genetics%20final%20helppppppp

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Feb 20, 2024

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1 . You mate an individual that is true breeding for long legs with an individual that is true breeding for short legs. All of the F1 offspring have short legs. Crosses between F1 individuals produce a total of 913 offspring. 1a. How many do you expect to have short legs if the trait is due to a single gene? Provide your answer to the nearest whole number. (3) 685 1b. If you were to do a testcross of an F1 individual, what would you expect the phenotypic ratio to be in the offspring? (3) 1:1 2. You have analyzed a cat at seven different genes and come up with the following genotype: Aa:Bb:cc:Dd:Ee:ff:Gg. You know that the capital letter allele for each locus is completely dominant over the lower case allele, and all loci assort independently. 2a. How many genetically distinct gametes will this cat be able to make? (3) 32 2b. If you crossed this cat with another cat with the genotype Aa:Bb:Cc:Dd:Ee:Ff:Gg , how many different genotypes would you expect in the offspring? (3) 972 2c. Referring to the cross in 2b, what proportion of the offspring would you expect to have the recessive phenotype at all seven loci? Provide your answer to 4 decimal places. (4) 0.0002 3. You are crossing two lines of mice that differ in tolerance to cold temperature and rate of oxygen consumption. Both traits show Mendelian expectations for single gene inheritance. When you cross a true breeding tolerant, low consumption mouse with a true breeding non-tolerant, high consumption

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mouse the F1 generation consists entirely of tolerant, low consumption mice. In the F2 generation you get the following data. Tolerant, low consumption = 568 Tolerant, high consumption = 194 Non-tolerant, low consumption = 183 Non-tolerant, high consumption = 61 Explain the inheritance of these two traits? (3) There is one gene for cold temperature tolerance and a second for rate of oxygen consumption. In the gene for cold temperature tolerance, tolerant is completely dominant to non-tolerant, and low consumption is completely dominant to high consumption for the gene determining oxygen consumption. Given the 9:3:3:1 ratio of phenotypes the two genes are independently assorting. 4. What cellular process results in two daughter cells that have the same number of chromosomes as the parent cell? (2) Mitosis 5. A diploid cell with N=6 has 13 picograms of DNA during G1 of interphase. How many picograms of DNA would it have during metaphase II of meiosis? (3) 13 pg 6. How many molecules of DNA are found on the average chromosome? (2) 1 7. Differentiate between homologous chromosomes and sister chromatids. (2) Homologous chromosomes share the same genes but are likely to have some different alleles since they originated from different parents. Sister chromatids, however, are identical copies of each other since they are the product of DNA replication of a chromosome. 8. List and define Mendel’s laws of inheritance and describe the cellular mechanism that results in each. (8)

The law of segregation states that the two copies of a single gene are split equally during gamete formation such that half of the gametes will contain one member of the pair while the other half of the gametes contains the other member of the pair. Segregation is the result of the separation of homologous chromosomes during Anaphase I of meiosis. Since the two copies (maternal and paternal) of a single gene reside on homologous chromosomes, when those homologues split into different cells the two copies have been split equally. The law of independent assortment states that the alleles of different genes segregate independently of each other such that all possible combinations of alleles will occur in the gametes at approximately equal frequency. Independent assortment occurs due to the fact that non-homologous chromosomes can align in different orientations during metaphase I of meiosis. Each orientation results in a different combination of alleles of the genes found on those non-homologous chromosomes and since each orientation has equal probability, all combinations of alleles will occur equally in the gametes. For example, in a cell with genotype Aa:Bb where genes A and B are located on different chromosomes, these chromosomes could align with A and B on the same side of the metaphase plate and a and b on the opposite side. That would result in A:B and a:b gametes. A second, and equally probable orientation would be for A and b to be on the same side of the metaphase plate while a and B are on the opposite side. This orientation will result in A:b and a:B gametes. Since both orientations are equally probable, each of those four combinations of alleles are all equally probable. 9. You cross rose capped roosters with a drab capped hens. All of the F1 offspring are rose capped. A cross between F1 hens and F1 roosters yields the following offspring: 47 rose capped roosters 26 rose capped hens 22 drab capped hens 9a. What do you hypothesize as the mode of inheritance of cap color in these chickens? (3) There is a single gene that determines cap color with rose capped completely dominant to drab capped. This gene is X-linked and males are homogametic while females are heterogametic. 9b. Design an experiment to test your hypothesis. What are the predicted outcomes if your hypothesis is correct? (4) I would conduct a reciprocal cross by mating a drab capped rooster with a rose capped hen. If my hypothesis is correct, the F1 generation should consist of ½ rose capped roosters and ½ drab capped hens. Crossing these would result in an F2 generation that consists of ¼ rose capped roosters, ¼ rose capped hens, ¼ drab capped roosters and ¼ drab capped hens. 10. You are conducting a cross between two individuals of the following genotypes: Aa : Bb x Aa : bb 10a. What phenotypic ratio do you expect if both A and B show complete dominance? (3)

3:3:1:1 10b. What phenotypic ratio do you expect if A shows incomplete dominance but B shows complete dominance? (3) 1:1:2:2:1:1 11. You cross two pure bred lines of birds with the following phenotypes: brown feathers with white tips x black feathers with yellow tips . The F1 all have black feathers with white tips. In the F2 you observe the following: Black feathers with white tips = 118 Black feathers with yellow tips = 59 Brown feathers with white tips = 54 Brown feathers with yellow tips = 3 11a. What do you predict is the cause of the above ratios? (3) There are two genes, one that determines the general feather color while the other determines the color of the tip. Black feathers are completely dominant to brown while white tips are completely dominant to yellow. These two genes are likely linked since the two parental combinations (brown with white and black with yellow) are present at higher proportions than expected while the two recombinant phenotypes are at lower frequencies than expected. The genes are acting in different pathways. 11b. Design an experiment to test your hypothesis. (4) A testcross of the F1 generation would be able to confirm linkage. If the genes are linked, then the testcross will show an overabundance of the Black:Yellow gamete and the Brown:white gamete, while the Black:White and the Brown:Yellow gametes will be at lower frequencies. This would lead to a recombinant frequency of <0.5. 12. Are different sex chromosomes homologous? Explain. (3)

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Different sex chromosomes are partially homologous in that they share some genes but not all and they pair during meiosis. 13. Genes A and B are known to be 18 m.u. apart on chromosome 2 while gene C is located on chromosome 4. The capital letter allele is completely dominant to the lower case letter allele for all three genes. You conduct a cross between parents of AA:BB:cc x aa:bb:CC genotypes to produce the F1. 13a. What gametes can be produced by the F1 and in what proportions? (3) A:B:C = 0.205 A:B:c = 0.205 A:b:C = 0.045 A:b:c = 0.045 a:B:C = 0.045 a:B:c = 0.045 a:b:C = 0.205 a:b:c = 0.205 13b. If two F1 individuals are crossed, what proportion of the offspring would be expected to have the genotype Aa:BB:cc? Provide your answer to the nearest 3 decimal places. (4) 0.018 14. You are studying the inheritance of fruit shape in cucumbers and have determined that there are two genes involved that interact in the following way:

Long ------------------------------> Long -------------------->Bulbous A B Both genes show complete dominance and are independently assorted. What phenotypic ratio would you expect if you were to cross the following individuals: aa:Bb x Aa:bb ?(3) 3 long : 1 bulbous 15. In breeding experiments involving true-breeding plants with red flowers you have found 7 different offspring with white flowers. In each of these 7 plants white is shown to be recessive to red and is the result of a single gene. You do a complementation test by making all pair-wise comparisons of the 7 white flowered offspring and the phenotypes of the resulting offspring are shown below (R= red flowers, W = white flowers). 1 2 3 4 5 6 7 1 W R W R R R W 2 W R W R R R 3 W R R R W 4 W R R R 5 W R R 6 W R 15a. What is the minimum number of genes involved in this trait? (3) 4 15b. Make up symbols for the genes involved in flower color and provide genotypes for all seven of the white flowered plants used in the experiment. (3) 1: aa:BB:CC:DD 5: AA:BB:cc:DD 2: AA:bb:CC:DD 6: AA:BB:CC:dd 3: aa:BB:CC:DD 7: aa:BB:CC:DD 4: AA:bb:CC:DD 16. In bees fertilized eggs develop into diploid female offspring, while males develop from unfertilized eggs and are haploid. Gene O determines eye color, with allele O+ resulting in golden eyes while O

results in opaque eyes. O+ is completely dominant to O. If you were to cross an opaque eyed female with a golden eyed male, what would the phenotypes of the male and female offspring be? (3) This would result in golden eyed female offspring and opaque eyed male offspring The nitrogenous bases of RNA and DNA nucleotides fall into two chemical categories. What are those categories and list which category each base belongs in. (4) Purines: Adenine and Guanine Pyrimidines: Cytosine, Thymine and Uracil 2.What are the components that make up each nucleotide? (3) A phosphate, a deoxyribose sugar (DNA) or ribose sugar (RNA) and a nitrogenous base (A, C, G or T for DNA, A, C, G or U for RNA) 3a. A mRNA molecule has the codon 5’-GCA. What is the anticodon found on the tRNA that complements this codon? (2) 3’ - CGU 3b. What amino acid would be attached to this tRNA? (2) Ala 4. The Central Dogma of Molecular Biology is used to describe the flow of genetic information within a cell. Draw a flow chart of the central dogma and identify the processes involved in each step. (3) DNA ---------------> RNA --------------------> Protein Transcription Translation 5. List two methods that can be used to isolate a specific gene and make millions of copies of it. (2) PCR and molecular cloning 6. List two methods that can be used to determine what traits are affected when a gene is absent from the genome or when it cannot be expressed. (2)

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RNAi and knockouts 7. You are studying two variants of the same protein which have the following sequences: Variant 1: Met Leu Pro His Asn Gly Tyr Glu Variant 2: Met Leu Leu Ile Thr Asp Met Lys 7a. What is the mRNA sequence for each of the variants? For any position in which more than one nucleotide is possible, use the designation “Pu” for any nucleotide that could be either Purine, “Py” for any nucleotide that could be either Pyrimidine, and “N” for any nucleotide that could be either a Purine or a Pyrimidine. Be sure to label the 5’ and 3’ ends. (4) 5’ – AUG PyUN CCN CAPy AAPy GGN UAPy GAPu – 3’ 5’ – AUG PyUN PyUN AUN ACN GAPy AUG AAPu – 3’ 7b. What kind of mutation resulted in the difference between the protein variants? (2) A deletion occurred in the third codon of the sequence in variant 2 (or an insertion in variant 1), losing the C in either the 7 th or 8 th position of the sequence. This resulted in a frameshift mutation. 8. List and describe the different types of mutations to chromosome structure. (8) Deletion – when a portion of a chromosome has been lost. Duplication – when a portion of a chromosome is repeated. Inversion – when a portion of a chromosome is in the opposite orientation. Translocation – when a chromosome contains a portion of a non-homologous chromosome. Can be either reciprocal, where two chromosomes have exchanged parts, or non-reciprocal, where one chromosome contains a portion of a second.

9. In what ways is a primary transcript modified in eukaryotic cells prior to translation? (4) A cap is added to the 5’ end, and it plus the next couple of nucleotides get methylated. The introns are removed and the exons spliced together into a single coding sequence. A sequence of A’s (poly-A tail) is added to the 3’ end. 10. How does a mRNA molecule in E. coli bind in the correct location of a ribosome? (3) The Shine-Dalgarno sequence in the 5’ UTR of the mRNA binds to a complementary sequence of a rRNA in the 30S subunit of the ribosome. 11. To which tRNA binding site on the ribosome does the initiator tRNA bind? (2) The P site 12. A tRNA moves through the binding sites of the ribosome in what order? (3) A  P  E 13. Describe the types of bonds that link nucleotides together in a DNA molecule, the relative strength of these bonds, and what role the differences in strength play in replication and transcription. (6) Adjacent nucleotides in the same strand of DNA are held together by covalent phosphodiester bonds between the phosphate and the sugar while adjacent nucleotides on different strands of the DNA molecule are bound by Hydrogen bonds between the bases. Covalent bonds are significantly stronger than are H bonds, so a DNA molecule is strongest within each of the individual strands, but the two strands of a DNA molecule can be easily separated. Since dsDNA must be broken into ssDNA for both replication of the molecule and transcription of a gene, the fact that the two strands are only weakly bound makes those processes easier to accomplish with less energy input. The strong covalent bonds that connect nucleotides of the same strand ensure that templates don’t randomly break easily.

14. Use the dsDNA sequence below to answer the following questions. CGTAATACGTTCCCTGAACCGTACTGGGCTGCCTTATCGAAGCTCAG GCATTATGCAAGGGACTTGGCATGACCCGACGGAATAGCTTCGAGTC 14a. During replication, the replication fork moves through this sequence from left to right and the complement to the bottom strand is synthesized in fragments. Where are the 5’ and 3’ ends of each strand. (2) Top – 3’ left, 5’ right Bottom – 5’ left, 3’ right 14b. This segment of DNA includes the entire coding region of a gene. Which is the template strand and which is the coding strand? (2) Top = template Bottom = coding 14c. What is the amino acid sequence of the protein encoded by this gene? (3) Met Gln Gly Thr Trp His Asp Pro Thr Glu 14d. Imagine there is a mutation in the 14 th base from the left where instead of a C top/ G bottom pair you now have an A top / T bottom pair. How will this change the protein encoded? (3) This will be a synonymous mutation and will not alter the amino acid encoded. 15a. You insert cDNA from the killifish Fundulus heteroclitus into expression vectors containing a bacterial promoter and then transform yeast cells with these recombinant vectors. You use an antibody for the killifish protein Superoxide Dismutase to screen the colonies. Would this work to find the colony transformed with the gene for Superoxide Dismutase? Why or why not? (3) No it will not. The RNA polymerase of eukaryotic yeast cells will not recognize the bacterial promoter. Therefore the cloned genes will never be transcribed and the proteins cannot be made. 15b. Would your answer change if you had transformed E. coli cells with these same recombinant vectors? Why or why not? (3) Yes, this would work. Since E. coli are bacteria, their RNA polymerase will recognize the promoter of the expression vector and the inserted genes will be transcribed. Since cDNA was used, the inserted

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genes will contain only coding sequence, so the bacteria will make the correct proteins, so the antibody should effectively find the superoxide dismutase. 16. Describe the process by which a prokaryotic cell makes a copy of its chromosome. (10) DNA replication in prokaryotes starts at a fixed origin known as OriC and proceeds in both directions until hitting the terminus on the opposite side of the circular chromosome. Replication is initiated with the binding of DnaA, which breaks a few H bonds on either side of OriC. The enzyme helicase (DnaB) enters the open area on either side of DnaA and moves in both directions away from OriC breaking the H bonds between complementary nucleotides on opposite strands. This forms a replication bubble around OriC where the DNA is single stranded and each strand can be effectively replicated by the addition of complementary nucleotides. Single strand binding proteins bind to the ssDNA in the replication bubble to keep the DNA from reannealing to form dsDNA, and topoisomerases break and reform phosphodiester bonds ahead of helicase to relieve torsional stress on the DNA as it is unwound. Since DNA polymerase III can only add nucleotides to other nucleotides, the enzyme primase adds a short RNA primer to the 3’ ends of each strand on both sides of the replication bubble. A DNA polymerase attaches to each primer/DNA complex and they begin adding nucleotides that are complementary to each of the original DNA strands. These nucleotides are in the form of deoxynucleotide triphosphates (dNTPs), and DNA Pol III catalyzes the reaction of forming a phosphodiester bond between the 3’ OH group on the deoxyribose of an already bound nucleotide and the first phosphate attached to the 5’ carbon on the deoxyribose of the incoming dNTP. The outer two phosphates are cleaved from the dNTP providing the energy for the polymerase reaction. Since DNA polymerase only catalyzes this reaction in the 5’ to 3’ direction, and the DNA strands are antiparallel to each other, one strand gets replicated in a continuous fashion and the other gets replicated in fragments. The strand that is continuously replicated is called the leading strand. On the leading strand, the DNA polymerase is adding nucleotides in the same direction as the replication fork is moving. The lagging strand is replicated in fragments, since the replication fork is moving in the direction opposite of which the DNA polymerase can add nucleotides. Therefore, an RNA primer is added to the lagging strand at the replication fork and DNA polymerase moves in the opposite direction adding complementary nucleotides until it hits double stranded DNA. The polymerase then detaches from the strand and rebinds near the replication fork on a new primer. It then adds complementary nucleotides until it reaches the beginning of the previously replicated fragment. DNA Polymerase I removes the RNA primers from the newly made strand utilizing its 5’ – 3’ exonuclease activity, and replaces them with DNA. Phosphodiester bonds are then formed between adjacent fragments by the enzyme DNA ligase. Once the entire chromosome has been replicated the DNA polymerases detach. This leaves two identical chromosomes.

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17. Below are the results of a series of experiments on fruit flies with two different phenotypes. Flies with phenotype 1 can digest rotting fruit without any problem while those with phenotype 2 have a drunken appearance and sometimes even die from eating rotting fruit. The researchers hypothesize that the two phenotypes are the result of differences in the enzyme alcohol dehydrogenase ( Adh ), which catabolizes the ethanol produced when fruits rot. The researchers decide to do three tests of this hypothesis. First, they do a northern blot for Adh in individuals of both phenotypes. Second, they do a western blot for Adh in individuals of both phenotypes. Finally, they sequence a portion of the 5’ end of the coding strand of the DNA of individuals of both phenotypes. The results of each are shown below. A. Interpret the results of the northern and western blots. B. Determine the coding sequences of the alleles of both phenotypes (label 5’ and 3’ ends). C. What do you hypothesize is the cause of the different phenotypes? D. What data would you need to generate to test your hypothesis? (12 points) A. The Northern blot shows that Adh is transcribed equally in both of the phenotypes and the resulting mRNA is the same size. The Western blot shows that while both phenotypes produce a version of Adh protein, the version being produced by phenotype 1 is larger or has a significantly different charge than the one produced by phenotype 2. Both are produced in approximately equal amounts though. B. 1: 5’ ATG GCC TGG CTA ACG AGG TGC AAG CAA CGA GAC 2: 5’ ATG GCC TGG CTA ACG AGG TGA AAG CAA CGA GAC C. Phenotype two has a nonsense base change in the 19 th nucleotide position of the coding sequence. This will result in a truncated protein, which explains the smaller protein being produced by phenotype 2. Truncated proteins are likely to be non-functional, so I hypothesize that the difference in the two phenotypes is due to phenotype 2 individuals having non-functional Adh. D. To test this hypothesis I would need to determine the activity of the Adh enzymes being produced in both phenotypes. If my hypothesis is correct, Adh in phenotype 1 will be functional but Adh in phenotype 2 will be non-functional BONUS: List the names of three people that were involved in determining the structure of DNA for 4 bonus points. Listing a fourth gets you 6. Erwin Chargaff

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Rosalind Franklin Francis Crick James Watson Maurice Wilkins https://drive.google.com/drive/mobile/folders/1LxSgprnKIdkgi9OCbTJz1qZp0oHeJmRT? usp=sharing_eip&ts=5eb95303

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