BIOL2323 S2024_Midterm 2 key

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Biology

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Apr 3, 2024

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1) What is the biological definition of a “female” organism? A) Female organisms always have two X chromosomes. B) Female organisms always give birth to live offspring. C) Female organisms are always more drab than males. D) Females produce larger gametes compared with males. (94%) E) There is no biological definition of a female organism. 2) Bees have a haploid number (n) of 16. Female bees are diploid, and male bees are haploid. How many chromosomes and chromatids does a male bee gamete (sperm) contain? A) 8 ; 8 B) 8 ; 16 C) 16 ; 16 (80%) D) 16 ; 32 E) 32 ; 32 3) Based on what you know about Drosophila sex determination, what sexual phenotype should a fly with two X chromosomes, two Y chromosomes, and two sets of autosomes display? A) It should have both male and female genitals. B) It should look completely female. (70%) C) It should look completely male. D) It should look completely male, but it will be sterile. E) There is a 50:50 chance that it will be male or female. 4) X0 individuals develop as _______ in Drosophila and ______ in humans. A) male ; male B) male ; female (79%) C) female ; female D) female ; male E) none of the above, because an XO genotype is developmentally lethal 5) You have a red dye that specifically binds to Barr bodies, and a blue dye that binds to all X chromosomes. You apply these two dyes to normal human male cheek cells. You should expect to see ___ red signal(s) and ___ blue signal(s) in the nucleus of each cell. A) 0 ; 0 B) 0 ; 1 (81%) C) 1 ; 1 D) 1 ; 2 E) 2; 2

6) Which of the following genotypes describes an individual that is hemizygous for all X-linked genes? A) 45, XO B) 46, XX C) 46, XY D) 47, XXY E) Both 45, XO and 46, XY (93%) 7) The Drosophila yellow gene is located on the X chromosome. Flies that are homozygous or hemizygous for a recessive y allele have yellow bodies. You cross a homozygous wild-type female to a yellow male. What should you expect to see in their offspring ? A) 100% wild type, regardless of sex (79%) B) 100% yellow, regardless of sex C) yellow males and wild-type females D) wild-type males and yellow females E) 50% wild type and 50% yellow, regardless of sex 8) The Drosophila miniature gene is located on the X chromosome. Flies that are homozygous or hemizygous for a recessive m allele are significantly smaller than wild-type flies. You cross a miniature female to a wild-type male. What should you expect to see in their offspring ? A) 100% normal size, regardless of sex B) 100% miniature, regardless of sex C) Miniature males and wild-type females (79% C or D) D) wild-type females and miniature males E) 50% wild-type and 50% miniature, regardless of sex 9) You set up a Drosophila parental cross involving a white-eyed female and a red-eyed male. You then crossed the resulting F1 siblings together. What proportions of phenotypes would you expect to see in the F2 generation ? A) 1/2 red females, 1/4 red males, and 1/4 white males B) 1/2 red males, 1/4 red females, and 1/4 white females C) 1/4 red females, 1/4 white females, 1/4 red males, and 1/4 white males (68%) D) 1/2 red males and 1/2 white females E) It is impossible to predict due to crossing over 10) Marfan Syndrome is an autosomal dominant disorder. Which of the following pedigrees is most consistent with the inheritance of Marfan Syndrome?

A) A (95%) B) B C) C D) B and C are equally likely E) None of these pedigrees is consistent with an autosomal recessive disorder 11) If individuals II-4 and II-5 had a daughter, what is the likelihood that this child would be affected by this condition? A) 0 B) 1/4 C) 4/9 D) 1/36 E) 4/36 (61%) 2/3 x 2/3 x ¼ = 4/36. The daughter part is irrelevant 12) Fragile X Syndrome is an X-linked dominant trait. A father with Fragile X syndrome should expect to pass this trait on to: A) none of his children. B) all of his children, regardless of sex. C) all of his sons and none of his daughters. D) all of his daughters and none of his sons. (91%) E) 50% of his children, regardless of sex. 13) The inability to taste phenylthiocarbamide is an autosomal recessive trait. People with this trait are called “nontasters,” and wild-type individuals are called “tasters.”

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A taster woman (who had a nontaster father) marries a taster man (who in a previous relationship fathered a nontaster daughter), and they are going to have a child. What is the probability that this child be a nontaster girl? A) 0 B) 1/18 C) 1/8 (72%) D) 1/6 E) 1/9 Both parents must be carriers, so all you have to do is calculate the two following probabilities and multiply them: P(two carriers have an affected offspring) = ¼ P(offspring will be a girl) = ½ 14) If individuals II-2 and II-4 were to have a child, what is the likelihood the child would be affected by this condition? A) 0 B) 1/9 C) 1/2 D) 1/4 (66%) E) 1 This is very similar to question 13. You can infer from the pedigree that both II-2 and II-4 must be carriers. If they had a child, then it’s simply the likelihood that two carriers would have an affected child. 15) Red-green colorblindness is a recessive X-linked condition. A woman who can see color, but who had a colorblind father, has several children with a man who is not colorblind. Their first child is a boy who can see color. They are expecting a second child, and they know it will also be a boy. What is the probability that their second son will be colorblind? A) 0% B) 25% C) 50% (72%) D) 66.7% E) 100%

From the question you should be able to infer that we are asking about a mating between a carrier female and a wild-type male. If they are going to have a son, then we know the father passed on his Y chromosome. The only unknown is whether the mother will pass on the normal or mutant X chromosome. So it’s a 50/50 chance. 16) Consider the following weird pedigree. Based on what you know about genetics and probability, what is the most likely genotype of individual III-5? Choose the best answer . A) AA B) Aa C) Aa (61%) D) It’s impossible to make any kind of educated guess E) Equally likely to be AA or Aa This was actually one of the quiz questions, and it’s asking you to think a little more deeply about probabilities. Individual III-6 passes on the recessive trait to all of her offspring. It is technically possible that II-5 is a carrier based on their parents. If so, they would pass on a recessive trait to 50% of their offspring, in which case you’d expect ~50% of generation IV should show the trait. But none of them do, and it’s not a small sample size. This would be like flipping a coin 7 times and getting 7 heads. Possible, but highly unlikely. The BEST and most likely answer, is that III-5 is not a carrier (aa). 17) Crossing over (recombination) occurs: A) between non-homologous chromosomes during mitosis B) between sister chromatids in a tetrad C) between non-sister chromatids in a tetrad (54%) D) only during meiosis in males E) all the time between closely linked genes 18) A chi-square value is a measure of the _______ between the observed and expected outcomes for a set of measurements. A) dominance

B) degrees of freedom C) null hypothesis D) phenotypic classes E) Deviation (96%) Consider the following scenario for questions 19-21 You are characterizing a new mutation in fruit flies that causes a blue-eye ( b ) phenotype. You have a true-breeding line that has blue eyes and curved wings, and when you cross these flies to wild-type flies, all the F1 progeny are wild type (red eyes and straight wings). You know that the curved wing trait is controlled by the c gene on the second chromosome. You cross your F1 dihybrids back to the true-breeding blue-eyed, curved wings strain and you obtain the following F2 progeny. 115 red-eyed, straight wings 80 blue-eyed, straight wings 85 red-eyed, curved wings 120 blue-eyed, curved wings 19) You decide to perform a chi-square analysis to determine whether these two genes are behaving in a Mendelian fashion. Your null hypothesis should be: A) The b and c genes assort independently, and we should observe a 1:1:1:1 phenotypic ratio in the F2. (76%) B) The b and c genes assort independently, and we should observe a 9:3:3:1 phenotypic ratio in the F2. C) The b and c genes do not assort independently, and we should observe a 1:1:1:1 phenotypic ratio in the F2. D) The b and c genes do not assort independently, and we should observe a 9:3:3:1 phenotypic ratio in the F2. E) The b and c genes do not assort independently, and thus we cannot make any predictions about F2 phenotypes. Choice A is what Mendel would have predicted, which should always be your null hypothesis. 20) Perform a chi-square analysis on the above dataset (using the chart located on the beginning of the exam). You should get a p value in the range of: a. 0.90 – 0.50 b. 0.50 – 0.20 c. 0.20 – 0.05 d. 0.05 – 0.01 e. 0.01 – 0.001 (62%) You should calculate a chi-square value of 12.5. With 3 degrees of freedom, that gives you a p value somewhere between 0.01 and 0.001. 21) Based on the results of your chi-square analysis in the previous question, and a threshold of p < 0.05, you should do which of the following: A) Fail to reject the null hypothesis and conclude that these genes are linked.

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B) Fail to reject the null hypothesis and conclude that these genes are unlinked. C) Reject the null hypothesis and conclude that these genes are linked. (43%) D) Reject the null hypothesis and conclude that these genes are unlinked. E) Neither, because the calculated p value was exactly 0.05 The p value is less than 0.05, so you must reject the null hypothesis (the genes are unliked or “assort independently”) and conclude that these genes are linked, which explains why your data don’t fit the 1:1:1:1 prediction. Consider the following scenario for questions 22-23 You are examining two traits in cucumbers. For flower color, the dominant allele of the w gene produces yellow flowers, and the recessive allele produces white flowers. For fruit texture, the dominant allele of the r gene produces smooth cucumbers, and the recessive allele produces rough cucumbers. You mate a dihybrid plant showing both dominant traits to plant showing both recessive traits. You obtain the following numbers of offspring: yellow, smooth 92 white, rough 93 yellow, rough 7 white, smooth 8 22) Based on the above information, what is most likely true concerning genes w and r ? A) Genes w and r are located on non-homologous chromosomes. B) Genes w and r are located close together on non-homologous chromosomes. C) Genes w and r are located on opposite ends of the same chromosome. D) Genes w and r are located close together on the same chromosome. (74%) E) We can’t infer anything about the physical relationship of these genes from this dataset. 23) In the above scenario, the map distance between genes w and r is: A) 0.75 mu B) 1.5 mu C) 7.5 mu (86%) D) 15 mu E) 30 mu Recombinants/total progeny * 100 = (7+8)/200 *100 = 7.5 24) What is the maximum possible distance between two genes that can be measured in a single experiment?

A) 0 cM B) 25 cM C) 50 cM (72%) D) 100 cM E) There is no upper limit 25) Which of the following is a valid way of describing a diploid organism that is heterozygous for recessive alleles of three genes located on the same chromosome? A) g h i + + + B) g / + ; h / + ; i / + C) g + + / + h i (85%) D) + / g ; h i / + + E) g g / h h / i i Consider the following scenario for questions 26 and 27 Genes t and u are located 40 map units apart. You perform a testcross on a Tu/tU dihybrid individual, and you observe 500 offspring. 26) How many offspring with a double-recessive ( tu ) phenotype should you expect to see? A) 20 B) 40 C) 100 (50%) D) 150 E) 200 A map distance of 40 units means that 40% of the offspring will be split between the two recombinant classes (each representing 20%). And you know the parental classes from the question, so: Parental 1 Tu/tu 30% Parental 2 tU/tu 30% Recom 1 TU/tu 20% Recom 2 tu/tu 20% double-recessive phenotype And 20% of 500 = 100 27) For the above scenario, which of the following could be the genotype of an offspring from one of the “parental” classes? A) tu / tu B) Tu / tU C) TU / TU D) TU / Tu E) Tu / tu (48%)

28) Genes A, B, and C are located on human chromosomes 19. You perform three linkage mapping experiments, and you determine that genes A and B are 25 m.u. apart, genes A and C are 10 m.u. apart, and genes B and C are 34 m.u. apart. Which of the following linkage maps is most consistent with your results? A) C–A–B (95%) B) A–C–B C) C–B–A D) B–C–A E) None of the above are consistent with the data. 29) The human ABO blood type system is controlled by three alleles of the I gene, which encodes an enzyme that attaches sugars to the H substance protein on the surface of red blood cells. What best explains why alleles I A and I B are codominant ? A) The i allele is due to a loss-of-function mutation. B) This gene is haploinsufficient. C) This gene is located on the X chromosome. D) The I A and I B alleles are both functional but encode different gene products. (92%) E) The I A and I B alleles inactivate one another. 30) In humans, albinism is a recessive autosomal phenotype. Imagine that two individuals with blood type AB and who are both carriers for albinism have a child together. What is the likelihood that their child will have blood type AB and display albinism? A) 3/16 B) 3/8 C) 1/16 D) 9/16 E) 1/8 (77%) P(two I A /I B parents have an AB child) = 1/2 P(two Aa individuals have an aa child) = 1/4 ½ * ¼ = 1/8 31) What blood types are theoretically possible to see in the offspring of a woman with type A blood and a man with type B blood? A) Only AB B) Only O C) A or B D) A, B, or AB E) A, B, AB, or O (74%) If both parents are hets (I A /i crossed to I B /i) then you can see all possible genotypes.

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32) In razorback hogs, coat color is under the control of multiple genes. For the red gene, the wild-type R allele (brown hair phenotype) is dominant over the r allele (red hair phenotype). A recessive allele of a second gene ( e ) is epistatic to the red gene: ee hogs always have white hair, regardless of the genotype of the red gene. A red hog and a white hog produce a litter or brown and white offspring in equal proportion. Which of the following are the most likely genotypes of the parents? A) RR; EE x rr; ee B) Rr; Ee x Rr; Ee C) rr; Ee x Rr; ee D) rr; Ee x RR; ee (60%) E) None of these are possible genotypes for the parents. This is completely analogous to the Labrador coat color example (replace phenotype black with brown, brown with red, and yellow with white). Answers C and D are both consistent with the phenotypes of the parents, but only D is consistent with the phenotypes of the offspring. C should yield ½ white, ¼ brown, and ¼ red. 33) You are a pea plant geneticist and you have isolated a new true-breeding recessive strain of dwarf pea plants. You also happen to have some of Mendel’s original stock of dwarf plants. For fun, you cross one of your dwarf plants to one of Mendel’s dwarf plants, and you find that all of their F1 offspring are tall! What is the most likely explanation for this miraculous result? A) Mendel was dead wrong. Genetics IS magic! B) These are codominant alleles of the same gene. C) You are witnessing epistasis between true-breeding and a heterozygous lines. D) You are witnessing complementation between true-breeding recessive lines for two different haplosufficient genes. (65%) E) You are witnessing hybrid vigor. 34) The genomes of identical twins are essentially 100% identical at the nucleotide level, meaning they share all the same single-nucleotide polymorphism (SNPs). The genomes of non-identical siblings will share: A) exactly 25% of SNPs B) on average ~25% of SNPs C) exactly 50% of SNPs D) on average ~50% of SNPs (68%) E) non-identical siblings will have none of the same SNPs 35) In chickens, there is a dominant condition known as “creeper,” in which the bird has very short legs and wings and appears to be “creeping” when it walks. The allele that causes the creeper phenotype is also recessive lethal. Therefore, if you were to cross two creeper chickens together, you should expect to see: A) 100% creepers B) No living offspring C) 2/3 creepers and 1/3 wild-type chickens (70%)

D) 3/4 creepers and 1/4 wild-type chickens E) 1/2 creepers and 1/2 wild-type chickens BONUS QUESTION (1 pt) Tortoiseshell cats are almost always female. However, there are rare examples of male tortoiseshell cats. What is a plausible explanation for why male tortoiseshell cats can exist? A) Trick question! All cats are female. B) These are XO males C) These are XXY males (87%) D) These are really females carrying a loss-of-function allele of the SRY gene E) These are males with spontaneous mutations in the O gene at the zygote stage.

BIOL2323 S2024_Midterm 2 key

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