BIS 101

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Sheila wants to fuse CAS9 to an activation domain called VPR. In the fusion polypeptide, she wants to have two or more Gly residues separating the CAS9 and VPR protein moieties. Which of the following fusions achieves Sheila's objective? Figure. A) Relevant sequences (5'->3') containing the CAS9 and VPR ORFs, as well as any additional codon. B) Example of fusion representation. C) Gly codons A

(pic)

Consider the data in Table 1. The following map fits the data best. (Symbols: *=0.5kb, **=1kb, ****=2kb, and so on. **B**=a 2kb fragment has a B site in the middle)

*X*********B****E**

Glucose is absent. Enter + for strong production, - for very low or no production. B-gal = beta-galactosidase. Perm = permease. Note (read if confused by Z and Y role): If you are wondering about the role of Z and Y in regulation, they have none. They are simply a readout of which operon is being transcribed. For example, assume that we have the following strain: chrom / F' lac I+, Oc, Z-, Y+ / I-, O+, Z+, Y- In the absence of lactose (non-inducing conditions), the strain above would express Y (permease), but not Z (beta-gal). This is because the Oc mutation would allow expression of the operon in the same phase while the repressor would sit on O+ and shut down the F'lac operon. This would tell us that Oc is not acting in trans, but is a cis element. If we provide lactose or IPTG, the F'lac operon would be activitated and the cell would express Y (permease) and Z (beta-gal) because both operons are active.

+, +, +, + Oc prevents repression. Lack of glucose stimulates expression via CAP. Expressed under all conditions. Z and Y are co-regulated.

5'ACCCTCGTGACCACCCTGACCTAC This sequence is taken from the middle of a gene. Ignoring the requirement for a Met codon, determine which reading frames are open (not interrupted) in the forward direction (left to right). Designate the frames by their order: +1 is ACC-CTC...; +2 is A-CCC-TCT... etc.

+1 and +3

5' GGCTGTGACAATCTAGTCAAAA The DNA sequence shown above is found in the middle of a gene. If this is the coding strand, which reading frame(s) is the most likely to be the one used for this gene? and what are the first three amino acids encoded by this frame? Hint: for the sequence 5'-GATTACA, frame +1 is GAT-TAC-..., +2 is ..G-ATT-ACA.. . For -1 you take the complementary strand and do the same.

+1, Gly Cys Asp Test all reading frame until you find one that is not interrupted by a STOP codon.

In each box enter + for strong production of beta-gal, - for very low or no production. For example, if you expect beta-gal expressed in all conditions you would choose answer +,+,+.

-, +, - This is the wild-type operon. Lac + glucose = Repressor off. CAP is not stimulating transcription. Therefore, very low expression* Lac, no glucose = Repressor off. CAP is stimulating transcription. Therefore, high expression No Lac, no glucose = Repressor on. CAP would be stimulating transcription. Therefore, no expression * It is debatable whether glucose+lactose conditions would turn off the lac operon. We choose to consider it off here. In fact, it varies from strain to strain. Some problems may imply some expression. Confusing? We promise that it will not be in the midterm.

Enter + for strong production of beta-gal, - for very low or no production

-, -, - No promoter! The operon is always OFF.

Glucose is absent. Enter + for strong production, - for very low or no production. B-gal = beta-galactosidase. Perm = permease.

-, -, -, - P- is in phase with Z+,Y+, resulting in no expression of the F operon. The chromosomal operon would be expressed but is Z-Y- . No beta-gal, no permease.

In I+,P+,O+,Z+,Y-/F' I+,P+,Oc,Z-,Y+ E.coli, lacZ and lacY protein expression in NO LACTOSE, NO GLUCOSE is respectively, ___,___ ; lacZ and lacY activity in LACTOSE, NO GLUCOSE is respectively, ___,___.

-,+ +,+

To practice restriction analysis you digest a PCR product with restriction enzymes B, E, and S in various combinations, then electrophorese the digested samples on a gel. Each dash (-) represents a distance of 1, and letters indicate cut sites in between the dashes. You get the fragment sizes shown in the table below. Digested With Fragment Lengths *analysis of E+S may be tricky, how could we cut twice and only see two fragments on our gel?

---B-E----S--

In a mutant mouse, the overall frequency of crossing over is doubled without changing their overall pattern. Assume you can make this mouse heterozygous at many loci and use it in a test cross, the following two statements best describe the genetic consequences.

--The distance in m.u. between any two genes on the same chromosome increases. --Any one gene is linked to fewer genes.

5'ACCCTCGTGACCACCCTGACCTAC This sequence is taken from the middle of a gene. Ignoring the requirement for a Met codon, determine which reading frames are open (not interrupted by any stop codon) in the backward direction, i.e. using the bottom strand 5' to 3' as coding strand. Designate the frames by their order: -1 is GTA...; -2 is TAG. etc.

-1 and -3 1. Set up the bottom strand: 5'GTAGGTCAGGGTGGTCACGAGGGT 2. Start with -1 frame: GTA-GGT-etc 3. -2 frame: TAG-GTC-ect

In a mink F2 population derived from two parents, on long haired, the other short haired, the frequency of individual hair length is distributed normally. The frequency of individuals phenotypically identical to parent 1 is 1/1000. The best estimate for the number of genes controlling this trait is:

5 1/(4^number of genes) will look like the recessive parent 4^5 = 1024 If it is one gene 1/4 of the offspring will look like the recessive parent. for two genes 1/(4^2) = 1/16 of the offspring will look like the recessive parent.

In Mendel's pea, T_ = tall, tt = short. A Tt F1 individual is selfed to make an F2 generation, which is in turn selfed to make an F3. What is the probability of heterozygosity among TALL F3 individuals?

0.4 Inbreeding generations = 2 Hets in the F3 = 0.5^2 = 0.25. Homs in the F3 = 0.75

Two normal parents have three kids. One kid is affected by Sicilian anemia, a fully penetrant condition. What is the probability that both of the normal kids are carriers?

0.44

Shire hobbits living in the villages of Bonning Gate and Newbiggin have curly hair on their feet rather than the normal smooth hair. The curly haired trait is controlled by different genes in the two families. Intervillage matings between the families resulted in all children (F1) having smooth hair. If such children were to intermarry, what would be the probability of their children (F2) having curly haired feet?

0.44 We are dealing with two loci because when the families intermate all F1s are "wild-type" or smooth. Thus, AAbb (curly) x aaBB (curly) create F1 AaBb (smooth). Intercrossing the F1 will result in a 9 : 3 : 3 : 1 genotypic ratio. However, the phenotypes will be different than 9 : 3 : 3 :1 because homozygous recessive bb creates the same phenotype as homozygous recessive aa. Therefore, the phenotypic ratio to be 9:7 smooth:curly, where curly can be: A_bb, aaB_, or aabb. Therefore the probability of the F2 having curly hair feet is 7/16, or .44. This is a case of double recessive epistasis.

A normal woman with no family history of genetic disease or albinism has a child with a man affected by albinism (recessive) and dwarfism (dominant, rare). The gene for albinism and the gene for dwarfism are linked and 20 m.u. apart. What is the chance that the child is dwarf and carries albinism?

0.5 A_ = normal pigment, aa=albino D_= dwarfism, dd=normal Mom is AAdd. Dad is aaDd. The question asks the P of a child with genotype aAdD. The P of getting Ad from mom is 1 because she is AAdd. Dad can only produce two types of gametes: aD and ad. Note that recombination does not affect the probability of gametes since the A locus is homozygous. So, 0.5 for aD gamete and aD/Ad child.

Consider the figure above. If you clone the 900 bp EcoRI DNA fragment of trpA in the EcoRI site of vector pUC21, what is the probability of making a trpA-GFP fusion protein? Hint: how many ways could it happen?

0.5 The figure shows that insertion of trpA in the right orientation would make two good junctions. The reading frame would be intact at both. Note, however, that this is not the only possible event.

This pedigree represents inheritance of a rare, fully penetrant disease in an extended family. If II-6 were to have married II-8, what would have been the probability of disease in their children?

0.5 boys, 0 girls

In a three point test cross, 15 of 1000 progeny had phenotypes indicating inheritance of a double cross-over recombinant chromosome. The distance between the three genes was 25 m.u and 15 m.u. Calculate cross-over interference between the genes.

0.6

Dark-green leaf is dominant over light-green. Two pure breeding varieties, one dark-green, the other light-green, are crossed. The F1 plant is selfed producing 147 dark-green and 53 light-green F2. The F2 generation is advanced by selfing producing the F3. The probability that a dark-green F3 plant breeds true is:

0.6 This question can be confusing Let's start: Dd, Dd = D_ = dark green dd = light green F1 = Dd F2 genotypes = 1:2:1 DD:Dd:dd F2 phenotypes = 3:1 dark:light What is the P that a dark green F2 will breed true? Dark green are 1DD:2Dd. Only DD breeds pure. So, in F2 1/3 breed pure. Now let's ask the same Q for F3s. There is a number of ways to solve this. One approach is to take a large family and carry it through. Suppose we have 300 dark green F2. Each is selfed and for simplicity we keep a single plant from the progeny F3. Start the F3 with 300 dark green F2s.....calculate the genotypic ratios expected....

You cross inbred pea, one drought resistant, the other drought sensitive. Equivalent numbers of individuals of the parental, F1, and F2 populations are phenotyped and scored from 0, sensitive, to 10, most resistant. Vp for P1, P2, and F1 is respectively, 3.8, 3.9 and 3.8. Vp for the F2 is 9.1. H2 is:

0.63

Paula and Jerome have a normal child, Sheila. Both Paula and Jerome have a mother who is affected by cystic fibrosis, an autosomal recessive condition. What is the probability that Sheila is a carrier?

0.66

Calculate the probability of finding at least one terminator codon in the +1 frame of a DNA segment 90bp long. Note: yes, this is a challenging probability problem. So, do not worry if it seems impenetrable. Look at the clues and see if you can work it out. The objective is for all to understand that in random DNA a STOP codon will pop up sooner or later.

0.76 For each codon the P of finding a stop is 3/64 because there are three stop codons out of 64. The P of NOT finding a stop codon is 61/64, or 1-(3/64) or 0.95. For a run of codons, it is easier to calculate the P of finding one or more stop codons by subtracting the P of NO STOP codon from 1. For example for two codons the P of finding at least one STOP = 1-(61/64)^2 = 0.89. For thirty codons, P of at least one STOP = 1-(61/64)^30 = 1-0.24 For each codon the P of finding a stop is 3/64 because there are three stop codons out of 64. To calculate the P of at least one codon for a run of codons it is easier to use the P of NOT finding a stop first. Since the total P of a system = 1, the P of NOT finding a stop = 1-(3/64). Consider now a two codon sequence. What is the P that neither of these two specifies a stop? Each codon can be considered independently. Therefore: P of NOT finding a stop^N, where N=number of codons.

A vector plasmid contains a single EcoRI site, a single BamHI site, a single HindIII site. After restriction enzyme digestion of the vector with HindIII, how many DNA fragments do you expect to see when you perform gel electrophoresis?

The expected number of centromeres per chromosome is:

In pea, purple flower (P) is dominant, white (p), recessive. Yellow seed color (Y) is dominant; green (y) is recessive. Tall stem (T) is dominant; dwarf (t) recessive. In a cross between a truebreeding (purple petal, yellow seeded, tall plant) and a truebreeding (white petal, green seeded, dwarf plant), what proportion of the offspring would be heterozygous for all three traits?

1 Both parents are homozygous. PPYYTT x ppyytt -> PpYyTt. All like parent 1.

In pumpkin: H_=large, hh=small O_=bright orange, oo=off white R_=resistant to caterpillars, rr=susceptible Map: O--5mu--H----10mu----R Interference=0.6 Cross: OhR//oHr x ohr//ohr. // = haplotypes How many plants producing bright orange, large and resistant pumpkins do you expect out of 1000 progeny?

1 This recombinant product results from a double crossover. The p of the double recombinant is 0.025x0.05=0.00125. Given 0.6 interference, coincidence is 0.4. An the resulting frequency should be 0.00125*0.4 =0.0005. 1 is the best estimate regardless of whether you consider interference or not.

In separate reactions, a purified DNA preparation of a certain plasmid is digested to completion with different combinations of restriction enzymes: Reaction 1: No enzyme (negative control) Reaction 2: BamH1 only Reaction 3: EcoR1 only Reaction 4: EcoR1 and BamH1 The diagram below shows the appearance of the original molecules and the digestion products in the four reactions after separation by electrophoresis in an agarose gel: How many restriction sites for BamH1 and EcoR1 does the plasmid have?

1 BamH1 site; 2 EcoR1 sites Starting with the conclusion that the plasmid is circular and 6 kb in length, we see that BamHI generates a single fragment. It must cut the circular form only once. The two fragments produced by EcoR1 digestion indicate two cuts. You can verify this using a toy necklace. One cut linearizes it to its total length, two cuts give two fragments that add up to the total length.

Phage Phi80 forms plaques on Pseudomonas F2 and F8. Mutant phages N,P,Q,R,S,T,U,V are only virulent on F2. When combinations of two mutants are applied to F8 at high MOI, P+T, P+U, and V+Q FAIL to form plaques. All other combinations do form plaques, i.e. in combination they become virulent on F8. You deduce that at least ____ genes are required for plaque formation on F8.

5 Get organized and start counting genes: Gene 1: P, T, U Gene 2: .... BY THE WAY: an early version of this problem had the virulence of the phages turned around (PTU do not complement, in the early version they did!). It was a mistake in writing the question and it did not make sense. The current version makes sense.

You hybridize a siamese to a javanese cat. The differences between the two (pure-breeding) breeds result from the action of four genes for which dominant and recessive alleles are segregating in the F1. The siamese is recessive at all loci. What is the P of getting a pure siamese in the F2?

1/256 Four genes: such as the cross between two hybrids AaBbCcDd  resulting in the following progeny -> aabbccdd

The following pedigree represents a rare autosomal disorder. Individual III-1 marries a woman whose family has no history of this disease. What is the probability that their first child will be male and affected?

1/4

A pea plant with the genotype AABBcc is crossed with a peaplant with the genotype AaBbCc. What is the probability of an offspring having the genotype AABbcc or AaBBcc?

1/4 AABBcc x AaBbCc. P of AABbcc = 1/2*1/2*1/2 = 1/8. P of AaBBcc = 1/2*1/2*1/2 = 1/8. The two events are alternative, not independent. Therefore, we add the two Ps.

All sneetches want their children to have stars on their bellies. Sneetches can be black or yellow, and star-bellied or starless. Two loci control these phenotypes: yellow is dominant to black, and star-bellied is dominant to starless. The combination of alleles that make a black, starless sneetch is lethal. If two heterozygous yellow, star-bellied sneetches mate, what is the likelihood their first child will be starless?

1/5 The question is at first sight a bit disconcerting. It deals with a lethal interaction; yyss is lethal. If two heterozygous parents are mated, what is the P of yyss? This genotype is taken out of the count. So what is left? You must factor both genes in the calculation.

A plant heterozygous for three unlinked loci is selfed. What proportion of the offspring will be recessive at all traits?

1/64 Consider a single locus. If a heterozygous parent is selfed, what fraction of the progeny will be homozygous recessive? If three loci are independent, what should you do with the individual probabilities?

The tinge of anatto color (a natural orange pigment) of the achiote plant has been attributed to the additive effect of the genotype at three independent and equivalent loci, W, X and Z. Upper case alleles contribute 2 units of red, while lower case allele contribute 0 units. A yellow wwxxzz plant is crossed to a red WWXXZZ plant. The F1 is selfed. What is the probability of F2 progeny with the yellow parent phenotype?

1/64 Consider the 6 allelic slots that determine the phenotype. The P of small vs large case allele for each is 1/2. Therefore, the probability of 6 x is (1/2)^6 = 1/64

Assume that III-6 and IV-1 marry and have a child. What is the probability that this child will be affected? Hint: Determine the type of inheritance pattern first.

1/8

If individual III-6 and IV-4 have a child together, what is the probability that the child will have the disorder?

1/8

Freckles and tasting of PTC, a bitter substance, are independent and dominant traits in human. Use T-t and F-f, respectively, for alleles. Mary, freckled and non-taster, and Pablo, non-freckled and taster, have a non-freckled and non-taster child. What is the probability that their next child will be female, freckled and taster?

1/8 Mary, freckled and non-taster: F_tt Pablo, non-freckled and taster: ffT_ first son: fftt, therefore both are heterozygous for the dominant trait.The cross is:Fftt x ffTt Note that this is a test cross with the added complication that the recessive for each trait is not the same. But, each phenotypic trait has 0.5 chances. For example Ff x ff -> 1:1 freckled to non-freckled. The P of each phenotype in the progeny, such as freckled and taster, is 0.5x0.5 = 0.25, or 1/4. Multiplied by the P of female (1/2), the P of the female, freckled and taster child is 1/8

The family tree above illustrates the inheritance of a rare disease condition. How many gametes are informative for LOD linkage analysis?

Three genes of corn, R, D, and Y, reside on chromosome 9 (see map, above). In a testcross RDY/rdy x rdy/rdy, how many RdY/rdy plants are expected in a progeny of 1000 if there was no interference?

10 Obtaining an RdY/rdy progeny requires a double cross-over event. Assuming no interference, in 1000 progeny, a double cross-over event should occur 1000 * (0.2) * (0.1) = 20 So, 20 progeny should have the double cross over phenotype. A double cross over event in the parent would produce recombinant gametes rDy or RdY in a 50:50 ratio. So P(RdY) = 20/2 = 0.01*1000=10

An LOD of 2 means that the likelihood of linkage for the given theta value is ______ more likely than the likelihood of independent assortment.

100 LOD is the log base 10 of the odds (likelihood of linkage / likelihood of independent assortment) 10^2 = 100

Pursuing the linkage of marker W5 to disease allele D, your team obtains the following LOD maximum at a theta of 0.03 for 4 independent family pedigrees: 2.2, 1.8, 1.3, 2.7. You conclude that linkage between marker alelle W5 and D is _______ fold more likely than no linkage and ___________ at ______ m.u.

100 million, strongly supported, 3

How many different mRNA sequences can encode a polypeptide chain with the amino acid sequence Met-Leu-Arg? (Be sure to include the stop codon.) Note: consider only the coding region of the mRNA

108

Matings between cousins can increase the rate of recessive genetic conditions in their progeny because cousins, are likely to share ___% of their DNA:

12.5 Start with siblings: how much DNA do they share? To make the calculation clear, assume that each parent had a completely different set of alleles. So, for each locus there are, so to say, four alleles: 1,2,3,4. Mom was 1,2; Dad 3,4. The two sibs could be 1,3 and 2,4 (Share no alleles at this locus), or (1,3 and 1,3, or 1,4 and 2,4 etc). If you play with the stats you see that they in average share 1 of 2 alleles. Now work out this calculation for the cousins. Genetic relatedness is explained in details in the Pedigree analysis page.

Consider the cross in Table 1. If the THREE genes considered were to be UNLINKED, the expected number of rrhhss progeny out of 10,000 is:

1250 Under independent assortment, SsTtQq -> 8 gamete types, each with P=1/8 In a testcross, genotypic progeny classes correspond to gametic classes. Therefore, stq gamete is 1/8*10,000 = 1250

Two mutations that affect plaque morphology in phages (a- and b-) have been isolated. Phages carrying both mutations (a- and b-) are mixed with wild-type phages (a+ b+) and added to a culture of bacterial cells. After infection and lysis, phage lysate are collected and cultured on bacteria cells to look at plaque morphology, obtaining the results, above. What is the frequency of recombination between the a and b genes

14% recombinants/(recombinants+parentals)

Consider the above figure, representing translation in a mammal. The distance between X and W along the mRNA is _____ bases. Hint: the same figure is used in another problem, which you should solve first.

If the % A content in one DNA sample is 35%, the % G content is:

15 A = T, so if A = 35%, then T = 35 %. A + T = 70% A + T + G + C = 100% and G = C So G + C = 100 - 70 = 30% and G = 30/2 = 15%

Based on the cross results in Table 1, the R-H and S-H distances in m.u. are, respectively __, __ . If you could count all missed double CO, the distances would be_____.

17, 3, longer Order and phase: RSH Recombinations between R and S = 0.1407 Recombinations between T and Q = 0.03 Observed double XOs = 7 Expected double XOs = 0.14*0.03 = 0.0042, in 10K -> 42 Coincidence = 7/42 = 0.16 Interference = 1-0.16 = 0.84 R-S distance 14 m.u. S-H distance 3 m.u. R-H distance 17 m.u.

Three ribosomes are catalyzing the synthesis of polypeptides. Consider polymer S, the best guess for the distance between X and W measured in nucleotides is:

Loci A and B are linked at a 15 m.u distance, with A-B alleles in coupling. The cross AaBb x AaBb produced 1000 progeny. How many aabb do you expect?

180 The cross is AB//ab x AB//ab. We wish to find the P of an ab//ab progeny. We need the P of ab gametes Each parent produces four types of gametes: Parental, cumulative P=0.85 AB, ab, each has P=0.85/2=0.425 Recombinant, cumulative P=0.15 Ab, aB, each has P=0.15/2=0.075 The probability that that two ab gametes form a zygote is Pab x Pab = 0.425x0.425 = 0.180

Morgan crosses a red-eyed male to a red-eyed female. In the resulting progeny there are 1/4 white-eyed flies. Sexing shows that in the progeny males are ____________, females are _______________:

1:1 red-eyed:white-eyed, red-eyed Female XwX+ x male X+Y --> half of male progeny are XwY. Female all inherit X+ from father.

A red-eyed fly and a white-eyed fly mate. From this progeny, which is all red-eyed, a female is mated to a white-eyed male. In the resulting progeny, females are ______, males are _______.

1:1 red:white eyed, 1:1 red:white eyed,

Considering epistatic yellow in retrievers. An aa dog is yellow regardless of the genotype at the B locus. In the presence of the dominant A allele, dogs are black or brown depending on the presence of the dominant B (black) allele or recessive b (brown) allele. If the mating of a brown and a yellow dogs has previously produced puppies of all three colors, what ratio do you expect for black:brown:yellow?

1:1:2

Investigating a suspicious gambling operation, you observe that the six-faced dice used yields 1 less frequently than expected. You measure its probability of "1" at 0.11. The expected odds are ____. The observed odds are ______.

1:5, 1:8 fair die: 1 chance of getting a "one" : five chances of not getting a "one" other die: 0.11:(1-0.11) = 0.11:0.89 = 1:8

In a pea plant the A and B genes are linked at a 10 m.u. distance. You genotype the pollen grains of this individual. What is the expected ratio of recombinant to parental gametes for the A-B genes?

1:9 Since the map distance between A and B is 10 m.u, you would expect recombination between A and B to occur 10% of the time. The answer 1:9 means that you expect one recombinant for every 9 parental gametes observed (or 10%).

Three plant lines each have an independent mutation that causes them to have a curled leaf phenotype. All are true breeding for this phenotype and the wild-type phenotype is straight leafed. When crossing these mutants, you observe the following: Plant 1 x 3 -> complementation Plant 1 x 2 -> no complementation Given these results, what phenotypes have you observed in the progeny?

1x3 = straight, 1x2 = curled

The S and Y genes are unlinked. In an individual with genotype SsYy, in the absence of crossing over, the following number of dsDNA molecules encoding the S allele are dragged to one pole at anaphase I.

A 7 bp insertion in the coding region of a gene could be reversed by a second, nearby insertion of ______ bp.

2 7 bp insertion disrupts the ORF of the gene causing a frameshift mutation. 7 + 2 = 9 9 is divisible by 3. Therefore, it restores the overall reading frame of the gene and could restore function.

Consider the genome map of Salmonella Kleiber-3 (isolated by the TAs from a toilette residue in the men restroom). The green arrows illustrate the position and orientation of one IS element present on F plasmid and chromosome. How many Hfr strains could be formed?

2 F plasmids predominantly use IS elements to insert into genome via homologous recombination.

A black ram and a white ewe from pure breeds are mated. The F1 progeny is white and when interbred they produce 40 white and 13 black F2 lambs. The number of different alleles of the wool color gene present in a G1 cell of an F1 individual is:

2 White is dominant. W_ = white, ww=black. P1 and P2 are pure breeding and thus ram is ww, ewe is WW. The F1 is heterozygous Ww. In G1 cells chromosomes are not replicated and two copies of the gene are present, each as a different allele.

A diploid organism at the following stages of meiosis has this number of chromosome sets per cell: prophase I - end of cytokinesis I - prophase II - end of cytokinesis II :

2 - 1 - 1 - 1

This scenario takes place in a liver cell nucleus. Most likely, this DNA contains:

2 genes, there is no information about introns The two RNA X-mas trees emanating from the DNA result from the expression of two genes. No information is available about introns.

Two pure breeding lines of pea, one tall and hairy, the other short and smooth, are crossed. The F1 hybrid is tall and hairy. The F2s assort in four phenotypic classes in a 9:3:3:1 ratio. Crossing a short, smooth plant to a short, hairy one will produce ____ phenotypic classes:

2 or 1 An organism with a dominant phenotype for a single gene trait, could be homozygous (AA) or heterozygous (Aa). What is the effect of genotype on the number of phenotypic classes it will produce in a testcross?

A wild-type fruit fly is heterozygous for genes A, B, and D. Genes A, B, and D are linked. Assuming no recombination, how many different gamete combinations of these genes can this fruit fly produce? How many could theoretically be produced assuming recombination?

2, 8 Regardless of the linkage phase, if there is no recombination between the loci, only two types of gametes can be produced. For instance if the linkage phase is AbC / aBc, then the only gametes produced would contain either AbC or aBc. With recombination, theoretically all different allelic combinations can be produced. 2 x 2 x 2 = 8 For example gametes could be ABC, abc, Abc, aBC, AbC, aBc, abC, ABc.

Flowers of Armenian cyclamens can have three types of petal color: solid dark-blue, red with yellow tips, or solid yellow. You purchase a blue flowered plant, self it, and obtain 230 blue, 64 red and yellow, and 13 yellow. Based on this, the most likely hypothesis is that ____ genes are involved and that the one conferring ___ color displays _______ epistasis.

2, blue, dominant

In a Pp heterozygous organisms, one of the chromosomes carrying the P locus is moving toward the pole at anaphase I of meiosis. In the absence of crossing overs, the P locus on this chromosome is represented by ____ dsDNA molecules with genotype ________ . (name every complete allelic copy)

2, either PP or pp

An XXX mouse is expected to have ____ Barr bodies and to be _____:

2, female

An XXXY mouse is expected to have ____ Barr bodies and to be _____:

2, male

In cow, horned is a recessive trait, polled (hornless) is dominant. You cross two polled animals that both had a horned mother. In their progeny, you pick a polled steer. What is the probability that this steer is heterozygous?

2/3

Where is the promoter for this gene, approximately? Choose a coordinate on the top ruler (basepairs of DNA fragment).

20 Figuratively, you grab the 5' end of the nascent RNA, stretch it out and align it to the dsDNA. The place where the RNA ends, is 25b (if eukaryotic) away to the right of the promoter. Where did transcription start?

Localizing the TATA box region of a eukaryotic promoter tells us that the RNA transcription initiation site is

20-30 bases 3' of the TATA sequence TATA box and -12 promoter region of prokaryotes are binding site for RNA polymerase. The distance from a TATA box is to transcription start is ~25b.

In mustard, branched stem is dominant over unbranched. Two pure breeding varieties, one branched, the other unbranched, are crossed. The F1 is selfed producing 600 branched and 220 unbranched F2 plants. How many of the branched plants are homozygous?

200 Consider the branched plants in the F2: what are the possible genotypes and what ratio do they display?

Consider the image above representing a cellular process in bacteria. . A promoter is most likely located closest to this nucleotide position

230

What is the distance in nucleotides between point X and W? Hint: how many bases does it take to add an amino acid to a growing polypeptide?

24 The polypeptides emerging from the two ribosomes differ by 8 amino acids in length. Therefore, there must be 8 codons between the two ribosomes.

In a generalized transduction system using P1 phage, the donor is leu+ thr- arg+ and the recipient is leu- thr+ arg-. The donor allele arg+ is initially selected for in recipients after transduction (all recipients will therefore be arg+), and 100 recipient colonies known to be arg+ are replica plated to determine the status of the leu and thr genes (see above). See above for the results of the replica plating experiment. What is the cotransduction frequency of a) arg and thr and b) arg and leu respectively?

24%, 36% Tip 1: Remember, all colonies are arg+! Tip 2: You are looking for cotransduction of arg+ with the other donor genes! Tip 3: Note that recipient is not "minus" for all markers! Tip 3: If arg+ and thr- are cotransduced from the donor into the recipient what will be the phenotype of the recipient? A potentially confusing fact about this cross is that the donor is leu+ thr- arg+ . That is thr minus. If the thr- allele is transduced into the recipient, the recipient will become thr-, i.e. incapable of growing on medium that is not supplemented with threonine. Scoring this phenotype is not a problem since we are replica-plating arg+ colonies. Therefore, thr+ recipient colonies HAVE NOT been transduced with the thr donor allele.

In a 100 million base pair genome, approximately how many times is the restriction site of the BamHI enzyme (5' GGATCC 3') by chance? Assume that the genome is random DNA sequence.

24,000 times

The image represent bacterial transcription. What is the most likely position for the promoter from which transcription is controlled?

240 W is the 5' end of the RNA. Lay down the Z-W molecule (RNA) on top of the DNA template. Its 5' end would reach ~ 230. That is the Transcription Start Site. The promoter is ~ -10b upstream, i.e. in the direction away from the transcription bubble.

Consider the design of Meselson and Stahl's experiment proving that DNA replicates in a semi conservative fashion. Starting with 15N (heavy) DNA, and after THREE generations in 14N medium, an E. coli cell should display the following percentage of heavy (15N), hybrid (15N/14N) and light (14N) DNA:

25% hybrid and 75% light DNA

5' TTACCCATGCTTGTATATGCGGATCATTGTTATTAG Which set of primers would you design in order to amplify as much as possible of this dsDNA by PCR? Only the top strand of the target is shown. Note: the primers shown are shorter than you would design in real life.

5'CTAATAACAATG and 5'TTACCCATGCTT

An mRNA has the codon 5' UAC 3'. What tRNA anticodon will bind to it?

5'GUA Pairing is antiparallel.

In parsley, tall is dominant to short and frilly to simple leaf. The tall and frilly genes assort independently. Crossing Tall frilly x short smooth yields this progeny: tall frilly=14, tall simple=7, short frilly=6, short simple=13. Assuming a two-factor testcross, chi squared for these data = _____. Therefore, chance alone ___ explain deviation from expected.

5, can This was a challenging question. 49% of the class got it right. A testcross is when an individual with a dominant phenotype, but unknown genotype is crossed to a recessive homozygous "tester" strain. Considering a single locus, if the dominant individual is homozygous, there will only be one type of progeny: phenotypically dominant. If heterozygous, two types will be found. Clearly, this is a testcross because: The question states so The phenotypically dominant parent is crossed to a double recessive Further, we conclude that the dominant parent is a double heterozygous because four phenotypic categories are seen in the progeny For two genes assorting independently (=unlinked), this testcross hypothesis involves an expected ratio of 1:1:1:1. Thus, we proceed as follows: We sum the progeny numbers = 40 Expected for each category = 10 Subtract from the observed, square, divide by observed, add result from all four = 5 From the table column for p=0.05 and 3 degrees of freedom, chi^2 of 5 is well below the threshold. We accept that chance could have resulted in the observed numbers when four equal sets are predicted.

Consider the complementation table below illustrating the analysis of 15 mutants, F to Z. All are recessive and homozygous. When crossed, the phenotype of the F1 is marked as 1 if complementation is seen (wild-type F1), or 0 if the F1 is mutant. How many genes are defined by this analysis and what alleles belong to what gene?

5, g1=FOQSZ, g2=MPY, g3=NRX, g4=TV, g5=UW

In the progeny of an F1 plant you expect four phenotypes resulting from the genetic action of dominant and recessive alleles at two independent genes. Out of 50 F2s, you count 27 double dominant, 7 and 9, respectively, single dominants, and 7 double recessive phenotypes. What Chi Square value do you calculate for this outcome and do you reject the above hypothesis?

5.5, no The (null) hypothesis is that the 9:3:3:1 ratio expected from a dihybrid cross will be observed. Out of 50, expected are 28.125: 9.375:9.375:3.125. The chi square is calculated by usual formula and results in 5.5. Looking at the column for 0.05 probability, the critical chi square value for 3 degrees of freedom is 7.8. Our lower number indicates that our P is higher (in fact, it is 0.141). There are 14% chances that random fluctuations will yield a phenotypic ratio as deviant as observed or more deviant. We accept the null hypothesis, i.e. that we are observing a 9:3:3:1 ratio and that any variation is due to chance.

The tinge of anatto color (a natural orange pigment) of the achiote plant has been attributed to the additive effect of the genotype at three independent and equivalent loci, W,X and Z. Upper case alleles contribute 2 units of red, while lower case allele contribute 0 units. A yellow wwxxzz plant is crossed to a red WWXXZZ plant. The F1 is selfed. What is the probability of progeny with exactly intermediate color?

5/16

A Tom cat and his kitten son share in average the following percent of alleles:

Consider the design of Meselson and Stahl's experiment proving that DNA replicates in a semi conservative fashion. Starting with 15N (heavy) DNA, and after TWO generations in 14N medium, an E. coli should contain this percentage of heavy (15N), hybrid (15N/14N) and light (14N) DNA?

50% hybrid and 50% light DNA

A mutation prevents splicing of a 600 bp intron positioned in the middle of a 3000 base long ORF. The resulting mRNA is 600b longer than the wild-type one. The following is the best estimate of the polypeptide length that could result from translation of this mRNA:

530

Table 2 represents progeny phenotypes from crosses between "miniature wing" fly mutants, each independent, recessive and single genic. 0=miniature. 1=wild-type. The data are consistent with the following number of genes.

6 Gene 1: F, M Gene 2: N, T Gene 3: O, Q Gene 4: P Gene 5: R Gene 6: S, U

6 kb

You expect an F1 plant with genotype MmTt to produce four phenotypic F2 classes (double-dominant, two different single dominant, and double-recessive). You count 80:18:36:10 F2 individuals, respectively. The chi-square = _____ . Your expectation is __________ .

6.1, accepted

The primary transcript of the T gene has the following characteristics: Primary transcript = 3000 bases Intron 1 = 300 Intron 2 = 300 Intron 3 = 300 5' UTR = 100 3' UTR = 200 Choose the closest estimate of T protein length in amino acids.

600 ORF = Primary transcript - introns - UTRs

You mate two mice with the following genotypes: MmKkLlPpQqRr x mmkkllppqqrr. Assume that each gene controls a different character and that the upper case allele is dominant, how many different genotypes and phenotypes do you expect?

64, 64

Three genes of corn, R, D, and Y, reside on chromosome 9 (see map, above). In a testcross RDY/rdy x rdy/rdy, how many RdY/rdy plants are you likely to get in a progeny of 1000 taking into account that interference = 0.3?

7 The frequency of recombinant gametes of that class is 0.1*0.2*COI (Coefficient of coincidence). The COI is 1-Interference = 0.7. So, recombinant gametes = 0.1*0.2*0.7*1000 = 14. This is for both types, RdY and rDy. RdY are half of 14, thus 7.

Which of these is added to the 5′ end of a pre-mRNA transcript shortly after the initiation of transcription in eukaryotes?

7-methylguanosine cap The molecule is methylated on the 7 position. it is attached to the 5' end of the mRNA.

What is the probability that a 7 pup mice litter has a gender ratio of 4:3 or 3:4 (i.e. it contains 4F or 3F)?

70/128 First, decide what is the P of any family? In this case, p and q are equal. Do not get confused by the fact that any family type has the same P. That is due to the fact that p=q. Second, how many ways are there to have three boys and four girls or four boys and three girls in a family? The binomial expansion should help here.

Gene E displays dominant epistasis on gene G. In the progeny of two EeGg individuals, the phenotypic effect of the G genotype cannot be determined in ______% of the progeny.

75 E_, __ : no G effect, P = 12/16 ee, G_ : G dominant effect visible, P = 3/16 ee, gg: g recessive effect visible, P = 1/16

Fig S1 displays a cellular event in an individual of a new rodent species. What is the somatic (in a cell such as found in the muscle) and gametic chromosome number of this species?

8, 4 This is anaphase I of meiosis. The cell still has all the chromosomes although it is about to split. I count 8. How many chromosomes would you expect a meiotic product to have? The number going to one pole: 4.

You prized hybrid cow has the following genotype: Qq,Rr,Tt. The Q, R and T loci are independent. She can produce ___ egg types, such as _______ .

8, qRt, qrT, Qrt, QRT

Coat colors of Labrador retrievers depend upon the action of at least two genes. An inhibitor of coat color pigment (ee) prevents the expression of color alleles at another independently assorting locus, producing golden coats. When the dominant condition exists at the inhibitor locus (E_), the alleles of the other locus may be expressed: E_B_ = black and E_bb = chocolate. When dihybrid (heterozgyous at both loci) black labs are mated together, what are the phenotypic proportions expected in the progeny?

9 black: 3 chocolate: 4 golden This is a typical example of recessive epistasis. Being ee creates golden coat color regardless of the alleles present at the other locus. A typical dihybrid cross produces progeny with the ee phenotype 4 times out of 16, so you would expect 4 golden coat colors.

Considering epistatic yellow in retrievers. An aa dog is yellow regardless of the genotype at the B locus. In the presence of the dominant A allele, dogs are black or brown depending on the presence of the dominant B (black) allele or recessive b (brown) allele. If the mating of two black dogs has previously produced puppies of all three colors, what ratio do you expect for black:brown:yellow?

9:3:4

Starting with a blue inbred line of violet, you isolate white recessive mutants that you classify in two complementation groups corresponding to independently assorting genes. If two of these white mutants are crossed and produce a violet F1, what ratio of violet:white plants do you expect in the F2?

9:7 Mutant 1 aaBB, Mutant 2 AAbb. Both A and B are needed for violet color formation. This works like double recessive epistasis: two genes, both needed, for wild-type phenotype. Only A_B_ is blue 9/16, all others (A___, __B_, aabb) are white. These classes add to 7/16.

Okay, let's practice a restriction enzyme map exercise. We start with a linear fragment, Sizes are in kb: No treatment: 15 XbaI: 8, 5, 2 BglII: 10, 5 PstI: 9, 6 BglII + PstI = 9, 5, 1 BglII + XbaI = 8, 3, 2, 2 XbaI + PstI = 8, 4, 2, 1 all three = 8, 3, 2, 1, 1

== X === B = P = X ========

Read the question above and provide the correct answer

A This is a wild-type operon except for permease (Y). No Lac, no glucose = Repressor on. CAP would be stimulating transcription. Therefore, no expression. Lac, no glucose = Repressor off. CAP is stimulating transcription. Therefore, high expression of beta-gal, but no permease. Lac + glucose = Repressor off. CAP is not stimulating transcription. Therefore, very low expression* * It is debatable whether glucose+lactose conditions would turn off the lac operon. We choose to consider it off here. In fact, it varies from strain to strain. Some problems may imply some expression. Confusing? We promise that it will not be in the midterm.

An mRNA molecule has the following percentages of bases: A = 23%, U = 42%, C = 21%, and G = 14%. What would be the corresponding percentages of bases in the template strand of the DNA that contains the information for this gene?

A = 42%, T = 23%, C = 14%, and G = 21% Remember that mRNA will be complementary to the template strand. Therefore, mRNA/Template A=T, G=C, U=A, C=G (read A = T here as A in mRNA = T in template strand, etc.)

A recombination frequency of 45% between the A and B loci suggests that:

A and B are nearly unlinked: they are distant on the same chromosome. 45% suggests they are on the same chromosome but sufficiently distant that C.O. scrambles them nearly as efficiently as being on different chromosomes.

Consider the above diagram of DNA molecules, each ~1000b long. Which molecule(s) could be a substrate for DNA polymerase III?

A and D

A eukaryotic transcription factor that represses gene activity could have the following activity on promoters:

A modified version of the CAS9 protein entails mutated endonuclease sites. By fusing the ORF of this CAS9 to that of a protein like GAL4, scientist have made chimeric CAS9-Activator protein. These proteins should work well for the following purpose:

A guide RNA can target CAS9-GAL4 to any promoter, resulting in activation of the target gene (if GAL80 is absent or inactive)

A mutant E. coli cell cannot transcribe the genes associated with a specific sigma factor, but expression of all other genes was not affected. What is the most likely explanation for this observation?

A mutation within the sigma factor gene caused it to no longer recognize and bind its target consensus sequence Sigma factors recognize specific promoters. A mutation in a specific sigma factor would only affect transcription of genes that contain the specific promoter(s) it recognizes. Mutation of all the consensus sequences is highly unlikely (lots of mutations would have to occur) and mutation in the core RNA polymerase should affect all gene transcription.

A 13b (b=base) intron is in the middle of the coding region of gene R5, which encodes a 400 amino acid protein. If this intron is not spliced out, what effect would you predict on the gene product?

A shorter protein is most likely, with a few different amino acids before truncation The reading frame will be changed (intron has a number of base non divisible by 3) and the P of an accidental stop codon is very high. The ribosome entering the intron will make different amino acids than expected even after it reenters the normal coding region because it will be in a different reading frame. To put this in a proper frame (no pun intended), consider a short polypeptide encoded by 20 codons. What happens if you add 12 bases to the middle of the mRNA? How different will be the encoded polypeptide? What if you add 13 bases? Remember that codons are 3b long. You will recognize that the outcome is a frameshift mutation. Once you frameshift the ORF, the P of finding a stop is very high as you progressively go through more random codons: (1-(3/64))^N, where N=number of codons considered

In a trip to Mexico you buy pure-breeding seeds of a tomato with green stripes over a red background, which you call Zebra. You cross zebra to a common red-fruited variety, Early Girl. The F1 is spotted green on red background. In the F2 you observe 33 solid red, 70 spotted, and 29 zebra. The mutation involves change(s) at ______. Individuals with spotted phenotype are _________ .

A single gene, heterozygous

Which of the following BEST describes alternative splicing?

A single pre-mRNA can be spliced in multiple ways that ultimately lead to the formation of several different proteins from the same gene sequence.

CRISPR-Cas9 is complexed with the target DNA forming an open DNA bubble.________ DNA strand(s) ____ based-paired to RNA; ____ endonuclease motif(s) cut(s) _________DNA strand(s):

A single, is, two, both

The following evidence would support genetic sex determination over environmental sex determination:

A small and a large chromosome pair during meiosis I. One or the other is found in gametes.

Which statement best describes a Holliday junction?

A structure between homologous DNA segments This is the only answer that accurately describes a feature of a Holiday junction.

You are studying the rII region of phage T4 just like Seymour Benzer did in the 1950's and 1960's You took four rII- strains carrying different mutations, identified as A, B, C and D. When mixed, infections are performed in E. coli strain K, and the following results are obtained (above). The conclusion most consistent with these results is:

A, B, and C have mutations in the same gene, D is in a different gene.

The following molecules are ~ 10,000b long. They are provided in a test tube with NTPs. Which molecule(s) could be a substrate for primase? PS. The question is not whether primase is needed, but whether primase could use the shown molecule for its enzymatic activity

A, C and D Primase can act anywhere there is an overhang, i.e. ssDNA, a template. This is because it can synthesize an RNA polymer without the need of a 3' end. Please review lagging strand synthesis to grasp this point.

Refer to the replication bubble above. Choose the answer that provides the correct 5'/3' polarity of the template strands (ends labeled A, B, C and D)?

A-3', B-5', C-5', D-3' The fork movement is clearly marked. You can deduce that the arrows represent the direction of DNA strand growth. DNA is always synthesized in the 5' to 3' direction, so it is easy to assign polarities to the newly synthesized DNA (see DNA strands labeled E,F): each arrow represents a 3' end. Since DNA is always anti-parallel, A and D must also be the 3' ends, and B and C the 5' ends.

Your friend from BIS101 performed transformation experiments in E. coli and proudly told you that genes A and B are closest to each other, followed by genes B and C, and A and C are farthest apart. Match his statement with the correct coinheritance frequencies, below. Explanation: A-B = 36% means that the frequency of cotransformation between A and B is 36%.

A-B = 36%, B-C = 24%, A-C = 12%

Loci A and B are linked at a 25 m.u distance. The cross AaBb x aabb produced 1000 progeny. Of this progeny, approximately _______ are expected to be recombinant (ignore cross-over interference). Alternatively, if the map distance is determined to be 40 m.u., ________ progeny are expected to be NON recombinant.

250, 600 0.25 * 1000 = 250 recombinant 0.4 * 1000 = 400 recombinant ---> 1000 - 400 = 600 NON recombinant

How many codons would be possible in a triplet code if only three bases (A, C, and U) were used?

The WLE gene of sweet orange has a transcript with the following characteristics: Primary transcript = 1637 bases Intron 1 = 152 Intron 2 = 248 5' UTR = 124 3' UTR = 237 (does not include stop codon) How many amino acid residues does the WLE protein have?

291 ORF = Primary transcript - introns - UTRs

Consider the epistatic problem of the piping vs spiny pineapple. Piping is epistatic on spiny. If the piping locus "P" has genotype pp, the spiny trait is manifested. P_ plants are always piping. ppS_ = spiny tip, ppss = spiny. Two inbred lines, one piping, one spiny are crossed and produce a piping F1. Selfing of the F1 produces the following F2s: 118 piping, 28 spiny tip and 12 spiny. What ratios, in that order, do you expect if you cross the F1 to the inbred spiny parent?

2:1:1

The genome of E. coli W contains IS51 at thee different sites. F plasmid M contains a single IS51. This means that E. coli W and F plasmid M can form this many different Hfr strains:

3 Each pair of homologous IS, one on the chromosome, the other on the F plasmid, can lead to a single type of Hfr integration. Three IS elements on the chromsoome --> 3 Hfr

A geneticist is working with a new bacteriophage called phage Y3 that infects E. coli. He has isolated eight mutant phages that fail to produce plaques when grown on E. coli strain K (similar to Benzer). To determine whether these mutations occur at the same functional gene, he simultaneously infects E. coli K cells with paired combinations of the mutants and looks to see whether plaques are formed (also like Benzer). He obtains the following results (see above). A plus sign means that plaques were formed on E. coli strain K; a minus sign means that no plaques were formed on E. coli strain K. How many functional genes (cistrons) do these mutations belong to?

3 This is equivalent to a complementation test with independent mutants in a diploid organism Solve this problem exactly like the complementation problems in the Allelic interactions module. Hint: start with mutant 1 and assign it to a cistron (gene) called M. Going down the corresponding column, which other mutants affect the same cistron? Once you have assigned all relevant mutants to M, make another gene (N) and assign to it the next mutant that is not in M. Etc etc...

A frameshift mutation can result from the following change (mark the WRONG statement): bp=base pair

3 bp insertion

Mendel test-crosses an F2 pea of unknown genotype obtaining 8 phenotypic categories. He concludes that this plant is heterozygous at ___ genes. The p of the fully recessive category in this testcross progeny is ___.

3, 1/8

Studying an F2 family, you suspect the action of two unlinked genes because the phenotypic ratio "looks like" 9:3:3:1. To determine whether your guess is reasonable, you subject your observed phenotypic counts to the chi square test obtaining a chi square value of 8.2. The degrees of freedom are _____. You conclude that your guess is ______.

3, unsupported

A test cross is performed on a Drosophila (fruit fly) presumed to be B/b, F/f. B = black body, b = brown body. F = forked bristles, f = unforked bristles. The resulting offspring has the following phenotypes: Black, forked = 230, black, unforked = 210, brown, forked = 240, and brown, unforked = 250. The Chi-square value of this data set is ____ and the degrees of freedoms are _________ . *Note on genotypic nomeclature: B/b is a equivalent to Bb *Use your textbook instructions, or an Internet site to calculate the chi square

3.75, 3

Two pure breeding lines of rice are crossed. One is dark-green and tall, the other light-green and short. The F1, dark-green and tall, is allowed to self. The probability of dark-green and short plants in the F2s is:

3/16

You breed a pigmented german shepherd heterozygous for recessive white coat color (Ww) to a white female. What is the probability that if they have four pups, two will be white and two pigmented?

3/8

What fraction of families with four children have an equal number of boys and girls?

3/8 First, decide what is the P of any family? In this case, p and q are equal. Do not get confused by the fact that any family type has the same P. That is due to the fact that p=q. Second, how many ways are there to have two boys and two girls in a family? The binomial expansion should help here.

The genome of a double stranded DNA virus contains 14% T (thymine). How much of this viral genome is composed of cytosine (C)?

36% Since T = 14 % and A = T, then A = 14 %, and A + T = 28 %. Since A + T + G + C = 100 %, then G + C = 100 - 28 = 72 % G = C, so G = 72 / 2 = 36%

Blue eyes are recessive to any other color. Mary and John have brown eyes, but each have a blue-eyed father. What is the probability that at least one of their three children is blue-eyed? Hint: sometime it is easier to calculate the P of the alternate event

37/64 B_ = brown-eyedbb = blue-eyedboth Mary and John are Bb. P of blue-eyed child is 1/4, P of brown-eyed is 3/4. The P of three children with brown eyes is 3/4^3. Subtract that from 1 and you get the P of the alternate event: one or more blue-eyed children.

In cow, horned is a recessive trait, polled (hornless) is dominant. You interbreed polled animals that you believe are heterozygous for the Polled gene. In their progeny you count 30 polled and 20 horned. What ratio of polled:horned animals is expected?

3:1 Pp x Pp -> 3 P_ : 1 pp

The codon(s) for a tRNA anticodon of sequence 5′-GGC-3′ is

3′-CCG-5′

A eukaryotic gene has 5 exons. It will have ___ introns.

Consider the following phases of meiosis: prophase I - telophase I - prophase II - telophase II. The content of DNA COPIES for a given LOCUS in a diploid cell undergoing these phases is as follows:

4 - 2 - 2 - 1 Four copies after S phase, split 2 and 2 at anaphase I, and again at anaphase II

The above map of a chromosomal region of Bacillus subtilis was derived by cotransformation mapping in the early 60s. The mapping cross used a wild-type donor and a recipient that was auxotrophic for tyrosine, histidine and tryptophan with genotype tyr1-, his2-, trp2-. Consider the molecular events leading to formation of these two transformants: i) trp2+, his2-, tyr1+ ; ii) trp2-, his2+, tyr1+. Most likely, the following number of crossing overs were involved, respectively:

4, 2

Two pure breeding lines of pea are crossed. One is purple-flowered and tall, the other white-flowered and short. The F1 is purple-flowered and tall. Crossing the F1 to the white and short parent results in ____ phenotypic and ____ genotypic classes.

4, 4 This is a testcross with two assorting loci. The het parent makes four gamete types. The homozygous recessive parent 1.

The figure below shows the relevant coding strands and reading frames for the wild type and for a frameshift mutation in gene G7. The best strategy to revert the mutation to wild-type is i) to use CRISPR-Cas9 to cut inside codon ____ of the mutant and ii) to use ________ .

4, the wild-type repair template

Testing the hypothesis that orange vs white color in koi carps is conferred by a single dominant locus, you cross two pure breeding lines, one orange, the other white, then intercross two F1s. In one hundred F2 hatchlings you observe 16 white and 84 oranges. The chi square of fit to single locus inheritance is _______ . Accordingly, you _______ the hypothesis .

4.3, reject

How many colonies are auxotrophic mutants?

A scientist conducted a hybridization experiment where she hybridized the DNA from a pig to mRNAs she purified from the same species and then examined the structure via electron microscope (note: as part of the hybridization procedure DNA was denatured into single stranded molecules). Which of the following observations would MOST likely be possible?

The portion of the DNA strands annealed to mRNA, but certain regions of DNA looped out. The intron regions of the DNA would not match with the spliced mRNA, and would loop out.

A 12b (b=base) intron is in the middle of the coding region of gene R5. If this intron is not spliced out, what effect would you predict on the gene product? Keep in mind that intron is essentially random, non-coding RNA.

The resulting protein will most likely be 4 amino acid residues longer The intron adds four codons. The P of a stop is 0.17. So, mostly likely a protein 4 amino acids longer than the expected one will result.

The following is not true about DNAseI Hypersensitive Sites (DHS):

The stronger the hypersensitivity the less likely that the connected gene will be expressed The formation of naked DNA regions next to a transcription start site, or in any other regulatory site (such as an enhancer), is highly predictive of impending activation.

Which of the following statement describes the 'wobble' rules correctly?

The third base pairing between the tRNA and mRNA is relaxed.

An RNA molecule has the following percentages of bases: A = 23%, U = 42%, C = 21%, and G = 14%. Which of the following statements is TRUE:

This RNA molecules is single stranded Since the A base composition of this strand does not equal that of T (and also G doesn't equal C) we must conclude this is a single stranded nucleotide sequence. It is RNA since it contains uracil bases. We do not have enough information to know if it is an mRNA (or any other type of RNA molecule).

ABO blood type in humans results from this type of genetic system:

Three alleles involving dominance and codominance

What is this?

Transcription: W=RNA, M,Q = DNA The melted bubble of dsDNA and an RNA strand "hanging out" is diagnostic.

Which of the following molecules is capable of targeting chromatin remodeling complexes to specific DNA sequences to modify chromatin structure and activate gene expression?

Transcriptional activator

Which step in the molecular cloning process directly generates a genetically modified organism (GMO)?

Transformation

Mr and Mrs Garcia obtained a karyotype of their unborn child. It displayed 22 autosomes and 2 X chromosomes. They bought baby girl outfits but when the baby is born they are surprised: he is male! The karyotype is confirmed. What is the best explanations?

Transposition of SRY to an X or an autosome

A transgenic organism always contains recombinant DNA in its genome

True

A scientist introduces into potato the coding region of gene A spliced to promoter from gene B. Both A and B are potato genes. Is this genetic engineering?

True Technically and according to our definition, this is genetic engineering. However, this does not use any DNA from another species and it may undergo a different regulatory treatment. The use of same species DNA in modifying an individual is called cisgenesis (no need to remember this term).

A tall plant was discovered in a field of pure breeding short sunflowers. Crossing the tall mutant to a true-breeding wild-type plant you obtain 16 tall and 10 short plants. What is the most likely genotype of the tall mutant and the inheritance pattern of the tall phenotype?

Tt, dominant

In the progeny of a white-eyed female and a red-eyed male, one of 500 flies is a white-eyed female. The gender of this exceptional fly is caused by __________ , and its eye color by _________ .

Two X + one Y, non-disjunction in the mother This is Bridges original experiment. Non disjunction in mother resulting in egg with two X, which is fertilized by sperm with Y. XXY is female in flies.

A single crossing over is well suited for integration of a plasmid into a chromosome, but it is not sufficient (actually, very deleterious) for insertion of a gene carried by an HFR conjugation strand?

Two circles are joined efficiently by a XO. However, a single XO between a linear DNA (HFR strand) and the chromosome will break the chromosome open.

Rosalind Franklin's picture 51 was obtained by X-ray diffraction through a DNA molecule. It provided the fundamental information that DNA's structure consisted of:

Two covalent chains wound in a double helix

Where along the length of a mature mRNA will one find the start codon?

Typically 30-70b from 5' end but could be 100s or even 1000b A common assumption by students is that an mRNA starts with the AUG. This is NEVER the case. The ribosome needs a landing pad and a 5' UTR (UnTranslated Region) is always present. Same, by the way, for the 3' UTR.

A TEMPLATE DNA strand reads 3' GATTACATTGGCC. The following could be an RNA transcribed from this DNA is: 5'...

UGUAACCGG 5'UGUAACCGG>3'GATTACATTGGCC

Which of the following codons codes a different amino acid from the rest? Hint: use the genetic code table

UUU

Siamese and Abyssinians are pure breeds of cats with high level of homozygosity. They are quite different both in appearance and genetically. Mating Siamese to Siamese and Abyssinian to Abyssinian will produce the following progeny type:

Uniform and highly homozygous

CRISPR-Cas9 can be utilized to knock out (make a null mutant) multiple genes at a time. Which answer best describes how this could be best accomplished?

Use multiple guide RNA molecules against targeted loci

Marker allele Y7 and allele D are linked in phase at 16 m.u. Consider an individual that has Y7-D/Y2-d haplotypes. The P of a Y7-d gamete is equal to the P of a ______ gamete, and is _________ .

Y2-D, 0.08 One chromosome has Y7--D linked in phase. The other has Y2--d. These correspond to the gametes with parental genotypes (Y7D, and Y2d). 0.16 is the fraction of all gametes that has undergone a CO event between Y and D forming the recombinant alleles,Y7d and Y2d. Calculate the frequency of all gamete types.

The CRISPR-Cas9 system has been obtained by engineering...:

a bacterial defense system against phages

The following is NOT required for discovery of DNAseI hypersensitive sites

a cell lysate capable of carrying out translation of exogenous RNA

A missense mutation is most likely to be deleterious when it occurs in:

a conserved protein motif

The recipient (female) cell is pro-, thr-, pur-. Formation of a pro+ thr- recipient requires the following molecular events:

a double crossing over between Hfr DNA and recipient chromosome

The process of recombination and dsDNA repair are connected in this way:

a dsDNA break initiates recombination dsDNA repair --> recombination. It is possible that a dsDNA repair could take place without resulting in recombination

A Holliday junction formed during meiosis is INITIATED by the following event:

a dsDNA cut in one of the two paired chromosomes

The homology search associated with dsDNA breaks repair requires:

a protein that binds ssDNA with 3' end dsDNA breaks are resected to free 3' protruding ends (ssDNA) that bind recA

The following is a substrate to which Telomerase could add nucleotides. Assume that the correct repeat is present in all:

a protruding ssDNA with a 3' end A protruding 3' is lengthened by telomerase: see module and lecture 13 notes.

In prokaryotic translation, the ribosome first binds this site on the mRNA:

a purine rich site close to the start codon

The following is a substrate to which DNA Pol III could add nucleotides. :

a recessed, fully base-paired 3' end

In most tissues, gene W is off due to a repressor binding a promoter DNA motif (like an operator) . When the hormone gastrin binds the repressor, it inactivates it, allowing W expression. Deleting the DNA binding motif will result in the following W expression pattern: with gastrin ______, without gastrin _______ .

activated, activated

Transposable elements are present as repeated sequences in the genome of many eukaryotes. A loss of function mutation in DNA methyltransferase would result most likely in:

activation and expression of transposable elements Most transposons in genomes are kept silent by DNA methylation and formation of repressive chromatin. The DNA methyltransferase is a key component of this repressive pathway.

You isolate yeast chromatin and break it down to fragments with one or two nucleosomes. Using immunological (antibodies, no need to know any detail), you selectively isolate nucleosomes that have acetylated histone tails. You extract DNA from this purified fraction and sequence it. Comparing the sequence you obtain to the annotated genome sequence, you expect to find the location of:

active promoters

Oak alleles T and t differ in transcription level and have an additive effect on tannin content (T high, t low). The two alleles differ only by a SNP 100 bp upstream of the gene TATA box. The best explanation for this trait is that the SNP:

affects binding of a transcription factor to DNA The different accumulation of tannins in this system is the likely result of different expression by each allele. The mutation in the promoter is a good candidate because it could have changed the binding of a transcription factor.

A cell loses telomerase. The cell will die:

after several divisions The chromosomes can stand some loss of terminal DNA without being negatively affected. Multiple divisions are needed for chromosomes to lose some of their genes and therefore to affect cell survival.

Consensus sequence motifs for promoters are derived by:

aligning different promoters and identifying shared bases

Which of the following scenarios will produce a "male" human?

all the other answers XYY XX with a copy of SRY on an autosome XXYY with one copy of SRY deleted XX with a copy of SRY on both X-chromosomes

Consider the Galactose regulatory factors GAL4, GAL80 and GAL3. Imagine having mutations that affect their function in different ways (complete loss of function, loss of binding to allosteric regulator, etc) and testing the different alleles using merodiploids (or diploids). Would you find that they work in cis or in trans?

all trans

A popular chromatin organization model involves loops. The insulator elements at the base of loops are most likely involved in this function: they

allow the loop chromatin to undergo a given conformation, e.g. open or close, and prevent that state from spilling over in the next loop

The typical ORF of a bacterial gene:

always clear, start to stop

An expanding replication bubble is depicted in Fig. 1. Consider the strands Z and Y. Within this structure and compared to Z, the number of polymerases of any type associated with Y is:

always higher Y represents the template for the lagging strand where multiple polymerases are at work: primase, DNA pol I and III.

An F' plasmid carrying the wild-type Trp gene most likely originated when:

an F integrated near Trp excised imperfectly

Inside a cell, a likely step toward replication of the following single DNA strand, 5'CCCCAGGGATTACGGG 3', will involve this :

an RNA primer made by primase A primer is needed to start replication. In cells, the primer is RNA and synthesized by primase.

An F' plasmid is produced when:

an integrated plasmid in a Hfr strain excises from the chromosome taking along a segment of host chromosome

Paired homologous chromosomes are separated at:

anaphase I

The S and Y genes are unlinked. In an individual with genotype SsYy, in the absence of crossing over, the most likely place for the S and s alleles to part company is

anaphase I

Crossing in peas involves the following manipulation of "female" and "male" individuals:

anthers are removed from flowers of chosen females, pollen from chosen males is applied to female stigma Each flower is hermaphroditic, i.e. it bears both male and female organs. Anthers, which cast pollen, are removed from the "female" to prevent self-pollination.

Mark the wrong statement about mitosis of haploid cells:

cannot occur Mitosis is cell division. Many haploid organisms exist, such as bread yeast.

A researcher uses an expression plasmid in which the cloning site EcoRI is within the ampicillin resistance gene. After cutting this vector with EcoRI she adds a DNA segment (S) that has EcoRI-complementary sticky ends (overhanging single stranded regions), ligates the mix, transforms E. coli and selects transformants on agar plates supplemented with ampicillin. The colonies that contain the S fragment cloned into the vector ____ be identified because ___________:

cannot, do not grow

The following organism(s) does (do) not have a life stage where one haploid cell forms two haploid cells:

cat gametes are the product of meiosis in all mammals and cannot divide further

An epigenetic trait is caused by:

changes in chromatin, but no change in DNA sequence

Analysis of mutations affecting a sequence motif in an enhancer site will display ____ properties:

cis enhancer sites act in cis

Wild-type T9 phage forms clear plaques. Two cloudy plaques mutant phages are grown separately, mixed 1:1, and infected at high or low M.O.I. on E. coli. High M.O.I. results in mostly clear plaques. Low M.O.I yields mostly _________ plaques because _______ . M.O.I. = Multiplicity of Infection

cloudy, complementation cannot take place The two phage strains are mutant in different genes because they complement when coinfected in the same cell (high MOI). Low MOI results mostly in single phage-infection events. Each phage type, alone, results in cloudy plaques.

Expression of blood serum albumin in bacteria requires the following:

coding region without introns, prokaryotic promoter, prokaryotic terminator

Two phage mutants display similar phenotypes: they both form very small but clear plaques. The wild-type phage makes much larger plaques. To determine whether two phage mutants can complement the following must be carried out:

coinfection of susceptible host followed by assay Coinfect permissive host: if the phages produced together the proteins necessary to make large plaques they complement.

HDAC4 is a histone deacetylase that stimulates female ovulation. HDAC4 binds DNA around the promoter of the NOV gene, whose expression inhibits ovulation. Likely, HDAC4 deacetylates histones, which leads to:

compact chromatin and NOV repression

Environmental variance, Ve, for a human quantitative trait can be best estimated by :

comparing twin pairs

The following mode of bacterial sex requires contact between live donor and live recipient:

conjugation

The family tree above illustrates the inheritance of a rare disease condition. Linkage between P locus and the disease locus is ________ with the data; likely, P5 is _________ with the disease allele.

consistent, in coupling Dominant condition resulting from allele T. Chelsea is Tt, P5P2 . Note that condition is associated with P5 . Linkage and the following phases make sense: T, P5//t, P2.

An Hfr strain is derived from a strain that:

contained an F plasmid Hfr strains are produced by the occasional integration of a sex-capable plasmid into the host chromosome.

In the presence of a homologous template DNA representing allele w1, cutting of a genomic site inside w2 allele of the W gene by CRISPR-Cas9 results in the following:

conversion of w2 to w1

Consider an F1 pea plant heterozygous at two independent loci, inflated vs constricted (Ii), and axial flowers vs terminal (Aa). From the selfed progeny you germinate and observe 16 F2s. The following can be said about the scored phenotypic ratios:

could be any ratio, but likely skewed toward the double dominant I_A_

A CRISPR-Cas9 contains a guide RNA complementary to a unique sequence in exon 5 of gene G. This CRISPR-Cas9 is expressed in fibroblast MCK cell line. MCK is heterozygous for G: it has a wild-type allele (G) and a mutant allele (gdel). In the latter, intron 4, exon 5 and intron 5 have been deleted. As a result of CRISPR-Cas9 expression, the MCK genome will be:

cut once at G, not a gdel

An XXX fruit fly is expected to be ____ because of _____:

dead, X:autosome imbalance

The Moderna and Pfizer COVID vaccines entails an RNA molecule in a lipid capsid. After injection in a patient, the RNA is taken up by cells causing expression of the COVID-19 surface antigen (a protein). A competitor's vaccine lacks the 5'-CAP of the mRNA. The expected outcome is:

decreased production of the protein product and efficacy of the vaccine

The function of tRNA is to:

deliver and guide amino acids for polymerization on the ribosome

A tRNA-Ser mutates its ANTICODON from 5'GGA to 5'GCA. The use of this mutant tRNA during translation of the +1 frame in RNA 5'GGUUGCCGU will yield:

gly-ser-arg Gly Cys Arg original peptide5' GGU UGC CGU The mutation on the Ser tRNA will enable the mutant anticodon to pair with UGC ACG-5'GGU UGC-3 CGU This will results in a : Gly-Ser-Arg peptide

Guanosine triphosphate (GTP) is incubated with polynucleotide phosphorylase, an enzyme that can polymerize NTPs without a template and a primer. The resulting polymer is incubated with an E. coli translation extract. Which labeled amino acid is retained by the Meselson and Stahl filter?

glycine An RNA made of Gs encodes only the GGG codon = Gly

A 5'-GGG RNA is mixed with an E. coli translation extract in the "triplet binding assay". Which labeled amino acid is retained by the Meselson and Stahl filter?

glycine The GGG codon = Gly

Gene Raf of the bacterium Pseudomonas encodes raffinase, an enzyme that hydrolyzes raffinose to simple sugars. You align the Pseudomonas raffinase sequence to that of multiple bacterial species and find strong conservation from amino acid 250 to 290. The rest of the protein varies considerably. A frameshift mutation caused by a single base pair insertion at nucleotide position 450 of the open reading frame, inactivates the Raf gene. At a later time, a two base pair insertion takes place at nucleotide 460. The probability that this second mutation could cause the Raf mutant allele to revert is:

good

A cystic fibrosis patient is homozygous for a 2 bp deletion in an exon encoding the CTFR protein. To cure this with CRISPR technology you target a _______ to a site _______ providing __________ .

guide RNA, very near the deletion, a wild-type copy of the affected exon

Cell size increases with DNA content. Yeast strain W has mean cell size of 40 microm^3. Yeast strain K has mean cell size of 80 microm^3. Rarely, two cells of strain W can fuse forming "new" strains that have the mean cell size of 80 microm^3. A plausible interpretation of these observations is that strain W is _____, strain K is ______.

haploid, diploid

In a large and consanguineous human pedigree for a dominant autosomal condition, there is no evidence for homozygous dominant individuals. This is most likely due to:

haploinsufficiency The mutant allele causing this condition is likely haploinsufficient. The homozygous mutant genotype result in lethality and is therefore never found.

Curly leaf of pepper is caused by a recessive allele at a single locus. A homozygous curly leaf plant is crossed to a heterozygous, normal plant. In their progeny you count 60 healthy and 15 affected individuals. The best explanation for these results is that the curly leaf trait

has incomplete penetrance

DNA highly compacted by nucleosomes and other regulatory proteins is called _________ . Promoter sequences in this DNA _______ accessible by RNA polymerase.

heterochromatin, are not

When overheated (heat shocked) all organisms express a specialized set of conserved proteins that help them overcome heat stress. Through the study of mutants in maize you identify a recessive mutation affecting a locus that is required for activation of the gene encoding heat shock protein 90 (HSP90). Homozygous mutants at this locus fail to activate HSP90 gene transcription. After several years of work, you determine that this locus is 70Kb from the HSP90 gene, acts in cis and corresponds to a DNAseI hypersensitive site. The most plausible hypothesis is that this locus:

is a enhancer

Comparing the DNA of T-Rex and human, their G+T % ______ because ____.

is the same and = 50%, of Chargaff's rule A basepairs with T, and G base pairs with C. Chargaff's rule states that A=T , G=C. Picking, for example, A = 49 %, it follows that T = 49 % and their total make up is 98 %. Therefore G + C = 2% and G=1%. Therefore G + T = 50%. Try this again, i.e. pick any A % between 0 and 50, and you will see that G + T always = 50%.

Homologous recombination is described by all of the following EXCEPT:

it involves chromosomes that do not normally pair with each other or DNA molecules that are not homologous. Homologous recombination by definition requires DNA sequences that match (are homologous) in order for recombination to occur.

The one gene, one enzyme hypothesis was an important advance in our understanding of genetics because:

it was the first hypothesis to link each gene with a single protein.

The image above represents translation in the cell of a mammal. The transcription bubble responsible for synthesis of this mRNA, is:

no longer connected, mRNA end is no longer base-paired to DNA In eukaryotes transcription and translation are separated.

You travel to the high Alpine valley from which prof Comai originated and measure heritability of IQ in the moderately inbred local population (they are all cousins of LC). The heritability you measure can now be applied to:

no other population

A healthy woman with no history of hemophilia in her family mates with a man with hemophilia, a recessive X-linked condition. Their first child is a girl with Turner Syndrome (XO sex chromosomes), who is also hemophiliac. Most likely, the child resulted from this egg __________, and this sperm ________:

no sex chromosomes from non disjunction at meiosis I or II, normal

A dog is heterozygous at the P locus (Pp), a gene on chromosome 2, produces a Pp egg. Assuming no crossing over, this would most likely result from:

non disjunction at anaphase I of meiosis

The DNA strand used as a template for transcription is termed the __________ strand.

non-coding Watch videos/read book if this is still confusing. It is called "non-coding" because its sequence is complementary to that of the RNA, which is by definition "coding".

A nonsense mutation occurs at codon 10 of a 230 codon ORF. The following change can likely restore the ORF to wild-type function:

none of the other answers works

Through genetic engineering you express the Cas9 protein and nothing else in a transgenic mouse. The probability that this protein will be cutting specific sites in the mouse chromosome is ___, because _______:

none, guide RNA is needed

Ferdinand is affected by hemophilia, a recessive x-linked condition. He mates with an unaffected woman with no history of hemophilia in her family. Their first child is a girl with Turner Syndrome (XO sex chromosomes), who is also hemophiliac. You infer that she resulted from the following sperm and egg (answers listed in the order of sperm, egg):

normal, non-disjunction at meiosis I

The DNA sequence T is contained within a larger fragment called S. You obtain the appropriate primers to amplify T. Your lab mate, however, does not give you the normal S dsDNA fragment. Instead he gives you only a preparation of the top strand (no bottom strand!). Your ability to amplify S from this template is:

not affected

"Freckled" and "Taster of PTC", a bitter substance, are independent and dominant traits in human. Use F-f and T-t, respectively, for alleles. Pablo, non-freckled and taster, lived with Sally and George, who were both non-freckled, non-taster. Sally has child who is freckled and taster. The father of this child could:

not be either man

Instead of using Taq DNA polymerase, from hot spring bacterium Thermus aquaticus, you decide to use E. coli DNA Polymerase I for your PCR reaction. This will ____ , _______.

not work, E. coli DNA pol will not survive the first cycle

The primer for RNA synthesis is provided by

nothing. It is not needed.

Consider Fig. 3. In the TATAA box region, the density of _________ is expected to be ______ due to the action of _________.

nucleosomes, low, chromatin remodeling complexes

Genes M and gene P of bacteriophage GammaDelta are required to infect E. coli UCD. Neither gene is necessary for GammaDelta to infect E. coli ChicoSU. Two unknown GammaDelta mutants, 1 and 2, grow on ChicoSU but not on UCD. You mix these two and infect UCD at high density. You obtain good infection and phage growth. A very likely scenario is that mutant 1 and 2 are:

one mutant in M, the other in P, no recombination needed for growth on UCD

In experiments involving bacterial sex, the transfer of DNA from donor to recipient is documented most easily by:

selecting for a gene present on donor DNA Out of all the answers presented here, this is the easiest to perform. If the donor is arg+ and the recipient arg-, see if the recipient becomes arg+ by growing it in the absence of arginine.

Absence of the "anhidrotic ectodermal dysplasia" (AED) gene product results in scaly skin, an X-linked condition. A man with AED has a daughter with mixed patches of normal and scaly skin. The best explanation is that:

she is heterozygous; skin patches express either the wild-type or the disease allele This is the human version of "calico". Woman inherited a normal allele from her mother, but a mutant allele from father. Following, X inactivation the defective allele (AED syndrome) is manifested in patches.

A homozygous loss of function mutation in the telomerase gene would result in the following phenotype:

shortening of chromosomes

A bacterial virus has a DNA genome composed of 10% A, 35% T, 35% G, and 20% C. Most likely, this genome is:

single stranded This genome violates Chargaff's rule: A=T, G=C. The best explanation is that it is ssDNA. ssDNA is not base-paired and therefore it does not follow the rule.

Which of the following proteins helps to hold the DNA strands apart while they are being replicated?

single-strand binding proteins

Repressors (also called silencers) that selectively regulate certain genes may function in multiple ways. A mode of action observed for certain repressors is to bind...

sites near to or also bound by an activator, physically preventing the activator from binding

During initiation, the _____subunit is the first part of the ribosome to associate with the mRNA.

small

Duchenne Muscular Dystrophy (DMD) is a recessive X-linked condition, where D = wild-type allele, d = defective allele. In affected families only males (X-d/Y) display the severe condition. Rarely, carriers (X-D/X-d) display mild form of dystrophy. A plausible explanation for this observation is that in these individuals:

some muscles are preponderately made up of cells in which X-D is inactivated

There are several tRNALys in kangaroo that are encoded by different tRNA genes. A mutation in one of these tRNA genes makes the anticodon of the encoded tRNALys complementary to one of the stop codons. As a result of this mutation:

some proteins will be longer and less functional These type of tRNA mutants are not uncommon. They result in the specific amino acid being inserted instead of translation termination.

A human female is produced when an X-containing egg is fertilized by an X-containing sperm. Inactivation of the X contributed by the sperm will occur _______ with _____ % probability:

sometime during embryo development, 50 (i.e. in half the cells)

Based on what you learned so far, what modification would you choose to express a bacterial gene product in a plant cell?

splice plant enhancer, promoter, 5'UTR, 3'UTR and terminator to bacterial ORF

A genomic DNA segment from prickly pear is hybridized to a mature mRNA purified from ripe fruit, obtaining the above image. The polymer in J is _____. ss: single stranded; ds: double stranded Note: All complementary nucleic acid polymers have fully hybridized

ssDNA, P = 3', R = 3' mRNA:DNA hybrid. The polymer in blue and with the bulges (introns) is ssDNA. The green polymer is ssRNA. The unhybridized portion of the green polymer is most likely a 3' polyA tail. The polarity of the DNA is opposite.

The poly(A) tail of the mRNA is primarily important for the

stability of the mRNA. Stabilization of the mRNA is the primary function of the poly-A tail.

A eukaryotic transcription factor that represses gene activity could have the following activities on promoters EXCEPT this one:

stabilize binding by RNA pol

A eukaryotic transcription factor that stimulates gene expression is most likely to have the following activity:

stabilize binding of RNA polymerase to the promoter

Pursuing the linkage of marker allele G3 to disease allele D, your team obtains the following theta_LOD values series. The first number is theta, the second, separated by _, is the corresponding LOD (some LOD are negative): 0.01_-4, 0.02_-2, 0.03_-1, 0.05_0, 0.07_1, 0.09_2, 0.11_5, 0.14_10, 0.17_15, 0.2_6, 0.25_3, 0.3_1, 0.35_0.2. You conclude that linkage is _______ at ___ m.u.

strongly supported, 17 Find the theta with the highest LOD score. Any LOD score that is greater than 3 is considered strong support for linkage.

The chromosome number of a cell is first reduced to half at:

telophase I

The total number of DNA molecules representing chromosomes is lowest at:

telophase II

You obtain the sequence for the mRNA transcribed from the actin gene of hammerhead shark. You identify the corresponding DNA template strand in the dsDNA of the actin locus. The 5' end of the template strand is closest to this DNA element.

terminator The synthesized RNA is made in the 5' to 3' direction, so the 3' end of this RNA will be near the terminator. Since the RNA is paired to the template strand, the 5' end of the template must be near the terminator.

Replica plating bacterial colonies from one type of medium to another is done to:

test if they grow on a different medium In multiple experiment types, bacteria are switched from complete (rich) and permissive medium to assorted minimal media to help discover auxotrophic mutants. Any mutant of interest will fail to grow on a specific auxotrophic medium (the test medium). For this reason, retrieval of the colony of interest is only possible if a the same colony is available on the rich medium. By comparing the replicated set of colonies, any colony that fails to grow on the test medium can be collected from the rich medium.

In Chinese cabbage, wrinkled leaf is dominant, smooth leaf is recessive. You would like to find out if a wrinkled plant is homozygous or heterozygous. What is your simplest and best strategy?

testcross

Two pure breeding lines of maize display light green leaves instead of the usual dark green of most varieties. Their F1 is dark green. The F2 consists of 461 dark green and 353 light green. You concludes that ____ gene(s) is/are responsible. Using A, B, etc designation for the gene(s), the dark green and light green F2s would have, respectively, the following generic genotypes _______, ________.

two, A_B_, anything else (aabb, aaB_, A_bb) The numbers resemble a 1:1 ratio, but the data fail the chi square test (chi-sq=14, well past the critical value for one degree of freedom). The numbers fit well a 9:7 ratio, consistent with two genes. There are 9 A_B_ genotypes, and 7 for all other combinations. The product of two genes is needed for wild type leaves. A_bb and aaB_ are equally defective.

The above map of a chromosomal region of Bacillus subtilis was derived by cotransformation mapping in the early 60s. The mapping cross used a wild-type donor and a recipient that was auxotrophic for tyrosine, histidine and tryptophan with genotype tyr1-, his2-, trp2-. Based on the map, select the least likely, but still possible genotype of transformants.

trp2+, his2-, tyr1+

Recombinant DNA technology makes precise changes to a targeted DNA sequence outside of cells , while genome editing makes precise changes to the DNA sequence inside of cells .

true

You have isolated cDNA 1, which contains the ORF for an important protein. You want to make the following fusion: NH2-GFP-cDNA1-COOH. Choose the correct junction. For simplicity, there is no hinge-coding region. Original stop codon and start codon are in upper case Start of cDNA 1: .... aaaaaagccATGtttaaa End of cDNA 1: .... ggggggtttTAGcccggg Start of GFP: .... ccccccattATGtttccc End of GFP: .... ttttttaaaTAGcccttt

ttttttaaaATGtttaaa

There are two questions on pineapple genetics. Start with this one first. The leaf edges of pineapple can be uniformly spiny spiny on the leaf tip piping, i.e. with nice smooth edges that resemble cloth piping folds Pineapple growers prefer varieties (Links to an external site.) that are piping because harvesting them is safer and easier. The following crosses were carried out using pure breeding varieties whose phenotype is indicated as piping, spiny tip, or spiny. Two pure bred parents are named as W and Z for later reference in the answers. Based on the experimental results, you must conclude that

two genes at work, Spiny is double recessive The results do not fit simple Mendelian genetics rules. At first sight they are confusing. Focus first on the F2 with the most phenotypic classes. There are two that have three classes. Epistasis could be at work! What ratios and phenotypes do you observe? When two gene are involved, you would expect 9:3:3:1. However, if the genes interact these ratios will change. If all fails, I suggest you look at this video: LinkLinks to an external site. Restating some of the explanation: There are three phenotypic classes in at least some of these crosses. This is the hallmark of epistasis, i.e. the interaction of two genes. In an epistatic dataset it is useful to: Focus on datasets which F2 have three phenotypic classes: you are looking at line 1 and 3 of data in table. Identify the ratio type: 12:3:1 or 9:3:4. This here looks like 12:3:1. Which phenotypic class is the one with frequency close to 1/16? The Spiny class. Which phenotypic class is the one with frequency close to 12/16? The Piping one. Which one is intermediate? The Spiny Tip one. Therefore, Piping 12 : Spiny Tip 3 : Spiny 1. Using simple A and B gene names, these correspond to A_,__ 12 : aaB_ 3 : aabb 1. A is epistatic on B. When A is present, the leaves are piping and the spiny edges are suppressed. When A is not present (aa), the spiny character is manifested. B_ = Spiny Tip, bb = Spiny. Now convert AA to PP for Piping, and BB to SS for Spiny. Check the other crosses.

This is image describes the method of Meselson and Stahl experiment. N15 is a heavy isotope of regular, lighter N14. Assume that DNA replicates by the dispersive model: the result for generation 1 is shown. Which pattern would one find for generation 2?

D New DNA is light. Generation 1 displays DNA with intermediate weight, like in the original Meselson and Stahl experiment. In this case, however, old and new DNA mix on the same strand. In the next replication, 1/4 of the each molecule is heavy, 3/4 light.

Two Hfr strains are available for Salmonella esparto. HfrC1 results in the following order of entry (each letter represents a gene): F-H-Z-L-K-A-T-M HfrD5 results in the following order of entry (each letter represents a gene): D-Y-U-W-M-T-A From these data you can assemble the following genetic map:

D-Y-U-W-M-T-A-K-L-Z-H-F The two Hfr have opposite transfer orientation. They overlap at MTA / ATM and thus the three genes can be used to join the two maps.

The image represents transcription. Please identify the nature of the molecular sites at H and W. They are, respectively:

DNA 5', RNA 5' The 5' end of the RNA is extruded from the transcription DNA bubble. It is base paired to an antiparallel DNA strand (the template). Z is the 3' end of the mRNA. The template strand of DNA is antiparallel. Therefore, H is the 5'.

Which of the following is a true statement?

DNA ligase joins a 5' phosphate to a 3' hydroxyl in the phosphate-sugar backbone of DNA

Which of the following statements is most correct when describing histone deacetylation and DNA methylation?

DNA methylation and histone deacetylation are associated with more tightly packaged chromatin which represses transcription.

The image shows some of the structures in a DNA replication bubble. For replication to proceed, the arrow tips of M, P and Q should ALL be associated with:

DNA polymerase III Growing chains of DNA are extended by the action of DNA Pol III

Which enzymes catalyzes DNA synthesis in the 5 → 3' direction?

DNA polymerases Primase is not the correct answer because it synthesizes RNA in the 5' to 3' direction

In which of the following processes is template DNA used?

DNA replication and RNA transcription Self explanatory assuming you understand the difference between replication, transcription and translation.

The transforming principle in Avery's Pneumococcus experiment could be destroyed by:

DNAse

A particular histone demethylase acts on histone methyl marks that actively repress gene expression. Loss of function mutation of this histone demethylase would most likely:

Decrease gene expression

F and G are two loci. Dominant=upper case letters, recessive=lower case letters. Hybrid FfGg is crossed to ffgg. The following phenotypic classes and progeny numbers are produced: F,G = 523; F,g = 45; f,G = 41; f,g = 529. Allele F is in coupling with allele _____. The map distance between the F and G loci is ___ m.u.:

G, 7.5 Parental gametes are FG, fg. Recombinant alleles are Fg and fG and have frequency 86/1138 = 0.0755, or 7.5 mu.

Based on the Gal regulatory pathway illustrated in this figure the regulatory pathway involving galactose, GAL80, GAL3 and GAL4 and leading to galactose catabolism requires the following element, _______, to bind directly the DNA of the ___________. When bound, this element is ________ activating transcription.

GAL4, enhancer, sometimes

A researcher creates random copolymers of three nucleotides by mixing polynucleotide phosphorylase with guanine and adenine nucleotides in a ratio of 5 guanine nucleotides to 1 adenine. Which of the following three nucleotide copolymers would MOST likely be most abundant?

GGG

As you increase the temperature of a solution, this dsRNA molecule will denature last: Note: only top strand shown, 5' to 3'. Just in case you wonder, RNA follows the same or very similar pairing rules as DNA.

GGUCUGCGCGG AT and AU bp employ two H-bondsGC employs three H-bondsThe more GC pairs, the stabler the dsRNA structure

Consider the sequence 5'GTAAAAATTTTTTAC. The following 5'->3' sequence could be a primer for its replication:

GUA

The dsDNA fragment is being replicated. Corresponding to P, a nucleic acid polymer is paired to the following Y strand sequence 5'AAC. The sequence of the P polymer is, 5' to 3'

GUU RNA primer is antiparallel to DNA with its 5' end is to the right. Y strand: 5' AAC RNA primer: 3' UUGor 5' GUU Therefore --UUG-5'

Identification of the transforming principle proved that:

Genetic information is contained in DNA

Because of an error in transcription, a tRNA gene changes the anticodon from 5'-UUU to 5'-UUG. As a result, some polypeptides in the affected cell will display the following amino acid residue change: Hint: draw tRNA paired to codon

Gln->Lys 5'AAA codon for lysine3'UUU anticodon for tRNA Lys5'UUU anticodon for tRNA Lys5'UUG mutated anticodon3'GUU same mutated anticodon5'CAA bound codon -> Gln, therefore Gln -> Lys The mutated tRNA will bind the Gln codon and bring Lys in that polypeptide position.

A geneticist conducts the experiment performed by Nirenberg and Matthaei in 1961 (see module for review), but this time she combines guanine nucleotides (instead of uracil) with polynucleotide phosphorylase. In which radioactively-labeled amino acid precursor tube should positive protein appear?

Gly G nuc ---NPase--> 5'GGGGGGG...GGG3' ---tRNAs and ribosome---> incorporates radioactive Gly RNA contains only one type of codon: GGG. Therefore, GGG triplet is recognized by tRNA-Gly.

A coding strand in a bacterial genome has the following base sequence. 5′-ATGATACTAAGGCCC-3′ What amino acid sequence will be encoded?

Met-Ile-Leu-Arg-Pro

A coding strand in the middle a bacterial genome has the following base sequence. 5′-GGATGACGACTGAA-3′ Choose the polypeptide product of the best ORF and the reading frame:

Met-Thr-Thr-Glu, +3 A simple matter of identifying the frame and translating the codons sequentially.

The image above shows a DNA molecule before the action of any enzyme, and then four subsequent DNA molecules (labeled A, B, C and D) after the action of one particular enzyme. Which DNA molecule would result from the action of DNA polymerase III?

Molecule C DNA polymerase III requires a 3' end and a template to synthesize DNA. It will extend any 3' end that has an available template. The C molecule is the only molecule where this process and only this process has occurred.

The number of eukaryotic proteins involved in regulating and carrying out transcription of the 10,000 to 25,000 genes present in the genome are encoded by the following number of genes:

More than 1,000

You wish to clone DNA fragments of a pathogen genome produced by digestion with the restriction enzyme HindIII. The plasmid vector has a unique HindIII restriction site, but does not have any type of insert screening system like blue-white screening. After a typical cloning procedure, the transformed E. coli that grow on appropriate medium can be screened for the cloned genome pieces by the following method:

Perform HindIII analysis on purified plasmid DNA: two bands are expected

You wish to clone DNA fragments of a pathogen genome produced by digestion with restriction enzyme HinDIII. The plasmid vector has a unique HinDIII restriction site. After a typical cloning procedure, the transformed E. coli cell that grow on appropriate medium can be screened for the cloned genome pieces by the following method:

Perform HindIII restriction analysis on purified plasmid DNA: two bands are expected

Pho, Met, Mal The orientation of each IS is important.

Consider an experiment where you have four rII- mutants. These mutants are all different, but are similar in that portions of the rII gene are deleted (called deletion mutations). Although these deletion mutants are unique, it is possible that some mutants overlap, i.e. have common missing DNA bases. If you carry out mixed infections of E. coli strain K with all possible pairs of mutants, which of the following would you expect?

Plaques can be formed if deletions don't overlap.

Think back to the prokaryotic gene expression regulation module. The Lac operon is an example of negative inducile control of gene expression. Choose the type of control illustrated by GAL4 in the control of genes for yeast galactose-metabolizing enzymes.

Positive inducible

In a Pp heterozygous organisms, one of the chromosomes carrying the P locus is moving toward the pole at anaphase I of meiosis. In this meiosis, this chromosome was previously involved in a single crossing-over with its homolog that took place between the centromere and the P locus. The P locus on this chromosome has genotype ________ . (name every allelic copy)

Why do you think DNA primer length is important when trying to amplify DNA (PCR) from genomic DNA? Pick the answer that is the most correct.

Primers of sufficient length are needed to ensure specificity for the target DNA

Interrupted mating experiments using multiple Hfr strains were performed. What is the correct gene map based on the data above? Note, answers make the chromosome look linear but E. coli chromosomes are supposed to be circular. Assume the first and last gene given in each answer are next to each other. Explanation: in this table, Hfr-14 A-B-C-D indicates that the time of entrance is shortest for A and longest for D.

Q - W - D - M - T - P - X - A - C - N - B Correct. Remember the F factor can insert anywhere, and in either direction. Here, the F factor inserted in the reverse direction for strains 2 and 4.

Human chromosomes are numbered 1 to 23 according to decreasing size (1 is largest). Trisomy 21 is a relatively common condition resulting in Down Syndrome. There is no syndrome described for trisomy 2. This is most likely because:

The many genes on chr. 2 result in lethal imbalance while the fewer genes on chr. 21 produce less, and thus viable, imbalance

A pure bred great dane dog is crossed to an F1 hybrid dog resulting from a "Great Dane x Boxer" cross. The progeny displays varying traits: size, skull shape, coat type, color, etc. Why?

The mixed breed gametes vary in allelic composition resulting in the variety of different phenotypic traits observed. Note that the Great Dane is homozygous for both dominant and recessive alleles. At least some of the boxer dominant alleles (present in the hybrid) are expected to have a dramatic effect.

Suppose that a mutation occurs in a small intron of a gene-encoding a protein. What could be the effect of this mutation on expression of the encoded protein?

The mutation may or may not affect the intron removal depending on whether nucleotides affecting splicing are altered. If the intron is not removed, the protein would be severely affected Splicing utilizes specific sequences within the introns (edges and middle). If a mutation affects these sequences, splicing may not occur. This would dramatically change the protein sequence made from this mRNA

Mr. T has blood type A. Mrs. T has blood type B. They have three kids, whose blood type is, respectively, A, B, and O. The best explanation for the blood phenotypes in this family is:

The parents are both heterozygous at the same blood type locus: Mr. T is IA/Io, Mrs. T is IB/Io.

A plasmid is conjugated from a donor to a recipient cell. The genes necessary to form the conjugation pilus are most likely encoded by:

The plasmid itself

NOTE: this was changed from an earlier version to make it clearer. A second question on the cross is available below. Generalized transducing phage P22 is used to transduce DNA from Salmonella Kleiber-3 into a lab strain that is aro-, gly-, tyr-. The results are shown in the table. Choose the most likely map ORDER for these markers.

tyr-aro-gly Organize you observations. You need to determine the order of these three. Look for the 4 crossing overs events. This is the least frequent event. One of the three should be -: it is the central one. This is because 4 crossing overs are needed to produce a + - + arrangement.

Transductants from this cross are selected for prototrophy for amino acid "m" (they must be m+). The gene order on the E.coli chromosome is:

v-s-r Compared to the parental (donor) genotype, the rarest class of recombinants (II) displays l as the discordant locus. Thus, s is in the middle.

Based on your overall understanding of gene interaction, cases of epistasis in human molecular and cellular functions are likely to be:

very common Cases of genes affecting the function of other genes are very common in every organism. Just think of transcriptional regulators.

Gene Raf of the bacterium Pseudomonas encodes raffinase, an enzyme that hydrolyzes raffinose to simple sugars. You align the Pseudomonas raffinase sequence to that of multiple bacterial species and find strong conservation from amino acid 250 to 290. The rest of the protein varies considerably. A nonsense mutation occurs at nucleotide position 150 of the open reading frame. This mutation is likely to be _______ for enzyme activity.

very deleterious

All of the following are required or very useful in a bacterial plasmid cloning vector EXCEPT -viral capsid proteins. -lacZ inactivation by insertion. -origin of replication. -selectable markers. -unique restriction enzyme sites.

viral capsid proteins.

The open reading frame of gene Z is 990 bases long. In mutant "m" a frameshift mutation occurs in nucleotide 120 of the ORF. A reversion mutation affecting nucleotide 142 fixes the frameshift and yields mutant m-rev2. In mutant "g" a missense mutation occurs at nucleotide 50 of the ORF. The probability that the polypeptide encoded by gene Zm-rev2 is active is _______ than in mutant Zg and depends also on ____________:

worse, evolutionary conservation of residues 41-47

Most sequence-specific Transcription Factors of eukaryotes cannot bind target DNA that is ___________ because the target sequence is hidden by __________.

wrapped around a nucleosome, histones

You cross two pure breeds of rabbit, one grey, the other white. Their F1 is grey. Crossing two F1 sibs yields 27 grey and 3 white. Does the grey coat color trait fit a single genic dominant trait? Choose yes or no and the chi square value.

yes, 3.6 The test is for 3:1 with total of 30 observations. Expected is 3/4*30 and 1/4*30. Relatively simple calculation (see lecture or Chi Squared page). For 1 degree of freedom critical Chi Squared value is ~3.8. This is still below it and we accept that chance could have caused the observed divergence.

Marker allele Y7 and allele D are linked at 16 m.u. An affected grandpa that has Y7-D/Y2-d haplotypes. The probability that a grandchild inherits D but not Y7 is: (Hint: this is a challenging problem. Make sure you consider all the possible event paths leading to the final outcome.

~0.07 There are two alternative scenarios for losing the linkage between D and Y7. A crossing over can happen in the first generation or in the second generation. This is explained better in the page dedicated to this event.

Alleles at the Y locus can be genotyped by a lab test and have different numbers (Y1, Y2, etc). An affected grandma has Y7Y2,Dd genotype. Assuming no linkage between Y and D, and that no other family member has these alleles, the probability that her grandchild inherits D but not Y7 is: (Hint: this is a challenging problem. Make sure you consider all the possible event paths leading to the final outcome)

~0.19

A plant that is heterozygous for gene A is self-fertilized, and all the resultant offsprings are continually self fertilized for another 5 generations. How many offsprings would you expect to be homozygous dominant for gene A in the 5th inbreeding generation?

~48% The formula to find the total homozygous fraction is 1-0.5^n, where n=number of inbreeding generations. The F2 has experienced 1 inbreeding generation. The F3 2, and so on. Homozygotes are either dominant or recessive. To find the fraction of either one divide by 2 the total.

A plant that is heterozygous for gene A is self-fertilized, and all the resultant offspring are continually self fertilized for another 4 generations. How many offsprings would you expect to be homozygous for gene A in the 4th inbreeding generation (a.k.a. the F5)?

~94% The formula to calculate fraction inbred from selfing generations is: Fraction inbred = 1-(0.5)^n where n=inbreeding generations. The formula means that half of the hets turn into homozygotes each generation

The human gene that determines maleness is on the ___ and encodes a ____:

Y, transcription factor

5'GGCTGTGACAATCTAGTCAAAA The DNA sequence shown above is found in the middle of a gene. If this is the template strand, which reading frame(s) is the most likely to be the one used for this gene? and what are the first three amino acids encoded by this frame? Hint: for the sequence 5'-GATTACA, frame -1 is TGT-AAT-...., -2 is T-GTA-...

-3, Leu Thr Arg It is easier to obtain the reverse complement and work with it. The reverse complement of 5' AAGC is 5' GCTT. Once you identify the reverse complement, note that the +1 frame GCT-T is actually the -1 frame for A-AGC. Test each frame for the presence of stops. Any stop-free frame is suitable. Assuming you gave up and need to see a solution here it is: 5'GGCTGTGACAATCTAGTCAAAA-3' the RNA templated from this would be: 5'-UUUUGACUAGAUUGUCACAGCC-3' ( or you could use the sense DNA strand),

Consider these three genes of the house fly: A gene (for antennae): A_=straight antennae, aa=curly antennaeM gene: M_=normal mouth, mm=small mouthC gene: C_=long combs, cc=short combs A female heterozygous at all three genes is testcrossed (AaMmCc x aammcc). The progeny has the following phenotypes and numbers. Based on the data, which of the following is the correct gene map?

-A--(10.3 m.u.)--C--(14.7 m.u)--M- Phase and order is A-c-M.Gamete, count, classificationΑcM: 382, parentalaCm: 379, parentalAcm: 69, C-M COaCM: 67, C-M COACm: 48, A-C COacM: 44, A-C COACM: 5, 2-COacm: 6, 2-COA---(92+11)/1000--C---(136+11)/1000---M

Which of the following is a true statement? -DNA ligase joins a 3' phosphate to a 5' hydroxyl in the phosphate-sugar backbone of DNA -DNA ligase forms hydrogen bonds between nucleotide bases. -DNA ligase joins a 5' phosphate to a 3' hydroxyl in the phosphate-sugar backbone of DNA -DNA ligase recognizes and cuts specific DNA sequences. -DNA ligase joins the 1' sugar hydroxyl to the imino group of the cognate base in DNA

-DNA ligase joins a 5' phosphate to a 3' hydroxyl in the phosphate-sugar backbone of DNA

Which of the following steps are carried out when conducting genetic engineering? Select all that apply: -Crossbreeding and selecting a plant for 10,000 years to make it more palatable -Identification of suitable genetic parts (promoter, coding region, terminator, vector) -Introduction of suitable genetic parts into a host cell -Gamma-irradiating tomato seeds to introduce random mutations -Assembly of suitable genetic parts parts outside of a cell -Crossbreeding and selecting a plant for 10 years to make it more palatable

-Identification of suitable genetic parts (promoter, coding region, terminator, vector) -Introduction of suitable genetic parts into a host cell -Assembly of suitable genetic parts parts outside of a cell -

Which of these activities fall under the definition of genetic engineering? Select all that apply: -Wolves becoming dogs through domestication -Inserting a gene encoding chymosin into the genome of a bacterium -Crossing and selecting (i.e., breeding) teosinte to form modern maize -Inserting a gene encoding herbicide resistance into the genome of a corn plant

-Inserting a gene encoding chymosin into the genome of a bacterium -Inserting a gene encoding herbicide resistance into the genome of a corn plant

A man with hypophosphatemia, an X-linked dominant trait causing low blood P, marries a normal woman. What proportion of their sons will have the condition?

An F' carries a gene that makes regular E. coli virulent on mice. After conjugation into the cell of a non virulent recipient strain, the minimum number of crossing overs between the plasmid donor DNA and the host chromosome required to confer virulence (is)are:

Generalized transducing phage P22 is used to transduce DNA from Salmonella Kleiber-3 into a lab strain that is "-" for all relevant markers. If a single phage infects each cell, the probability of transducing the leu and phe markers in the same event:

A bacterial genome has two identical IS closely flanking a met+ gene. The IS' are oriented in opposite direction. An Hfr is formed by insertion of an F-plasmid into one of the two genomic IS'. What is the probability of forming an F'met+ via IS recombination?

0 Note that IS orientation has an effect on the type of recombination product they generate. When two IS are close, F' formation is possible. The F plasmid can recombine with one to form an Hfr. Then, the F plasmid can excise by recombination with the other IS. Unless the IS are pointing in the same direction, however, the recombination event cannot lead to excision. Try this yourself in the simplified, 2-IS situation below. Recombination does not excises a circle, it simply flips around the region between the IS.

In a previously unreported experiment, Morgan chose an F1 male fly produced from the original mating of a red eyed female x white eyed male flies. This male was crossed back to its mother. The probability of white eyed progeny in this backcross is the following: __% females, __% males.

0, 0 X+ = wild type allele (red) Xw = white eyed allele All males produced in the original cross are X+Y. Mother was X+X+. Therefore, from the X+Y x X+X+ backcross, P=0 for both sexes.

Consider the dataset illustrated above. Each dot represents the phenotypic trait values for apple size of parents and progeny. The regression line is shown for each. Choose the best estimate of narrow sense heritability for each population (shown in order)

0, 0.5, 1

An XO mouse is expected to have ____ Barr bodies and to be _____:

0, female

You are testing the hypothesis that a disease allele is linked to marker 1. Assuming theta = 0.2, what is the probability of the above pedigree?

0.0001

Consider the above pedigree. It depicts inheritance of a dominant condition. Affected allele, R. Normal allele, r. The numbers refer to alleles of a marker that may be linked. To explore this possibility, derive a probability for theta = 0.1. Choose the best answer.

0.00092

Two incompletely dominant genes affect the same trait: M and T, are linked at 7 m.u. M and T alleles add 10 units each to the trait, m and t add 0. If you self F1 hybrid Mt/mT (20 trait value), the frequency of high trait value (40) plants in the F2 is expected to be:

0.0012

F and G loci assort INDEPENDENTLY. An F1 with genotype FfGg is selfed for multiple generations to produce an F5 generation. The probability to find an FfGg F5 individual is (choose the closest number):

0.004 This is probably the most difficult question in the exam. To solve this you have to: Treat F and G locus as independent Calculate the fraction for each and then...

Freckles and tasting of PTC, a bitter substance, are independent and dominant traits in human. Marla and Pablo are heterozygous at both loci and have a non-freckled and non-taster child. If Marla and Pablo have five more children, what is the probability that half the kids are freckled tasters, and the other half non-freckled, non-tasters? Hint: Consider what you know at the start before calculating the answer.

0.007 Here are some pointers: 1. Determine the P of having a double recessive phenotype child (fftt). Let's call that q. 2. Determine the P of having a double dominant phenotype child (F_T_). Let's call that p. 3. Since we know that the first child is double recessive, we only need to calculate the P of having a family with 2 double recessive kids, and three double dominant, which together with the first child would give us a 50:50 family. 4. That can be calculated using the binomial expansion. See the video...

A plant heterozygous at five independent loci is selfed. What fraction of its progeny will be heterozygous at all loci?

0.031

The gene for albinism (recessive) and the gene for dwarfism (dominant, rare) are linked and 20 m.u. apart. Mating between Meg, a normal woman with no family history of albinism, and Francisco, an albino dwarf, produced a dwarf son, Diego. Diego mates with Rhonda, a normal woman whose father was albino. What is the chance that their child will be albino and not dwarf?

0.05 A_=normal, aa=albino B_=dwarf, bb=normal This question requires you to follow genotypes throughout the generations. Meg has a genotypes AAbb and she has a dominant allele for A in phase with the recessive b allele (effectively a repulsion; Ab/Ab). Francisco has genotype aaBb and has the linkage phase aB/ab Diego has genotype AaBb and the linkage phase Ab/aB. Rhonda has the genotype Aabb with linkage phase Ab/ab. The question asks the probability that Diego and Rhonda have a baby with the genotype aabb (ab/ab). In order for this to occur, Rhonda must contribute the parental gamete ab. Recombination here has no effect on the gametes. Therefore, ab = ?% Diego must contribute ab as well, which requires a recombination event. Therefore, P(aabb) =

Blue eyes are recessive to any other color. Mary and John have brown eyes, but each have a blue-eyed father. They have seven children. The probability that three are brown-eyed and four are blue-eyed (in any order) is:

0.06

Sally's grandmother is albino, a single-genic recessive trait. Sally wants to marry her cousin, who has the same grandmother. The probability that their first child is albino is:

0.0625 a=albino condition Each grandmother is aa and contributes an a to progeny. P that Sally and cousin are Aa is 0.5 for each. P of aa from two carriers = 0.25 Multiply all Ps = 0.5*0.5*0.25

H_ = hairy leaves, hh = smooth leaves. An F1 plant with genotype Hh is self-fertilized, and all the resultant offspring are continually self fertilized until you derive the F5 generation. If you pick a hairy plant, what is the probability that it will be heterozygous?

0.118 Fraction homozygous = 1-0.5^N, where N = number of inbreeding generations Heterozygosity + homozygosity = 1 Fraction Heterozygous = ? F5 = 4 inbreeding generations Fraction with dominant phenotype = Fraction homozygous/2 + Fraction heterozygous

The family tree above illustrates the inheritance of a rare dominant disease condition manifested after age 40. The ? marks individuals too young to be diagnosed. The P locus is at 30 m.u. from the disease locus and the causal allele is phased with P9. What is the probability that Rupert is affected and carries allele P9?

0.12 P of parental gamete T-P9 = 0.7/2=0.35 P of two generation transmission = 0.35^2 = 0.1225

Yellow seed is dominant over green. A pea plant heterozygous for this trait is selfed resulting in a pod with 6 seeds. What is the probability that the pod contains 3 yellow seeds and 3 green seeds in any order?

0.132

A disease allele, D, is linked to marker M3 with theta = 0.2. Starting with a grandfather with genotype DM3//dM4, what is the probability that a grandchild will inherit the D-M3 haplotype? Assume that M3 is not found outside the grandfather lineage. // : indicates haplotype. E.g. AB//ab has A-B and a-b coupling.

0.16

The pedigree illustrates inheritance of disease X. If individuals II-1 and IV-1 mate, what is the probability that their first child is affected?

0.17

You are testing the hypothesis that a disease allele is linked to marker 1. Assuming theta = 0.2, calculate the LOD score.

0.21

In horses the Red gene (R or r) displays recessive epistasis on the Bay gene (B or b). Horses with the rr genotype are red regardless of their Bay genotype (B_ or bb). Recessive Bay (bb) are black if R_. Crossing a black stallion to a red mare has produced three colts: one bay, one red and one black. What is the P that the next colt from the same pair will be Bay?

0.25

Alleles at the Y locus (no relation to Y chromosome!) can be genotyped by a lab test and are defined by different numbers (Y1, Y2, Y3, etc). A mother affected by a dominant condition (D) has Y7Y2,Dd genotype. She marries a man with a Y5Y5,dd genotype. Assuming no linkage between Y and D, the probability that her child inherits D but not Y7 is:

0.25 P(kid gets D allele) = 0.5 P(kid gets Y2 allele) = 0.5 0.5*0.5 = 0.25

0.274

In German shepherds, long coat is a recessive trait. Two heterozygous dogs are mated. What is the probability that their litter consists of 2 long-coated and 4 wild-type puppies in any order?

0.296 P of long coated = 0.25 = q P of normal coat = 0.75 = p A specific litter in a given order would have total P of: How many different litters (orders of birth) are possible with 6 choose 2?

You compare tracking ability in a large number of parents and progeny bloodhounds. You plot mean progeny ability vs. mean parent ability and obtain a regression coefficient of 0.3. Narrow sense heritability is:

0.3

A translocation between two chromosomes joins gene A and gene B within an intron as follows: junction -------GENE-A>>>>] ----(intron)---|----(intron)---[GENE-B>>>>> ------- The genes are in the same orientation. What is the probability that a fusion protein representing the respective portions of both gene products is made?

0.33

In your greenhouse populations you observe 2 mutant plant lines which you call "biteme" mutants. Plants homozygous for a recessive allele you call 'a' are vulnerable to zombie mite bites. The same is true for plants homozygous for another recessive allele you call 'b'. You cross individuals homozygous for each of the mutants and observe that the entire F1 generation remains vulnerable to zombie mite bites. How many loci are manifested by this analysis and what are the genotypes of the parental generation.

1 locus, a and b are two alleles of the same gene) This question is testing whether you understand how complementation works. If the mutation occurs in the same gene, crossing two mutants together would yield F1 progeny that still display the vulnerable-to-zombie-bites phenotype. This is because each parent donates a mutant allele, and no wild-type allele is present. If mutation b were in a separate gene, then each parent would contribute a bad allele of one gene and a functional copy of the other. Having one functional copy of each gene would produce the wild-type phenotype in the F1. Each mutant is homozygous recessive. Since complementation does not occur, we know the mutations are in the same gene: we are dealing with one locus. The parental cross is aa x bb and the F1 then has the genotype ab. The F1 genotype has two mutated alleles at this locus and is thus vulnerable to zombie bites. See the question about mice and hairlessness as an example of what happens when complementation occurs.

What is the most likely event leading to a frameshift mutation?

1 or 2 bp insertion or deletion

In snapdragons, a true breeding red flowered plant (AA) crossed with a true breeding white flowered plant (aa) produces pink flowered F1 progeny. What phenotypic ratios do you expect from selfing the F1 generation?

1 red: 2 pink: 1 white

In black-eye pea, the T and P genes are linked and 20 m.u. and P_ = pennate leaves, pp = oval leaves. T_ = thick pods, tt = narrow pods. A PpTt plant with the dominant alleles in REPULSION, is SELFED. The probability of an oval-leafed, narrow podded plant in the progeny is:

1% Rephrasing, the problem asks this: from Pt/pT -(probability)-> pt/pt Two pt gametes must fuse to get pptt. P of pt = 0.2/2 = 0.1 (there are two recombinant products, PT and pt). P of pt x pt = 0.01

The following pedigree represents a rare autosomal disorder. What is the respective probability that individuals III-4 and IV-4 are carriers?

1, 0.67 This is a recessive autosomal condition: dd = disease. The left branch is obviously inheriting d from I-2. The right side inherited either from the I-3,4 couple, or from II-8. III-3 and III-4 must be carriers. III-4 inherited the disease allele from either II-8 or II-9. Therefore, P of carrier for III-4 = 1. Consider now IV-4. Any child in the IV family could be either dd, Dd, or DD as both parents are heterozygotes. IV-4 is not dd or he would be affected. This is a classical case of conditional probability: there are only two possible genotypes: Dd and dd, in a 2:1 ratio. Thus P of being a carrier = 2/3.

A starving E. coli in a GLUCOSE-FREE, but LACTOSE-CONTAINING medium has genotype Is,O+,Z+,Y+/I+,O+,Z+,Y+. Expression of beta-Gal is expected to be _____ and the expression of lac-permease is expected to be _____:

1, 1

A starving E. coli in a LACTOSE-FREE and GLUCOSE-FREE medium has genotype I+,O+,Z+,Y+/I-,O+,Z+,Y+. Expression of beta-Gal is expected to be _____ and the expression of lac-permease is expected to be _____:

1, 1

A starving E. coli in a LACTOSE-FREE and GLUCOSE-FREE medium has genotype I+,Oc,Z-,Y+/Is,O+,Z+,Y-. Expression of beta-Gal is expected to be _____ and the expression of lac-permease is expected to be _____:

1, 100

Two pure breeding lines of pepper, one dark green and round, the other light green and oval, are crossed. The F1 hybrid is dark green and round. The F2s assort in four phenotypic classes in a 9:3:3:1 ratio. Crossing a light green, oval F2 plant to a dark green, round one will produce ____ phenotypic classes:

1, 2, or 4 G_ = dark green, R_ = round Light green is double recessive: grr. Dark and oval F2 can have four genotypes, listed below with the expected phenotypic classes in the cross. GgRr -> four classes GgRR or GGRr -> two classes GGRR -> one class

This pedigree represents the inheritance of a RARE condition in humans. Assuming full penetrance and considering the most likely way of inheritance, the individual probability that III-6 and IV-1 are carriers is, respectively:

1, 2/3

Two T4 phages carry each a different mutation in the rII-A gene. Their co-infection at high density on permissive strain B results in rare wild-type phages. Most likely, the recombination event leading to the formation of such a wild-type phage genome involved ______ crossing-over(s) and produced ____ phage(s)

1, one wild-type and one double-mutantvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv

Non disjunction at anaphase II of meiosis affects chromosome 10 of maize (corn). Considering ALL cells produced by this meiosis, the copies of chromosome 10 are as follows (1,1,1,1 would indicate that all 4 cells have one copy):

1,1,2,0

The expected number of chromatids per chromosome during G1, G2, prophase, anaphase (both of mitosis) is:

1,2,2,1

The expected number of chromatids per chromosome during G1, prophase I, anaphase I, anaphase II is:

1,2,2,1

Maize seeds can be purple (Y) or yellow (y), plants can be normal (D) or dwarf (d). A plant heterozygous for both traits was self-fertilized and produces the following progeny: purple, normal= 87, yellow, normal = 33, purple, dwarf = 27, and yellow, dwarf = 13. Assuming independent assortment, what is the Chi-square value for these data (round to nearest tenth decimal)?

1.6

A pea plant with the genotype AABbCcDdeeFf is self-pollinated. What is the probability of an offspring having the genotype AAbbCcDDeeFF?

1/128 You must calculate individual Ps for each locus. What then? Are each independent? Then you must (multiply OR add)?

In the cross AaBbCcDd x AaBbCcDD, what proportion of offspring are expected to be heterozygous for all four genes?

1/16

The probability of II-3 and II-4 being a carrier is, respectively:

1/2, 1

Inbred wild and domesticated beans are crossed. Four genes are involved in bean domestication. For all four genes, the wild bean has dominant alleles, while the domesticated bean has recessive alleles. What is the frequency of the fully domesticated phenotype among the F2s?

1/256

In the progeny of two grey herons, 14 birds are grey and 5 are sandy. If each bird in this clutch is mated to a sandy bird, the most likely number that will produce only grey birds is:

5 If you work through all possible parental genotypes, you will find that for the grey herons to produce two types of progeny, both must be heterozygous. Grey is dominant. Sandy is recessive. Thus: G_ = grey gg = sandy Both parents are Gg. The progeny is 1GG:2Gg:1gg. For. grey x sandy cross to produce only grey birds, the genotype must be GG x gg. Of the 14 grey birds, one third are expected to be GG and two thirds Gg. The answer closest to one third is "5" (14/3 = 4.7).

Forbes' parakeet and the red-crowned parakeet are pure breeding animals. An excaped pair hybridized producing birds of mixed appearance. These F1s intercrossed producing ~1000 birds of ranging phenotypes. Among these, one looks like a pure-bred Forbes and another like a pure-bred red-crowned parakeet. You conclude that ______ genes are likely to control the differences between these "species".

5 The progeny with the phenotype of interest (one or the other race) appears in 1/1000 F2s. What kind of ratio is this? What would you have concluded if each race type appears in a 1/4 ratio? and what about 1/16? ..and 1/64?

What is the sequence of the DNA template strand at position +1 to +8

5"GCAATATG You have to flip the answer sequence around (3-5') and see what pairs to the mRNA remember pairing rules and antiparallel arrangement of dsRNA, dsDNA, and DNA-RNA hybrid

5'ACCGGTCGCCACCATGGTGAGCAAGGGCGAGGAGCTGTTCA The sequence shown is part of a gene that could encode at least 9 amino acids. What mRNA sequence could be transcribed from this gene if this is the coding strand?

5' ACCGGUCGCCACCAUGGUGAGCAAGGGCGAGGAGCUGUUCA

5'-ATAGGCGATGCCA3'-TATCCGCTACGGT <--template strand The above sequence represents DNA that is part of the RNA-coding sequence of a transcribed gene. Give the sequence found on the RNA molecule transcribed from this DNA and identify the 5′ and 3′ ends of the RNA.

5' AUAGGCGAUGCCA 3' the mRNA sequence is always similar to the nontemplate (aka coding strand) except that the T are replaced with U.

Inside a cell, a likely step toward replication of the following single DNA strand, 5'CCCCAGGGATTACGGG 3', by DNA polymerase will involve previous base pairing by this nucleic acid :

5' CCCGUAA

Choose the double stranded DNA molecule that will melt (denature) at the highest temperature (top strand is shown only):

5' CCGCAGGG Highest content of GC base pairs, which are more stable than AT pairs because they form three H-bonds.

The TEMPLATE strand of a gene has the following sequence 5' AGCTTGCACCG 3' ( or can be viewed as 3' GCCACGTTCGA 5' ). What would the corresponding mRNA transcript look like?

5' CGGUGCAAGCU 3' The TEMPLATE strand is the DNA strand the polymerase binds to and is used to template (provide a base pair guide for) the RNA transcript. Therefore, it should be the reverse complement of the transcript. We call it the reverse complement because the polarity is reversed (3' to 5' becomes 5' to 3') and it is complementary to the template. E.g. if template is 3'-TAC-5', mRNA will 5'-AUG-3'

The following single DNA strand, 5'CCCTTAAG 3', will form a proper and complete double stranded DNA when paired to this DNA (choose the best answer):

5' CTTAAGGG

Choose the single stranded DNA molecule that has the best potential to form dsDNA without an enzyme:

5' GATTACACCTGTAATC DNA strand is self-complementary and forms hairpin dsDNA

The following single DNA strand, 5'AGGGATTAC 3', will form a proper double stranded DNA when paired to this DNA (choose best answer):

5' GTAATCCCT The 5' end of one piece of DNA must match the 3' end of the other: 5' AGGGATTAC 3' (sequence in the question) 3' TCCCTAATG 5' (sequence in the answer read backwards!)

Pairing and extension by DNA pol III of this primer, 5'CCCUUGG 3', in association with this DNA, 5' GATTACACCAAGGGACA, will result in formation of the following new sequence (primer omitted from sequence):

5' TGTAATC

Pairing and extension by DNA pol III of this primer, 5'CCCUUGG 3', to this DNA, 5' GATTACACCAAGGGACA, will result in formation of the following NEW sequence (primer omitted from sequence):

5' TGTAATC

Which component of an mRNA sequence would be most useful to use for identification of the promoter controlling the expression of a gene of interest? UTR = untranslated

5' UTR of mRNA 5' UTR of the mRNA is essentially the region just downstream of the TSS. This region is directly downstream of the promoter.

DNA polymerase III carries out the majority of DNA replication. Compared to DNA polymerase III, DNA polymerase I has this unique _______ enzymatic activity, which acts on ______.

5' to 3' exonuclease, RNA DNA pol I extends an Okazaki fragment, digesting the RNA primer, 5' to 3', of the adjacent Okazaki primer. This activity is also called nick translation.

RNA polymerase is in the middle of transcribing gene W. In a bubble of melted dsDNA it is polymerizing ribonucleotides onto the nascent RNA chain in the ______ direction . The promoter of this gene would be most easily reached if RNA polymerase began walking towards the _______ region on the __________. (Hint: remember what strand RNA polymerase actively transcribes RNA from)

5' to 3', 3', template strand Draw out the process of transcription with the nascent RNA strand. What end is growning?

In Q, a nucleic acid polymer is paired to CTA. What is the paired sequence on this polymer?

5'-GAU An RNA primer is antiparallel to DNA with its 5' end is to the left. The shown sequence is 3'-CTA. Therefore 5'-GAU

Consider the ssDNA 5'-CGTTAC. The following sequence is perfectly complementary to it:

5'-GTAACG Complementary strand must pair in antiparallel orientation

The following ssDNA could form the best hairpin: Hairpin = dsDNA stem and small ssDNA loop

5'-GTAACGAAACGTTAC The two complementary section of the strand must pair in antiparallel orientation: 5'-GTAACGAAACGTTACFolding this at AAA5'-GTAACG|3'-CATTGC|

What feature of eukaryotic mRNA is first recognized by the ribosome?

5'CAP

Carrying out PCR on a DNA sample and gel electrophoretic analysis of products will provide information on the following:

Both presence and size of target

You grow sweet peas for a living and know that purple pigment is dominant over white. You also know that two genes are involved in producing pea pigment and display complementary gene interaction (duplicative recessive epistasis). One of your clients wants purple sweet peas for his wedding. Which two sweet pea lines should you cross in order to maximize the number of purple sweet peas in the next generation?

AAbb x aaBB Any pea plant that is homozygous recessive for EITHER gene A or B results in white pea plants (based on your knowledge this is duplicate recessive epistasis). AAbb x aaBB crosses yield all purple plants (AaBb). No other answer will give you all purple flowers in the F1 generation.

Choose the best 4 base primers to PCR amplify the shown target in such a way that the final major product has EXACTLY and only the bracketed DNA . 5'-AATT[ACCATTGCATCTGATTACAATTC]ACGG All primers are listed 5' - 3'.

ACCA, GAAT

Genes A,B, and C are linked. You cross an AaBbCc plant with an aabbcc plant and get offspring listed in the table above. The heterozygous parent has the following alleles in the following order on ONE of its chromosomes (note phase AND gene order):

ACb

The following sequence could form a dsDNA:

AGCTAATTAGCT

The enzyme responsible for synthesizing the polyA tail of eukaryotic mRNA requires:

ATP, a polymer providing a 3'OH end,

Chromatin remodeling complexes hydrolyze _____ and _________ in chromatin.

ATP, reposition nucleosomes

A TEMPLATE DNA strand reads 3' GTTTATGCTGGAAAT. The following could be an RNA transcribed from this DNA is: 5'...

AUACGACCUUUA

Transcriptional activator and repressor proteins often affect the level of transcription of eukaryotic genes in the following way

Activators and repressors bind to cis elements of promoters and enhancers and promote or prevent, respectively, the assembly of the complete transcription apparatus.

The peptide sequence Met-Arg-His can be encoded by the DNA sequence: -5'CGATGCGGCATTGAG -5'TCATGCGACACTAGC -5'CTATGAGACACTGAG -All of the other answers are correct 5-'CATGAGGCACTAATT

All of the other answers are correct

What statement best describes a test cross?

An individual of unknown genotype displaying dominant traits is crossed with an individual displaying recessive traits for the same characters Yes. It is called test cross because it is the most suitable cross to test genotype. If the individual to be tested is heterozygous, recessive genotypes will appear at 50% frequency.

Nettie Stevens model of sex determination in mealworms involved the following:

An unusual chromosome pair results in two types of sperm: those with a small chromosome, and those with a large chromosome. The first form male zygotes, the latter female zygotes.

What phase of cell cycle is this cell involved in?

Anaphase I of meiosis

Why is it important that gene mapping experiments in bacteria require the recipient to be resistant to an antibiotic and the donor to be sensitive to the same antibiotic?

Antibiotic selection ensures that only recipient cells are analyzed for gene mapping.

Phage T18 has linear dsDNA genome. Strains K and L are both wild-type and differ by SNPs that have no effect on phenotype. Upon coinfection of K and L on a permissive bacterial cell, the two phage genomes recombine. What unique number of crossing-over is compatible with survival of both recombination partners? The answers imply that other numbers are not compatible.

Any number Draw possible recombination events. Which ones result in intact genomes?

Because of an error in transcription, a tRNA changes the anticodon from 5'-CCA to 5'-CCG.. As a result, some polypeptides in the affected cell will display the following amino acid residue change: Hints and notes: Easy to go wrong here. We recommend drawing the tRNA anticodon paired to codon. Keep 5' and 3' straight. The change in tRNA sequence does not change the aminoacyl-tRNA-synthetase's specificity. Finally, draw an example RNA and test the change effect. Also, mind your time on this Q.

Arg->Trp

What is the most likely mode of inheritance displayed by this pedigree?

Autosomal dominant

Read the question above and provide the correct answer

B CAP is mutant and it is not stimulating transcription. Therefore, very low expression* * It is debatable whether lack of CAP conditions would turn off the lac operon. We choose to consider it off here. In fact, it varies from strain to strain. Some problems may imply some expression. Confusing? We promise that it will not be in the midterm.

In a cross between AaBbCc and aabbcc, the most frequent gamete classes produced by the hybrid are ABc and abC. The least frequent are aBc and AbC. The parental haplotype and gene order is:

BAc and baC

Assume that this cell has genotype Bb for a gene located on the chromosome marked by G. The allele(s) associated with G could be:

BB, Bb or bb, depending on occurrence of crossing over

Consider the above image representing transcription and translation inside a prokaryote. You should be able to identify the important components. Where did the ribosomes first associate with the mRNA?

Between Z and W The AUG must be between Z and W. The ribosome at W has a 9 amino acid peptide, indicating that the AUG is 27b away toward the 5' of the mRNA. Z is the 5' since the mRNA is growing at its 3' end (LL). Consider an mRNA 5' to 3'. Where would you land a ribosome?

How is the structure of an RNA nucleotide similar to that of DNA? How is it different?

Both contain a sugar, phosphate, and base, but only RNA has a 2′ OH.

Red-green colorblindness is X-linked recessive. Polydactyly (extra fingers and toes) is autosomal dominant. Mary and her mother are normal, her father is color blind and polydactylous. Bob has normal vision and is polydactylous. His mother is normal. If Mary and Bob have children, what types of children will they produce?

Both sexes can be polydactylous but only boys can be colorblind XC = normal vision, Xc = recessive allele -> colorblind P_ = polydactylous / = mark separates alleles and is sometime used for clarity. So, pp and p/p represent the same genotype Mary is XC/Xc, pp Bob is XC/y, Pp P of polydactyly = 0.5, P of colorblind male = 0.5, all females get XC from Bob.

Read the question above and provide the correct answer

C C+, wild-type CAP, will stimulate transcription only in the absence of glucose. In the first operon, Oc is in phase with P-. This operon cannot be expressed because it lacks an active promoter. I- in the second operon is complemented by the I+ in the first, which acts in trans. The permease gene, Y+, will be expressed under induced, no glucose conditions. The Z+ on the first operon is not expressed because the promoter is defective

In an alternative universe, Stahleson and Mels carry out their classical DNA replication study starting with N14 and transferring the culture to N15. Their observation for generation 1 is shown. Which pattern will they find for generation 2? Note: DNA replicates the same way in their universe as in ours. They use first light, then heavy N! N15 is a heavy isotope of regular, lighter N14.

C New DNA is heavy. Generation 1 displays DNA with intermediate weight, like in the original Meselson and Stahl experiment. One strand light, the complementary one heavy. Heavy-Heavy-Heavy-Heavy (new DNA)|| ||| ||| ||Light-Light-Light-Light In the next replication, all new DNA is heavy. Half the DNA molecules will be of intermediate weight, the other half will be heavy. Heavy-Heavy-Heavy-Heavy (new DNA)|| ||| ||| ||Light-Light-Light-Light (original light strand)Heavy-Heavy-Heavy-Heavy (new DNA)|| ||| ||| ||Heavy-Heavy-Heavy-Heavy (from Generation 1)

Consider the above DNA molecules . Which molecule(s) could be a substrate for telomerase?

C and D Telomerase polymerizes nucleotides if a 3' overhang is present. It essentially extends the 3' overhang and is used in DNA replication to ensure the ends of linear chromosomes don't shorten with every subsequent DNA replication cycle.

Consider this single stranded DNA molecule: 5'-CTTG-3'. A perfectly complementary DNA will have the following sequence, 5' to 3'.

CAAG

This tRNA sequence in the tRNA-Met binds the starting codon of typical mRNAs (listed 5' to 3'):

CAU See the paring between tRNA and mRNA:tRNA 3'-UAC-5'mRNA 5'-AUG-3'Therefore, the anticodon is 5'CAU

Which component of the CRISPR-Cas9 system is responsible for sequence specificity?

CRISPR RNA (crRNA)

The Moderna and Pfizer COVID vaccines entail an mRNA molecule that is expressed in the cells of the patient. Based on these properties, the RNA molecule in the vaccine is likely to incorporate the following elements, from 5' end to 3' end: Note: ORF here means a sequence of codons starting with AUG and ending with UAA or UGA or UAG. After injection in a patient, the RNA is taken up by cells causing expression of the COVID-19 surface antigen (a protein).

Cap-5'UTR-ORF-3'UTR-PolyA

What is necessary for integration of an F plasmid into the host chromosome?

Chromosome and plasmid share a region of homology, typically an Insertion Element (IS) IS elements jump around (they can transpose). This means that they are ubiquitous and therefore well suited to provide regions of homology that can cause recombination between two genetic elements. Further, IS are present in different chromosomal loci and can mediate insertion of a plasmid, such as F, in different sites.

Circular

Using a filtered U tube, a glass tube with two compartments separated by a bacteriological filter, would prevent this type of sexual transfer between two bacterial cultures:

Conjugation Contact between cells is required to transfer DNA through a pilus.

Consider the cross in Table 1. Coeff. of Coincidence is ___, interference is ____:

Consider the cross in Table 1. Coeff. of Coincidence is ___, interference is ____: --Order and phase: RSH --Recombinations between R and S = 0.1407 --Recombinations between T and Q = 0.03 --Observed double XOs = 7 --Expected double XOs = 0.14*0.03 = 0.0042, in 10K -> 42 --Coincidence = 7/42 = 0.16 --Interference = 1-0.16 = 0.84

Covid-19 omicron may be the product of recombination between Covid-19 delta (CoΔ) and an unknown coronavirus (CoU). Likely, such a recombinant was formed when:

CoΔ and CoU infected the same cell

What role do restriction enzymes normally play in bacteria?

Cutting foreign DNA as a defensive mechanism

What role do restriction enzymes play normally in bacteria?

Cutting foreign DNA as a defensive mechanism

In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen (N15) and then transferred them to a medium containing "light" nitrogen (N14). Which of the results in the figure above would be expected after ONE round of DNA replication in the presence of N14?

D After one round, you should expect two dsDNA molecules, both contains one light and one heavy nitrogen DNA strands.

Read the question above and provide the correct answer

D Cc, constitutively active CAP, is actually easy to account for: it will stimulate transcription regardless of glucose. Is will shut down the second operon, which has wild-type operator. The permease gene (Y) is the only wild-type in that operon, but it will not be expressed. Oc will protect the first operon from Is and result in transcription under all conditions. The Z and Y genes in phase with Oc are, respectively, + and -. Therefore, there will be no permease expression.

This scene takes place in a bacterium and, although not shown explicitly, it includes protein synthesis. Which marked position is likely associated with the LONGEST polypeptide being synthesized is:

A male mink affected by the rare hunchback condition mates with several normal females producing 98 healthy pups, 45 affected females and 53 affected males. The affected pups display light to very severe symptoms. Matings between the healthy pups produce only healthy progeny. The best explanation for this syndrome is:

Dominant, variable expressivity The question provides important hints: The affected male mates with a presumably healthy female and they have affected and healthy progeny -> dominant Progeny ratio -> father is heterozygous Strong 1:1 -> consistent with full penetrance Healthy pups produce only healthy progeny -> fully penetrant or some of these would produce affected progeny Severity varies -> variable expressivity

In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen (N15) and then transferred them to a medium containing "light" nitrogen (N14). Which of the results in the figure above would be expected after TWO rounds of DNA replication in the presence of N14?

E After two rounds of semi conservative replication you would expect 4 molecules of dsDNA. Two having one heavy and one light nitrogen chain, while the other two contain two light nitrogen chains

Which element below is NOT required for a polymerase chain reaction (PCR)?

E. coli cells

In horses, E/_ is dominant black and e/e is red. If you want to engineer a black horse with small red spots, the BEST strategy is to inject an ___ zygote with ________.

E/e, CRISPR-Cas9 targeted to E and acting relatively late in embryonic development Inactivating E during embryo development results in red spots (clusters of cells with no E activity). The later the inactivation, the smaller the clusters and the corresponding spots.

Chargaff rule proved crucial in identifying the DNA structure by suggesting that :

Each T formed H-bonds with A, and each G with C If T and A are paired together via hydrogen bonds, the number of T's must be equal to the number of A's in double stranded DNA. Same goes for G-C pairs

A chocolate male lab mates with a black female lab. The resulting puppies include two golden and one chocolate. Based on the genetics of Q.2 (see also below), the male and female genotypes could be, respectively: _______ , _________. from Q.2: Coat colors of Labrador retrievers depend on two genes: one, the inhibitor of coat color pigment (ee) prevents the expression of color alleles at another independently assorting locus, producing a golden coat color. When the dominant condition exists at the inhibitor locus (E_), the alleles of the other locus may be expressed, E_B_ producing black and E_bb producing chocolate.

Eebb, EeBb The golden puppy must be ee__, and we know that at least one E is required for brown and black, so we know both parents must be Ee. Since the mother is black we know she must be B_ and the brown father is bb. Since they have a chocolate pup, we know the mother must also carry a b.

Concentration of beneficial Ca++ in banana is directly related to the expression of gene CHX1. To increase the level of CHX1 the following gene could be transformed in papaya

Either (1) or (3)

You mate an Abyssinian to a Siamese and obtain generation F1. The fastest method to homozygosity is to cross an F1 cat to: (mark the most complete answer)

Either a parent or a sib

Pairing of sex chromosomes is necessary to:

Ensure production of two gamete types

You are given a tube of DNA and a restriction map of a linear piece of DNA from the BIS101 staff. They assure you that the DNA you received is definitely the one described in the restriction map but for some reason you are skeptical of their claims. You run two experiments. In experiment one you do not cut the DNA, and in experiment two you digest the DNA with both BamHI and EcoRI. You perform gel electrophoresis on the DNA from your experiments (with other samples). Which lanes are consistent with the claimed structure of the DNA samples from each experiment?

Experiment 1 --> Lane C, Experiment 2 --> :Lane A

Genetic variance, Vg, can be estimated by comparing total phenotypic variance, Vp, of:

F1 and F2

The Arabidopsis gene FLD encodes a histone deacetylase enzyme. The FLD protein binds DNA around the FLOWER LOCUS C (FLC) gene promoter, and expression of the FLD gene stimulates flowering. FLC expression inhibits flowering. Which of the following is TRUE?

FLD deacetylates histones in nucleosomes that normally form open chromatin around FLC. This compacts the chromatin and represses FLC transcription It helps to lay out a pathway: FLC --| FLC --| Flowering So, FLC is a repressor of a repressor. This is consistent with the hypothesis that FLD opposes expression of FLC. We know that acetylated histones are associated with increased expression. Then, the activity of FLD makes sense, right?

A genome-edited organism always contains recombinant DNA in its genome

False

A white-eyed male fly is mated to a homozygous red-eyed female. Sons and daughters are mated. What phenotypic ratios do you expect in in the "grandchildren"?

Females: 100% red-eyed, Males: 50% red-eyed, 50% white-eyed

Vermillion eye color (a very bright red) in fruit flies is X-linked and recessive to wild-type red, which appears darker and can easily be distinguished. A red-eyed heterozygous female is mated to a red eye male. What genotypic and phenotypic ratios do you expect?

Females: all red-eyed, half being heteroygous. Males: half red-eyed, half vermillion-eyed

Freckles and tasting of PTC, a bitter substance, are independent and dominant traits in human. Use F-f and T-t, respectively, for alleles. Mary, freckled and non-taster, and Pablo, non-freckled and taster, have a non-freckled and non-taster child. What genotype do Mary and Pablo have, respectively?

Fftt, ffTt FFtt x ffTT -> 100% F_T_, freckled and tasters. FFtt x ffTt -> 100% F_, freckled Fftt x ffTT -> 100% T_, tasters The only genotype combination that yields recessive progeny for both traits is Fftt x ffTt

What is the consequence of methylation of DNA sequences in a promoter?

Formation of compacted chromatin and transcriptional repression Highly methylated DNA is recognized by nuclear factors (proteins) that favor formation of heterochromatin.

Penetrance refers to:

Frequency of affected individuals among those with the relevant genotype

The figure illustrates the distribution of beta-carotene content in peas. P1 and P2 are inbred lines. F1 is their hybrid and F2 the filial generation derived by selfing the F1. Consider the distribution of F2. If you select and self individuals that are the very extreme of that distribution, you chances of obtaining progeny that inherit the extreme parental phenotype is _____. Vd = variance due to allelic dominance Vi = variance due to gene interaction Va = variance due to allelic additivity Ve = variance due to environment

Good if Va high, Vi and Vd low Va, the additive component of genetic variance, is predictable because each genotype's value is known from the phenotype. Therefore, high Va is promising. Vd and Vi are tricky variances (Vi more than Vd) because the same phenotype can result from multiple different genotypes and their behavior in a cross cannot be predicted with accuracy.

The Q and H genes segregate independently. During prophase I of meiosis of an individual with QqHh genotype, the following alleles are closest:

H and h

Comparing narrow sense heritability, h2, to broad sense heritability, H2, the following is wrong:

H2 predicts breeding success Breeding success is predicted by h2, the additive part of Vg.

You cross a hairless mouse (aaBB genotype) to a mouse with curly hair (AAbb genotype). All of the F1s have straight hair. You cross two of the F1 mice together. In the F2 generation 18 mice have straight hair, 8 mice are hairless, and 6 have curly hair. What is the phenotype of an aabb mouse in the F2 generation?

Hairless The F1 is AaBb and straight haired (i.e. wild-type). In the F2 we expect the following ratio: A_B_ : A_bb : aaB_ : aabb in a 9:3:3:1. aaB_ mice are hairless, A_bb mice have curly hair. The 9:4:3 ratio suggests recessive epistasis with 9 straight haired, 4 hairless, and 3 curly. From this ratio and the aaB_ phenotype, we infer that aabb must be hairless.

All of the following organisms have been genetically modified, in the sense that human intervention altered their genetic makeup. However, only some were genetically modified through genetic engineering. Of the organisms listed here, which were the product of genetic engineering? Select all that apply: -Herbicide-resistant "superweeds" obtained via repeated herbicide application to a farmer's field -Modern-day tomatoes, obtained through cross-pollination and selection -Herbicide-resistant corn, made so by scientists inserting a gene that encodes resistance into the genome of a corn plant -A plant near Chernobyl with radiation-induced mutations

Herbicide-resistant corn, made so by scientists inserting a gene that encodes resistance into the genome of a corn plant

_________ DNA refers to a structure resulting when a single strand of one homologous chromosome basepairs with the complementary strand of DNA of the other homologous chromosome. This occurs during the formation of a Holliday junction.

Heteroduplex This is the definition of heteroduplex DNA. Note: recombinant DNA sounds related but it is not. Recombinant DNA is formed when two pieces of DNA from different sources are combined by attaching one ds cut end to another ds cut end, splicing together different pieces.

Which of the following statements about histone and gene expression is correct?

Highly condensed chromatin represses gene expression.

Choose the best answer: When self-fertilizing plants, why does homozygosity increase following generations of successive inbreeding?

Homozygous plants will always produce homozygous offspring, whereas heterozygous plants will produce homozygous offspring in a 1:1 ratio

Watson used to play with card board nitrogenous bases on his desk. What was he trying to determine?

How the bases interact

A new mutation in the lac operon prevents expression of Z and Y under any condition. To test whether it acts in cis or in trans, you introduce in the mutant this F'lac plasmid __________ . If the new mutation acts in trans, in the presence of lactose the expression level of Z will be _____.

I+,O+,Z+,Y+, 1

The DNA in chromatin can be analyzed by treatment with DNAseI, which can easily cut DNA regions that are free of nucleosomes but not DNA regions associated with nucleosomes. The cut sites can be identified by sequencing and are located on the genome of any sequenced eukaryote. You would expect to find the most sensitive sites (hypersensitive) at the following positions:

Immediately upstream of transcription start sites Regulatory DNA at promoters needs to be exposed for the proper transcription factors to recognize it.

The CRISPR-Cas9 system in bacteria is akin to our body's:

Immune system

Which of the following statements about translation is correct?

In eukaryotes, the 5 cap and the 3 poly(A) tail are involved in translation initiation.

Compare and contrast how regulation of genes for yeast galactose-metabolizing enzymes is different from regulation of E. coli lactose-utilizing enzymes. Choose the INCORRECT answer:

In yeast, the GAL expression system requires stabilization of RNA polymerase at the promoter region to express galactose metabolizing enzymes. In E. coli, no stabilization of the RNA polymerase is required for expression of lactose metabolizing enzymes.

All of the following are required or very useful in a bacterial plasmid cloning vector EXCEPT

Insulin Insulin, as a gene, is not useful or used in cloning technology per se. The plasmid typically has two genes: a gene for resistance to antibiotic for selection of transformants and the lac Z-alpha gene for insertional inactivation.

You are interested in expressing a plant gene in bacteria. You have determined the sequence of the transcribed region. The gene has 6 exons. You have made PCR primers that recognize the 5' and 3' UTR, as well as each exon boundaries. The BEST approach is to: 1) .......... . 2) Insert into a plasmid that provides a strong promoter in the right orientation (check orientation). 3) Transform bacteria, screen, etc.

Isolate mRNA, make cDNA, PCR amplify with 5' and 3' UTR primers Expressing a plant gene with exons and introns is not possible in bacteria. You must use a continuous complete ORF as found in mRNA. The cDNA is the best approach because it provides exactly that.

The following is not true about mapping functions. RF = recombination frequency CO = crossing over

Mapping function allows precise measurement of RF between any two loci The mapping function attempts to correct for double CO events, but it cannot predict whether they happen and how frequently. It is based on theoretical assumptions. It uses RF as an input. Very small corrections are applied to closely linked genes, but increasingly larger to distantly linked genes.

Why was it important for Mendel to start his work with different types of pure bred (or true breeding) lines?

It ensured that all starting parents used were homozygous for all genes of interest When homozygous pea is selfed, or crossed to an individual with the same genotype, all progeny is uniform and matches the parental phenotype. This provides a necessary "basal expectation": the line is stable and its progeny is predictable. Many of our domesticated species fit this quality: wheat, rice, beans, poodles, brown Swiss cows, etc., all matings between individuals of the same breed produce progeny nearly identical to the parents.

What would be the most likely effect of moving the AAUAAA consensus sequence within the 3′ UTR ten nucleotides upstream?

It will result in a shorter mRNA transcript. AAUAAA is the sequence recognized in the 3' UTR, where the protein for cleaveage are recruited to cut the mRNA tail prior to polyadenylation.

Three ribosomes are catalyzing the synthesis of polypeptides. In this scenario, an RNA polymerase is most likely found nearest to:

K In prokaryotes transcription and translation are coupled and an RNA polymerase is expected in the transcription bubble M

Using pedigree data, linkage evaluation (Fig. LOD above) was carried out between a dominant disease locus K and three molecular marker loci: T, W, and R. The best genetic map showing m.u. distance BETWEEN closest loci is:

K--2--R--3--T--10--W The maximum of each curve identifies the optimal recombination frequency (RF) with respect to locus K. The distance of each marker from K is RF x 100.

The following is false about LOD:

LOD = log of the number of inferred crossovers in pedigree correct answer, i.e. false: LOD = log of the number of inferred crossovers in pedigree

In a human family pedigree, linkage between a trait and a genotyped locus is best analyzed by the following method:

LOD analysis using many potential recombination frequencies

You are given a tube of DNA and a restriction map of a linear piece of DNA from the BIS101 staff. They assure you that the DNA you received is definitely the one described in the restriction map but for some reason you are skeptical of their claims. If you digested the DNA with restriction enzyme EcoRI only and ran the DNA on a gel (gel electrophoresis) with other samples. Which lane would make you believe their claims concerning the identify of the DNA in the tube?

Lane B The map indicates that Eco R1 cuts the DNA 6kb from one end (2+4) and 5kb from the other. Its digestion generates two fragments: 5 and 6kb. Those are found in lane B.

An EcoR1 fragment contains the ORF for the flu virus coat protein. To make a vaccine, you mix these fragments with a vector plasmid cut with EcoR1. No promoter is near this vector site and no transcription takes place. To obtain expression, you include in the ligation mix another EcoRI fragment contains the POtet. After ligation, transformation in E.coli followed by selection, and screening, you identify colonies that carry both fragments in the same plasmid (coat protein ORF + OPtet). The fragments can ligate in any order and orientation since they have equal ends. How many of these colonies will express in the absence of the tet repressor?

Less than 1/4 You clone two fragments into a vector. All ends are equal and so each fragment can go in forward or backward orientation. There are four possible arrangements: -PO-> -----ORF----> -PO-> <----FRO----- <-OP- <----FRO----- <-OP- -----ORF----> Only the top one works for expressing the ORF: 1/4

Which step in the molecular cloning process directly generates recombinant DNA?

Ligation

Consider the above figure. Of the positions indicated, the one associated with the LONGEST polypeptide being synthesized is:

In an individual with genotype Mm, a single nondisjunction event at meiosis II affects the chromosome carrying this locus. No crossing over is involved. The resulting gametes are: ( 0 = no allele present)

M, M, mm, 0

In this scene, site ___ is most likely bound by an enzyme with 5'-->3' ________ activity.

M, exonuclease The thicker lines are obviously primers. M displays a ss nick between DNA 3' on the left and the RNA primer 5' on the right. DNA Pol I will recognize this site and elongate the DNA while its 5'->3' exonuclease digests the RNA primer. Ligation of this site would result in a strand made of DNA and RNA. Ligase will act once the primer has been completely digested and both ends flanking the nick are DNA.

5'ACCGGTCGCCACCATGGTGAGCAAGGGCGAGGAGCTGTTCA The sequence shown is part of a gene that could encode at least 9 amino acids and it is the same as the previous question. What is the best predicted polypeptide encoded by this sequence? Use single letter amino acid codes. Codes for amino acids GGlycineGly PProlinePro AAlanineAla VValineVal LLeucineLeu IIsoleucineIle MMethionineMet CCysteineCys FPhenylalaninePhe YTyrosineTyr WTryptophanTrp HHistidineHis KLysineLys RArginineArg QGlutamineGln NAsparagineAsn EGlutamic AcidGlu DAspartic AcidAsp SSerineSer TThreonineThr

MVSKGEELF You must find an ORF, i.e. an ATG (AUG in RNA language) followed by a series of triplets that encode a polypeptide sequence. The ORF will end when a stop is found. This exercise uses a very very short ORF for simplicity. The scope of this exercise is to make sure that you can manually "translate" a gene (i.e. apply ORF analysis). Of course, a gene must be first transcribed and the RNA is translated. The previous question makes you derive the RNA. However, when analyzing gene function, it is also possible to apply the genetic code directly to DNA by quickly converting an AUG to an ATG (for example). You do not have to memorize the code. It will be provided in any test.

In the X-Y and Z-W systems of sex determination, which sex is the heterogametic sex, male or female?

Male in X-Y, Female in Z-W

The following abnormality COULD NOT contribute to formation of an XXX zygote:

Male nondisjunction meiosis I

Which of the following scenarios occurring simultaneously could produce an XXXY individual?

Male nondisjunction meiosis I, female non disjunction meiosis I or II This individual most likely results from an XX egg fertilized by an XY sperm. Both gametes result from nondisjunction (ND). In the case of the egg, XX can be produced either by ND at meiosis I, or meiosis II. In the case of the sperm, ND at meiosis II would have produced either a YY or XX sperm. Therefore, it must have been at meiosis I. For the record, the four products of that single meiosis were XY, XY, __, __.

Which of the following scenarios would NOT produce a phenotypically male human?

Male sex-determination requires the presence of the SRY region carrying the TDF gene.

Red-green colorblindess is X-linked recessive. Roger is color blind and affected by Klinefelter syndrome (XXY). His mother and father have normal color vision, while his maternal grandfather is colorblind. Which of Roger parents contributed an abnormal gamete due to nondisjunction? Assuming no crossing over, in which meiotic division did the nondisjunction event take place?

Mother, meiosis II Get organized as follows: 1. Give wild-type dominant allele and the condition recessive allele suitable names. I suggest C and c, respectively. Where is the C gene located? 2. What is Roger's genotype? Remember, he has Klinefelter syndrome and is color blind. 3. Roger inherited the color-blind condition. Who could have transmitted it to him? 4. Roger has two conditions: what could be the relationship? Hint: see answer to 2 above.

What criteria must be met in order for a complementation test to be valid?

Mutant lines must be true breeding This is the only answer that makes sense. Unless they are true breeding you cannot interpret the results. "Mutant lines must show different phenotypes" is not correct because complementation tests are done to determine whether a single mutant phenotype is caused by independent mutations in the same or in different genes. "Mutant lines must have mutations in different genes" is wrong because we are testing whether mutant lines possess mutations in the same or different genes. It could be either. "Mutant lines must have mutations in the same gene" is wrong because see explanation above "Mutant lines must have been isolated together" is wrong because the complementation test is often used on independently discovered mutants.

What would be the MOST likely effect of base-pair change mutation in the DNA 15 bases downstream (in the direction of transcription) of the start site of transcription of an E. coli gene?

No effect on transcription is likely, but the resulting RNA will display a base change A mutation downstream of the TSS is unlikely to affect transcription. Transcription initiation is determined by the promoter region, which is always upstream of the TSS. A mutation downstream of the TSS is going to be reflected in the RNA that is made. If this mRNA is translated into protein, this mutation may change the amino acid sequence which can alter protein function. Typically, at least 50 bases of RNA are present before a start codon is found. So, an effect on the encoded protein is unlikely.

For the experiment performed by Nirenberg and Mattaei in 1961 (see module for review), could RNA polymerase been substituted for polynucleotide phosphorylase without otherwise modifying the experiment?

No, because RNA polymerase requires a DNA template

A female white-eyed female fly is crossed to a red-eyed male. In their progeny, about 1/25 females is white eyed. The best explanation for this is:

Non-disjunction at AI or AII in the female, which is XXY This is the classical observation explored by Calvin Bridges. It is ND in the XXY female.

Sepia eyes are produced by a recessive allele of the autosomal red eye gene. Miniature wings result from a recessive allele of an X-linked gene. A female fly with miniature wings and sepia eyes is crossed with a red eye homozygous male with normal wings. What phenotypic ratios are expected in the F1 generation?

Normal wing, red eyed females, and miniature winged, red eyed males Xm = miniature, XM = wild type, S_ = red, ss = sepia Female is Xm/Xm, ss. Male is XM/y, SS. All progeny inherit the S allele from father. All female inherit XM from father. All males inherit only Xm from mother,.

Consider the above figure. In a merodiploid that has TWO wild-type copies of the tet operon, you isolate a single mutation that causes expression of tetT and tetW under any condition. Most likely, this mutation affected:

Refer to the replication bubble above. The strands labeled E are called _____________ and comprise the _____________.

Okazaki fragments...lagging strands

pCAMBIA plasmids are an example of a plant transformation vector. Their use entails two steps. The first is transformation of a very special bacterium called Agrobacterium. Agrobacterium is then incubated with plant cells, in which it conjugates a section of the plasmid, resulting in plant transformants. Many pCAMBIA have two genes that encode resistance to the same antibiotic: kanamycin. They encode the same protein, but had different promoter and terminator. Given the use of these vectors, what is the most likely reason that they contain two genes for kanamycin resistance?

One copy is expressed in plants, while the other is expressed in bacteria Bacterial promoters do not function in plants or in any eukaryotes because they are not recognized by the eukaryotic RNA polymerase. Viceversa for plant promoters. Therefore, if kanamycin selection is planned in bacteria and then in plants, two different genes must be used, one with a bacterial promoter, the other with a plant promoter.

A 200 amino acid protein is encoded by a mature mRNA that is 800b long. Examination of this mRNA sequence and ORF analysis should reveal:

One major ORF 600b long

Compare restriction enzymes to the CRISPR system. Consider a bacterium deploying these defense tools. Which system involves an adaptive response?

Only CRISPR

Fig S1 displays a cellular event in an individual whose genotype at locus F is Ff. If the structure(s) in position K contain(s) both allele F and f, then the most likely place to find another set of F and f alleles is

P P and K mark structures of homologous chromosomes. You can tell they are homologous because they have similar structure (centromere position, arms length). This is anaphase I (duplicated chromosomes present, pairs are separated). If one homolog has Ff, the other must too.

Consider the figure. The cell in ____ represents a phase of mitosis in a haploid organism with X = ____.

P, 5

The image shows some of the structures in a DNA replication bubble. DNA ligase will be needed soon in the following structures:

P, Q Okazaki fragments must be ligated into a single continuous strand.

Bonus Question: The CRISPR array in the genome of Streptococcus bacteria contains DNA sequences that are transcribed to make guide RNA, which are loaded onto the CAS9 protein. Why doesn't this complex cut the CRISPR array in the Streptococcus genome?

PAM (NGG sequence) is absent in the array

To demonstrate its activity, a small segment of chromosome _____ was introduced into a ___ embryonic cells resulting in a _______ mouse:

Y, XX, male

In a diploid organism with a somatic chromosome number of 10 (x=5), which of the cells in this figure is in anaphase I of meiosis? Choose the best answer.

R R represents anaphase I in a diploid cell with chr. number = 2X = 10 P: anaphase with chromatid splitting. No homologous pair is visible: each splitting chromosome is different from all others. Could be mitosis (X=5 in a haploid) or anaphase II (X=5 in a diploid with 2X=10). Q: anaphase with chromatid splitting. Homologous pair are visible: for each splitting chromosome there is an identical one: therefore, not meiosis II. Mitosis in a diploid (2X=10) . R: anaphase with homologous chromosomes splitting, therefore in meiosis I. There are 5 pairs (X=5, in this diploid 2X=10) S: anaphase with homologous chromosome pair splitting, therefore in meiosis I. There are 10 pairs, in this diploid 2X=20. T: anaphase with duplicated chromosomes moving to poles, therefore it should be meiosis I. However, the splitting does not involve matched members of a pair. This is an aberrant cell and not expected in a regular meiosis. In this diploid, 2X = 10.

This replication fork advances, finally reaching the chromosome end. Most likely, which nucleic acid end will be directly elongated by telomerase?

R T is 5'. Therefore, R is 3'. R will protrude because any RNA primer templated from the end of the R strand will be digested and will leave the R end as ssDNA. Telomerase will extend R providing sufficient template space for primase. See lecture on telomerase.

Gene order problem. Consider the cross in Table 1. The gene order (no phase implied) is:

R-S-H Gene order problem. Consider the cross in Table 1. The gene order (no phase implied) is:

Consider CRISPR-Cas9. While clearly different in function and origin, the following complex resembles it in several key features: This material was not covered in 2021. The correct answer is RISC.

RISC, i.e. complex with argonaute

Chose the statement that is FALSE when describing prokaryotic or eukaryotic transcription:

RNA polymerase catalyzes RNA polymerization in the 3' to 5' direction Violates the principle that polynucleotides are polymerized on the 3' end.

Telomerase is an enzyme made of ____ and protein. It acts on a _____ substrate providing both a _______ and _________ activity:

RNA, DNA, template strand, polymerase

Francisco and Margo have no history of disease in their families. Yet, two of their five children have a severe skin flaking disorder. A reasonable hypothesis is that the allele responsible for this condition is ______ and that __________ for the disease allele.

Recessive, both parents are heterozygous

The presence of 3' to 5' exonuclease activity enables DNA polymerase to:

Remove a mispaired nucleotide just after adding it

How do the two basic types of terminators found in bacterial cells differ?

Rho-independent terminators do not require a protein to aid in termination, while rho-dependent terminators do.

An individual is heterozygous at the R and G genes (unlinked): its genotype is RrGg. The following gametic genotype is not possible:

Rr Both alleles of R are present in this gamete. This breaks the first law of Mendel: alleles segregate. In meiosis of an Rr individual, R and r move apart (=segregate). Gametes are either R or r.

Consider an Rr heterozygote. Nondisjunction at meiosis I affecting this chromosome could result in the following set(s) of sperms. Note: Crossing may (or not) have affected this chromosome. Each shown set consists of four sperms. A "-" means that neither R or r is present.

Rr, Rr, -, -; or RR, rr, -, -. Replicated chromosomes start meiosis with RR or rr conformation. They exit anaphase I with what possible conformations?

Two pure breeding lines of snapdragon, one red-flowered and tall, the other white-flowered and short, are crossed. The F1 hybrid is red and tall: you name the dominant alleles R and T. The F2s assort in four phenotypic classes in a 9:3:3:1 ratio. Crossing a red, tall F2 plant called Sam to a white, tall F2 called Andrea produces 4 phenotypic classes. The genotype of Sam and Andrea is, respectively:

RrTt, rrTt Sam must be het at both loci, or it would make 100% red and tall plants. Andrea is obviously homozygous recessive for color (rr). It must be het for height, or all progeny would be tall.

Two pure breeding lines of snapdragon, one red-flowered and tall, the other white-flowered and short, are crossed. The F1 hybrid is red and tall: you name the dominant alleles R and T. The F2s assort in four phenotypic classes in a 9:3:3:1 ratio. Crossing a red, short F2 plant called Rupert to a white, short one produces 2 phenotypic classes. The genotype of Rupert is:

Rrtt

In this scene, the type of polymerase associated with J is also likely associated with:

S DNA Pol III is elongating both the leading strand J and the actively growing Okazaki strand S. L is an elongating primer due to the action of primase.

The S and Y genes are unlinked. An individual with genotype SsYy can produce the following normal gametes:

SY, Sy, sY, sy

In the bacteria where CRISPR-Cas9 is a defense system, the Cas9 protein is expressed together with multiple guide RNAs encoded in the CRISPR array. Given that the bacterium encodes the guide RNA in its own genome (as DNA), why does it not cut its own genome and commit suicide?

The guide RNAs in the CRISPR array do not contain the PAM site

Which one of the following is not associated with the process of transcription?

The leading strand of DNA

Gene mapping by timing of entry and interrupted conjugation determines gene order of a bacterial chromosome but does not provide a detailed map (gene distances are not determined). In the above image, the gene order has been determined and the donor is pur+, thr+, pro+ while the recipient is pur-, thr- pro-. Which statement best describes an experimental design for how a researcher could create a more accurate gene map using timing of entry data and interrupted conjugation?

Select for last gene (pro+) to enter recipient and use replica plating to determine presence of donor alleles in recipient (thr+, pur+). In this case, there is not really a decidedly better method. In other words, you could probably pick first or middle gene and still get results. The choice of the last (most distal gene) makes better sense and is chosen here as the best answer. Obviously, this would not make a great midterm question, but it is an excellent thinking question.

Example of a looped structure obtained by hybridizing nucleic acid strands that have partial identity. The bubble sequence is not present in the facing strand. A geneticist isolates a gene that contains eight exons. He then isolates the mature mRNA produced by this gene. After making the DNA single stranded, he mixes the single stranded DNA and RNA. Some of the single-stranded DNA hybridizes (base pairs) with the complementary mRNA. Assuming that the mRNA containes all eight exons, how many circular loops would you expect to observe from the hybridized structure?

Seven Draw this to visualize why it's 7. The figure could represent an mRNA-DNA hybrid for a single intron gene: DNA = Exon--Intron--Exon RNA = Exon-Exon For 2 exons, you have 1 intron. The number of introns will always be 1 less the number of exons.

A strain of bacteria possesses a temperature-sensitive mutation in the gene that encodes the rho protein. At high temperatures, rho is not functional. When these bacteria are raised at elevated temperatures, which of the following effects would you expect to see?

Some RNA molecules are longer than normal. To terminate the transcription, a palindromic sequence at the end of the mRNA forms a hairpin that makes the RNA pol stall and finally fall off, thereby effectively terminating transcription. However, for some transcripts, this hairpin structure is not adequate to terminate the transcription. A helper protein (called Rho) is need to stop the transcription. If Rho protein has a mutation that prevents its function, then some genes' transcription will not terminate at the right time. The transcripts from these genes are going to be longer than normal.

Which of the following statements about introns and exons is wrong?

Some RNAs start (from their 5' end) with an exon, others start with an intron. An RNA cannot start with an intron because it could not be spliced. 5'-INTRON-|-EXON-3' cannot be processed!

Typically, a recessive X-linked condition will appear in an affected 4-generation pedigree as follows:

Some males display it. Females usually unaffected.

A orange cat is impregnated by a black male. One of her kittens is a male calico. The phenotype of the kitten results from:

Sperm resulted from non disjunction of sex-chromosomes, egg was normal

The above LOD score obtained for linkage of condition D to marker allele M1. This data indicates that the linkage of D-M is:

Supported at 20 m.u.

The genetic map of chromosome 1 of arabidopsis is 120 m.u. with the centromere located approximately in the middle (see below). The T and W genes are located at the two opposite ends of the left arm, R in the center of the left arm. The most likely linkage status between these genes as determined in a genetic cross is:

T-R linked, T-W unlinked The left arm of the chromosome appears to be 60 m.u. The two genes are opposite end of the arm and should appear unlinked because the distance is longer than 50 m.u. (50 m.u. would also appear unlinked)

Consider this figure. Which cell is displaying non-disjunction and at which stage is it?

T. Anaphase I.

Plasmid vector pExpTet carries the promoter-operator region of the Tet operon next to a Shine-Dalgarno site, the MCS, and a transcriptional terminator as follows: POtet - SD - MCS - ter You cut the MCS with EcoRI and ligate in an EcoRI fragment containing the ORF for GFP. The GFP protein fluoresces green when excited with blue light. Among transformed E. coli, you identify 40 colonies that carry the GFP ORF. Under what conditions could they express GFP, and how many of the colonies would you expect to do so?

Tetracycline, ~20

The difference between a double Holliday Junction (DHJ) and a crossingover (CO) is:

The DHJ involves entanglement of one strand per dsDNA molecule, the CO involves whole dsDNA swaps across molecules the a crossingover is a complete swap between two dsDNA molecules, while the double Holliday Junction is an intermediate structure that in which only one strand per dsDNA molecule is entangled with the other.

A human baby with XY karyotype appears female. Which of the following explanations is most plausible?

The TDF gene on the baby Y is mutated to loss of function (it does not produce an active protein)

A three-D map of the genome in the nucleus has shown the following

The base of chromatin loops are often defined by the interaction of sequence motifs that position certain DNA binding factors in a convergent orientation

Which of the following is TRUE regarding RNA?

The basepairs in single stranded RNA molecules can interact via hydrogen bonds RNA can form secondary structures as is observed in hairpins seen in tRNA molecules and during transcription termination. Transcription requires a template, it can be double stranded (dsRNA viruses), and RNA can be catalytic without the assistance of proteins.

There are two questions on pineapple genetics. Answer the other one first. The leaf edges of pineapple can be uniformly spiny spiny on the leaf tip piping, i.e. with nice smooth edges that resemble cloth piping folds Pineapple growers prefer varieties (Links to an external site.) that are piping because harvesting them is safer and easier. The following crosses were carried out using pure breeding varieties whose phenotype is indicated is indicated as piping, spiny tip, or spiny. Two pure bred parents are named as W and Z for later reference in the answers. You conclude that in regard to the two genes controlling the leaf edge phenotype, variety W and Variety Z have the following genetic relationship.

W is dominant for both genes, Z only for one P = piping gene, P dominant on pS = spiny tip gene, S dominant on s P_,__ = Pipingpp,S_ = Spiny tippp,ss = Spiny Variety W = PPSSVariety Z = PPss There are three phenotypic classes in at least some of these crosses. This is the hallmark of epistasis, i.e. the interaction of two genes. In an epistatic dataset it is useful to: Focus on datasets which F2 have three phenotypic classes: you are looking at line 1 and 3 of data in table. Identify the ratio type: 12:3:1 or 9:3:4. This here looks like 12:3:1. Which phenotypic class is the one with frequency close to 1/16? The Spiny class. Which phenotypic class is the one with frequency close to 12/16? The Piping one. Which one is intermediate? The Spiny Tip one. Therefore, Piping 12 : Spiny Tip 3 : Spiny 1. Using simple A and B gene names, these correspond to A_,__ 12 : aaB_ 3 : aabb 1. A is epistatic on B. When A is present, the leaves are piping and the spiny edges are suppressed. When A is not present (aa), the spiny character is manifested. B_ = Spiny Tip, bb = Spiny. Now convert A to P and B to S. P_, __ : Piping etc etc

A genomic DNA segment from the chicken is hybridized to a mature mRNA purified from chicken ovaries, obtaining the above image. The mRNA is the molecule marked by the letter ______. The structure marked by G is _________ , and the one marked by M is _____. b = bases, nucleotides

W, the polyA tail, an intron

The Q and W genes are on located on different chromosomes. Consider the image above derived from a organism with QqWw genotype. If structure G carries the W and w alleles, structure H carries. (name every allelic copy)

W,w

In Campanula plants, blue (W) flower color is dominant to white (w). A cross between a blue plant and a white plant results in a large number of exclusively blue progeny. What is the genotype of the blue parent?

In Morgan's experiment where a red-eyed wild-type female was crossed to the original white-eyed male mutant, what was the key observation made regarding the F2 generation that suggested an X-linked pattern of inheritance?

White-eyed flies were only found among males

In German shepherds dogs, white coat (ww) and long hair (ll) are recessive traits controlled by independent genes. The corresponding wild-type alleles are dominant and result in pigmented coat and (medium-)short hair. A pigmented, short coated male is crossed to a pigmented, long hair female. Their pups are as following: 3 pigmented and short-haired 4 white and short-haired 2 pigmented and long-haired 1 white and long-haired. What are the genotypes of the parents (male x female):

WwLl x Wwll All dominant and recessive traits are manifested in the pups. So, both parents must contribute recessive alleles at both genes. Ratios are not important because the number are small. No gene can be homozygous dominant in either parent or all pups would display the corresponding dominant trait. The female must be ll because of her hair length.

You are interested in a particular segment of rhinoceros DNA and would like to clone it into a cloning plasmid. You have the following restriction map of the rhino DNA region that includes the DNA segment of interest (box). You also have the restriction map of the plasmid. Which restriction enzymes would you choose to clone the entire DNA of interest into the cloning vector?

X and N

The next movement of W will be toward ____. The polarity of Z is _____.

X, 5' Z is the end of the mRNA being translated. Note that the opposite end is obviously in a transcription "bubble". That end must be 3' since it is being acted upon by RNA Pol. The direction of the ribosome is indicated by the mRNA orientation and which ribosomes are associated with the longest protein chain.

What is the most likely mode of inheritance displayed by this pedigree?

X-linked recessive

What mode of inheritance does this pedigree most likely display?

X-linked recessive

What mode of inheritance fits this pedigree best?

X-linked recessive

Two normal looking fruit flies were crossed, and in their progeny, there were 202 females and 98 males. Which answers provides the best genetic explanation for this anomaly.

X-linked recessive lethal allele

This pedigree represents the inheritance of a rare, fully-penetrant condition in humans. The best explanation for its genetic mode of inheritance is _______. The P that individuals III-6 is a carrier is ___.

X-linked recessive, 0 Must be recessive because it skips II-2. If I-1 is rr, II-2 is Rr. Sons have p=0.5 of inheriting condition. Recessive autosomal is possible, but it requires two unrelated carriers (I-2, and either I-3 or -4), which is improbable.

What are X and Z?

X=RNA pol, Z=5' end of RNA

Mark the wrong statement about crossing overs (XO):

XO between non-homologous chromosomes ensure independent assortment of unlinked loci

The dsDNA fragment in Fig. 1 is being replicated in a test tube. All enzyme activities and factors needed to expand the replication bubble, including processing and ligating the Okazaki fragments, are present, but nothing else. By the end of the process shown, which template segment is unlikely to be copied into a DNA strand?

Y Y points to the 3' end of a template strand. During replication, an RNA primer will be templated from the Y sequence. Once the RNA is digested, there is no way to make DNA using the Y sequence as template. Telomerase is needed to extend the Y providing sufficient template space for primase. See lecture on telomerase.

Phage T19 has a circular dsDNA genome. Strains M and P are both wild-type and differ by SNPs that have no effect on phenotype. Upon coinfection of M and P on a permissive bacterial cell, the two phage genomes recombine. What unique number of crossing-over is certainly compatible with independent survival of the two recombination partners? The answers imply that other numbers could be or could not be compatible.

any even number Any odd number results in fusion of the two circles to form a single two-genome circle. It is unclear what would happen to such structure. What is clear, is recombination events involving two, four, or another even number (as unlikely as higher number may be) will still result in separate circles and thus guarantee survival of both partners.

In a recessive allele ORF, compared to the dominant, wild-type allele, you might expect to find:

any mutation in the other answers Recessivity is connected to loss of function, which could be caused by any of these mutation types.

Enhancer elements can be as distant as 10 or 100kb (kb=kilobase pairs) from a target promoter. The best explanation for the ability to act at this distance is that enhancers...

are brought into close contact with the target promoter by looping of the DNA

Gene F is located on the X-specific region of the X chromosome. In a G2 epidermal cell of a woman, there _____ of this gene, ____ located in the Barr body.

are four copies, two are

Alternative splicing refers to choice of different:

assorted donor and acceptors sites of introns

The function of aminoacyl-tRNA synthetases is to:

attach appropriate amino acids to corresponding tRNAs.

The figure illustrates the distribution of beta-carotene content in peas. P1 and P2 are inbred lines. F1 is their hybrid and F2 the filial generation derived by selfing the F1. Consider the distribution of P1, P2 and F1. If you select and self individuals that are the very extreme of each distribution (such as very low or very high in beta-carotene compared to the rest of their sibs), you chances of obtaining progeny that inherit the extreme parental phenotype is:

bad for any of these populations good for all Individuals in each population (P1, P2, and F1) are genetically homogeneous. I.e. they have the same genotype. Selecting one at the extreme of the distribution or in the middle gives us the same genotype. They are different because of the environment. No selection gain can be made.

F'virH+ carries the virulence determining gene virH+ that makes a particular strain of E.coli highly pathogenic. An F- E.coli carrying allele virH-, a loss of function mutant allele, receives F'virH+ through conjugation. The resulting cell will _____ pathogenic _____.

be, in any case If a plasmid is introduced into a new, compatible cell what happens? It starts replicating, is maintained independently of the main chromosome and it expresses its genes.

Consider the dataset illustrated above. Each dot represents the phenotypic trait values for apple size of parents and progeny. The regression line is shown for each. If you wanted to breed for increased apple size, which population(s) would be best and which worst?

best III, worst I The slope of the regression line is h2. Response to selection (R) is a function of h2. Specifically, given a certain selection value (S): R = h2*S Therefore, the population with the highest h2 is where breeding or evolutionary progress can be made most readily. Specifically, when h2 = 1, it tells us that if we select as parents individuals that deviates 10 size units from the mean of the population we should expect progeny that displays 10 size units deviation from the mean. I.e. we make great progress.

The following is not a quantitative trait:

blood type

Which phenotypic outcome from each of these three genes fits the qualitative definition?

both M and R The phenotypic classes resulting from any genotype of M and R are distinct and show no overlap. For example, you could classify the resulting leaves in narrow, medium and wide and assign the proper genotypes without errors.

Completely unaffected parents have an affected boy. Assuming full penetrance, the following is possible:

both autosomal recessive and x-linked recessive An earlier version of this GPQ featured the answer "dominant, low expressivity" as possible explanation. The term low expressivity is not always clear. In this context, it was unclear.

Pure breeds of dogs can have either floppy or stiff, erect ears. This character is controlled by a single gene. Two mutts with floppy ears mate and have four puppies, two with floppy ears and two with stiff ears. Choose the correct statement about the ear gene genotype in the parents:

both dogs are heterozygous

Two normal german shepherds have three puppies. One puppy has very short legs. Assume a single fully penetrant gene is responsible for the short leg condition. The best hypothesis for this scenario is that the parents are ______, The condition is _______.

both heterozygous, autosomal recessive

Genes M and P of bacteriophage GammaDelta are required to infect E. coli UCD. Without one or both of them, GammaDelta can still infect E. coli ChicoSU. Two unknown GammaDelta mutants, 5 and 6, grow on ChicoSU but not on UCD. When you mix either of these two with a known M mutant and infect UCD at high density, you obtain good infection and phage growth. A very likely scenario is that 5 and 6 are

both mutant in P This looks like a complementation test. How does it work? If in doubt, see the video. wt --[ M ]--[ P ]---- m- --[ X ]--[ P ]---- Draw the possible mutations in 1 and 2. Match them to m-. Do both proteins get made?

A CRISPR-Cas9 targets the middle of the gal4 gene in yeast. In a yeast cell that is haploid is at the G1 phase of cell cycle has a null mutation for an enzyme required in Non Homologous End Joining repair the CRISPR-Cas9 cuts into the gal4 DNA The most likely outcome would be:

breakage and loss of a chromosomal segment, possibly death

Identify the single wrong statement. A eukaryotic promoter..

can give high expression simply by having an optimal TATA box promoter element and no other factor binding site

In certain viruses, DNA is found as a single stranded molecule. In one such virus, the % A+T is 30. The expected G is:

cannot be estimated Since it is single stranded the rules that A = T and G = C do not apply. There is no complementary strand base-paired to the single strand. Thus, even though we know A + T = 30%, this doesn't tell us the specific G content. It does tell us that G + C = 70%, which means that G is between 0 and 70%.

The first demonstration that DNA was the genetic material was based on conversion of the rough, non pathogenic streptococcus form into a smooth, pathogenic one. The critical experiment entailed:

digestion of purified transforming factor with DNAse The key experiment involved treatment of the transforming factor with different hydrolytic enzymes. DNAase destroyed the factor's efficacy. The P32 labeling would not have worked. It was used by Hershey and Chase to demonstrate that phages inject their DNA and not proteins into bacteria. Phages are efficient injectors. Transformation is a very low efficiency phenomenon and uptake from a medium is very non specific. Bacteria would absorb any potentially nutrient substance, such as proteins or DNA.

A typical transcription factor is found as a ______ and binds an __________ sequence:

dimer, inverted repeat

W and K and unlinked. The F2 progeny of F1 WwKk will display the ratio characteristic of ___________.

dominant epistasis Genotype: phenotype, frequency W___ : white, 9/16 wwK_: black, 3/16 wwkk: red, 1,16

The Ziblets, a family of fairies, display through generations the occasional Pink Nose. This syndrome was first recorded in great-great-grandfather Zibbo. Worried about genetic disease they have since taken great precautions to marry into separate families that have no history of Pink Nose. Alas, Pink Nose pops up with a certain probability among descendants of Zibbo. Children with an affected parent are often normal, but their children can display Pink Nose. The following explanation of the condition is consistent with the observations above:

dominant, incompletely penetrant.

Norway spruce can have stiff or droopy branches. You cross two pure breeding varieties, one stiff the other droopy obtaining stiff F1s. Selfing one of these you obtain 178 stiff and 144 droopy F2. The best explanation is two independent loci are involved in

double recessive epistasis

Chromosomes in the cells from Normal human and from Patient after DNA damage. The following chain of events is the best explanation for the Patient's chromosome structure:

dsDNA break > homology search > Holliday junctions > crossover

Bloom syndrome is caused by the loss of a helicase that opposes resolution of Holliday junctions toward the crossing over outcome. In this syndrome, accidental...

dsDNA break could lead to translocations involving homologous DNA sequences on different chromosomes

Crossing overs are initiated by ____ and are formed typically between _______ :

dsDNA breaks, chromatids of homologous chromosomes

The above progeny were obtained in a testcross of a female Drosophila heterozygous for three linked genes. Construct a genetic map with the data.

e - 2.5cM - j - 10cM - w

The above progeny were obtained in a testcross of a female Drosophila heterozygous for three linked genes. What genes are in coupling phase in the parental haplotypes?

e+ and w+ The nonrecombinants are the most frequent haplotypes (e+w+j , ewj+). They reflect the parental chromosomes. The + alleles of e and w are in coupling, and both are in repulsion with j. Restated: The combinations that are most frequent, i.e. the parental ones, represent the phase. These are e+,w+,j and e,w,j+. Therefore, based on the table and identification of parental chromosomes, the phase is e+,w+,j // e,w,j+..

The following tool is well suited to test dominance of one bacterial allele over another:

either an F' or a specialized transducing phage

The occurrence of a double stranded DNA break can be __________ .

either an accident or a programmed feature of meiosis dsDNA break occur as accidents in somatic cells. In prophase meiotic cells, dsDNA breaks are programmed and lead to recombination.

Using PCR and primers specific for VNTR locus Y, you genotype 1000 UC Davis students. You measure their blood pressure and find the relationship shown above. You conclude that locus Y is ______ a gene affecting the blood pressure QT.

either linked to or inside The three groups correspond to YY, Yy, yy. Y : large VNTR band, y: small VNTR band. Heterozygote has both alleles. The strong association between blood pressure and the genotype at this locus indicates close linkage and potentially colocation with a gene affecting the trait .

An enhancer can cause increased expression of a gene even from which it is separated by as many as 50,000 base pairs because:

enhancer gets close to the promoter by formation of a chromatin loop

A eukaryotic DNA sequence that positively regulates transcription regardless of orientation and relative position to the promoter is called a(n)

enhancer.

PCR provided a quantum jump in the ability of identifying criminals from samples collected at crime scenes for the following reason:

exponential amplification from diluted templates

Crossing over interference results in:

fewer double crossing overs Interference represses the occurrence of a second CO nearby the first CO

You treat chromatin with DNAse I and determine that DNA regions near certain genes are highly sensitive to the action of this enzyme. This sensitivity to DNA is most likely an indication that those regions are

free of nucleosomes and transcriptionally active. Sensitivity to nuclease -Naked DNA: highest -DNA highly complexed by DNA-binding protein: least

A double stranded DNA break in a young mouse is repaired by non-homologous end joining (NHEJ). A possible outcome of NHEJ is

frequent DNA changes such as 1 base insertion in the sealed break

Galactose utilization genes in yeast are regulated by the concerted action of three proteins: gal4, gal80, and gal3 (Figure 4). Loss of function mutation(s) in gene(s) _______ would be required to activate galactose genes in the absence of galactose:

gal80

Based on the Gal regulatory pathway illustrated in this figure, the regulatory pathway involving galactose, GAL80, GAL3 and GAL4 and leading to galactose catabolism is best summarized as follows:

galactose --> GAL3 --| GAL80 --| GAL4 --> transcription of catabolic genes

Stable acquisition by a recipient cell of an allele via bacterial Hfr conjugation... (mark the single FALSE statement)

involves pairing and recombination of two F plasmids The only role of a plasmid in Hfr conjugation is to provide the origin of transfer. So, recombination between two F plasmids makes no sense.

You have obtained two true-breeding strains of mice, each homozygous for an independently discovered recessive mutation that prevents the formation of hair on the body. The discoverer of one of the mutant strains calls his mutation naked, and the other researcher calls her strain hairless. You cross naked and hairless mice with each other and all the offspring are phenotypically wild-type. What were the genotypes of the parents?

hhNN x HHnn This question is testing whether you understand how complementation works. If the mutation occurs in the same gene, crossing two mutants together would yield F1 progeny that still display the mutant hairless trait. This is because each parent donates a mutant allele, and no wild-type allele is present. If the mutation is in a separate gene, then the parent will each contribute one bad allele of one gene and a functional copy of the other. Having one functional copy of each gene produces the wild-type phenotype in the F1. We know each mutant breeds true for the mutation, so they must be homozygous recessive for their respective mutation. Since complementation occurs, we know the mutations are in different genes. If we designate genes H and N, one parent must be HHnn (the naked mutant) and the other hhNN (the hairless). The F1 progeny is then a HhNn heterozygote and shows the wild-type phenotype because a good copy of each gene is present. See the question about plants and zombie bites as an example of what happens with complementation does not occur.

The figure illustrates the distribution of beta-carotene content in peas. P1 and P2 are inbred lines. F1 is their hybrid and F2 the filial generation derived by selfing the F1. You estimate that Vg (genetic variance) is _____ , based on comparison of ____________ .

high, P1,P2,F1 vs F2

Consider the above figure. The tetRS mutation resembles lacIS. Your E. coli has the following tet genotype: R+, P+, O+, T-, W+/ F': RS, P+, OC, T+, W-. In the absence of tetracycline, the activity of tetT and tetW is, respectively: .

high, none

Gene F is off until petal formation, and during petal formation it is actively transcribed. DNAseI sensitivity of gene F's promoter region will be ______ in chromatin isolated from _______ :

highest, petals

F'his+ strS F- his- strR A cross between the two strains above is plated on minimal medium with streptomycin. Most growing colonies will have this genotype: strR, ____ Nomenclature: strS= streptomycin sensitive, strR=streptomycin resistant his-: one his- allele is present his-, his+: two his alleles present

his+, his-

Haploinsufficient gene M in the genome of your germ line undergoes a dsDNA break. The best repair strategy for your progeny fitness is:

homologous recombination

Fidelity of DNA replication directly involves the following processes (choose all and only the relevant ones): i) Selectivity of DNA polymerase for the correctly based-paired nucleotide ii) Processing of RNA primer by 5'->3' exonuclease of DNA Polymerase I iii) Activity of 3'->5' exonuclease of DNA Polymerase III iv) Nucleotide Mismatch Repair v) Recognition of the origin of replication

i, iii, iv i) Selectivity of DNA polymerase for the correctly based-paired nucleotide: yes, 10^3 factor ii) Processing of RNA primer by 5'->3' exonuclease of DNA Polymerase I: no, it has no direct effect on fidelity iii) Activity of 3'->5' exonuclease of DNA Polymerase III: yes, this is the proofreading activity that removes mispaired nucleotides, 10^3 factor iv) Nucleotide Mismatch Repair: yes, it recognizes mismatched base pairs that escaped i and iii adding another 10^3 clean up factor v) Recognition of the origin of replication: no, it has no direct effect on fidelity Together, i, iii, and v provide a 10^9 specificity. I.e. only one/10^9 polymerized nucleotides are wrong.

Identify the sequence of events during double stranded DNA break repair. i) strand invasion ii) resection iii) DNA synthesis iv) resolution v) homology search

ii, v, i, iii, iv

In an XX mouse X inactivation fails and both X are active in all cells. The following result is most plausible:

imbalance of X-encoded products has severe and most likely lethal consequences

In the absence of a homologous template DNA, cutting of a genomic site inside a gene by CRISPR-Cas9 results in the following:

imprecise repair and likely gene inactivation

The enzyme telomerase is needed [mark the single wrong statement] -once the replication fork has reached the chromosome end -during embryo development -in the cells of a fully developed brain -in bone marrow

in the cells of a fully developed brain Telomerase is only needed in dividing cells and at the termini of a fully replicated chromosome.

You are looking for the ORF in a eukaryotic gene that has many introns. The ORF will be detectable:

in the coding strand, if the introns are removed and the exons are joined Any translated exon will have an uninterrupted reading frame that typically lacks the AUG unless it represents the start or accidentally happens to have an AUG in that position. The AUG---Stop reading frame will only be evident once the exons are stitched together.

Two calico cats, Svetlana and Lucia, differ in the size of their orange and black patches. Svetlana has 4 patches, Lucia has more than 100. The best explanation for this difference is that:

inactivation of X was earlier during Svetlana's embryogenesis Early inactivation in growing embryo -> few large patches Late inactivation in growing embryo -> many small patches See X-inactivation video

Two calico cats, Marlene and Lily, differ in the size of their orange and black patches. Marlene has 100 patches, Lily has 7. The best explanation for this difference is that:

inactivation of X was earlier during embryogenesis for Lily

BrDU staining of chromosomal DNA. The following is a good explanation for the response of patient to irradiation ( = DNA damage): the patient displays _____________ due to the deficiency of this enzyme ___________.

increased crossovers, helicase The staining patters indicates many crossovers between the chromatids in patient. This is an effect of Bloom syndrome: during dsDNA repair Holliday Junctions are normally forced to resolve as non-crossover. The helicase necessary for this function is defective and missin in Bloom Syndrome.

The open reading frame of gene Z is 990 bases long. In mutant "m" a frameshift mutation occurs in nucleotide 120 of the ORF. A reversion mutation affecting nucleotide 132 fixes the frameshift and yields mutant m-rev1. The reversion mutation most likely involves:

insertion or deletion of 1 or 2 base pairs

Two pure breeding lines of snapdragon, one red-flowered and tall, the other white-flowered and short, are crossed. The F1 hybrid is red and tall: you name the dominant alleles R and T. The F2s assort in four phenotypic classes in a 9:3:3:1 ratio. How should you determine the genotype of a red, tall F2? Cross it to...

itself This is a feasible strategy. If the plant is heterozygous, some progeny (25% for each trait) will be white and short (in any combination). The best way to test this plant is to cross it to a double recessive (white and short) plant. This is called a test cross and it is better than selfing because the frequency of each recessive trait will be 50% in the progeny.

The vector plasmid shown above has the kanamycin resistance gene (kanR) and a HindIII site in the lacZ-alpha gene. You ligate Your Favorite Gene (YFG) in the HindIII site. After transformation of an E. coli strain that expresses the omega portion of lacZ (for alpha complementation), cells containing the cloned YFG can be selected with __________ , ____________ express beta-gal activity, and appear __ on X-gal.

kanamycin, do not, white

The strands labeled F are called_____________and are synthesized by___________.

leading strands...DNA polymerase

Selfing a pea plant involves:

leaving the flower alone

less then 150, 330

No individuals homozygous for BRCA2 mutation have been found. This observation is consistent with:

lethality

In fruit flies, both X chromosomes of a female are active. The following strategy would work well to solve the male-female X-imbalance. Males:

make an X-specific regulator that doubles expression of all X-genes

Consider both a human and a fly whose sex chromosomes are XXY. Their respective sex (or condition) is:

male, female

An XO fruit fly is expected to be _____ because of ______:

male, single X

Roger is affected by haemophilia-A, a recessive X-linked condition. If he were to mate with a carrier (a person not affected, but carrying the disease allele), their progeny would have the following probabilities of being haemophiliac:

males 0.5, females 0.5

A calico cat mates with an orange cat. The kittens will be:

males 1 black : 1 orange, females 1 calico : 1 orange

A fur color gene located on the X chromosome controls orange (O) and black (+) color. An orange female mates with a black male. The kittens will be:

males all orange, females all calico

Calvin Bridges discovered non-disjunction by observing that a cross between a white-eyed female and a red-eyed male resulted in rare _______, that had ___ eyes instead of the expected ___ eyes.

males, red, white

A haploid cell cannot successfully complete the following process:

meiosis

Homologous recombination is limited to DNA molecules that are nearly identical by the following mechanism:

melting of DNA at recombination sites and formation of heteroduplex DNA

The surface oligosaccharides of Toad red blood cells are determined by locus S, which encodes an enzyme necessary for oligosaccharide synthesis in a manner similar to IA and IB of humans. Homozygotes for alleles SR or SW display type R or type W oligosaccharides, respectively. The heterozygote displays both types. If the SR allele arose by mutation of the SW allele, the mutation is most likely to be of this type:

missense A single amino acid change is the most likely cause, therefore missense mutation. The other mutations are too severe.

A mutant allele of the catabolite activator protein gene encodes an otherwise normal protein that cannot bind cAMP. This mutation is most likely a ________ and affects a ________ amino acid residue:

missense, conserved

Gene M displays recessive epistasis on unlinked gene F. MmFf is selfed. Among the F2s, a phenotypic class occurring with a frequency of 1/4 will have the following genotypic formula ("_"=any allele):

mm__ mm _ _ = epistatic phenotype, 4/16 M_F_ = phenotype F dominant, 9/16 M_ff = phenotype F recessive, 3/16

What hypothesis is based on the histone code? That...

modification of gene expression is a consequence of chemical alterations on the tails of histone proteins

The orange and black coat colors of cats are controlled by an X-linked gene with two alleles: O = orange and o+ = black. Oo+ females are "calico" (black and orange patches). Two cats mate and have 2 orange males, 2 black males, 2 orange females and 3 calico females. The parents' genotypes are (X and Y are omitted for simplicity):

mom is Oo+, dad is O 2 orange males: mom must carry O 2 black males: mom must carry o+ 2 orange females: dad must carry O 3 calico females: got o+ from mom and O from dad

After purifying the DNA of the F factor and its F'trp derivative, you digest both with restriction enzyme HaeIII (cuts GGCC, which is fairly common). Digestion of the F factor with HaeIII yields 120 fragments adding to 80 kb. The most PLAUSIBLE gel pattern for the F'trp will include ______ of the fragments from the original F plasmid _____________ .

most, plus a new set of fragments. F' consist of the original F plus a segment of chromosomal DNA. A somewhat larger plasmid is expected. Most of the original F restriction products should be present. The Trp region added may range between 10 and 30kb.

Sites in eukaryotic chromatin that are preferentially digested with DNAseI, correspond to DNA regions that are:

mostly free of histones

You obtain a male and a female dog, which you intend to breed. While healthy, they are heterozygous for multiple deleterious recessive alleles. The affected loci are unlinked. You have plenty of resources and time and you want to use these two individuals to produce a (inbred) dog breed that has no deleterious traits. What is your P of success?

nearly 1 You can make approximate calculations to see what it would take. The question is whether it is possible, not whether it is easy or difficult. Given large resources, is it possible?

In the late 1990s, researchers developed golden rice, a rice plant that was genetically engineered to produce high levels of beta-carotene in rice grains. Two genes were introduced: one a plant enzyme called phytoene synthase (Psy), the other a bacterial enzyme called carotene desaturase (Crt). In addition, a selectable marker gene, Pmi, was introduced to select transformed rice cells on medium containing mannose. You have the following promoters at your disposal: pGlu: Rice promoter that is active only in rice grain (endosperm) pUbi: Rice promoter that is active in all rice tissues (leaves, roots, stems, grains) pPmi: E. coli promoter that is active in the presence of mannose Which combination of promoters and transgenes has the best chance of producing a rice plant with beta-carotene in the grains?

pGlu:Psy; pGlu:Crt; pUbi:Pmi

Select the single feat that is not possible for F-type plasmids:

package its DNA in a phage particle This is the only answer that is false. Viruses package their DNA in phage particles.

The first exon in a eukaryotic gene will certainly contain:

part or the whole 5' UTR

Activation of a eukaryotic promoter cannot involve the following:

pausing of the ribosome over codons because the corresponding tRNA is depleted

Using pedigree data, linkage evaluation (LOD figure above) was carried out between a dominant disease locus K and three molecular marker loci: T, W, and R. The curved lines are derived by comparing:

pedigree likelihood for different linkages vs pedigree likelihood for independent assortment In each case, a different marker is genotyped in the pedigree. The P of each pedigree is evaluated under different recombination frequency hypotheses and the LOD is calculated.

The recipient (female) cell is pro-, thr-, pur-. Formation of a pro+ , thr-, pur+ recipient is:

possible but rare, requires four crossing overs

Genetically engineered CAS9 and guideRNA can mimic fairly well the general function of all these proteins, EXCEPT this one:

primase All proteins except primase have activities (binding a given motif and potentially cutting it) that are found in Cas9 and its mutant derivatives. Primase is a polymerase.

This polymerase ______, which never needs a primer, is likely active at this site____:

primase, L At site L primase is polymerizing an RNA primer. Primase does not need a primer.

The ____________ on a gene is the DNA sequence that the transcription apparatus recognizes and binds to initiate transcription.

promoter

Breaking and rejoining of chromatids occurs normally at:

prophase I of meiosis

In the following phases of meiosis chromosomal crossovers will be present:

prophase I, metaphase I

Bacteriophage molecules can be labeled with either 35-S or 32-P radioactive precursors. The 35-S and 32-P isotopes label, respectively:

protein, DNA Proteins are composed of amino acids, some of which contain sulfur. DNA doesn't have sulfur as part of its structure. DNA's structure includes a phosphate backbone. Some proteins include phosphate in their molecular structure, typically by postranslation modification of amino acids Tyr and Thr. In any case, the amount of P present in most proteins is very low compared to that of DNA. See DNA video in module.

Using this Salmonella, the TAs isolate Hfr strains. Hfr-Erin transfers Pyr+, Gal+, Leu+ into an F- strain that is "-" for all these markers, but ampicillin resistant. You select exconjugants progeny on a medium supplemented with ampicillin, glucose and pyrimidines but lacking leucine. The least likely progeny is:

pyr+, gal-, leu+ Remember how many crossing overs are needed and possible. What is the least likely pattern?

The ______ protein forms a _________ that recognizes homologous, non-denatured dsDNA:

recA, complex with ssDNA recA and ssDNA form a specialized structure capable to search dsDNA.

Provide the best answer. The condition displayed by the affected individuals in this pedigree is ________. The genotype of individual II-9 is _________.

recessive and X-linked, XdXD

Fig S2 displays a family pedigree in which some members are affected by a fully penetrant condition. The best explanation for this condition is:

recessive, X-linked Skips generations: recessive. Multiple outcrossings to presumed non carriers (rare) make it improbable for the condition to be autosomal.

Fig S3 displays a family pedigree in which some members are affected by a fully penetrant condition. The best explanation for this condition is:

recessive, autosomal This is a classical case: two affected progenitors must have transmitted disease alleles to the parents of the affected progeny. Obviously recessive and autosomal. X-linked inconsistent with the affected woman in the progeny and transmission through III-4.

In E.coli, next to a gene encoding restriction enzyme EcoRI there is a second, related gene encoding a methyltransferase. A mutation causing loss of function in the methyltransferase gene results in death of the bacterium. This is most likely because the EcoRI protein:

recognizes and cuts non-methylated target sites

Consider this cross and the shown progeny. What type of gamete did parent 1 contribute? (CO=crossing over)

recombinant gamete, CO between W and the translocation

Identify the false statement. Introns are

removed in the process of DNA replication

Which of the following are all trans-acting factors involved in regulation of the Lac operon?

repressor, CAP, RNA polymerase

A and B = chromosomal loci. Dominant and recessive alleles = upper and lower case letters, respectively. The hybrid AaBb is crossed to aabb. The following phenotypic classes and progeny numbers are produce: AB, 25; Ab, 450; aB, 410; ab, 30. Alleles for genes A and B are in _____________ phase. The best estimate of map distance between the A and B loci is _______ m.u.

repulsion, 6 The genes must be in repulsion phase because the most observed progeny represent the parental class, and in this example, the parental classes are gametes Ab or aB. Map distance is estimated by recombinants/total offspring = (25 + 30) / (25 + 30 + 410 + 450) = 0.06 or 6% or 6 M.U.

Meiosis will always.. (choose the wrong statement):

result in direct gamete production

Plasmid vector pExpPtet carries the PO region of the Tet operon next to a Shine-Delgarno site (SD) (ribosome binding), the multiple cloning site (MCS), and a transcriptional terminator sequence (Ter) as follows: POtet-SD-MCS-Ter You cut the MCS with EcoR1 and insert (ligate) into it an EcoR1 DNA fragment carrying the ORF for Green Fluorescent Protein (GFP). GFP, when excited with blue light, fluoresces green. Among transformed E. coli colonies you identify forty that carry the GFP ORF. Under what conditions could they express GFP and how many would you expect to do so?

tetracycline, ~20 The ORF fragment can be cloned into the MCS in two possible orientation. Forward, i.e. with 5' end of coding strand proximal to the promoter, or reverse, with the 3' end of coding strand proximal to the promoter. Only the forward orientation can be expressed. Consider how the ORF can be cloned into the MCS.

The image shows some of the structures in a DNA replication bubble. Ribonucleotides are likely to be found at these places:

the arrow tails of M, P, and Q M is a leading strandOkazaki fragments are represented by the P and Q arrows. The arrow is made of a tip and a tail. We use the arrow symbol because the tip represents the growing end and the tail the start. Each synthesized strand starts with an RNA primer. Therefore the tails consist of RNA primers, in which the ribonucleotides are present. The RNA is eventually digested away by the exonuclease function of DNA pol I. Deoxyribonucleotides (DNA) are present at the tip and through most of the arrow, but they are not ribonucleotides (RNA).

Comparison, i.e. alignment of the amino acid sequence of RNA polymerase from multiple plants and animal species is likely to show the following:

the consensus motif for their active site

Insertion of DNA in the MCS region of the lacZα gene of common bacterial plasmids used for cloning results in the following useful outcome:

the lacZα protein is inactivated

The nucleotide U is associated with this feature of a replication fork:

the lagging strand

An advantage of measuring a quantitative phenotypic trait using recombinant inbred lines (RIL) compared to an F2 generation is that with RIL:

the phenotype of multiple individuals with identical genotypes can be tested

Wild-type phage M4 infects both E. coli K and E. coli B. Mutants F and G infect only E. coli B. You prepare lysates by high MOI infection of E. coli B and get the following results. The F and G mutations are: Note: MOI=multiplicity of infection. The mix of F and G is made by mixing independent F and G lysates. For coinfection, a mix of F and G is used to infect E. coli B, and the the resulting phages are used in the infection of E. coli K.

the same gene High MOI shows that F and G cannot complement. Therefore they are mutant in the same gene. The high MOI lysate from coli B enabled recombination. Relatively rare single crossovers in the affected gene produce a WT allele that gives clear plaques.

This is the second part of Q.5. The old Q.5 was broken down in two questions to make matter clearer. Generalized transducing phage P22 is used to transduce DNA from Salmonella Kleiber-3 into a lab strain that is thi-, aro-, gly-, tyr-, pur-. After selection for thi+, the colonies are replica-plated to determine their genotype at the other markers. The results are shown in the table. Choose the most likely map ORDER for thi, aro, gly and tyr markers.

thi---gly-aro-tyr

A chromosomal translocation involves:

transfer of one segment bewteen chromosomes of different types

The presence of the enzyme DNAse in a medium would hinder the following mode of bacterial sex:

transformation This answer tests whether you understand the modes of DNA transfer in bacteria. Transformation is the only type of DNA transfer that involves naked DNA in the outside environment (not inside some protective envelope). DNAse would destroy DNA in the medium, but would not necessarily affect DNA packaged in a phage or hidden inside a conjugation pilus.

RecA protein binds single stranded DNA and favors formation of heteroduplexes with complementary DNA. It is essential for recombination. Nevertheless, loss of function mutation of the recA gene in E.coli is compatible with survival, even though a number of processes are compromised. Mark the one process that should not be compromised.

transformation with an F plasmid All other processes require homologous recombination (HR) and are compromised in the mutant. Transformation with a plasmid does not require HR and it is not affected. The plasmid replicates autonomously (by itself). In fact, the most common bacteria used for genetic engineering are recA-: they support plasmid transformation and replication, but do not allow homologous recombination.

During eukaryotic transcription, an RNA-DNA heteroduplex is formed:

transiently in the melted DNA region The newly formed RNA is made by base pairing it to the DNA template strand. As the RNA lengthens the 5' end falls off of the DNA template allowing the transcription bubble to close.

A man heterozygous at Z locus of chromosome 19 (Zz) produces a Zz sperm. f this sperm fertilizes a normal egg the resulting fetus would display this condition:

trisomy 19

Consider mice E, G and K, whose VNTR genotype at locus D33 (a 5bp tandem repeat) is analyzed in this gel. If mice E and G are mated, their pup K's genotype could be explained by:

trisomy from non disjunction at anaphase 1 (in the absence of recombination) Each band displays a D33 allele. In most individuals, two bands are expected, such as D33-8 and D33-12. Each parent should donate a band to the progeny, If the progeny has three bands, it must have received two from one parent. This is expected if nondisjunction resulted in a disomic gamete carrying both alleles from one parent, If recombination is ruled out, the two alleles would normally separate at anaphase I. Therefore ND at Anaphase I is implicated.

Based on the growth phenotypes of this replica plate experiment, colony R is .

trp-, tyr-, arg+

The above map of a chromosomal region of Bacillus subtilis was derived by cotransformation mapping in the early 60s. The mapping cross used a wild-type donor and a recipient that was auxotrophic for tyrosine, histidine and tryptophan with genotype tyr1-, his2-, trp2-. Based on the map, select the probability of coinheriterance of the given alleles in the transformants when the primary selection was for growth without tyrosine.

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