Genetics Quiz 2
) A man has attached earlobes, whereas his wife has free earlobes. Their first child, a boy, has attached earlobes. A) If the phenotypic difference is assumed to be due to two alleles of a single gene, is it possible that the gene is X-linked? B) Assuming the trait could be autosomal or X-linked, is it possible to decide if attached earlobes are dominant or recessive?
) a. Sons inherit their X chromosome from their mother. The mother has free earlobes, the son does not. If the allele for free earlobes is dominant and the allele for attached earlobes recessive, then the mother could be heterozygous for this trait and the gene could be X-linked. b. It is not possible from the data given to decide which allele is dominant. If attached earlobes is autosomal dominant, then the father would be heterozygous and the son would have a 50 percent chance of inheriting the dominant attached earlobes allele. If attached earlobes is recessive, then the trait could be autosomal or X-linked, but in either case, the mother would be heterozygous.
) In humans, PKU (phenylketonuria) is a disease caused by an enzyme inefficiency at step A in the following simplified reaction sequence, and AKU (alkaptonuria) is due to an enzyme inefficiency in one of the steps summarized as step B here: phenylalanine —A—> tyrosine —B—> CO2 + H2O A person with PKU marries a person with AKU. What phenotypes do you expect for their children? All normal, all having PKU only, all having AKU only, all having BOTH PKU and AKU, or some having AKU and some having PKU? Discuss any assumptions you made.
Assuming homozygosity for the normal gene, the mating A/A · b/b x a/a · B/B would only result in A/a · B/b heterozygous offspring. All of their children would be normal, assuming both loci are haplosufficient (which is true in both diseases).
) A certain type of deafness in humans is inherited as an X-linked recessive trait. A man who suffers from this type of deafness married a normal woman, and they are expecting a child. They find out that they are distantly related. Part of the family tree is shown here. How would you advise the parents about the probability of their child's being a deaf boy, a deaf girl, a hearing-enabled boy, or a hearing-enabled girl? Be sure to state any assumptions that you make.
Because the disorder is X-linked recessive, the affected male had to have received the allele, Xa, from the female common ancestor in the first generation. The probability that the affected man's wife also carries the a allele is the probability that she also received it from the female common ancestor. That probability is 1/8. The probability that the couple will have an affected boy is: p(father donates Y) * p(the mother has Xa) * p(mother donates Xa) 1/2 * 1/8 * 1/2 = 1/32 The probability that the couple will have an affected girl is: p(father donates Xa) * p(the mother has Xa) * p(mother donates Xa) 1/2 * 1/8 * 1/2 = 1/32 The probability of normal children is: = 1 - p(affected children) p(affected children) = affected male OR affected female = 1/32 + 1/32 = 1/16 = 1 - 1/16 = 15/16 Half the normal children will be boys, with a probability of 15/32, and half the normal children will be girls, with a probability of 15/32.
In corn, purple kernels are dominant over yellow kernels, and full kernels are dominant over shrunken kernels. A corn plant having purple and full kernels is crossed with a plant having yellow and shrunken kernels, and the following progeny are obtained:purple, full 112 purple, shrunken 103 yellow, full 91 yellow, shrunken 94 What are the most likely genotypes of the parents and progeny? Test your hypothesis with a chi-square test.
Given P (purple); p (yellow) and F (full); f (shrunken): The yellow shrunken parent must be pp ff. The purple full parent could be homozygous (PPFF) or heterozygous (PpFf) at either locus. If the purple full parent were homozygous (PPFF) at either locus, there would be no recessive progeny. Because there are 94 yellow shrunken progeny, the purple full parent must be PpFf. Therefore, the parental cross looks like this: PpFf x ppff Given that cross, you EXPECT the progeny to look like this: 1/2 purple; 1/2 yellow 1/2 full; 1/2 shrunken 1/2 full = 1/4 purple full 1/2 purple —> 1/2 shrunken = 1/4 purple shrunken 1/2 full = 1/4 yellow full 1/2 yellow —> 1/2 shrunken = 1/4 yellow shrunken So given a total of 400 progeny, we expect: 100 purple full 100 purple shrunken 100 yellow full 100 yellow shrunken remember: �2 = (112 - 100)2 + (103 - 100)2 + (91 - 100)2 + (94 - 100)2 = 144/100 + 9/100 + 81/100 + 36/100 100 100 100 100 = 1.44 + 0.09 + 0.81 + 0.36 = 2.7 Degrees of freedom = number of phenotypic classes (4) - 1 = 3 Using the chart in your book, the p value is between 0.2 and 0.5. This falls in the range "Fail to Reject Null Hypothesis." Therefore, we can conclude that the progeny DO appear in the 1:1:1:1 ratio predicted by our genetic explanation.
) A curious polymorphism in human populations has to do with the ability to curl up the sides of the tongue to make a trough ("tongue rolling"). Some people can do this trick, and others simply cannot. Hence, it is an example of a dimorphism. Its significance is a complete mystery. In one family, a boy was unable to roll his tongue but, to his great chagrin, his sister could. Furthermore, both his parents were rollers, and so were both grandfathers, one paternal uncle, and one paternal aunt. One paternal aunt, one paternal uncle, and one maternal uncle could not roll their tongues. A) Draw the pedigree for this family, defining your symbols clearly, and deduce the genotypes of as many individual members as possible (use a dash where an allele is ambiguous). B) The pedigree that you drew is typical of the inheritance of tongue rolling and led geneticists to come up with the inheritance mechanism that no doubt you came up with. However, in a study of 33 pairs of identical twins, both members of 18 pairs could roll, neither member of 8 pairs could roll, and one of the twins in 7 pairs could roll but the other could not. Because identical twins are derived from the splitting of one fertilized egg into two embryos, the members of a pair must be genetically identical. How can the existence of the seven discordant pairs be reconciled with your genetic explanation of the pedigree?
In order to draw this pedigree, you should realize that if an individual's status is not mentioned, then there is no way to assign a genotype to that person. The parents of the point in question has a phenotype (and genotype) that differed from his. Therefore, both parents were heterozygous and the boy, who is a non-roller, is homozygous recessive. Let R stand for the ability to roll the tongue and r stand for the inability to roll the tongue. The pedigree becomes:Assuming the twins are identical, it is likely that the R allele is not fully penetrant and that some genotype "rollers" do not express the phenotype. In this case, the reported data suggest that the R allele is 65% penetrant.
A pure-breeding strain of squash that produced disc-shaped fruits was crossed with a pure-breeding strain having long fruits. The F1 had disc fruits, but the F2 showed a new phenotype, sphere, and was composed of the following proportions: long 32 sphere 102 disc 378 Propose an explanation for these results, and show the genotypes of all three generations.
Note that the F2 are in an approximate 12:3:1 ratio. This suggests a dihybrid cross in which Dominant Epistasis is occurring. Let the two genes in the pathway be represented by A and B. Where the pure-breeding disk is AA BB and the pure-breeding long is aa bb. P - A/A ; B/B (disc) x a/a ; b/b (long) F1 - A/a ; B/b (disc) F2 9 A/-;B/- (disc) 3 A/- ; b/b (disc) 3 a/a ; B/- (sphere) 1 a/a ; b/b (long)
The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are available. Similarly, wings are normally long, but there are strains with short wings. A female from a pure line with brown eyes and short wings is crossed with a male from a normal pure line. The F1 consists of normal females and short-winged males. An F2 is then produced by intercrossing the F1. Both sexes of F2 flies show phenotypes as follows: 3/8 red eyes, long wings 3/8 red eyes, short wings 1/8 brown eyes, long wings 1/8 brown eyes, short wings Deduce the inheritance of these phenotypes. State the genotypes of all three generations and the genotypic proportions of the F1 and F2.
Red-eyed is dominant to brown-eyed and since both females and males are red-eyed, this gene is autosomal. Since males differ from females in their genotype with regard to wing length, this trait is sex-linked. Knowing that Drosophila females are XX and males are XY, the long-winged females tells us that long is dominant to short and that the gene is X-linked. Let B = red, b = brown, XS = long and Xs = short. The cross can be rewritten as follows. P: bb XsXs x BB XSY F1: 50% female: Bb XSXs 50% male: Bb XsYF2: 1/16 BB XSXs - red, long, ♀ 1/16 BB XsXs - red, short ♀ 1/8 Bb XSXs - red, long, ♀ 1/8 Bb XsXs - red, short ♀ 1/16 bb XSXs - brown, long ♀ 1/16 bb XsXs - brown, short ♀ 1/16 BB X SY - red, long, ♂ 1/16 BB X sY - red, short ♂ 1/8 Bb X SY - red, long, ♂ 1/8 Bb X sY - red, short, ♂ 1/16 bb X SY - brown, long ♂ 1/16 bb X sY - brown, short, ♂ The final phenotypic ratio is: 3/8 red, long 3/8 red, short 1/8 brown, long 1/8, brown, short with equal ♂ and ♀ in all classes
A dominant allele H reduces the number of body bristles that Drosophila flies have, giving rise to the "hairless" phenotype. In the homozygous condition, H is lethal. An independently assorting dominant allele S has no effect on bristle number except in the presence of H, in which case a single dose of S suppresses the hairless phenotype, thus restoring the hairy phenotype. However, S also is lethal in the homozygous (S/S) condition. A) What ratio of hairy:hairless flies would you find in the live progeny of a cross between two hairy flies both carrying H in the suppressed condition? B) When the hairless progeny are backcrossed with a parental hairy fly, what phenotypic ratio would you expect to find among their live progeny?
The cross is Hh Ss x Hh Ss. Because this is a typical dihybrid cross, the expected ratio is 9:3:3:1. However, the problem can't be worked in this simple fashion because of the epistatic relationship of these two genes. Therefore, the following approach should be used. For the H gene, you should expect 1 (HH): 2 (Hh): 1 (hh). For the S gene, you should expect 1 (SS): 2 (Ss): 1 (ss). To get the final ratios, use the forked method: A) 1/4 HH All progeny dead regardless of S. THESE ARE NOT COUNTED 1/4 SS 1/8 - Dead THESE ARE NOT COUNTED 1/2 Hh 1/2 Ss 1/4 - hairy (H is suppressed by S) 1/4 ss 1/8 - hairless 1/4 SS 1/16 - Dead THESE ARE NOT COUNTED 1/4 hh 1/2 Ss 1/8 - hairy 1/4 ss 1/16 - hair Of the living progeny (9/16), the ratio is 7 hairy: 2 hairless. B) 1/4 HH All progeny dead regardless of S. THESE ARE NOT COUNTED 1/2 Hh 1/2 Ss 1/4 - hairy (H is suppressed by S) 1/2 ss 1/4 - hairless 1/4 hh 1/2 Ss 1/8 - hairy 1/2 ss 1/8 - hair Of the living progeny (12/16), the ratio is 2 hairy: 1 hairless.
Black, sepia, cream, and albino are coat colors of guinea pigs. Individual animals (not necessarily from pure lines) showing these colors were intercrossed; the results are tabulated as follows, where the abbreviations A (albino), B (black), C (cream), and S (sepia), represent the phenotypes. A) Deduce the inheritance of these coat colors. Show all parent and progeny genotypes. B) If the black animals in crosses 5 and 6 are crossed, what progeny proportions can you predict by using your model?
The cross would be Cb/Ca x Cb/C- . Therefore, the progeny would be 3 Black to 1 Cream or Albino (depending on identity of the "-"). - look at answer sheet
From a presumed testcross A/a x a/a, in which A represents red and a represents white, use the chi-squared test to find out which of the following possible results would fit the expectations: A) 120 red, 100 white B) 5000 red, 5400 white C) 500 red, 540 white D) 50 red, 54 white6) In a certain plant, the flower petals are normally purple. Two recessive mutations arise in separate plants and are found to be on different chromosomes. Mutation 1 (m1) gives blue petals when h
The hypothesis is that the organism being tested is a heterozygote and that the A/a and a/a progeny are of equal viability. The expected values would be that phenotypes occur with equal frequency. There are two genotypes in each case, so there is one degree of freedom. � 2 = ∑ (observed-expected)2 /expected a. � 2 = [(120-110)2 + (100-110)2 ]/110 = 1.818; 0.05 < p < 0.2, nonsignificant; hypothesis cannot be rejected b. � 2 = [(5000-5200)2 + (5400-5200)2 ]/5200 = 15.385; p < 0.001, significant; hypothesis must be rejected c. � 2 = [(500-520)2 + (540-520)2 ]/520 = 1.538; 0.2 < p < 0.5, nonsignificant; hypothesis cannot be rejected d. � 2 = [(50-52)2 + (54-52)2 ]/52 = 0.154; 0.5 < p < 0.9 , nonsignificant; hypothesis cannot be rejected 6) a. If enzyme A was defective or missing ( m2/m2), red pigment would still be made, and the petals would be red. b. Purple, because it has a wild-type allele for each gene, and you are told that the mutations are recessive.
In cats, blood type A results from an allele (IA) that is dominant over an allele (IB) that produces blood type B. There is no O blood type. The blood types of male and female cats that were mated and the blood types of their kittens follow. Give the most likely genotypes for the parents of each litter. A) ♂ A x ♀ B = 4 A and 3 B B) ♂ B x ♀ B = 6 B C) ♂ B x ♀ A = 8 A D) ♂ A x ♀ A = 7 A and 2 B E) ♂ A x ♀ A = 10 A F) ♂ A x ♀ B = 4 A and 1 B
a) The female parent must have the genotype IBIB. The male parent must have the genotype IAIB. b) Both parents must be homozygous for the recessive allele (IBIB). c) The male must be IBIB. A female with type A blood could be either IAIA or IAIB. The fact that all offspring have type A blood suggests that the female is heterozygous dominant. She could be IAIB, but it is unlikely that chance alone would have produced eight kittens with blood type A. d) Both parents must be heterozygous e) Either both parents are homozygous for blood type A or one parent is homozygous for blood type A and the other parent is heterozygous. The blood types of the offspring will not allow us to determine the precise genotype of either parent. f) The female parent is homozygous B and the male parent is heterozyg
The pedigree below was obtained for a rare kidney disease. note: Interpret rare to mean that anyone unlreated who marries into this family would be assumed to not be carrying the allele (example: II 1, II 5, III 5 and III 6). look at pedigree A) Deduce the inheritance of this condition, stating your reasons. B) If persons 1 and 2 marry, what is the probability that their first child will have the kidney disease?
a. Autosomal recessive: affected individuals inherited the trait from unaffected parents and a daughter inherited the trait from an unaffected father (therefore not sex-linked). b. Both parents must be heterozygous to have a 1/4 chance of having an affected child. Parent 2 is heterozygous (her father was homozygous recessive, but she has unaffected). Parent 1 has a 50% chance of being heterozygous dominant and a 50% chance of being homozygous dominant (her father is heterozygous due to his affected mother; the allele is very rare and therefore it can be assumed that her mother was homozygous dominant). In order for the child to have the disease, both Parent 1 and Parent 2 would need to be heterozygous (there is a 50% chance of this for Parent 1 and a 100% chance of this for Parent 2) AND then, there is a 25% chance that their first child will be affected: 50% * 100% * 25% = 12.5% also: 1/2 * 1 * 1/4 = 1/8
In a certain plant, the flower petals are normally purple. Two recessive mutations arise in separate plants and are found to be on different chromosomes. Mutation 1 (m1) gives blue petals when homozygous (m1/m1). Mutation 2 (m2) gives red petals when homozygous (m2/m2). Biochemists working on the synthesis of flower pigments in this species have already described the following pathway: A) Which mutant would you expect to be deficient in enzyme A activity? B) A plant has the genotype (M1/m1; M2/m2). What would you expect its phenotype to be? C) If the plant in part b is selfed, what colors of progeny would you expect and in what proportions? D) Why are these mutations recessive? E) Are these genes epistatic to each other?
a. If enzyme A was defective or missing ( m2/m2), red pigment would still be made, and the petals would be red. b. Purple, because it has a wild-type allele for each gene, and you are told that the mutations are recessive. c. 9 M1/-;M2/- purple 3 m1/m1 ;M2/- blue 3 M1/- ; m2/m2 red 1 m1/m1 ; m2/m2 white d. The mutant alleles do not produce functional enzyme. However, enough functional enzyme must be produced by the single wild-type allele of each gene to synthesize normal levels of pigment. e) The 9:3:3:1 gene ratio suggests no genetic or biochemical in
To understand the genetic basis of locomotion in C. elegans, recessive mutations were obtained, all making the worm "wiggle" ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F1 hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F1 hybrid was wild type (gliding) and the "w" means that the hybrid wiggled. A) Explain what this experiment was designed to test. B) Use this reasoning to assign genotypes to all 12 mutants C) Explain why the phenotype of the F1 hybrids between mutants 1 and 2 differed from that of the hybrids between mutants 1 and 5.
a. Intercrossing mutant strains that all share a common recessive phenotype is the basis of the complementation test. This test is designed to identify the number of different genes that can mutate to a particular phenotype. In this problem, if the progeny of a given cross still express the wiggle phenotype, the mutations fail to complement and are considered alleles of the same gene; if the progeny are wild type, the mutations complement and the two strains carry mutant alleles of separate genes. b. From the data, 1 and 5 fail to complement - gene A 2, 6, 8, and 10 fail to complement - gene B 3 and 4 fail to complement - gene C 7, 11, and 12 fail to complement - gene D 9 complements all others - gene E There are five complementation groups (genes) identified by this data. c. mutant 1: a 1 /a 1 · b + /b + · c + /c + · d + /d + · e + /e + (although, only the mutant alleles are usually listed) mutant 2: a + /a + · b 2 /b 2 · c + /c + · d + /d + · e + /e + mutant 5: a 5 /a5 · b + /b + · c + /c + · d + /d + · e + /e + 1/5 hybrid: a 1 /a5 · b + /b + · c + /c + · d + /d + · e + /e + phenotype: wiggles Conclusion: 1 and 5 are both mutant for gene A. 2/5 hybrid: a + /a5 ,· b + /b 2 · c + /c + · d + /d + · e + /e + phenotype: wild type Conclusion: 2 and 5 are mutant for different genes.
) In the fruit fly Drosophila melanogaster, red and yellow eye pigments are synthesized through the pterin pathway. These, plus the brown pigment synthesized through the ommochrome pathway, produce the dark red Drosophila eye color. A part of the pterin pathway (simplified) is shown below. Consider the genes encoding Enzyme 1 ("e1"), Enzyme 2 ("e2") and Enzyme 3 ("e3"), respectively. What are their predicted genetic interactions? Which gene(s) is(are) epistatic to which gene(s)? Are they recessively or dominantly epistatic? Assume each enzyme is haplosufficient.
e1 is recessively epistatic to e2 and e3. Assuming e1 is haplosufficient, flies homozygous mutant for e1 will have colorless (white) eyes regardless of the genotypes at e2 and e3. In other words, recessive alleles at e1 mask the phenotypes associated with e2 and
