Genetics 371B, Autumn 1999 Answer key, Problem set 2

1.	When you are faced with a multiple-trait problem such as this one, rather than trying to understand the entire cross at once, it makes sense to break the problem down and solve it in parts. One thing to keep in mind you cannot assume that wildtype is dominant.
	(a)	The question here is about star-eyes alone, so you can look at that phenotype for now and ignore the others. The cross is: star eye x star eye --> 118 star eye : 42 normal clearly, a 3:1 ratio of star eye : normal; the parents must both be heterozygous for star eye . Furthermore, the star eye phenotype must be dominant over normal (s+s x s+s giving a 3:1 ratio of s_ and s+s+). Therefore, the normal progeny are homozygous s+s+; we should not see star-eyed progeny if we cross the normal flies with each other.
	(b)	For bent wing and forked bristle, we can again look at each trait in isolation and ask if we can deduce anything about their modes of inheritance. Looking first at bw, we see that normal males, when crossed to bent wing females, gave normal F1 progeny. Therefore, bw must be recessive to normal (bw+), and the normal male parent cannot have been heterozygous bw+ bw -- else his progeny would have consisted of bent wing and normal wing flies in 1:1 proportions. Likewise, forked bristle males had normal progeny, so forked bristle must be recessive to normal bristle. Again, the progeny were exclusively normal with respect to bristle, so the normal-bristle female parent must have been homozygous. (Another way of thinking about it so far if this were a pedigree, we'd say that both bw and f phenotypes skip generations, so neither one can be dominant.) Now let's look at the progeny, again starting with bw vs bw+. Ignoring the presence or absence of forked bristle phenotype, the F2 progeny are: Females 1212 normal : 398 bent wing Males 1192 normal : 398 bent wing In both females and males, there's a 3:1 ratio of normal : bent wing progeny. We can see no evidence of sex-specific ratios in either F1 or F2, so the bent wing trait must be autosomal recessive. Note also that the F2 ratios make sense: if the original parents are bw+ bw+ homozygous normal (male) and bw bw homozygous recessive (female), the F1 progeny should be heterozygous bw+ bw so the F2 progeny should be normal and bent wing in 3:1 ratio, and it's gratifying that they are indeed so. Presumably, the genotype ratio is 1 bw+ bw+ : 2 bw+ bw : 1 bw bw Moving on to forked bristle -- here, the F2 progeny are: Females 1610 normal Males 805 normal : 785 forked bristle There is clearly a difference between males and females. Furthermore, the overall ratio is: 2 normal female : 1 normal male : 1 affected male a ratio we've seen before with the X-linked white gene. Because the trait is not seen in F1 flies, it cannot be Y-linked; we can tentatively assign X-linked recessive as mode of inheritance for forked bristle. The cross with respect to forked bristle is:
	(c)	To do the chi-square test, we need to predict the phenotype classes in the F2 as well as the numbers of progeny of each phenotype. The easiest way to proceed is to draw a Punnett square -- The chi-square table for this expected phenotype ratio (for 3200 progeny total) is shown: Categories Expected Observed (O-E)2/E normal female 6/16 x 3200 = 1200 1212 0.12 bent wing female 2/16 x 3200 = 400 398 0.01 normal male 3/16 x 3200 = 600 602 0.0067 forked bristle male 3/16 x 3200 = 600 590 0.167 bent wing male 1/16 x 3200 = 200 203 0.045 bent wing, forked bristle male 1/16 x 3200 = 200 195 0.125 Chi-square value = 0.473 There are six categories, so the number of degrees of freedom (df) = 5. The P value corresponding to the observed chi-square value of 0.473 and df of 5 is 0.975 < P < 0.995. Therefore, the probability of getting the observed deviation from expected values is high (between 97.5% and 99.5%); the null hypothesis that the observed values are not significantly different from expected cannot be rejected.

2.

The girl's mother of course has inherited one complete set of chromosomes from the maternal grandmother and one from the grandfather. So the question is, what is the probability that the mother will transmit only the grandmother's chromosomes and not the grandfather's. For each chromosome, the probability that the grandmother's chromosome will be transmitted = 1/2. Therefore, the probability that all 23 grandmotherly chromosomes will be transmitted = (1/2)23. For the boy, the probability is slightly different we already know that he inherited the Y chromosome from his paternal grandfather. So we only have to consider the remaining 22 chromosomes; the probability that he inherited a complete set from his grandfather is (1/2)22.

3.	(a)	Assigning genotypes for Pedigree 1 becomes straightforward if we keep two things in mind -- affected women must be HbHb (because the trait is recessive in females; if a woman has even one Hh allele she will not show the balding phenotype), and unaffected men must be HhHh (if they had even one Hb allele they would show the balding phenotype). Keeping this information in mind, we can fill in the genotypes as shown: I-1 and I-2 must both be heterozygous because they have an affected daughter (II-2); II-3 must be heterozygous because she is herself unaffected but has an affected son (III-2). The son (III-2) cannot have got the trait from his father (II-4), because the father is not affected. III-3 and III-4 are each Hh_ , with a 50% probability of being HhHb.
	(b)	(i) The trait is not autosomal recessive, because unaffected children (IV-2 and IV-3) are born to an affected couple (III-4 and III-5). (ii) The pedigree is consistent with autosomal dominance about half the offspring of affected individuals are affected, the trait does not skip generations, and an affected couple have unaffected children. If the trait is autosomal dominant, I-1 must be a heterozygote (because she has unaffected children). (iii) It is not X-linked recessive, because affected women (e.g., I-1) have unaffected sons (e.g., II-4). (iv) It is not X-linked dominant, because affected men (e.g., II-2) have unaffected daughters (e.g,. III-1). (v) It is not Y-linked, because women are affected also, and affected men have unaffected sons. (vi) It is not sex-limited, because men and women are affected. (vii) It is not sex-influenced, dominant in males/recessive in females, because affected women have unaffected sons. (viii) It is not sex-influenced, dominant in females/recessive in males, because affected men have unaffected daughters. An aside: Achoo syndrome is a real phenotype, not something I made up for this homework; it is really believed to be autosomal dominant.
	(c)	The baldness allele can come only from the woman (III-4, Pedigree 1); the sneezing allele can come only from the man (IV-1, Pedigree 2); and sex of the child is determined by whether the man transmits his X chromosome or his Y (probability of each = 1/2). We have already seen that the probability that the woman carries the baldness allele is 1/2. What is the genotype of the man (IV-1)? His parents are both heterozygotes (they each have an unaffected parent); if A = dominant, sneezing, and a = recessive, non-sneezing, IV-1 must be either AA (probability = 1/3) or Aa (probability = 2/3). [Where do these numbers come from? We know that his parents are heterozygotes, so their children could each be AA or Aa or aa with a 1:2:1 ratio of relative probabilities. In his case, we already know that he is not aa because he is affected so he is either AA (1/3 chance) or Aa (2/3 chance). If you are still confused, draw the mating of IV-1's parents as a Punnett square, eliminate the homozygous recesssive, and count the remaining possibilities.] (i) For the the first child to be a bald, sneezing male, three things have to happen -- the mother has to transmit a baldness allele, the father has to transmit a sneezing allele, and the father has to transmit a Y chromosome. The probability of all three events happening is the product of the individual probabilities; in addition we have to factor in the probabilities of the parental genotypes. p(Hb allele being transmitted) = (1/2)(1/2) because there's a 1/2 chance that the mother is heterozygous, and a half chance that she will transmit the correct allele. p(allele A being transmitted by the father): Here, it might be easier to figure out the probability that he will transmit the recessive allele (a); the probability that he will transmit the dominant allele will then be 1 - p(a). The father has a 2/3 chance of being Aa, so there is a 1/3 chance that he will transmit allele a; therefore, there is a 2/3 chance that he will transmit allele A. (Note that this calculation can also be done as a sum rule if he is (1/3)AA there is a 1/3 chance of transmitting allele A; if he is (2/3)Aa, there is likewise a 1/3 chance of transmitting allele A; so the overall chance of transmitting A = 1/3 + 1/3 = 2/3.) p(Y chromosome from father) = 1/2 So the overall probability of a bald, sneezing male = (1/4)(2/3)(1/2) = 1/12. (ii) For a daughter to show the baldness phenotype, she will have to inherit the baldness allele from both parents. Since the father does not carry this allele, the chance that she will get the allele from both parents = 0; the probability of a bald, sneezing daughter = 0. (iii) The probability that the son will be hairy depends on whether the mother transmits the baldness allele or not. As we have seen, the probability that she will transmit the baldness allele = 1/4. Therefore, the probability that she will transmit the other allele Hh = 3/4. So the probability of a hairy, sneezing son = (3/4)(2/3)(1/2) = 1/4.

4.	(i)	The trait is not autosomal recessive because unaffected children are born to parents who are both affected (II-4 and II-5).
	(ii)	It is not autosomal dominant, because affected children e.g., III-1 and III-2) are born to unaffected parents.
	(iii)	It is not X-linked recessive an affected woman (III-1) has an unaffected father (II-2).
	(iv)	It is not X-linked dominant, because an affected father (II-5) has unaffected daughters.
	(v)	It is not Y-linked because women are affected as well as men.
	(vi)	...so it is not sex-limited either.
	(vii)	It is not dominant in males and recessive in females, because an unaffected man (II-2) has an affected daughter.
	(viii)	In the same family, an unaffected woman (II-1) has an affected son, so the trait cannot be dominant in females and recessive in males.
	So how do we explain this strange pedigree? A number of explanations come to mind. For example: reduced expressivity or penetrance will complicate analysis of the pedigree incorrect diagnosis by the physician, or incorrect recollection of family history on the part of the patient when the genetic counselor was constructing the pedigree sporadic occurrence of the phenotype, e.g., because of a new mutation, or because the symptoms can arise for more than one reason false paternity

5.

In Cross 1, homozygous slow Oreo-lovers are mated with homozygous slow Oreo-haters so the progeny of Cross 1 will be homozygous slow and heterozygous with respect to cookie love/hate. In Cross 2, homozygous fast Oreo-lovers are mated with homozygous slow Oreo-haters so the progeny should be fully heterozygous. To determine the mode of inheritance for these two traits, we can look at each trait by itself. With respect to speed, the cross between the F1 progeny is homozygous slow x heterozygous --> 80 fast : 80 slow (i.e., a 1:1 ratio) Therefore, slow must be recessive else the progeny would all be slow. With respect to Oreos, the cross is heterozygote x heterozygote --> 119 haters : 41 lovers (i.e., a 3:1 ratio of haters : lovers) Clearly, Oreo-hating must be dominant to Oreo-loving. So if F = fast and L = hater, the crosses can be depicted as:

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