Practice DAT from ADA bio question.

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amino

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1. Assuming no linkages, how many different kinds of gametes can be produced by an organism with the genotype AaBbcc? answer:4

can someone explain this answer?

2. In watermelons, the unlinked genes for green color G and for short length S are dominant over alleles for the striped color g and long length s. Predict the phenotypes and their ratios for the cross Ggss x ggSs.

answer: 1:1:1:1 green short:striped short: green long: striped long

Whats the best way to figure this out?

thanks
a

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amino said:
1. Assuming no linkages, how many different kinds of gametes can be produced by an organism with the genotype AaBbcc? answer:4

can someone explain this answer?

2. In watermelons, the unlinked genes for green color G and for short length S are dominant over alleles for the striped color g and long length s. Predict the phenotypes and their ratios for the cross Ggss x ggSs.

answer: 1:1:1:1 green short:striped short: green long: striped long

Whats the best way to figure this out?

thanks
a

1. AaBbcc has three components. The last component (cc) is homogenous, so you don't have to consider this one because all of the gametes will contain c. AaBb will segregate into gametes with AB, Ab, aB, and ab, resulting in 4 gametes.

2. For this question, I find it easier to tackle each allele individually and find the probability of getting each phenotype.
- Gg x gg will result in 1 green: 1 stripped ratio. (probability of green is 1/2, probability of stripped is 1/2)
- ss x Ss will result in 1 short: 1 long ratio. (probability of short is 1/2, probability of long is 1/2)

Then you find the probability of both phenotypes.
- Probability of Green and short is (1/2 * 1/2 = 1/4).
- Probability of Green long is (1/2 * 1/2 = 1/4).
- Probability of Stripped short is (1/2 * 1/2 = 1/4).
- Probability of Stripped long is (1/2 * 1/2 = 1/4).

This results in 1:1:1:1 ratio.

I hope this helps
 
amino said:
1. Assuming no linkages, how many different kinds of gametes can be produced by an organism with the genotype AaBbcc? answer:4

can someone explain this answer?

2. In watermelons, the unlinked genes for green color G and for short length S are dominant over alleles for the striped color g and long length s. Predict the phenotypes and their ratios for the cross Ggss x ggSs.

answer: 1:1:1:1 green short:striped short: green long: striped long

Whats the best way to figure this out?

thanks
a


for the first one, make the gametes and put them in the pundit squre, you will find that the total number of different gametes is only 4, ABc, Abc, aBc, abc. got that?

pundit square. these are kiddy problems. try not to get too much stressed about these things. good luck.
 
786mine said:
for the first one, make the gametes and put them in the pundit squre, you will find that the total number of different gametes is only 4, ABc, Abc, aBc, abc. got that?

pundit square. these are kiddy problems. try not to get too much stressed about these things. good luck.

pundit square... :laugh: :laugh: :laugh: I do apologize, but I can't help it. After studying for about ten hours at this point that is hilarious. My favorite SDN misspelling was from the pre-med forum though...someone spelled 'phlebotomist' as 'phlebotanist' made me think of maple syrup. :laugh:
 
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