{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

CLAS_Final_MCDB_1A

CLAS_Final_MCDB_1A - Greg Neel — C LAS — FINAL REVIEW...

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Greg Neel —- C LAS — FINAL REVIEW Sources for pmctice Emblems: General tips: —Practice a lot of problems. \/ Your lVIidterm 1+2 —Try not to cram. \/ The posted Midterm 1+2 -Don’t miss ANY questions that are from that list (6) \/ My midterm 1+2 —Yes, the final is cumulative. (50+50+150) / The posted Final -Yes, there will probably be some curve, but that shouldn’t ‘/ This practice Final change how you study anyways. \/ Online quiz problems —Read my emails. \/ CLAS Packets/ Quizzes 1) Which will give a 1:1 ratio of phenotypes? a) AABB x aabb b) AaBb x AaBb c) AaBb x aabb d) AaBB x ”BB e) AAbb x aaBB 2) Label the PHENOTYPIC ratios the following would produce. Genes are unlinked. Hint: Know how to get to the genotypes, from the words A test question may not put the genotypes into the letters for you; sometimes you may have to infer what the question is getting at. (choices: 100% dominant, 913:3:1, 1:1:1:1, 3:1, 3:3:1:1) Two genes segregating but one of the parents is homozygous recessive for one of the genes. (AaBb x Aabb) :: Two genes segregating but one of the parents is homozygous recessive for both of the genes. (AaBb x aabb) :: One gene segregating and both parents were heterozygous (Aa x Aa) :: Two genes segregating and both parents heterozygous for both genes. (AaBb x AaBb) 2: Two genes segregating, one of the parents homozygous dominant for both genes. (AaBb x AABB) :: 3) When a monohybrid is “selfed” (ie Aa x Aa), one sees a # : # phenotypic ratio. 4) The __ tells the “actual” alleles present in an individual. 5) A(n) ____is a specific version of a gene. 6) What F2 phenotypic ratio shows codominance? 7) What F2 phenotypic ratio shows typical dominance? 8) A Turner syndrome female (X0) could be the result of nondisjunction in either Meiosis 1 OR Meiosis 2, in the mother OR father (any of the four combos. Memorize this!) True/False 9) XYY individuals most likely arise from nondisjunction at the second meiotic division of the father. (Hint: think of the only way you can get an extra Y-chromosome. Ifneeded, review the nondisjunction videos I emailed) True/False 10) XXX individuals can arise from nondisjunction at the 1St or 2nd meiotic division of the mother, or the second in the father. 11) Crossing over between homologous chromosomes occurs during: 12) A male’s X chromosome contains genes derived from his maternal grandfatherAND grandmother. (Hint: think of where the male got his Y—chromosome from. Then think of where his X—chromosome came from. His mother had 2 X—chromosomes, right? What do these X—chromosomes do during Meiosis?) True/False 13) Suppose the F and G genes are linked and separated by 20 map units. What fraction of the progeny from the cross FG/fg x fg/fg would be phenotypically FG? 14) By what process is it “decided” or “determined” which X—chromosome in a female becomes the Barr body? A) DNA methylation in early development activates the Xist gene (random selection) Greg Neel — CLAS — FINAL REVIEW B) The paternal X chromosome is always the Barr body 15) Ifa woman displays a sex-linked recessive trait/disease: a) all of her sons will have it. b) half of her daughters will be carriers of it. c) both of these 16) The dominant autosomal trait, Jeez disease, causes brain swelling at about age 45. A female whose mother has Jeez disease, marries a male who has no familial history. Because of the late onset, it is not known if the female has the disease. What is the probability their next child will not be afflicted with Jeez disease? 17) True—breeding Albino mice (aacc) crossed with true-breeding Agouti mice (AACC). All ofl'spring are of the Agouti coloration. When these F1 (agouti) mice interbreed, the next generation (F2) is in a ratio of 9 agouti: 3: black : 4 albino. This illustrates the principle of: 18) When a virus infects a bacteria and integrates into the genome, so it can be replicated as part of the host’s genome, this type of viral life cycle is called: 19) When the influenza virus leaves from the host animal cell, it does so by budding of the plasma membrane, forming an envelope. True/False 20) Influenza virus genome replication occurs in the : 21) Catabolite repression of the lac operon is sometimes referred to as the “glucose effect,” because in E.Coli, glucose is a preferable carbon source (over lactose). Thus, glucose presence will suppress (read: NOT enhance) the expression of the Beta-galactosidase genes (which metabolize lactose). This system of catabolite repression involves a cascade with Glucose/cAMP/CRP. CRP stands for CAMP Receptor Protein. Remember that ACTIVE CRP-CAMP complex binds the DNA to enhance transcription. 22) It is found that a particular enzyme is synthesized whenever Gregtose (an alternative sugar) is added to E. coli cells. This enzyme is responsible for metabolizing/breaking down Gregtose. This is likely gene regulation. (Think lac operon. . .) 23) A woman who is heterozygous for an X—linked recessive lethal allele marries a normal man. The lethal gene kills hemizygous or homozygous individuals before they are born. What is the expected sex ratio of their offspring? (Hint: draw a Punnet square and cross out the deceased) 24) A man who is heterozygous for an autosomal recessive lethal allele marries a normal woman. What is the expected sex ratio of their offspring? 25) The genes G/ g and N/n are linked with 20 map units between. The G/ g and N/n alleles show simple dominant/ recessive characteristics. Two parental strains GGNN and gun were crossed. The heterozygous offspring from this parental cross, were testcrossed. Ifyou had 1000 progeny from this testcross, how many would you expect to have the phenotype GN ?? 26) What fraction of offspring of the cross (ERrSsTt x (lgRrSsTT are expected to have the same phenotype as their parents assuming that all 4 genes assort independently? (hint: look at each locus individually. Draw out each punnet square if you need to) Give answer as fraction of /64. 27) An agouti mouse that is homozygous dominant at the agouti and albino loci (AACC) is mated to an albino mouse (aacc). lVIice without the dominant A allele are black, if theyre not albino (cc). The progeny from this cross breed. What proportion of the F2 progeny do you expect to be albino? 28) Nondisjunction may result in a human sperm that carries two sex chromosomes. True/False 29) In regards to X—linked traits, males are bemizygous, While females are either homozygous or heterozygous. True/False Greg Neel ~ CLAS — FINAL REVIEW 30) In garden peas, the tallness trait is dominant over the dwarf condition. Further, the round-seed trait is dominant over the wrinkled-seed trait. These genes are unlinked. Two plants that are heterozygous for both conditions (dihybridsll) are crossed and 160 offspring are produced. Approximately how many will be tall with round seeds? 31) In the lac operon, whether the repressor is active (bound) or not depends upon the presence of lactose only. Therefore, lactose determines whether transcription will be ON or OFF. ( Remember, these genes are for metabolizing lactose, and further, glucose is a preferable sugar.) Catabolite repression refers to the other “side” of the control mechanism: “Enhanced” or not. IfGLUCOSE is present, CAMP levels decrease, and it follows that there is decreased CRP—cAlVIP complex. Normally, transcription is thus never enhanced if glucose is present. If glucose is NOT PRESENT, there are increased levels of CAMP, increased levels of active cAlVIP—CRP complex, and thus CRP can bind the DNA and ENHANCE transcription. 32) If an individual has 1 Y—chromosome or more, they are always male (XXY, XY, XYY). Ifan individual is X0 or is XXX, they are female. 33) Differences in RFLP banding patterns indicate that the two DNA's being tested possess different base pairs (ie different alleles). This is because particular restriction enzyme recognition sites are either present or missing. True/False 34) The mouse autosomal genes C and T are linked and 38 map units apart. Genotypes CT/CT and ct/ct are intercrossed, and then the F1 is testcrossed. What is the F1 genotype? The proportion of Cctt progeny will be what? Write as a decimal. 35) In recombinant DNA technology, the target DNA is attracted from the source, cut with a restriction enzyme, and then ligated to another DNA molecule which is termed the “vector.” Transformation refers to the process of inducing a host cell to “take up” the recombinant plasmid. True/False 36) In order to locate and identify a human disease gene by positional cloning, W to vgio' us moleculg mgkers need to be detegnrn' ed. True/False 38) Among the shown crosses (below), circle any which will give a 1:1:1:1 phenotypic ratio. a) AABB x aabb b) AaBb x AaBb c) AaBb x Aabb d) AaBB x aaBB e) Aabb x aaBb 39) When a cross is made and a trait disappears in the F1 generation, then reappears in the F2, the trait is likely. dominant or recessive? 40) What F2 phenotype ratio would indicate complete dominance? 41) The separation of homologous chromosomes during Meiosis I is the main aspect of Meiosis responsible for the 1:1 allele ratio of the products. True/False 42) Ifyou are a male, your X chromosome contains genes derived from BOTH your paternal grandfather AND grandmother. True/False 43) A certain dominant trait, when displayed in men, is transmitted to half his sons and half his daughters. Is this gene Greg Neel — CLAS — FINAL REVIEW on an autosome 01' a SCX Chromosome? 44) Suppose the A and B genes are linked by 36 map units. What percent of the progeny from the cross Ab/aB x ab/ ab would be phenotypically AB? 45) Coffee syndrome is caused by a sex—linked dominant allele. Ifa lady has Coffee syndrome: A) her father must have Coffee syndrome B) her mother must have Coffee syndrome. C) all her sons will have Coffee syndrome. D) all of her daughters will have Coffee syndrome E) none of these 46) What monk wrote about the “particulate nature of inheritance” and fundamental genetic inheritance/ transmission patterns? 47) A disease, which causes death by age 3, is caused by a simple autosomal recessive mutation. A couple loses its first three children to the disease. If they decide to have a fourth child, what is the probability that the child will have the disease? (Hint: be careful) 48)The following recombination frequencies were observed in dihybrid testcrosses between three diflerent pairs of three genes. Gene Pair Observed RF A—B 30% B—C 10% A—C 20% What is the correct order of these three genes on the genetic map? 49) When a dihybrid is selfed, it will characteristically produce progeny phenotypes in the ratio: 50) Iftwo genes are very tightly linked and thus never show recombination, at selfed dihybrid with the genotype Fg/fG, would produce progeny phenotypes in the ratio A) 3FG :1 fg B) 9FG:3Fg:3fG:1fg C) 1FG:1Fg:1fG:1fg D) 1Fg:2FG:1fG E) none of the above 51) lfthese genes are tightly linked and never show recombination, a selfed dihybrid with the genotype BC/bc, would produce progeny phenotypes in the ratio: A) 3BC :1 be B) 9BC:3Bc:3bC:1bc C) lBCleczle:1bc D) 1Bc:2BC:1bC 52) Ifyou then crossed the dihybrid F1 agouti mice (AaCc) with the true-breeding albino strain (aacc), what phenotypic ratio would you expect among the progeny? 53) The type of viral life cycle, particularly prominent in bacteriophages, which causes the cell to burst after replicating itself many times is called: 54) For influenza to enter animal cells, it first binds RECEPTORS on the cell surface. Immediately after influenza viral RNA enters the host cell, it is copied into RNA and packaged into new virions in the CYTOPLASM. It is never copied into DNA. HIV, however, is copied from RNA9 DNA via “reverse uanscriptase,” which it brings into the cell. The DNA copy then enters the nucleus and is integrated into the host genome, forming a provirus. Reference viral life cycles in packet, if needed. True/False 55) Catabolite repression of the lac operon prevents (glucose/lactose) from being metabolized in the presence of (glucose/lactose). Greg Neel — C LAS FINAL REVIEW 56) It is found that a certain enzyme is responsible for the synthesis of phenylalanine. When phenylalanine is provided in the media that E. coli grows in, less of the enzyme is present. The gene for this enzyme is most likely regulated by gene regulation. (Hint: this is the type of gene regulation usually used when you need to maintain a camtarzt level of sometbing. Think trp operon) 57) A particular mutation in the operator of the trp operon prevents the repressor protein from ever binding to the operator region. Under what circumstances will transcription of this mutant trp operon be turned ofi‘? In the trp operon, does this have anything to do with glucose? 58) A mutation in the OPERATOR of the lac operon prevents the repressor protein from binding to the operator. When will transcription of this mutant operon be off? (Think of/ Circle 2 scenarios in which it would be off. Hint: Consider ‘un—enhanced’ to be ofi) +/— lactose +/— glucose +/— lactose +/- glucose 59) Calico cats (black and orange patches) are heterozygous for two coat color alleles, found on the X—chromosomes (one for black and the other for Orange). The patchy coloration is formed because early in development, each cell imprints an X-chromosome for inactivation. This is propagated through mitotic divisions forming the patch—pattern you observe, but the activation/inactivation “choice” is “reset” during meiosis. Because these genes are on X— chromosomes, male cats can only be calico if they are abnormally m. True/False 60) What fraction of offspring of the cross AaeeDd x AaEedd are expected to have the same phenotype as either one of their parents assuming no linkage? 61) When a dihybrid red, long—winged bee is crossed to a double—recessive white, whack—winged bee, the frequency at which red whack—winged and white long—winged bees are seen in the progeny is called the (meiotic/ mutation/ recombination) frequency. (circle) (Hint: assign letters fust) 62) The number of Barr Bodies is ALWAYS ONE FEWER than the total number of X—chromosomes. (eg XXX 2 barr bodies, XXY one barr body, XO zero barr bodies) True/False 63) Suppose the A and B genes are on the same chromosome but separated by 100 map units. What fraction of the progeny from the cross AB/ab x ab/ ab would be phenotypically Ab (genotype = A_bb)? (Hint: I know we didn’t do any problems with map units greater than 40—something. Ifrecombination frequencies are 50 or larger, the genes behave as if they are on separate chromosomes. Ie parentals=recombinants) A) 10% B) 25% C) 50% D) 75% E) 100% 64) Ifthere were a mutation in the CRP protein so that it could no longer bind cAlVIP and change conformation, could a cell carrying this mutation ever allow full expression of the operon? Yes/No 65) A human male carrying an allele for a trait on the X chromosome is called (hemi/mono)—zygous. (circle) 66) Exvivo method involves inserting the “desired” gene into cultured cells which were isolated from the patient. In vivo method involves inserting the “desired” gene into cells in the body of the patient. Germ—lines cells should NEVER be used because of ethical reasons. Somatic cells are thus used for all current gene therapy approaches. True/False 67) The mouse autosomal genes A and F are linked and 26 map units apart. AF/AF and af/af are crossed, and the F1 is testcrossed to af/af. The proportion of aaF f progeny will be what??? (give as a decimal). (Hint for these types of problems: While solving, do so without looking at the possible answers on the test. When you get an answer, THEN find it on the test. Don’t let other, attractive answers fool you) Greg Neel —- CLAS — FINAL REVIEW 68) After electrophoresis, which separates DNA fragments based on size, one does not know which of the fragments contains the sequence af interest. This is solved by the “Southern Blot,” named after its inventor. This transfers the bands of DNA to a filter. Then, labeled DNA probes can be “washed” over the membrane, which bind complementary sequence and signal the sequence presence for the scientist. True/False 69) The probability of having a crossing—over event between two genes is direcfly proportional to the map distance between them. (Increased map distance means increased recombination frequency. Any map distance of 50 map units (or more) means that the genes behave as if they are unlinked. Ie Parentals=recombinants.) True/False 70) Triplet repeat expansion causes Fragile—X. Individuals with a “premutated” allele are more likely to have a child with Fragile—X than those carrying the normal allele, likely because the “premutated” allele is much more likely to expand during DNA replication than the “normal” allele, due to slippage of DNA polymerase. True/False 71) The Y/y and Z/z genes are linked and 20 map units apart. In the cross YZ/yz x YZ/yz, what fraction of the progeny will be phenotypically Y2 (Y _zz).> A.. 8 96 B.. 9 % C.. 20 % (Hint: this is one of the trickier linkage problems, most are easier, but one of these could show up. If you know how to do the easier problems, move onto these. Notice it is not a test cross, but most of the questions you have seen are. Memorize how to do these.. ..See if this helps you figure this one out: One parent’s gametes in left column with their associated probability, the other parent’s gametes in the right column) .4 YZ YZ .4 .4 yzxyz .4 Y2 phenotype can be achieved through: Yz —Yz, Yz—yz, and likewise, yz —Yz. .1 Y2 Y2 .1 (.1x.1 + .4x.1 + .4X.l)= 9% .1 yZ yZ .1 72) Genes QZq and W/w are linked and known to be 20 map units apart. Individuals heterozygous for both of these genes (genotypes: QW/qw) were mated with each other. Ifthere are 2000 offspring from this cross, how many of the offspring would you expect to show the qW phenotype combination? (Hint: this is another one of the tricky ones! Memorize how to do these. . .so they’ re quick work during the test.) First step should be writing down the parental and recombinant gamete genotypes. Solution: (one parent’s gametes left column, the other parent’s gametes right column) PQW .4 .4QW qu .4 4qw RQV .10 8.10% RqW .1 . OqW So, how can we “ma ” the qW phenotype?? qW and qW(.10x.10) + qw and qW or (.4x.10) and likewise, qW and qw (.10x.4) = 9%, 2000x.09=180) 73) The slow block of polyspermy during fertilization is associated with the lifting and hardening of the fertilization envelope. True/False 74) A frog momla is a solid ball of undifferentiated blastomeres. Here’s a nice wiki excerpt: In humans, blastomere formation begins immediately following fertilization and continues through the first week of embryonic development. About 90 minutes after fertilization, the zygote divides into two cells. These mitotic divisions continue and result in a grouping of cells called blastomeres. During this process, the total size of the embryo does not increase, so each division results in smaller and smaller cells. When the zygote contains 16 to 32 blastomeres it is referred to as a "momla." Greg Neel — CLAS — FINAL REVIEW 75) In spermatogenesis, a diploid spermatogonium divides by mitosis to give a diploid primary spermatocyte and anotber spennatogonimn. True/False 76) Put these events into the correct order: 1) the acrosomal reaction 2) lifting and hardening of the fertilization envelope 3) the conical vesicle reaction i Qi .7"); This person has a child with an individual who does not have the disease and is known to be homozygous for the short DNA fragment. Southern blot analysis of the child indicated the presence of both the long and short DNA fragment (heterozygous for the fragment lengths). What is the likelihood that this child will have this disease? *3; : Asa“ kblo ’“WWW‘Qh This is a pedigree for a rare human genetic disease. -ST—é—h—O What is the probability that the newborn question mark will be affected? \ “if 7 A woman who is phenotypically normal has a brother who has hairy fingers (X—linked recessiv...
View Full Document

{[ snackBarMessage ]}