C1015 '03 Answers to Problem Set 4
4-1. A. The 2 people are double heterozygotes. Genotypes = IA IB Br bl, where Br = brown and bl = blue.
Note: To avoid confusion, it is customary to use different symbols for phenotype
and genotype when discussing the ABO blood type (phenotype) and the alleles of
the ABO gene (genotype). It is customary to call the blood types (phenotypes) A,
B and AB while the alleles (genotype) of the blood type gene are called IA
and IB. You can call the alleles A and B or anything else you
prefer, but the IA IB terminology is standard, and helps
you to avoid mixing up genotype and phenotype. (The ABO gene actually has three
alleles, but only 2 appear in this problem.)
You also have to be careful here not to confuse B meaning "type B" with B
meaning "brown." That's why we used bl and Br as symbols for the alleles
of the eye color gene, instead of B and b. .
B. 3/4 brown, 1/4 blue.
C. 1/4 A, 1/2 AB, 1/4 B.
D. 1/4 each of IA Br, IA bl, IB Br, and IB bl. Either allele for blood type is equally likely to go with either allele for eye color.
E. All combinations of blood type and eye color are expected. Blood type A can go with blue or brown eyes and ditto for blood type B or AB. So there are (overall) 6 possible phenotypes. For those who like to know, proportions expected are:
3/16 brown eyes, blood type A 3/16 brown eyes, blood type B 3/8
(6/16) brown eyes, blood type AB
1/16 blue eyes, blood type A 1/16 blue eyes, blood
type B 1/8 (2/16) blue eyes, blood type AB
You can figure this out from a large Punnett square, using the gametes in part D. Alternatively, you can calculate it using probability, if you assume the chance of picking any particular eye color genotype is independent of picking any particular blood type genotype.(Chance of brown eyes, blood type A = product of the two probabilities, etc.)
F. 1/4. (The eye color doesn't matter -- being blue-eyed doesn't predispose you toward or against any blood type. In other words, eye color and blood type are determined independently.) Note the question is "Given blue, what is probability of type A?" not "What is probability of being blue and type A?"
G. 2 traits (trait = characteristic = eye color or blood type), 2 genes, 2 alleles of each gene (4 total), 6 phenotypes and 9 genotypes. (Note: ABO gene has 3 alleles, but only 2 appear in this problem. Also note that determination of eye color involves more than 1 gene, so blue and brown aren't always inherited in the simple manner described here.)
4-2. A. All IB Br and IA bl. Symbols are same as in
previous problem.
B. You will get equal numbers of IA Br and IB bl from crossing over. If the genes are closely linked, the number of gametes with IA Br or IB bl formed by crossing over will be relatively small, and most gametes will be the type that you started with, that is, IB Br and IA bl.
C. Blood type A. We are assuming that the blood type gene and the eye color gene are closely linked (for this problem only). Since the woman had a blue-eyed type A parent, the allele for blood type A and the allele for blue eye color should be on the same chromosome and should be inherited together from the woman. (The man should pass on a chromosome with the blue allele of the eye color gene and the IA allele of the ABO gene.)
4-3. A. The genes for brown/blonde hair color and green/blue eye color are
closely linked on chromosome 19; brown and green are dominant. The cross here is
GB/gb X gb/gb.
Note on symbols: what's written before the slash mark = alleles on one homolog; what's written after the slash = alleles on the other homolog. How do we know which alleles are together? Whether it is GB/gb or Gb/gB? Woman has one chromosome with 2 dom. alleles from her mom = GB and one chromosome with 2 rec. alleles from her father = gb, so her genotype is best written GB/gb. Husband is double recessive or gb/gb.
Most of the children should be GB/gb (double dom. pheno) or gb/gb (recessive). These kids are "parental" types -- what you get if there is no crossing over. If crossing over does occur (during meiosis in the mom), a kid could get a Gb or gB chromosomes from his/her mom. These "recombinant" types should be rare -- since the genes involved are close to each other, crossing over between them (during meiosis) should be infrequent. The very rare recombinant kids would be blonde haired and green eyed or brown haired and blue eyed.
B. The genes responsible for albinism and PKU are unlinked; they are on chromosomes 11 and 12 respectively. So the cross here is AaPp X AaPp where a = recessive albino allele and p = recessive PKU allele. It doesn't matter which side of the family the recessive alleles came from, since the albino and PKU genes/alleles are not linked and are passed on independently. There is no sense in writing the genotype of a carrier as AP/ap or Ap/aP -- the "a's" and "p's" are not connected. The chance of a child with both conditions is 1/4 X 1/4 = 1/16. (From Aa X Aa, chance is 1/4 of getting aa. From Pp X Pp, chance is 1/4 of getting pp. Chance of getting both aa and pp is the 1/4 X 1/4 = 1/16). You don't have to be able to calculate the exact numbers here, but you should realize that the chances of getting either disease are independent. There is no tendency for the disease alleles to travel together (or to stay apart), so once you inherit one disease, you are no more likely than anyone else to inherit the other.
4-4. A. Call the generations (counting from the top) I, II, etc. Number the
people in the pedigree in each generation starting from the left. Then V-3, V-5
& IV-1 must be A hd / C HD, V-4 must be A hd /A hd or A hd / C hd, IV-2 must be A
hd / B hd and so on. (The two symbols before the slash go on one chromosome; the 2
symbols after the slash go on the other homolog.)
B. VI-5 may be a recombinant; it depends on the genotype of V-4. If V-4 was
AA, then VI-5 must have received a recombinant gamete (carrying C and hd) from
his mom. If V-4 was AC, then there is no reason to think VI-5 is a recombinant
-- he could have got a C (with hd) from his dad and an A (with hd) from his mom
without any crossing over. Note that individuals can have a chromosome with a C
and a normal allele of hd, but that all individuals who inherit a C allele
(haplotype) from
I-1 carry an HD allele. How is this so?
In the population at large, any one of the A, B, or C alleles
(really haplotypes) can go with
either allele of the HD gene (HD or hd). Whatever combination(s) we start with
will be passed on most of the time -- the combinations can be reshuffled by
crossing over, but reshuffling will be rare since the A/B/C marker and the HD gene
are closely linked. In this family, the original ancestor with HD (I-1 or
before) had a chromosome with a C allele (haplotype). That chromosome acquired the HD allele
because of a mutation of the normal, recessive hd allele. The chromosome
carrying the HD and C alleles was then passed on from generation to generation;
anyone who got it developed HD. Other ancestors had chromosomes with A, B, or C
and the normal allele (hd) of the HD gene. Some of these people passed on
chromosomes that carried the C and hd combination.
C. AA or AB - no, AC yes. The chromosome with the C carries the allele for HD.
D. If kid gets C (= chromosome with C + HD) from dad, she will have HD. If kid is BB or AB, we know kid did not get C from dad and will be okay. If kid is AC we know kid will get HD, because the A must come from mom and so C must come from dad. If kid is BC we can't be sure because we don't know if C came from mom or dad. We know mom has one B, but we don't know what her other allele is. (If they've already had an AB or AC kid, then we know mom is AB and C is from dad and kid will get HD.) All answers here assume no crossing over, because crossing over is very rare between the A/B/C marker and the HD gene. If crossing over is taken into account, then the predictions here are about 95% certain instead of 100%.
4-5. A. Yes. Parent #1 is aaBB (missing enzyme 1, encoded by allele A) and parent 2 is AAbb (missing enzyme encoded by allele B). Kids should be AaBb and able to make both enzymes (since they have one A and one B).
B. (1). Test which enzyme is missing (which step is blocked) in each parent. If different steps are blocked, then parents have defects in different genes and kids will be okay. Parents are then probably AAbb and aaBB and kids will be AaBb (as in part A). If same step is blocked, then parents are homozygous recessive for the same gene and kids will also have the defect. In this case, parents are both aa BB (or both AAbb) and all kids will be aa (or bb) and unable to make serine because of a block in the pathway.
(2). Probably okay. If child #1 is normal, child should be AaBb, so other kids should be too. (In other words, parents were AAbb and aaBB as in part A.) When 2 people with blocks in the pathway have kids, either all children should be normal (if parents were AAbb and aaBB) or none of the kids should be normal (if both parents were homozyous for the same recessive allele -- both aa or both bb).
4-6. A. Genetic equilibrium can be reached at different points, with different frequencies (p & q) for the two alleles (M & N). At any time, p + q =1, but at equilibrium the frequency of the genotypes (MM, MN, & NN) must be p2, 2pq and q2, respectively, and p2 + 2pq + q2 must = 1. For example, in one population p can = .1 and q= .9 while in a different population p can = .5 and q = .5. The frequencies of MM, MN, & NN will be different in each population, but in both cases, the frequency of MM will be p2, the frequency of MN will be 2pq, etc.
B. The usual factors responsible for differences in p & q values between populations are genetic drift (including the Founder effect -- see below) and/or selection. When two groups split off from each other or from the ancestral population, they usually start with the same values of p & q. However, if the groups are small, either at the start or at a later time, then significant fluctuations in p & q may occur due to chance alone. These chance changes in p & q that occur in a small population are known as genetic drift. For example, if the starting population is small, there may easily be more (or less) M's present in the starting group than in the ancestral population. In other words, the starting group may not be a statistically representative sample of the group from which it came, simply by accident. (This is called the Founder effect.) Once a small group settles down, and starts to reproduce, the number of M's passed on to the next generation may be larger (or smaller) than the proportion that is statistically expected. Sometimes the fluctuations cancel each other out -- there are too many M's passed on one generation, but too few the next, and so on. But sometimes by chance the fluctuations occur in the same direction for several generations. This leads to significant changes in p & q and t
his is what is meant by genetic drift.