Quiz 6 Question Pool Spring 2023

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Quiz 6 Question Pool Spring 2023 ALWAYS SHOW YOUR WORK! Avoid the use of ambiguous symbols for dominant vs. recessive alleles. Hint: you may have to consult your textbook (Chapter 7.5 for metabolic pathways) for some of the answers! Questions on metabolic pathways : 1) What are the nutritional needs of wildtype Neurospora (excluding water)? (3 points) -suitable carbon source -some inorganic salts -the vitamin biotin 2) What are hyphae & mycelia? (2 points) (*mycelium is made up of hyphae) Mycelium is the vegetative part of a fungus or fungus-like bacterial colony - it consists of hyphae (a mass of branching, thread-like) 3) Does Neurospora spend most of its life as a haploid or as a diploid? (1 point) haploid 4) How can mutations in two different genes give rise to the same phenotype? (2 points) the product of the blocked synthesis is being added to the minimal medium so even though they all can grow, each mutation maps at a unique site and a different step in the ordered biosynthesis of a product (ex. arginine) is affected in each arg- strain we use 5) Name two precursors in the biosynthesis of arginine. (2 points) orginine ---> citrulline ---> arginine 6) What kind of spores does Neurospora produce asexually? (1 point) conidia 7) Precursor-------------->Succinate--------------->Fumarate------------->Malate---------------->Oxaloacetate Ketoglutaric Succinic Fumarase Malate Dehydrogenase dehydrogenase dehydrogenase a. If a Neurospora mutant can grow when given fumarate or malate but not succinate, which enzyme has been mutated? (1 point) this is a fumurate-deficient strain: the enzyme mutated is succinic dehyrogenase b. Can this mutant grow if given oxaloacetate? (1 point) yes Problem 4-9 (4-23 in 6th edition): Somatic cells of chimpanzees contain 48 chromosomes.
How many chromatids and chromosomes are present at (a) anaphase of mitosis, (b) anaphase I of meiosis, (c) anaphase II of meiosis, (d) G 1 prior to mitosis, (e) G 2 prior to mitosis, (f) G 1 prior to meiosis I, and (g) prophase of meiosis I? (a) anaphase of mitosis: During this phase sister chromatids split. We would have 48 chromosomes in each pole and 48 chromatids. (b) anaphase I of meiosis: During this phase homologous chromosomes split, being a reductional division. In each pole we will have half the chromosomes we had after DNA replication. This is 24 chromosomes but 48 chromatids (remember they will split during anaphase II). (c) anaphase II of meiosis: This is an equational division, we will have 24 chromosomes in each pole and 24 chromatids. Each chromatid is considered a chromosome. (d) G1 prior to mitosis: During this phase DNA has not replicated yet and it is not condensed either. This formed is called chromatin. We will assign one chromatid for each chromosome. This is a somatic cell, so: 48 chromosomes and 48 chromatids. (e) G2 prior to mitosis: After S phase, we have duplicated all chromosomes. We will assign two chromatids per chromosome: this is 96 chromatids and 48 chromosomes. (f) G1 prior to meiosis: Before DNA duplication, 48 chromosomes, 48 chromatids. (g) Prophase of meiosis I: After DNA replication, condensation of the chromatin takes place: 48 chromosomes, 96 chromatids. How many chromatids and chromosomes are present in (h) an oogonial cell prior to S phase, (i) a spermatid, (j) a primary oocyte arrested prior to ovulation, (k) a secondary oocyte arrested prior to fertilization, (l) a second polar body, and (m) a chimpanzee sperm? (1 point each) (h) An original cell prior to S phase: This is G1 phase, 48 chromosomes. (i) A spermatid: This is the male haploid gamete, after meiosis: 24 chromosomes and 24 chromatids. (j) A primary oocyte arrested prior to ovulation: They are arrested at prophase I of meiosis. This means their DNA is still duplicated and chromatids have not divided yet. 48 chromosomes and 96 chromatids. (k) A secondary oocyte arrested prior to fertilization: They are halted at metaphase II of meiosis, meaning they have half the chromosomes (24) but chromatids are still together (48).
(l) A second polar body: They suffered the same process as the mature ovum but remained with little cytoplasm. They have 24 chromosomes and 24 chromatids. (m) A chimpanzee sperm: They have completed both meiosis as well, they have 24 chromosomes and 24 chromatids.   Problem 4-12 (the Jurassic Park problem!): In a certain strain of turkeys, unfertilized eggs sometimes develop parthenogenetically to produce diploid offspring. (Females have ZW and males have ZZ sex chromosomes. Assume that WW cells are inviable). What distribution of sexes would you expect to see among the parthenogenetic offspring according to each of the following models for how parthenogenesis occurs? (2 points each) a. The eggs develop from oogonial cells that never undergo meiosis. ZW   female eggs can give rise only to   ZW   females. b. The eggs go all the way through meiosis and then duplicate their chromosomes to become diploid. After eggs go through meiosis, there can be   Z   or   W   chromosome in the oocyte. After duplication we receive:   ZZ   and   WW . Since   WW   genotype is not vialble, there can be only males   ZZ . c. The eggs go through meiosis I, and the chromatids separate to create diploidy. After eggs go through meiosis I, if the chromatids separate and duplicate, we receive  ZZ  and  WW  genotypes. The last one is not viable, so there are only  ZZ  males d. The egg goes all the way through meiosis and then fuses at random with one of its three polar bodies (this scenario assumes the first polar body goes through meiosis II). The genotypes of the offspring are the following: 2 ZZ  - 2/10, or 1/5 males; 8 ZW  - 8/10, or 4/5 females; 2 WW  - not viable. Problem 4-13 (4-27 in 6th edition): Imagine you have two pure-breeding lines of canaries, one with yellow feathers and one with brown feathers. In crosses between these two strains, yellow female x brown male gives only brown sons and daughters, while brown female x yellow male gives only brown sons and yellow daughters. Propose a hypothesis to explain these results. (3 points) In birds, ZW= female, ZZ= male; alleles are on Z chromosome and Z^B > Z^Y Problem 4-15 (4-29 in 6th edition):
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In Drosophila , the autosomal recessive brown eye color mutation displays interactions with both the X- linked recessive vermilion mutation and the autosomal recessive scarlet mutation. Flies homozygous for brown and simultaneously hemizygous or homozygous for vermilion have white eyes. Flies simultaneously homozygous for both the brown and scarlet mutations also have white eyes. Predict the F 1 and F 2 progeny of the following true-breeding parents: (3 points each) a. vermilion females x brown males F1: XvY b+b – vermillion males; and XvXv+b+b – WT females F2:  both males and females have the same ratios: 1/8 white, 3/8 WT, 1/8 brown and   b. brown females x vermilion males F1 =  wild type females (XvXv+b+b) and WT males (Xv+Yb+b)   F2: females ¾ v+_b+_ (WT): ¼ v+_bb (brown) c. scarlet females x brown males F1: sc/sc+; b+/b – all are WT F2: no x-linked alleles in this cross. 9 WT: 3 scarlet: 3 brown: 1 white d. brown females x scarlet males F1 b+/b ; sc+/sc   F2 is the same as cross #3. Problem 4-18: In 1919, Calvin Bridges began studying an X-linked recessive mutation causing eosin-colored (pink) eyes in Drosophila . Within an otherwise true-breeding culture of eosin-eyed flies, he noticed rare variants that had much lighter cream-colored eyes. By intercrossing these variants, he was able to make a true-breeding cream-eyed stock. Bridges now crossed males from this cream-eyed stock with true- breeding wild-type females. All the F 1 progeny had red (wild-type) eyes. When F 1 flies were intercrossed, the F 2 progeny were 104 females with red eyes, 52 males with red eyes, 44 males with eosin eyes, and 14 males with cream eyes. Assume these numbers represent an 8: 4: 3: 1 ratio. a. Formulate a hypothesis to explain the F 1 and F 2 results, assigning eye colors to all possible genotypes. (3 points) b. What do you predict in the F 1 and F 2 generations if the parental cross is between true-breeding eosin- eyed males and true-breeding cream-eyed females? (2 points) c. What do you predict in the F 1 and F 2 generations if the parental cross is between true-breeding eosin- eyed females and true-breeding cream-eyed males? (2 points) Problem 4-20: As we learned in this chapter, the white mutation of Drosophila studied by Thomas Hunt Morgan is X- linked and recessive to wild type. When true-breeding white-eyed males carrying this mutation were
crossed with true-breeding purple-eyed females, all the F 1 progeny had wild-type (brownish-red) eyes. When the F 1 progeny were intercrossed, the F 2 progeny emerged in the ratio 3/8 wild-type females: 1/4 white-eyed males: 3/16 wild-type males: 1/8 purple-eyed females: 1/16 purple-eyed males. a. Formulate a hypothesis to explain the inheritance of these eye colors, assigning eye colors to all possible genotypes. (3 points) b. Predict the F 1 and F 2 progeny if the parental cross was reversed (that is, if the parental cross was between true-breeding white-eyed females and true-breeding purple-eyed males). (3 points) Problem 4-24 (4-37 in 6th edition): Each of the four pedigrees shown here represents a human family within which a genetic disease is segregating. Affected individuals are indicated by filled-in symbols. One of the diseases is transmitted as an autosomal recessive condition, one as an X-linked recessive, one as an autosomal dominant, and one as an X-linked dominant. Assume all four traits are rare in the population. a. Indicate which pedigree represents which mode of inheritance, and explain how you know . (4 points) b. For each pedigree, how would you advise the parents of the chance that their child (indicated by the hexagon shape) will have the condition? (1 point each) Problem 4-33 (4-46 in 6th edition): The ancestry of a white female tiger bred in a city zoo is depicted in the following pedigree. White tigers are indicated with unshaded symbols. (As you can see, there was considerable inbreeding in this lineage. For example, the white tiger Mohan was mated with his daughter). In answering the following questions, assume that “white” is determined by allelic differences at a single gene and that the trait is fully penetrant. Explain your answers by citing the relevant information in the pedigree. Could white coat color be caused by a: (2 points each) a. Y-linked allele? b. dominant X-linked allele? c. dominant autosomal allele? d. recessive X-linked allele? e. recessive autosomal allele?
Problem 4-31: Consider the following pedigrees from human families containing a male with Klinefelter syndrome (a set of abnormalities seen in XXY individuals; indicated with shaded boxes). In each, A and B refer to codominant alleles of the X-linked G6PD gene. The phenotypes of each individual (A, B, or AB) are shown on the pedigree. Indicate if non-disjunction occurred in the mother or father of the son with Klinefelter syndrome for each of the three examples. Can you tell if the nondisjunction was in the first or second meiotic division? (2 points each) Problem 5-8: CC DD and cc dd individuals were crossed to each other, and the F 1 generation was backcrossed to the cc dd parent. 903 Cc Dd , 897 cc dd , 98 Cc dd , and 102 cc Dd offspring resulted. a. How far apart are the c and d loci? (1 point) b. What progeny and in what proportions would you expect to result from testcrossing the F 1 generation from a CC dd x cc DD cross to cc dd ? (3 points)
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Problem 5-9: In mice, the autosomal locus coding for the β-globin chain of hemoglobin is 1 m.u. from the albino locus. Assume for the moment that the same is true in humans. The disease sickle-cell anemia is the result of homozygosity for a particular mutation in the β-globin gene. a. A son is born to an albino man and a woman with sickle-cell anemia. What kinds of gametes will the son form, and in what proportions? (2 points) b. A daughter is born to a normal man and a woman who has both albinism and sickle-cell anemia. What kinds of gametes will the daughter form, and in what proportions? (2 points) c. If the son in part a grows up and marries the daughter in part b, what is the probability that a child of theirs will be an albino with sickle-cell anemia? (1 point) Problem 5-11: Albino rabbits (lacking pigment) are homozygous for the recessive c allele ( C allows pigment formation). Rabbits homozygous for the recessive b allele make brown pigment, while those with at least one copy of B make black pigment. True-breeding brown rabbits were crossed to albinos, which were BB . F 1 rabbits, which were all black, were crossed to the double recessive ( bb cc ). The progeny obtained were 34 black, 66 brown, and 100 albino. a. What phenotypic proportions would have been expected if the b and c loci were unlinked? (2 points) b. How far apart are the two loci? (3 points) Problem 5-12: In corn, the allele A allows the deposition of anthocyanin (blue) pigment in the kernels (seeds), while aa plants have yellow kernels. At a second gene, W- produces smooth kernels, while ww kernels are wrinkled. A plant with blue smooth kernels was crossed to a plant with yellow wrinkled kernels. The progeny consisted of 1447 blue smooth, 169 blue wrinkled, 186 yellow smooth, and 1510 yellow wrinkled. a. Are the a and w loci linked? If so, how far apart are they? (1 point) b. What was the genotype of the blue smooth parent? Include the chromosome arrangement of alleles. (2 points) c. If a plant grown from a blue wrinkled progeny seed is crossed to a plant grown from a yellow smooth F 1 (progeny) seed, what kinds of kernels would be expected, and in what proportions? (3 points) Problem 5-13: If the a and b loci are 40 cM apart, and an AA BB individual and an aa bb individual mate: a. What gametes will the F 1 individuals produce, and in what proportions? What phenotypic classes in what proportions are expected in the F 2 generation (assuming complete dominance for both genes)? (3 points) b. If the original cross was AA bb x aa BB , what gametic proportions would emerge from the F 1 ? What would be the result in the F 2 generation? (2 points) Problem 5-22: In the tubular flowers of foxgloves, wild-type coloration is red while a mutation called white produces white flowers. Another mutation called peloria causes the flowers at the apex of the stem to be huge.
Yet another mutation called dwarf affects stem length. You cross a white-flowered plant (otherwise phenotypically wild-type) to a plant that is dwarf and peloria but has wild-type red flower color. All of the F 1 plants are tall with white, normal-sized flowers. You cross an F 1 back to the dwarf and peloria parent, and you see the 543 progeny shown below (only mutant traits are noted): dwarf, peloria 172 white 162 dwarf, peloria, white 56 wild-type 48 dwarf, white 51 peloria 43 dwarf 6 peloria, white 5 a. Which alleles are dominant? (1 point) b. What were the genotypes of the parents in the original cross? (1 point) c. Draw a map showing the linkage relationships of these three loci. (3 points) d. Is there interference? If so, calculate the coefficient of coincidence and the interference value. (2 points) Problem 5-26: Drosophila females heterozygous for each of the three recessive autosomal mutations with independent phenotypic effects (thread antennae [ th ], hairy body [ h ], and scarlet eyes [ st ]) were testcrossed to males showing all three mutant phenotypes. The 1000 progeny of this testcross were: thread, hairy, scarlet 432 wild type 429 thread, hairy 37 thread, scarlet 35 hairy 34 scarlet 33 a. Show the arrangement of alleles on the relevant chromosomes in the triply heterozygous females. (1 point) b. Draw the best genetic map that explains these data. (3 points) c. Calculate any relevant interference values. (1 point) Problem 5-27: Male Drosophila expressing the recessive mutations sc ( scute ), ec ( echinus ), cv ( crossveinless ), and b ( black ) were crossed to phenotypically wild-type females, and the 3288 progeny listed here were obtained. 653 black, scute, echinus, crossveinless 670 scute, echinus, crossveinless 675 wild type 655 black 71 black, scute
73 scute 73 black, echinus, crossveinless 74 echinus, crossveinless 87 black, scute, echinus 84 scute, echinus 86 black, crossveinless 83 crossveinless 1 black, scute, crossveinless 1 scute, crossveinless 1 black, echinus 1 echinus a. Diagram the genotype of the female parent. (1 point) b. Map these loci. (3 points) c. Is there evidence of interference? Justify your answer with numbers. (1 point) Problem 5-29: A true-breeding strain of Virginia tobacco has dominant alleles determining leaf morphology ( M ), leaf color ( C ), and leaf size ( S ). A Carolina strain is homozygous for the recessive alleles of these three genes. These genes are found on the same chromosome as follows: M C S 6 m.u. 17 m.u. An F 1 hybrid between the two strains is now backcrossed to the Carolina strain. Assuming no interference: a. What proportion of the backcross progeny will resemble the Virginia strain for all three traits? (1 point) b. What proportion of the backcross progeny will resemble the Carolina strain for all three traits? (1 point) c. What proportion of the backcross progeny will have the leaf morphology and leaf size of the Virginia strain but the leaf color of the Carolina strain? (2 points) d. What proportion of the backcross progeny will have the leaf morphology and leaf color of the Virginia strain but the leaf size of the Carolina strain? (1 point) Problem 5-31 (shortened): In Drosophila , the dwarp mutation leads to flies with small bodies and warped wings, the rumpled mutation causes bristles to be deranged, the pallid mutation makes wings pale, while raven leads to dark eyes and bodies. Females heterozygous for all 4 genes are crossed to males from a true-breeding dwarp rumpled pallid raven stock. The 1000 progeny obtained were as follows: pallid 3 pallid, raven 428 pallid, raven, rumpled 48 pallid, rumpled 23 dwarp, raven 22 dwarp, raven, rumpled 2
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dwarp, rumpled 427 dwarp 47 Indicate the best map for these four genes. Calculate interference values where appropriate. (5 points) Problem 5-34 (shortened): Do the data that Mendel obtained fit his hypothesis? For example, Mendel obtained 315 yellow round, 101 yellow wrinkled, 108 green round, and 32 green wrinkled seeds from the selfing of Yy Rr individuals (a total of 556). Use the chi-square test to determine whether Mendel’s data are significantly different from what he predicted. (3 points) Problem 5-35: Two genes control color in corn snakes as follows: O- B- snakes are brown, O- bb are orange, oo B- are black, and oo bb are albino. An orange snake was mated to a black snake, and a large number of F 1 progeny were obtained, all of which were brown. When the F 1 snakes were mated to one another, they produced 100 brown offspring, 25 orange, 22 black, and 13 albino. a. What are the genotypes of the F 1 snakes? (1 point) b. What proportions of the different colors would have been expected among the F 2 snakes if the two loci assort independently? (1 point) c. Do the observed results differ significantly from what was expected, assuming independent assortment is occurring? What is the probability that differences this great between observed and expected values would happen by chance? (3 points)