Bruce P. answered • 10/11/17
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20+ year college biology/genetics teacher; I want you to understand.
Hello, Sonia!
The question here starts with 'what is independent assortment'. It's a fancy term for the claim that what happens in the distribution of one locus (a locus is the 'home' or location for a gene) is not related to what happens at another locus. So here, you are being asked to use a Punnett square in support of the proposition "What happens with the alleles A1 and A2 does not determine/is not influenced by what happens with alleles B1 and B2"
Note that a Punnett square is just a way of 'tracking' what happens when the parents are creating gametes (sperm and eggs). The row labels (top) are the genotypes of gametes (sperm, eggs) that one parent makes; the column labels are the gametes the other parent creates.
The way you create the Punnett square represents certain assumptions, and the generated genotypes are a form of prediction. So if the assortment of A and B is independent, then you would predict that no matter which A allele gets put into a gamete, there would be equal likelihood of getting B1 or B2. In other words, you predict that the 4 possible gamete genotypes generated from a heterozygote (genotype A1/A2, B1/B2 =>A1B1, A1B2, A2B1, and A2B2) would each be equally likely.
So you would create a Punnett square with those genotypes as the row and column 'header' contents. The rest of the Punnett square represents the possible outcomes if the gametes 'meet'--so if A1B1 'meets' A1B1, you generate an offspring A1/A1,B1/B1
IF independent assortment applies here, THEN the likelihood of each of the gametes is equal, and each resulting diploid square is equally likely. So your Punnett square reveals the PREDICTED OFFSPRING from such a cross.
You then compare the PREDICTIONs of the "random assortment" model in the table to a TEST (actual cross performed between individuals of genotype A1/A2,B1/B2 x A1/A2, B1/B2 and see if the predictions of the Punnett square are met by the outcome in the real world.
IF the loci are NOT independently assorting, then there will NOT be an equal distribution of gametes. If, for example, the parental chromosomes were
----------A1------B1------
and
----------A2------B2------
(i.e. A gene and B gene are 'linked') then there would be a disproportionately HIGH number of gametes that got A1B1 or A2B2, whereas gametes A1B2 and A2B1 would be relatively rare--because they require a crossing over event in the small space between the A and B genes.
Bruce P.
Aha!!! Well done, Sonia! This looks like a well-executed answer and solid reasoning. As to the symbolism and co-dominance--YES, that's the usual usage and meaning... but it's always best to make sure that's what the instructor/writer of the question intended. But genotypically speaking, it looks like you've generated sensible outcome from the Punnett square (note that if this is a case where you are supposed to 'show your work' then the Punnett square should be part of the formal answer).
Nice job!
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10/11/17
Sonia B.
Hello :)
Can u help me with this?
Among the traits that Mendel studied in peas, he determined that the round allele for shape (W) was dominant to the allele for wrinkled (w) peas. Thus, Ww plants produce round peas. You have planted a mix of pea plants from seeds you obtained at the gardening store. You plant 100 plants and after they make pods you look at the peas contained in those pods. From your 100 plants, 25 have wrinkled peas and 75 have round peas. Of those round pea plants, how many do you expect to have a WW genotype? Show your working.
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10/25/17
Bruce P.
Hi, Sonia! This question is asking for two skills.
1) Making a Punnett square. Remember, a Punnett square is simply a way of keep track of the GAMETES each parent can make (or more specifically, what the genotypes of the gametes are). This is what goes in the top row and left column. The rest of the table predicts what happens when each combination of gametes meet (i.e. when sperm fertilizes egg). So if a parent is Ww, then it can make W gametes as well as w gametes; these would be the OUTSIDE of the Punnett
2) From the definition of dominance, we know that both WW and Ww peas genotypes will produce the W phenotype (ROUND phenotype).
This problem is a little weird in that it IMPLIES, but does not STATE that the peas you picked up at the store were generated by a cross of Ww x Ww. So it is either a trick question (you can't tell bc you don't know what the parents were; maybe some of the peas have different parents than others) OR you are supposed to assume the cross is Ww x Ww.
My guess is you should assume Ww x Ww.
Make your Punnett square for this cross; hopefully you will see that 3 of the 4 squares would have round (W) phenotype. The problem is asking "Of the ones with round phenotype, how many are round because their genotype is WW and how many round because genotype is Ww?"
Cheers
Bruce
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10/25/17
Sonia B.
100 progeny- 25 wrinkled, 75 round
Of those 75; 50 will be heterozygous and 25 will be homozygous dominant?
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10/25/17
Bruce P.
Looking good! Nice work!
(again, assuming we're correct in thinking these peas came from Ww x Ww parents)
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10/25/17
Sonia B.
Hello :) thnks for that
can u help me with this one?
If a population comprised 52 AA, 114 Aa and 34 aa individuals and the ‘A’ allele was dominant, what would be the
allele, genotype and phenotype frequencies (i.e., as a percentage)?
Hint: Since we have given you the genotypes, you can see that this is not a Hardy-Weinberg situation.
2.
allele, genotype and phenotype frequencies (i.e., as a percentage)?
Hint: Since we have given you the genotypes, you can see that this is not a Hardy-Weinberg situation.
2.
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10/25/17
Bruce P.
Hi, Sonia! This one is just asking you to process the information and convert to convent format...
So, recalling that in all cases, percent is (number of item divided by total items times 100)
--you are told the 3 genotypes, so you just need to figure out how many TOTAL individuals there are, and then what % are AA, etc.
For PHENOTYPES, you'll need to convert genotypes to phenotypes; this is the reverse of what you just did (AA and Aa both give 'A' phenotype)
For ALLELES, you'll need to total up the number of A (dominant allele) and a (recessive allele). Of the 52 AA individuals, how many A would this be? How many A are in the 114 Aa individuals? etc.
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10/25/17
Sonia B.
52 AA, 114Aa, 34aa
AAA --> 52+52+ 114/ 200 = 1.09 = p
aaa --> 114+ 34 + 34/ 200 = 0.91 = q
= p2 + 2pq + q2
= 1.18 + 1.98 + 0.82
Should I divide it by 200?
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10/25/17
Bruce P.
total alleles is the correct denominator, since you are putting alleles in the numerator.
So you should divide by (52 * 2) + (114 * 2) + (34 * 2)
The other clue is that the FRACTIONS should add up to 100% or 1.0
Since your fractions (1.09 + .91) add up to 2, you are indeed missing a factor of 2, and your proposal is a good one!!
Note that in the problem description, you are told that your population is NOT in hardy-weinberg equilibrium, so you're not required to do the p2 + 2pq + q2 check, but it is good that you are doing it, and you should look and see what you get when you adjusting your numbers!
Also, note that the question wants you to answer using %; your answer is certainly a correct format, but it's not the specific format being asked for!
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10/25/17
Sonia B.
(52 * 2) + (114 * 2) + (34 * 2)
okay I am still confused what formulas do I need to work out
allele frequency: (%)
genotype frequency (%)
phenotype frequency (%)
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10/25/17
Sonia B.
also can u assist me with this one
you put 5 male and 5 female wild type (vg+/vg+) flies, with 5 male & 5 female vestigial winged (vg-/vg-)
flies into your control bottle.
Assuming equal fitness, what would you expect the phenotype ratio to be (wild type to vestigial)?
flies into your control bottle.
Assuming equal fitness, what would you expect the phenotype ratio to be (wild type to vestigial)?
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10/25/17
Sonia B.
If the two genes assort independently the offspring of this cross will be:
1/4 A1A1 B1B1
1/4 A1A1 B1B2
1/4 A1A2 B1B1
1/4 A1A2 B1B2
If you find the four types of offspring in a ratio that is not 1:1:1:1, then the two loci are not independently assorting, and are linked.
10/11/17