Biology Forum Genetics HELP! Dihybrid crossing – not sure I understand??

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• #5734
MariaKsk88
Participant

I’m studying grade 12 biology on my own at home since I just had my son… My teacher told me to read the chapter on genetics and she gave me a bunch of exercises to do, and I did the ones on dihybrid crossing the best I could, but I’m not sure I understood it right. For example, the first exercise reads the following:

"A hereditary disease exists among humans called phenylketonuria, which is caused by an autosomal recessive mutant allele. Two parents are heterozygous for this condition. One parent has blood group O and the other has blood group AB. Answer the following:
a) Their first child is a son with blood group A and phenylketonuria. What was the probability for that to occur?
b) What was the probability that his future siblings were to be a brother and a sister who are carriers of the disease and have blood group B?
c) What is the probability for the family to receive a completely healthy male member with blood group O?"

I’m supposed to use Punnetts grid to explain so I tried to draw one but this is where I’m not sure I’m right. What I did was that, since the parents genotypes are Ppii and Pp(IA)(IB) (if you know what I mean with the (IA) and (IB)).. I drew a grid with 16 squares, put Pi, pi, Pi and pi on one side, and P(IA), P(IB), p(IA) and p(IB) on the other and then did the math.. Is that how you are supposed to do it?

If that’s right, I suppose my answers are right too…
a) probability 1/8
b) 1/4 chance for ONE child, so 1/16 for a brother AND a sister
c) They can’t have a child with the blood group O

—————–

Now there’s the next exercise which took me days to understand… It reads:

"Some plants of the white clover Trifolium repends produce hydrogen cyanide (HCN) when damaged and are sai to be cyanogenic. Acyanogenic plants do not produce HCN. Cyanogenesis depends on the presence in the leaves of certain glucosides, which can be broken down to release HCN. The presence of these glucosides is controlled by the dominant allele G. The genotype of the acyanogenic plants is gg. Cyanogenesis also depends on the presence of an enzyme, linamarase, which breaks down the glucosides. The presence of linamarase is controlled by the dominant allele L. Plants with the genotype ll produce no linamarase. In an experiment, 1600 plants were collected, being the progeny of crosses among all AaLl plants.
a) How many plants would be expected to be cyanogenic and how many acyanogenic?
b) Leaves taken from the acyanogenic plants were individually crushed, and purified glucosides were added. How many of these plants would now produce HCN?
c) Leaves taken from the acyanogenic plants were again individually crushed and the purified enzyme linamarase was added. How many of these plants would now produce HCN?
d) How many plants would be expected to remain acyanogenic at each treatment mentioned above?"

I tried to do the Punnett grid by writing LL, Ll, lL and ll on one side and GG, Gg, gG and gg at the other.. Is that correct? Then I found these answers:
a) 900 plants are cyanogenic (those with both G and L in their genotype) and 700 are acyanogenic.
b) 300 plants would produce HCN if you added glucosides.
c) 300 plants would produce HCN if you added linamarase.
d) 100 plants (the ggll) would remain acyanogenic, but they would produce HCN if you added BOTH glucosides and linamarase.

So… Sorry this was so long, but if anyone could let me know if I understood it correctly, and help me explain if I didn’t, I would reallllllly appreciate it! I study social sciences so I’m not too good at stuff like this hehe…

/Daphne

• #55005
sdekivit
Participant

for the first question reagrding phenylketonuria and the bloodgroup system ABO:

a) probability for A = 1/2 (and thus type Ai) and probability for PKU = 1/4

–> thus indeed 1/2 * 1/4 = 1/8

b) also probability for B = 1/2 (Bi) and to be a carrier = 1/2 ( = 2/4)

–> probability brother and sister are carrier + B = 2 * (1/2 * 1/2) = 1/2 (note that these are separate probabilities that you must add, not multiply)

c) completely healthy = PP = probability = 1/4 blood type O can only be found in the second generation (due to the AB of one parent). Note that your answer is incorrect. They can get a perfectly healthy member with blood type O but it can’t be a child of the given parent. The children are always Ai or Bi so someone outside the family who marries genotype that has the i allele can give a child in the second generation that has ii ( = O)

• #55055
MariaKsk88
Participant

Oh, thank you! I understand the c question but not the b… I suupose they want the probability of having first a brother, and then a sister? In that case, why shouldn’t I multiply 1/4 by itself? If you add the two probabilities together it will make it more probably for the family to get a brother AND a sister, than just one child.. That doesn’t make sense to me. ðŸ˜•

• #55174
sdekivit
Participant
quote MariaKsk88:

I suupose they want the probability of having first a brother, and then a sister? If you add the two probabilities together it will make it more probably for the family to get a brother AND a sister, than just one child

no the probability to occur the traits is higher when more children are born from the same parents.

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