Answer 41

basically, you need more information to find the genotype

Answer 42

This was the question and why I am confused: You breed a green-eyed female fly with a green-eyed male fly. This results in 50 offspring as follows: 13 red-eyed and 37 green-eyed. Please use the letters "G" and "g" to denote the genotypes. a) what is the genotype of the female fly? b) what is the genotype of the male fly? c)which color eye is dominant?

Answer 43

Ah, now we can actually solve this.

Answer 44

Great Please help me to understand this.

Answer 45

Well what are the possibilities for the parents? Remember, they can either be homozygous dominant, heterozygous or, homozygous recessive. GG, Gg, or gg. Three possibilities for each of the parents. The easiest way to find which parents genotypes correspond to the offspring distribution given is to test each one (There should be 9 choices, since 3 possibilities for father and 3 possibilities for the mother):

Father = GG; Mother = GG
Offspring = GG, GG, GG, GG

Father = Gg; Mother = GG
Offspring = GG, GG, Gg, Gg

Father = gg; Mother = GG
Offspring = Gg, Gg, Gg, Gg

Father = GG; Mother = Gg
Offspring = GG, Gg, GG, Gg

Father = Gg; Mother = Gg
Offspring = GG, Gg, Gg, gg

Father = gg; Mother = Gg
Offspring = Gg, gg, Gg, gg

Father = GG; Mother = gg
Offspring = Gg, Gg, Gg, Gg

Father = Gg; Mother = gg
Offspring = Gg, Gg, gg, gg

Father = gg; Mother = gg
Offspring = gg, gg, gg, gg

Now, just calculate the probabilities for each one, and find the one that matches the offspring distribution.

Answer 46

If you look at the offspring distribution its 13 red-eyed vs. 37 green-eyed, which is approximately 25% red-eyed and 75% green-eyed. So we need to find a distribution that shows a similar distribution to this.

Answer 47

I mean genotype combination that shows a similar distribution to this.

Answer 48

To save time, I'll give you a hint the father's genotype should be Gg

Answer 49

so would that than make the mother gg

Answer 50

Gg and gg make a 50/50 distribution we are looking for 25/75 or 75/25. Try again.

Answer 51

Gg

Answer 52

Yep, thats right

Answer 53

Gg and Gg give you 3 dominant expressed phenotypes and 1 recessive expressed phenotype

Answer 54

so to figure the dominant eye color how would you figure that out

Answer 55

So compare it to our distribution. There are 75% green-eyed and 25% red-eyed. We have 75% dominant and 25% recessive, so which correspond with which?

Answer 56

So than the dominant color is green

Answer 57

Yep, and now you've solved the problem

Answer 58

So summarizing, you find the genotype of the parents that yields a offspring distribution as given in the problem, and then compare to find dominant/recessive

Answer 59

You are the absolute best and a life saver!!!!

Answer 60

Glad to help