Exam 3 Potential Questions

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Exam 3 Potential Questions 1. Two organisms heterozygous for gene A are bred together. Two alleles of gene A show complete dominance . What would be the ratio of phenotypes in the offspring? a. 1:1 b. 1:2:1 c. 3:1 d. 9:3:3:1 e. All offspring are the same 2. Two organisms heterozygous for gene A are bred together. Two alleles of gene A show incomplete dominance . What would be the ratio of phenotypes in the offspring? a. 1:1 b. 1:2:1 c. 3:1 d. 9:3:3:1 e. All offspring are the same 3. You breed two pink snapdragons together. What phenotypic ratio would you expect in the offspring? a. 1 red: 1 white b. 1 red: 2 pink: 1 white c. 3 red: 1 white d. 3 pink: 1 white e. All pink 4. Which of the following is an accurate paraphrasing of Mendel’s Law of Segregation? a. Genes exist as pairs, and in gamete formation, each gamete gets one member of each pair b. In meiosis I, non-homologous chromosomes will line up at segregates positions c. During gamete formation, the way ne pair of alleles assort into gametes does not affect any other pair of alleles d. Dominant and recessive alleles are kept segregated from one another in mitosis e. In meiosis I, the two members of pair of genes will line up in metaphase at independent positions 5. Which of the following is an accurate paraphrasing of Mendel’s Law of Independent Assortment? a. Genes exist as pairs and in gamete formation, each gamete gets one member of each pair b. During gamete formation, the two members of a pair of alleles can assort into gametes in any manner c. During gamete formation, the way on pair of alleles assort into gametes does not affect any other pair of alleles d. The dominant alleles in a pair will always assort together e. During gamete formation, the two members if a pair of genes will line up in metaphase at independent positions 6. A testcross is performed by breeding an individual of dominant phenotypes/unknown genotype with a … a. Sibling

b. Parent c. Heterozygote d. Homozygous dominant e. Homozygous recessive 7. A testcross is performed on a plant derived from a round, yellow pea. The offspring from this testcross are 50% round green, 50% round yellow. What is the genotype of the parent plant? W = round, w = wrinkled, G = yellow, g = green a. WWGG b. WwGg c. WWGg d. WwGG e. wwGg 8. A testcross is performed on a plant derived from a round, yellow pea. The offspring from this testcross are 50% wrinkled yellow, 50% round yellow. What is the genotype of the parent plant? W = round, w = wrinkled, G = yellow, g = green a. WWGG b. WwGg c. WWGg d. WwGG e. wwGg 9. Genetically speaking, an organism that is true-breeding is … a. Hybrid b. Homozygous c. Heterozygous d. Hemizygous e. Dominant 10. A woman is affected by an X-linked recessive mutation. She mates with a wildtype male and has a son. This son will be _____ for the mutation a. Carrier b. Hemizygous c. Homozygous dominant d. Homozygous recessive e. heterozygous 11. Mendel’s wrinkly peas were created when a(n) ______ inserted into the gene SBEI, causing the pea to be unable to create amylopectin from starch a. Virus b. Transposable element c. Intron d. Centromere 12. Gregor Mendel chose to study traits in his pea plants that are diametrical and mutually exclusive. A trait with these properties usually … a. Is controlled by multiple genes b. Results from aneuploidy c. Exhibits incomplete dominance

d. Is controlled by a single gene e. Is X-linked 13. A pea plant of genotype Gg is self-crossed. You find a pod on this plant that contains three peas. What is the probability of finding three green peas inside the pod? G=yellow, g=green. a. ⅛ b. 1/12 c. ¼ d. 1/32 e. 1/64 14. A pea plant of genotype Gg is self-crossed. You find a pod on this plant that contains two peas. What is the probability of finding two green peas inside the pod? G=yello, g=green. a. ⅛ b. 1/16 c. ¼ d. 1/32 e. 9/16 15. Colorblindness is caused by a recessive X-linked mutation. If a wildtype man mates with a carrier female, what is the probability that the offspring will have normal vision? a. 0 b. ¼ c. ½ d. ¾ e. 1 16. Colorblindness is caused by a recessive X-linked mutation. If a wildtype man mates with a colorblind female, what is the probability that the offspring will have normal vision? a. 0 b. ¼ c. ½ d. ¾ e. 1 17. You are studying four linked genes. In the F2 generation of dihybrid crosses, you observe the recombination frequencies detailed below. What is the order of genes, A, B, C, and D on their chromosome? (Remember, the order you find might be backward from the options listed below) C-D = 7%, C-A=6%, D-B=5%, A-B=8% a. ACBD b. CBAD c. BCDA d. DACB e. ABCD 18. You are studying four linked genes. In the F2 generation of dihybrid crosses, you observe the recombination frequencies detailed below. What is the order of genes, A, B, C, and D on their chromosome? (Remember, the order you find might be backward from the options listed below) B-C=11%; A-C=6%;A-D=9%;B-D=8% a. ACBD b. CBAD c. BDCA d. DACB e. ABCD

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19. A couple has had three girls in a row. What is the probability that their fourth child will be a girl? a. ⅛ b. ¾ c. 1/32 d. 1/16 e. ½ 20. Because all mutations are rare, when looking at a pedigree, we can assume that people breeding into the family are… a. Wildtype b. Heterozygous c. Mutant d. Carriers e. Recessive 21. You are making a genetic map, and it happens that there is a dead zone for crossovers (fewer than average crossovers occur here) in between two of your genes. This will… a. Make it impossible to estimate the genetic distance between the two genes b. Not be relevant to the physical distance between the two genes c. Cause you to overestimate the distance between the two genes d. Cause you to underestimate the distance between the two genes e. Make the genes completely linked 22. You are making a genetic map, and it happens that there is a hot spot for crossovers (more than average crossovers occur here) in between two of your genes. This will … a. Make it impossible to estimate the genetic distance between the two genes b. Not be relevant to the physical distance between the two genes c. Cause you to overestimate the distance between the two genes d. Cause you to underestimate the distance between the two genes e. Make the genes completely linked 23. You perform a tetrahybrid cross (AaBcCcDd x AaBbCcDd). What is the probability of getting offspring that is AaBBCcDd a. ⅛ b. 1/16 c. 9/16 d. 1/64 e. 1/32 24. You perform a tetrahybrid cross (AaBbCcDd x AaBbCcDd). What is the probability of getting offspring that is AaBBCcdd a. ⅛ b. 1/16 c. 9/16 d. 1/64 e. 1/32 25. Below you see a mock-up of an electrophoresis gel (wells at top) analyzing a restriction fragment length polymorphism. The creation of the recessive ‘a’ allele most likely involved….

a. A SNP that created a new restriction site b. A SNP that destroyed a restriction site c. Expansion of an SSR d. Increased copy number of a CNV e. Decreased copy number of a CNV 26. Below you see a mock-up of an electrophoresis gel (wells at top) analyzing a restriction fragment length polymorphism. The creation of the recessive ‘a’ allele from the wildtype ‘A’ most likely involved…. a. A SNP that created a new restriction site b. A SNP that destroyed a restriction site c. Expansion of an SSR d. Increased copy number of a CNV e. Decreased copy number of a CNV 27. A mutation causes mice to have a stumpy tail. Some mice with the same mutation will have stumpier tails than others. What concept is likely responsible for this? a. Codominance b. X-inactivation c. Gene conversion d. Splicing e. Variable expressivity 28. A mutation causes mice to have a stumpy tail. Some mice with the same mutation witll have completely normal tails. No intermediate phenotype of partially tumpy tails is seen. What concept is likely responsible for this? a. Codominance b. X-inactivation

c. Incomplete dominance d. Incomplete penetrance e. Variable expressivity 29. What technology is used by companies like 23andMe or AncestryDNA to give an estimate of a person’s ancestry? a. Genome sequencing b. Restriction mapping c. Southern blot d. SNP chips e. Electron microscopy 30. If two genes on the same chromosome are far enough apart for multiple crossovers to regularly fall in between them, what is the result? a. The parental types are preferentially formed b. The recombinant types are preferentially formed c. The genes become lightly linked d. The genes appear to assort independently e. Parental and recombinant types are switched 31. Below is a Holliday juntion that might be found in a cis hybrid. If the Holliday junction- resolving enzyme made a horizontal cut, what would be the result? a. A crossover with a short heteroduplex region b. No crossover with a short heteroduplex region c. A crossover with no heteroduplex d. No crossover and no heteroduplex 32. Below is a Holliday juntion that might be found in a cis hybrid. If the Holliday junction- resolving enzyme made a vertical cut, what would be the result? a. A crossover with a short heteroduplex region b. No crossover with a short heteroduplex region c. A crossover with no heteroduplex

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d. No crossover and no heteroduplex 33. If a person is heterozygous for a SNP, and one of their gametes contains a heteroduplex at this locus, the two strands of DNA will not be completely complementary. The mismatch repair system will often try to fix this section of DNA, sometimes changing the genotypes of the gamete. This is called… a. Holliday Junction resolution b. Gene Conversion c. Epistasis d. Restriction Cutting e. Transformation 34. An organism has the genotype AABbccDdEe. How many types of gamete can it produce with respect to these five genes? a. 8 b. 16 c. 32 d. 64 e. 5 35. An organism has the genotype aaBbCcDdEe. How many types of gamete can it produce with respect to these five genes? a. 8 b. 16 c. 32 d. 64 e. 5 36. In a cis dihybrid with linked genes A and B, what is one of the most common genotypes (parental type) for a gamete? a. AB b. Aa c. BB d. AaBb e. Ab 37. In a trans dihybrid with linked genes A and B, what is one of the most common genotypes (parental type) for a gamete? a. AB b. Aa c. BB d. AaBb e. Ab 38. Two genes in humans regulate the response to COVID-19 (coronavirus). Gene R regulates the immune system, and a rare recessive mutation weakens the immune system, leaving the human more susceptible to infection (R=resistant, r=susceptible). Gene P regulates rational common sense. A common dominant mutation in Gene P induces irrational paranoia, causing the human to check the news for daily COVID-19 updates and go to the grocery store and buy up all the milk, eggs, and hand sanitizer they can carry, as if this will help somehow (P=paranid, p=calm). A susceptible, paranoid (Pp) person breeds with a resistant (Rr), calm person. What would be the phenotypic ratio among their offspring? Rp rp

rP RrPp rrPp rp Rrpp rrpp 1 (resist paranoid) : 1 (susceptible paranoid) : 1 (resist calm) : 1 (susceptible calm) 39. Two genes control coat color in labradors. Gene B controls pigment color with complete dominance (B=black, b = chocolate). Gene I is epistatic to gene B and encodes an enzyme that inserts the pigment into the fur (I = pigment fur, i = unpigmented, yellow lab). You cross a chocolate lab (bbIi) with a black lab (BbIi). What is the phenotypic ratio of the offspring? BI Bi bI bi bI BbII BbIi bbII bbIi bi BbIi Bbii bbIi bbii 3 Black : 3 Chocolate : 2 Yellow 40. Two genes control feather color in budgies. Gene Y adds yellow pigment and Gene B adds blue pigment. Recessive alleles do not produce pigment. If a dominant yellow and blue alleles are both present, the feathers are green. If no dominant alleles are present, the feathers are white. You cross a green budgie (BbYy) with a blue budgie (Bbyy). What would be the phenotypic ratio of the offspring? BY By bY by By BBYy BByy BbYy Bbyy by BbYy Bbyy bbYy Bbyy 3 green : 3 blue : 1 yellow : 1 white 41. In squirtles, gene T controls growth of hair on the tail with complete dominance (T=hairy tail, t=bald), and gene L controls shell development with incomplete dominance (recessive=normal shell, heterozygote = thich shell, dominant = spouts form on the shell). You cross a bald tailed, thick shelled squirtle with a hairy tailed (homozygous), thick shelled squirtle. What would be the phenotypic ratio of the offspring? TL Tl tL TtLL (hairy TtLl (hairy

sprouts) thick) tl TtLl (hairy thick) Ttll (hairy normal) 1 hairy tail, sprouts shell : 2 hairy tail, thick shell : 1 hairy tail, normal shell 1. In squirtles, gene T controls growth of hair on the tail with complete dominance (T=hairy tail, t=bald), and gene L controls shell development with incomplete dominance (recessive=normal shell, heterozygote = thich shell, dominant = spouts form on the shell). You cross a hairy tailed (Tt), thick shelled squirtle with a different hairy tailed (TT), thick shelled squirtle. What would be the phenotypic ratio of the offspring? TL Tl tL tl TL TTLL (hair sprout) TTLl (hairy thick) TtLL (hairy sprout) TtLl (hairy thick) Tl TTLl (hairy thick) TTll (hairy norm) TtLl (hairy thick) Ttll (hair norm 2 hairy sprout : 4 hairy thick : 2 hairy normal 1. Prior to Gregor Mendel, the Blending Hypothesis was the leading idea on how inheritance functions, where heritable factors mix and change through the generations. The observations of breeding on a red snapdragon with a white snapdragon to yield pink offspring supported the idea of blending. How did Gregor Mendel refute this hypothesis? Use Punnett squares in your answer. a. Blending was “replaced” with incomplete dominance (or possibly codominance depending on the situation). This thought of blending was proven wrong because further generations could be Red or White even when coming from pink parents. b. If blending was the case, pink (Rr) flowers would always be Rr and would not be able to separate into R and r R R r Rr Rr r Rr Rr * Rr = pink R r R RR Rr r Rr rr * RR = red, Rr = pink, rr = white

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1. A certain type of beetle can be black, spottes, or gray. Allele B confers a blacksheel to the beetle and is completely dominant to b, which confers gray shell. Allele T confers black spots to the shell, and is completely dominant to t, which is un-spotted. The black spots can only be apparent when the shell os gray (black spots are not visible when the whole shell is black), making Gene B epistatic to Gene T. if you crossed a BbTt beetle with a Bbtt beetle, what would be the phenotypic ratio in the offspring? BT Bt bT bt Bt BBTt BBtt BbTt Bbtt bt BbTt Bbtt bbTt bbtt 6 black : 1 gray with spots : 1 gray 2. You have a trans dihybrid organism for linked genes A and B. Genes A and B are known to be 4 centimorgans apart. What types of gametes would this organism produce with respect to genes A and B, AND in what proportions? TRANS CIS Recombinant AaBb 2% Parental AaBb 48% Aabb 2% aabb 48% Parental Aabb 48% Recombinant Aabb 2% aaBb 48% aaBb 2% 3. An autosomal recessive mutation causes a disorder. Two heterozygotes mate. In a sibship of 2, what is the probability of having one affected and one unaffected offspring? Show your work. Multiplication vs. Addition Affected ¼ Unaffected ¾ 1) ⅓ * ¾ = 3/16 2) ¾ * ¼ = 3/16 3/16 + 3/16 = 6/16 = ⅜ A vs. a

a. What is the most likely mode of inheritance for this mutation? Autosomal recessive b. What is the likely genotype of individual III-2? Aa 2. a. What is the most likely mode of inheritance for this mutation? X-linked dominant b. What is the likely genotype of individual II-2? X A y 1.

a. What is the most likely mode of inheritance? X-linked dominant b. What is the likely genotype of individual II-2? X A X a 1. You have an organism of genotypes Ab/aB. Genes A and B are 14 cM apart on their chromosome. What types of gamete will this organism produce with respect to genes A and B, and in what proportions ? TRANS Recombinant AaBb 7% Aabb 7% Parental Aabb 43% aaBb 43% 2. In snapdragons, flower color is autosomal, shows incomplete dominance, and can be red, pink, or white. You mate two pink snapdragons, and your flower makes three seeds. What is the probability of getting two pink and one red snapdragons from your mating. Show your work as much as possible. a. RR = red (Rr) 1/2 * (Rr) 1/2 * (RR)1/4 = 1/16 Rr = pink (Rr) 1/2 * (RR) 1/4 * (Rr) 1/2 = 1/16 rr = white (RR) 1/4 * (Rr) 1/2 * (Rr) 1/2 = 1/16 1/16 + 1/16 + 1/16 = 3/16 R r R RR Rr r Rr rr

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