C2005/F2401 ’10Answers to Review Questions for Exam #4
Each answer and explanation was worth 2 pts unless it says otherwise. The explanations given below are much longer and more detailed than expected on the exam.
Key to Problem 1
A. Answers: A-1. ¼; A-2. 3/32.
Explanation: Cross is Rh+/Rh- MN IA i X Rh+/Rh- MM IB i.
You can figure out the results for each gene separately, using a Punnett square or other method.
For A-1, Chance of Rh- child is ¼.
For A-2, Chance of child who is Rh+ MN and blood type A
= (chance of Rh+) X (chance of MN) X (chance of blood type A) = ¾ X ½ X 1/4. = 3/32.
More details:
All 3 genes are on autosomes and so assort independently. (The autosomes are always numbered 1 to 22, and the X and Y are not given numbers.)
The Rh gene has only two alleles, Rh+ & Rh-, and Rh+ is dominant to Rh-. (See problems 14-2 & 14-4.) Since the2 Rh+ parents had a Rh- child, you can figure out that they must be heterozygous, and that Rh+ is dominant to Rh-.
The parents are also heterozygous for ABO blood type (IA i & IB i) since they had an O child.
The MN blood group gene has only 2 co-dominant alleles, M and N. There is no null allele equivalent to i. You know the genotypes for the MN blood group, because the two alleles are co-dominant, and phenotype = genotype. Phenotypes M, MN and N correspond to genotypes MM, MN and NN respectively. (The MN blood group was discussed at length in class when demonstrating the Hardy-Weinberg Law. See handout 23 B and problem 9-2.)
B. Answers: B-1. 1/16; B-2. 2 dif. Genotypes.
Explanation: Child can be homozygous Rh+ or homozygous Rh-, but there is only one possible homozygous choice for the other two genes – MM & blood type O (ii). So you need to calculate the chance the child will be homozygous Rh+ or Rh-, MM, and ii.
Homozygous Rh+ = 1/4; same for homozygous Rh-. Chance of one or the other = sum = 1/4 + 1/4 = ½. Chance of MM = ½.
Chance of ii = ¼.
Chance of triple homozygote = product of all three chances = ½ X ½ X ¼ = 1/16.
C-1 & C-2. Answers: C-1. Person #1 is mother; C-2. results support father’s refusal to pay.
Explanation: Child is N phenotype, meaning NN genotype. Only person #1 has an N, so if one of them is the mother, it must be #1. (Person #2 cannot pass on an N allele to his/her offspring.) Where did the child’s second N allele come from? It couldn’t be from person #2. Therefore the ‘father’ is not the second parent, and some other male must have donated the N allele.
C-3. Answers: (a) Neither phenotypic test is useful. (b) If child is genotype IAIA that will reinforce the father’s case.
Explanation (2 pts each part):
For (a). Any Rh or ABO phenotype is possible with those two parents. Whatever you find in the child would be consistent with the parents’ alleles. Nothing you could find would rule out the father, or prove he was the actual father.
An inclusion, or match with the father’s genotype (for Rh or ABO), is not meaningful as there are plenty of other men who are the proper ABO and/or Rh phenotype. If the father and child matched at many variable loci, that might mean something. Matching at one or two is not meaningful. (If someone of type A and Rh+ committed a crime, and you are type A and Rh+, does that mean you did it??) However, a failure to match at one or two loci (an exclusion) is a strong argument that person #2 is not the father. (If someone of type A and Rh+ committed a crime, and you are type B and Rh-, are you a suspect?)
For (b) see next page.
Explanation for Problem 1, part C-3, cont.
For (b). Person #2 could be the parent of any phenotype or genotype except homozygous A. (Homozygous B is not a possibility here, as the mother has no IB allele. ) If the child is homozygous A then some male with an IA allele must be the father. Since person #2 has no IA allele, this is an exclusion – person #2 is ruled out. You could argue there is some chance of a rare event like a mutation to explain a lack of a match at one locus, but if two independent loci like MN and ABO do not match, the best explanation is that person #2 is not the father.
Note that even if child inherited an i or IBallele from his/her male parent, which is consistent with person #2 as the father, it does not mean that person #2 should pay up. It just means that some male with a IB allele or i allele is the father. (This is an inclusion; see above.)
Key to Problem 2.
A. Answers: A-1. 3 bands; A-2. >6 bands.
Explanation: A trisomic individual has 3 copies of the polymorphic region, so there can be three copies with different versions of the repeats. Each different version (6 to 23 repeats) gives a PCR product of a different length. If there are different numbers of repeats on each of the 3 chromosomes, there will be 3 bands on the gel.
The trisomic individual received two chromosomes from one of his/her parents. These could be the same, derived from sister chromatids, if ND occurred at 2nd division. In that case the trisomic individual would have only two bands. However, the two copies from one parent do not have to be the same. They could be different if the parent in which ND occurred was heterozygous, and ND occurred at 1st division (see C-2 below).
Note there is no cutting by restriction enzymes here. The variation (polymorphism) is detected by amplifying the repeat region with PCR and measuring the length of the amplified part. (Primers are to sequences on each side of the repeats – these sequences are the same in virtually everybody.)
B. Answers: 6/10 or 6/12 (1 pt each).
Explanation: Each parent contributes one chromosome carrying the polymorphic region. Both of the father’s chromosomes contain 6 repeats, so the child has to get a one chromosome with 6 repeats. The child can get either of his mother’s chromosomes – the one with 10 repeats or the one with 12. So the child’s haplotype, designated by the number of repeats on each chromosome, is either 6/10 or 6/12. (Note that in terms of haplotype, 6/10 is the same as 10/6.)
C-1. Answer: 2. The number of bands and their position will be the same as in a normal child, but their intensities will be different.
Explanation: The father is homozygous for the repeats on chromosome 16. If ND occurs to produce a disomic gamete with two copies of chromosome 16, both copies will have the same haplotype (6 repeats), whether ND occurred at 1st or 2nd division (see C-2). The mother is heterozygous, and produces normal gametes with one copy of chromosome 16. That chromosome has either one haplotype or the other -- 10 or 12 repeats. If the polymorphic region of the DNA from the child (really a spontaneous abortus) is amplified using PCR, there will be two bands on the gel – one corresponding to an amplification of the region with 6 repeats and one corresponding to an amplification of the region with either 10 or 12 repeats.The number of bands and their position will be the same as in a normal individual.
The number of bands will be the same as normal, but the intensity of the bands will be different. That’s because there are more copies of the DNA with 6 repeats in the DNA of a trisomic than in the DNA of normal diploid.A normal individual has one copy of the DNA with 6 repeats, but a trisomic has two. (Both have one copy of the DNA with 10 or 12 repeats.) The amount of PCR product is proportional to the amount of DNA template available. Therefore, PCR of DNA from a normal person will yield two bands of equal intensity, but PCR of DNA from a trisomic individual will not. PCR of a trisomic will yield one band at the normal intensity (from the one chromosome with 10 or 12 repeats) and one band of double the normal intensity (from the two chromosomes with 6 repeats).
Note – there are no probes here. You are looking at the amount of PCR product, not at its ability to hybridize to a probe.
Key to Problem 2, cont.
C-2. Answer: The mother. That’s because the mother is heterozygous for the polymorphism, but the father is not.
Explanation (4 pts): You need to be able to tell whether the two copies of chromosome 16 (resulting from ND) came from different homologs (ND at 1st div.) or only from one homolog (ND at 2nd div.). If the two copies of chromosome 16 in the parent are the same, as in the father, you can’t tell the difference. In either case, the two copies in the gamete will have the same haplotype (6 repeats.) If the two copies of chromosome 16 are different, as in the mother, you can tell the difference. In that case, if homologs fail to separate in 1st division, there will be a chromatid from each homolog in the gamete, and the two copies in the gamete will have different haplotypes (10 copies on one; 12 on the other). If homologs separate properly at 1st division, but sister chromatids fail to separate at 2nd division, there will be two chromatids from the same homolog in the gamete, and both copies will have the same haplotype (both will have 10 repeats or both will have 12).
D-1. 47 chromosomes and 47 chromatids.
Explanation (not required): A trisomic cell has 2N + 1 chromosomes = 2 X 23 +1 (for humans). The number of chromosomes is the same before and after mitosis, but the number of chromatids changes -- the # of chromatids/chromosome goes from two (before mitosis) to one (after mitosis). The # of chromatids/chromosome doubles before mitosis so each daughter cell can get one copy of each chromosome.
D-2. Answer: DNA in a triploid in G-2 is 6c. Note that question asks about the amount of DNA, not the number of chromatids.
Explanation: Triploid means 3N or 3 of each homolog. It is not the same as trisomic which is 2N +1. So a triploid human cell has 3N = 69 chromosomes.
In G-2, DNA is doubled, and there are two chromatids per chromosome. So you need to figure out the amount of DNA in a 3N cell with doubled chromosomes = amount of DNA in 138 chromatids.
‘c’ is the amount of DNA in an N cell before DNA replication (23 chromatids); 2 X 3c is the amount of DNA in a 3N cell after DNA replication (138 chromatids).
Details: ‘c’ is the amount of DNA in a haploid (N) cell with 1 chromatid/chromosome
= DNA in N (unduplicated) chromosomes = DNA in N cell in G-1 (before DNA replication)
= DNA in one haploid set of chromatids = DNA in 23 chromatids (for humans)
DNA content in a haploid (N) cell will be c before DNA replication (in G-1) and 2c after DNA replication (in G-2).
DNA content in 3N cell will be 3 times that – 3c before DNA replication (in G-1) and 6c after DNA replication (in G-2).
A DNA content of 6c corresponds to 138 (6N) chromatids, but the question asks about the DNA content, not the # of chromatids. The amount of DNA in 138 chromatids is 6c.
Key to Problem 3
A-1. (6 pts for answers and explanation.) Answer: Most common to least: #1 > #4 > #3 > #2 > #5.
Short explanation: #1 (parental); #4 (single recombinant -- one crossover in longer interval); #3 (single recombinant – one crossover in shorter interval); #2 (double recombinant – 2 crossovers required);
#5 (impossible in gametes of these parents).
Detailed explanation: (** refers to note at** on next page, after #5.)
#3 & #4. A and B are 15 cM (centiMorgans or map units) apart, which corresponds to a RF of 15%. That means that recombinants for A and B make up 15% of the gametes from either parent. If you start with A-1 B-1/A-2 B-2, 15% of gametes will carry A-1 B-2 or A-2 B-1. Any particular single recombinant (between A & B) will be about 7.5%. ** Similarly, B and D are 22 cM apart, so 22% of gametes will be recombinant for B and D, and any one single recombinant (between B & D) will be about 11%. Recombinants for B & D will be more common than recombinants for A & B. Therefore #4 – about 11% expected, should be more common than #3 – about 7.5%.
#2. A double recombinant with one crossover between A & B and one between B & D will occur 15% of 22% or about 3% of the time. Therefore #2 is expected about 1.5% of the time. **
#1. Parentals will be more common than recombinants – that’s what linkage (RF < 50%) means, and why #1 is most frequent. (About 40% will be recombinant for either interval or both, leaving 60% parental, or #1 at about 30% **).
Key to Problem 3, cont.
#5. This combination is impossible in the gametes of either parent. It requires a recombination between a chromosome in the mother and one in the father. These chromosomes are not in the same person, so no crossover between them can occur during meiosis in either parent. Their children get one maternal chromosome and one paternal chromosome, so this combination could be produced by crossing over during meiosis in the children, but not in the parents.The children could not inherit this combination, but the grandchildren could.
** Note: Since crossing over occurs by cutting and rejoining of DNA molecules, any time you generate one recombinant chromosome, you generate another, reciprocal one. Therefore the frequency of say, A-1 B-2 D-2 should equal the frequency of A-2 B-1 D-1, etc. (You were not expected to calculate the exact proportions of each gamete – the numbers are included to make the answers clearer and more concrete.)
A-2. Both. A-2 B-1 D-1 is the reciprocal of #3; it should be the same frequency. (See ** above.)
A-4 B-3 D-3 is a single recombinant with a crossover in the A to B interval. The distance and frequency of crossing over should be the same for A to B whether it is in the mother or father. So A-4 B-3 D-3 should be the same frequency as the other two. (2/4 for correct explanation of either one but not both.)
B. Answer: > 25% but < 37%. These distances are so large that RF values are not strictly proportional to distance, and so the distances (& RF values) are not strictly additive. (2/4 for adding the distances to 37% or considering double crossovers but getting the effect backwards, and choosing the answer >37% but <50%. See detailed explanation below.)
Explanation: RF values are proportional to distance and additive in the right range, which is the range in which multiple crossovers between the markers or genes are infrequent. In this range, which is usually taken to be RF < 25%, the effects of multiple crossovers can be ignored. When crossovers are rare, the number of recombinants you detect reflects the actual number of crossover events.
In all cases, the number of actual cut and rejoin events is proportional to distance. However, if distance gets too large, not all the crossover events will yield detectable recombinants. When the distance is too big, the number of multiple crossovers becomes significant, and the number of recombinants that you count is an underestimate of the number of actual cut and rejoin events. For longer distances, such as the distance from A to D, some of your recombinants will not be detected, because they will have multiple crossovers between A & D, and these will put the A & D haplotypes back to the way they started. So the RF will be smaller than expected from the actual distance. That’s why, in this case, a direct measurement of the RF between A & D will be less than 37%. The shorter distances (derived from smaller values of RF) are more accurate, as they involve less double crossovers. Adding them up gives a good estimate of the actual distance, and actual real number of crossover events. However if you measure the A-D RF directly (instead of adding up shorter distances), the RF you get will be an underestimate – because you will not detect all multiple crossovers. Multiple crossovers don’t make the RF come out larger – they make the RF value smaller. (That’s why the maximum value of RF plateaus.)
Another way to see this is to look at the graph of RF vs. distance on the class handout. At small distances, RF and distance are proportional. As distances gets larger, the RF approaches a limit – it doesn’t keep increasing. If you double the distance, you don’t double the RF. At large distances, RF is an underestimate of distance, and an underestimate of the real frequency of crossing over. No matter how large the real distance is, the RF never gets larger than 50%.It’s the RF that isn’t an accurate measure of crossing over and distance, not the other way around. To correct for double crossovers you have to correct the RF, not the distance.
Key to Problem 4.
A-1. This condition is dominant. Why? In short, the pattern of inheritance is consistent with a rare dominant but not a rare recessive.
Arguments why it isn’t recessive:
(1). When affected people have children with normal partners, they keep having affected kids. If condition were recessive, this would only happen if many of the partners of normal phenotype were carriers, which seems extremely unlikely, because the condition is so rare.(Only about 50 people with this condition have been documented since the middle ages!) For example, the first affected female in generation III had multiple partners, and had affected kids with 3 out of 4 partners.
(2) is on next page.
Key to Problem 4, cont. Why condition isn’t recessive, cont.
(2). Normal descendants in this family don’t pass it on. If it were recessive, two carriers of normal phenotype could have an affected kid.
Arguments why it IS dominant:
(1). Everyone with the condition has a parent with it.
(2). The affected people have both normal and affected kids.
This is what you would expect from a rare dominant – all affected individuals are heterozygous, and have one affected parent. When the affected people have children with a normal person, they have a mix of normal and affected kids. (The affected kids from the normal X affected is about ½ and ½, which is expected. This pedigree is large enough that using proportions is not out of the question.)
Another approach:
Once you decide the gene is on the X, the pattern of inheritance is incompatible with recessive inheritance, and fits nicely with dominant inheritance.