Answer key to practice problems--2000 week 1
| 1. | If B is the allele for brown eyes and b is blue, the genotype of the blue-eyed man is bb. His wife must be a Bb heterozygote--she must have a B allele (she has brown eyes), but she must also have inherited a b allele from her blue-eyed mother (whose genotype was bb). Therefore, the "cross" is Bb x bb; half the children from such a cross are expected to show the dominant (brown-eyed) phenotype. (Draw a Punnett squares diagram if you're not convinced.) |
| 2. | The brown-eyed man's here could either be homozygous BB or heterozygous Bb. If he were heterozygous, we'd expect to see a 1:1 ratio of brown- and blue-eyed children (because his wife is homozygous recessive, bb). However, all that says is that each child has a 50% chance of being brown-eyed--because each child is an independent event, it is possible to get 10 brown-eyed children in a row even if the father is heterozygous (just as it is possible to flip a coin and get ten 'heads' in a row). So at this point, we cannot say with certainty determine the man's genotype.
However--the 11th child is blue-eyed, and therefore must have received a recessive allele from her/his father. Therefore (barring some aberrant event), the father must be heterozygous (Bb). |
| 3. |
You can tell right away if an animal has the dominant allele--it'll have yellow fur. Therefore if you simply remove the yellow-furred animals, the animals left behind (the black-furred ones) must all be homozygous recessive. [Removing the black-furred ones would not work--there could be heterozygotes left behind.] |
| 4. |
The Manx allele (M) is lethal when homozygous. and gives a dominant (tail-less) phenotype:
The cross Mm x Mm is expected to give a 1:2:1 ratio of MM : Mm : mm animals. The MM ones die, giving the observed 2:1 ratio. |
| 5. | (a) | Two crosses show that red must be the dominant phenotype. In the first cross, there are yellow progeny of red parents, so the yellow allele must have been present in the red parents (hence, yellow must be recessive). In the third cross, red and yellow parents have only red progeny, again indicating that that the red phenotype must be dominant. |
| (b) | Using R for the allele for the dominant (red) phenotype and r for the recessive (yellow) --
Cross 1: red x red The progeny show the dominant (red) and recessive (yellow) phenotypes in 3:1 ratio. Therefore, the parents must both be heterozygotes -- Rr. The progeny must be RR:Rr:rr in 1:2:1 ratio; the red progeny are R_ and the yellow progeny are rr. Cross 2: red x red The progeny are all red, so both parents must be homozygous dominant (RR), in which case all the progeny are also homozygous dominant, or one parent is homozygous dominant and the other is heterozygous (Rr), in which case the progeny are also RR and Rr in 1:1 ratio. |
| 6. | (a) | Brown is dominant, white is recessive -- because the F1 progeny were all brown. |
| (b) | If B = brown and b = white, the parents must have been BB (brown) and bb (white). We know that the white parent must be homozygous recessive (else it wouldn't have been white); the brown parent must also have been homozygous, otherwise we would expect to see a 1:1 ratio of brown and white F1 progeny. | |
| (c) | The third cross is the only one. Let's consider each cross --
P brown x F1 brown: This cross is BB x Bb --> the progeny will be BB and Bb in equal proportions; they are all brown, so we can't tell what their genotypes are without doing further tests. F1 brown x F1 brown: Here, the cross is Bb x Bb and the progeny are BB:Bb:bb in 1:2:1 ratio. We know that the whites must be bb, but again, the browns could be either BB or Bb. P white x F1 brown: The cross is bb x Bb --> the progeny are bb (white) and Bb (brown) in equal proportions. Here, we can determine the genotype of each progeny -- the white ones are all bb, and the brown ones are all Bb. |
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| (d) | The F2 whites are bb (otherwise they wouldn't be white); therefore, this cross is just like the 3rd cross (i.e., Bb x bb) and will give similar results -- bb (white) and Bb (brown) in equal proportions. | |
| (e) | Since it is brown, it must be either BB or Bb. A testcross will reveal its genotype -- cross it to white marmot; if the progeny are all brown, the unknown must be BB, and if the progeny are white and brown in equal proportions, the unknown must be a heterozygote. |
| 7. | Since this is double-stranded DNA (A basepaired to T) A content = T content = 30% each; A + T = 60%. Therefore, G + C = 40%; G = 20%, or 0.2 * 650000 * 2 = 130000.
Note the correction (the " * 2" factor) -- since this is double-stranded DNA, each strand has 650,000 bases so the total number of bases is 650,000 * 2. |
| 8. | (a) | Based on the rules of base pairing, if these were normal, double-stranded nucleic acids, we would expect the number of A bases to equal the number of U's and the number of C's to equal the number of G's. Virus (i) does follow this pattern, but in (ii), the number of A bases does not equal the number of U bases, and C &Mac173; G. |
| (b) | This viral RNA cannot be double-stranded -- it must be a single-stranded RNA virus. |
| 9. | AO x BO --> progeny could be A, B, AB, or O. |
| 10. | Since no two babies had the same blood type, the B x O parents must have had twin babies of two different blood types. This outcome is possible only if the B parent is heterozygous (BO), as the O parent is homozygous (OO). Therefore, these parents are BO and OO; the twins are BO (blood type B) and OO (blood type O) also.
The AB baby could only have been born to the A x B parents (this is the only parent pair where one parent has the A allele and one has the B allele). However, we cannot with certainty tell what the genotypes of the parents are -- the A parent is AA or AO, and the B parent is BB or BO. That leaves the baby with blood type A; it must belong to the AB x O parents, whose genotypes are AB and OO. The baby's genotype must be AO. |
| 11. | The simplest approach is a trial-and-error method: interpret each cross one at a time, and see if your interpretation is consistent with the interpretation of the previous crosses. To begin with, it is clear that there are three phenotypes, so just for simplicity, I am going to assign them 3 allele designations (R, B, W, for Red, Blue, and White) and assume that they are alleles of the same determinant. I may have to revise this initial hypothesis later on--e.g., this may be a case of incomplete dominance between two alleles--but at least for starters, I'm going to assume simple dominant/recessive interactions.
Cross (a) -- Red #1 selfed -- yields a 3:1 ratio of red and blue-flowered plants in the progeny. This looks like a typical heterozygous F1 cross, with R being dominant and B recessive. So I'm tentatively assigning Red #1 a genotype of RW. Cross (b) -- Red #2 selfed -- similarly suggests that R is dominant over W; the genotype would be RW. Cross (c) -- Blue selfed -- gives a 3:1 ratio of blue:white; blue must be dominant over white and the genotype of the blue-flowered plant must be BW. At this point, we have a hypothesis for all of the genotypes:
We are now in a position to predict the results of the remaining crosses, and seeing if our predictions are met. Cross (d) -- Red #1 x Red #2 = RB x RW:
Cross (e) -- Red #1 x Blue -- should be RB x BW, which should give a 1:1 ratio of red:blue (draw Punnett squares if you're uncertain about this). Again, that's what we see. Cross (f) -- BW x WW should give 1:1 blue and white Cross (g) -- WW x WW gives only white-flowered progeny. So our initial hypothesis appears to be sound as far as we can tell from the data provided. We can predict the results of cross (h): Red #2 x blue = RW x BW:
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