genetics final (all quiz questions)

Part II: The figure below shows the pedigree of a family in which a completely penetrant, autosomal dominant disease is transmitted through three generations. The disease allele, which is rare, has a single origin. The alleles for a SNP locus are shown (A or C). The likelihood ratio for linkage is _______________________ , which means we ______________ sufficient data to support linkage with statistical confidence.

(0.9^9 ) x (0.1^1) x (2^10); don't have

Part I: A wild-type (denoted by +) female mouse is crossed to a male mouse with apricot eyes (ap) and gray body (gy). The F1 mice are all wild-type for both traits. When the F1 are interbred, 200 F2 progeny are distributed as follows: Females: All wild-type = 100 Males: wild-type = 46 apricot = 6 gray = 4 apricot, gray = 44 What is the genotype (with correct arrangement of alleles) of the F1 females? ap gy / ap gy + + / + + + gy / ap + + + / ap gy

+ + / ap gy

Hemophilia is caused by an X-linked recessive mutation in humans. A woman whose sister is a hemophiliac marries a man whose brother is a hemophiliac, but the couple does not show symptoms of hemophilia. What is the probability that a daughter they are expecting will have hemophilia? (Assume that the mutant allele is rare and completely penetrant, and carriers do not show any symptoms.) 0 1/2 1/4 1 1/8

0

In 2008, Margaret Binkele was murdered in her Georgia home. The number of repeats at three CODIS simple sequence repeat (SSR) loci were determined using the crime scene sample and blood from three suspects. The results of genotyping three SSR loci are summarized in the table. Allele frequencies for each locus are as follows: Locus 1, allele 10: 0.1 Locus 2, allele 11: 0.2 Locus 3, allele 6: 0.2; Locus 3, allele 9: 0.3 What is the probability of matching the genotype identified from the crime scene sample? 0.000048 0.17 0.0012 0.0006 0.000024

0.000048

If the parents of a family already have two boys, what is the probability that the next two children will be a girl and a boy? Assume that boys and girls are equally likely. (Hint: The order is unspecified.) 0.125 1 0.0625 0.5 0.25

0.5; There are two options: a girl first then a boy, and a boy first then a girl.

Two alleles of gene C control hair color in horses: C¹ and C². Horses homozygous for C¹ allele are red, heterozygotes are yellow, and C² homozygotes are cream. In the offspring of mating between a cream horse and a yellow horse, what phenotypic ratio is expected? 1 red: 1 cream 1 cream: 2 red: 1 yellow 1 cream: 1 yellow 3 yellow: 1 cream 3 yellow: 1 red 1 red: 2 cream: 1 yellow 1 red: 1 yellow 1 red: 2 yellow: 1 cream

1 cream: 1 yellow

Red-green color blindness is controlled by an X-linked gene in humans. The allele that causes color blindness is recessive to the allele for normal vision. A man and a woman both with normal vision are married as first cousins. The mother of the woman and the father of the man are siblings and they are both colorblind. What is the probability that a child from this consanguineous mating will be colorblind? (Assume that the mutant allele is rare and completely penetrant, and carriers have normal vision.) 1/2 1/8 1 0 1/4

1/4

In corn, having ligules (L) is dominant to liguleless (l), and green leaves (G) is dominant to white leaves (g). If a testcross is performed with a dihybrid plant (i.e. heterozygous for two genes), what proportion of the progeny would be green and liguleless? 1/16 1/2 1/4 0 9/16 3/16

1/4; P (green) = 1/2 and P (liguleless) = 1/2 in this testcross (a cross involving a recessive homozygote).

Part I: Sickle cell anemia is a recessive disease in humans. Two carriers of the sickle cell allele (but without the disease) marry and have three children. What is the probability that all three children will be diseased? 1/8 1/4 3/4 1/64 27/64 1/2

1/64

Part III (Do Part II first): Below is a pedigree of a human genetic disease in which solid color indicates affected individuals. Assume that the disease is caused by a gene that has two different alleles A and a, and the disease allele is rare. If the female 4 marries an affected man (exhibiting the disease) whose parents are both heterozygotes, what is the probability that they will have a normal child? (Hint: First, consider the probability that the affected male has the genotype needed for the couple to have a normal child.) 1/4 1/3 1/8 1/2 1/6

1/6; For the couple to have a normal child (aa), they need to be both heterozygous affected (Aa). Then they would have 1/4 chance of having a normal child, but this is contingent upon the affected male having the right genotype. P (a normal child from two affected parents) = P (both parents = Aa) x P (each parent contributes a allele)

Part III (Do Part II first): A wild-type (denoted by +) female mouse is crossed to a male mouse with apricot eyes (ap) and gray body (gy). The F1 mice are all wild-type for both traits. Now, you backcross one of these F1 female mice to the male parent mouse with apricot eyes (ap) and gray body (gy). Among 100 female F2 progeny, how many females are expected to have either apricot eyes or gray body (but not both)? Females: wild-type apricot gray apricot, gray total = 100 Males: wild-type = 46 apricot = 6 gray = 4 apricot, gray = 44 Note: Since this is a different cross, you would expect to see four different phenotypic classes among the female F2 progeny, and the map distance between the genes should still be the same. 10 25 20 100 50

10

Part I: The figure below shows the pedigree of a family in which a completely penetrant, autosomal dominant disease is transmitted through three generations. The disease allele, which is rare, has a single origin. The alleles for a SNP locus are shown (A or C). What is the genetic distance between the two loci? 0 m.u. 20 m.u. 30 m.u. 10 m.u. 50 m.u. 40 m.u.

10 m.u.

Part II (Do Part I first): In humans, the genes for red-green color blindness (C = normal, c = colorblind) and hemophilia A (H = normal, h = hemophiliac) are both X-linked and only 3 map units apart. A normal woman whose mother is colorblind (with normal blood clotting) and whose father is hemophiliac (with normal vision) marries a normal man. If she were to have 100 sons and 100 daughters, how many of her children are expected to be normal? Note: RF is calculated by scoring only those progenies that can be identified as parental vs. recombinant. All daughters will have the same phenotype whether they're parental or recombinant type (indistinguishable). 3 103 200 101.5 (101-102 children) 1.5 (1-2 children) 100

101.5 (101-102 children)

Had Hershey and Chase performed their experiment designed to determine the molecule of heredity using the 16N radioisotope, what would have been their results? 16N will label both DNA and protein, so radioactivity would be found within bacterial cells and on phage ghosts 16N will label protein only, so radioactivity would be found on phage ghosts 16N will label DNA only, so radioactivity would be found within bacterial cells 16N will label neither DNA nor protein, and phages would not be radiolabeled as a consequence

16N will label both DNA and protein, so radioactivity would be found within bacterial cells and on phage ghosts

Part II: After you submit the aforementioned sequence (887 nt) in ORF Finder, select the longest ORF in the list and do BLAST. (Do not change any setting in the new BLAST window. The program selected by default should be blastp.) For the protein with the third highest score, how many amino acids are identical between the query (your sequence) and the subject (from the database)? (Hint: Click the appropriate protein under 'Description' to go to the alignment.)

197 (Find the number in the 'identities' column.)

Achondroplasia is a form of dwarfism in humans. It is caused by a mutant allele of the fibroblast growth factor receptor (FGFR) gene that produces an overactive protein. Having one copy of the mutant allele results in dwarfism. Having two copies of the mutant allele results in stillbirth. (Infant dies in the womb.) If two people with achondroplasia have a child together, what is the probability that their child will also have achondroplasia? (Note that only live progeny are considered.) 1 0 1/3 2/3 3/4 1/2 1/4

2/3

Suppose a protein contains 100 amino acids. How many unique sequences could it potentially represent? 2,000 100^4 400 100^20 4^100 20^100

20^100

The R/r and S/s genes are unlinked. In the cross Rs / rS x rs / rs, what fraction of the progeny will be Rs / rs? 25% 10% 50% 100% 20%

25%; Rs is one of two parental gamete types made by Rs / rS heterozygote. RF is 50% when two genes are unlinked.

Part I: You want to PCR amplify the trinucleotide repeat region of the Huntington disease (HD) locus and analyze the PCR product by gel electrophoresis. The 5' end of one primer is located 70 nucleotides upstream of the first CAG repeat, and the 5' end of the other primer is located 60 nucleotides downstream of the last CAG repeat. Normal alleles for HD gene contain up to 35 CAG repeats, while disease-causing alleles carry 36 or more. What would be the size of the PCR product in base pairs for an allele with 50 CAG repeats?

280

A plasmid vector contains a single EcoRI site in the polylinker (multiple cloning site). An insert is flanked by two EcoRI sites, one at each end. You cut the vector and the DNA containing your insert with EcoRI and ligate the two. How many EcoRI site(s) are present in the recombinant DNA molecule and how many band(s) would you see when you run your EcoRI digest of the recombinant DNA on a gel? (There is no internal EcoRI site in the insert DNA.) The recombinant DNA contains ["0", "2", "1"] EcoRI site(s), and there would be ["0", "1", "2"] band(s) when you run the recombinant DNA digested with EcoRI on a gel.

2;2

The following picture shows ethidium bromide-stained DNA bands revealed by gel electrophoresis of the same circular DNA digested in three different tubes containing the following restriction enzymes: EcoRI only, BamHI only, or a mixture of the two enzymes. The arrow represents the direction of electrophoresis. (FYI, a restriction enzyme digests DNA by cutting both strands of a specific, short DNA sequence it recognizes.) How many EcoRI and BamHI site(s) are present in this circular DNA? Select two that apply. (You don't need to analyze the last lane on the right.) 3 EcoRI sites 2 BamHI sites 2 EcoRI sites 3 BamHI sites 1 BamHI site 1 EcoRI site

3 EcoRI sites 2 BamHI sites

You find eight different fly strains (A-H) that show the recessive white-eye trait. Pairwise crosses are performed and summarized as follows. W indicates white-eyed progeny, and R indicates wild-type red eyes. Based on these crosses, how many different genes are present? 1 gene 2 genes 4 genes 3 genes More than 4 genes

3 genes (As you read the results of complementation tests, recall that two homozygous mutant parents will produce a mutant F1 if mutation occurs in the same gene. two homozygous mutant parents will produce a wild-type F1 if mutation occurs in different genes.)

You performed a Sanger sequencing reaction and obtained the following chromatogram (a computer-generated trace of the intensity of each color's fluorescence). In this figure, A = green, C = purple, G = black, T = red. The height of the peaks is unimportant. The 5' end of the sequence is at the left of the trace. What is the sequence of the template DNA used for this sequencing reaction? (You can still answer the question even if you cannot distinguish the different colors.) 5' TTTGCTTTGTGAGCGGATAACAA 3' 3' TTTGCTTTGTGAGCGGATAACAA 5' 5' AAACGAAACACTCGCCTATTGTT 3' 3' AAACGAAACACTCGCCTATTGTT 5' 5' TTTGCTTTGTGAGCGGATAACAA 3'

3' AAACGAAACACTCGCCTATTGTT 5'

Given the partial RNA sequence for a gene, what is the sequence of the template strand? 5' AACUGGGUAA 3' 5' AACUGGGUAA 3' 5' AACTGGGTAA 3' 3' TTGACCCATT 5' 5' TTGACCCATT 3' 3' AACTGGGTAA 5'

3' TTGACCCATT 5'

If 20% of the bases in a region of the mouse genome are cytosine, what percentage (%) in that region are adenine?

30

Part III (Do Part II first): In humans, the genes for red-green color blindness (C = normal, c = colorblind) and hemophilia A (H = normal, h = hemophiliac) are both X-linked and only 3 map units apart. A normal woman whose mother is colorblind (with normal blood clotting) and whose father is hemophiliac (with normal vision) marries a man who is both colorblind and hemophiliac. If she were to have 100 sons and 100 daughters, how many of her children are expected to be normal? Note: For this mating which can be considered a testcross, the egg of the normal woman must be of a particular genotype to have a normal child, whether it's a son or a daughter. 1.5 (1-2 children) 101.5 (101-102 children) 200 3 100 103

3; The normal woman transmits C H (a recombinant type) to her normal sons and daughters.

How many of the listed events are post-transcriptional processes that occur after transcription starts but before translation begins in eukaryotes? Addition of a poly-A tail on the 3' end mRNA export to the cytoplasm Scanning down the 5' end of mRNA to locate the first AUG Addition of a 5' cap Promoter binding of RNA polymerase Base pairing between the initiator tRNA anticodon and the start codon Peptide bond formation Binding of the large subunit with the small subunit of ribosome Splicing to remove introns and join exons together 1 3 2 4

4

Part II (Do Part I first): Below is a pedigree of a human genetic disease in which solid color indicates affected individuals. Assume that the disease is caused by a gene that has two different alleles A and a, and the disease allele is rare. In generations II and III combined, how many individuals are heterozygotes for sure (100% chance)? Assume complete penetrance. (All individuals with the disease-causing genotype show the expected phenotype.)

4

Given that the trait displayed in this pedigree is *rare and X-linked dominant, what is the highest penetrance if I-1 female is heterozygous? (*Assume fewest genetically unrelated individuals have the rare allele. I.e., people marrying into the family with the trait history do not normally carry the rare allele unless there is evidence to suggest otherwise.) 40% 50% 100% 67% 33%

40%; Penetrance = #people showing the trait ÷ #people with the trait-associated genotype × 100 The numerator is 2. How many additional people have the genotype associated with this X-linked dominant trait?

What is the average size of fragments in base pairs (bp) when a random sequence of DNA is digested with a restriction enzyme that has a 6-base recognition site? (Note: Assume equal ratio of each nucleotide.) 4096 bp 1024 bp 4 bp 64 bp 256 bp 16 bp

4096 bp

In Drosophila, the genes y (yellow body) and car (carnation eyes) are located at opposite ends of the X chromosome. In doubly heterozygous females (y+ y car+ car), a single chiasma is observed somewhere along the X chromosome in 90% of the examined germ cells. No X chromosomes with multiple chiasmata are observed. What percentage of the male progeny produced from such a female would be recombinant for y and car? 40% 45% 90% 80% 10% 20%

45%

The R/r and S/s genes are linked and 10 map units apart. In the cross Rs / rS x rs / rs, what fraction of the progeny will be Rs / rs? 5% 10% 50% 20% 45% 90%

45%; Rs is one of two parental gamete types made by Rs / rS heterozygote. RF is 10%.

Which of the following sets of primers could you use to amplify the entire target DNA sequence below, which is part of the last protein-coding exon of the CFTR gene? 5'GGCTAAGATCTGAATTTTCCGAG...TTGGGCAATAATGTAGCGCCTT 3' 3'CCGATTCTAGACTTAAAAGGCTC...AACCCGTTATTACATCGCGGAA 5' 5' GGCTAAGATC 3' and 3' CATCGCGGAA 5' 5' GTAGCGCCTT 3' and 3' CATCGCGGAA 5' 5' GTAGCGCCTT 3' and 3' CCGATTCTAG 5' 5' GGCTAAGATC 3' and 3' CCGATTCTAG 5'

5' GGCTAAGATC 3' and 3' CATCGCGGAA 5'

Adenine in the wobble position of a tRNA is always modified to inosine (I), which can base-pair with U, C, or A. Amino acid Ile can be specified by AUU, AUC, and AUA codons (reading from 5' to 3'). What is the anticodon of tRNA that carries Ile? 5' AUI 3' 5' UAI 3' 5' IAU 3' 5' IUA 3'

5' IAU 3'

You performed a Sanger sequencing reaction and obtained the following chromatogram (a computer-generated trace of the intensity of each color's fluorescence). In this figure, A = green, C = purple, G = black, T = red. The height of the peaks is unimportant. The 5' end of the sequence is at the left of the trace. What is the sequence of the smallest DNA molecule that is synthesized in the sequencing reaction and contains dideoxyC (ddC)? (You can still answer the question even if you cannot distinguish the different colors.) 5' TTTGCTTTGTGAGCGGATAACAA 3' 5' primer-TTTGC 3' 3' primer-AAC 5' 5' primer-TTTGCT 3' 3' primer-AACA 5'

5' primer-TTTGC 3' (DNA is synthesized in a 5'-to-3' direction. Here, the first nucleotide added to the 3' end of a primer is T.)

Part I: In some genetically engineered corn plants, a Bt gene was added to a chromosome. The Bt gene specifies a protein called Bt that is lethal to certain flying insect pests that eat the corn plants. If a corn plant is heterozygous for the Bt gene (one homolog has the introduced Bt gene and the other does not), what proportion of its pollen (sperm) would carry the Bt gene? 25% 75% all pollen (100%) 50% no pollen (0%)

50%

Part II: When you go to UCSC Genome Browser and search for the aforementioned gene, what is the "sum" of the numbers of exons and introns? (Note: If there are multiple transcripts, work with the highlighted one that corresponds to the RefSeq.)

51 OR 133 (Find the numbers of exons and introns by moving your cursor close to the horizontal and vertical lines on the annotated gene.)

Part I: Go to ORF Finder: https://www.ncbi.nlm.nih.gov/orffinder/ Links to an external site.. In the space below "Enter Query Sequence," copy and paste the following sequence (887 nt). Under "Choose Search Parameters," select any sense codon and submit. (Do not change other search parameters.) TGAAATGAATAAGGCCTTTATTAGCCAGAGAAAAGAAAACAATATTGAAACTAAACATAAGAAAGTGAGGGCTGTAAGTTATCGTAAAAAGGAGCATCTAGGTAGGTCTTTGTAGCCAATGTTACCCGATTGTCCTACAGCTTTGTCCAGTGGCTGTAGCGGTCCCGTTGCTGCGGTGAGCTGGCTGCGTTGATGGGCGGTAAGTGGCCTAGCTGGTGCTCCATTCTTGAGTGTGTGGCTTTCGTACAGTCATCCCTGTACAACCTGTTGTCCAGTTGCACTTCGCTGCAGAGTACCGAAGCGGGATCTGCGGGAAGCAAACTGCAATTCTTCGGCAGCATCTTCGCCTTCCGACGAGGTCGATACTTATAATTCGGGTATTTCTCTCTGTGCATGGCCTGTAATTTCTGTGCCTCCTGGAAGAATGGCCATTTTTCGGCTTCAGTAAGCATTTTCCACTGGTATCCCAGCTGCTTGCTGATCTCTGAGTTTCGCATTCTGGGATTCTCTAGAGCCATCTTGCGCCTCTGATCGCGAGACCACACGATGAATGCGTTCATGGGTCGCTTCACTCTATCCTGGACGTTGCCTTTACTGTTTTCTCCCGTTTCACACTGATACTTAGAGTTACAGCTTTCAGTGCAAAGGAAGGAAGAGCTTCTCCGGAGAGCGGGAATATTCTCTTGCACAGCTGGACTGTAATCATCGCTGTTGAATACGCTTAACATAGCAGAAGCATATGATTGCATTGTCAAAAACAAGGAGAGTGCGACAAAATTGAAAGGTGCCAGAGTTCGAAACTTATTTTACTATCCAAAACTCACTTCTACCAGATTCTTTGTTACGTTAACTTTTGTAATGAAACTTGCATTTCTCCGCCCTCAACA There are _____ ORF(s) larger than 200 nt, and the longest ORF codes for ________________ .

5; more than 200 amino acids

Part II: Sickle cell anemia is a recessive disease in humans. Two carriers of the sickle cell allele (but without the disease) marry and have three children. What is the probability that at least one child will be normal? (Hint: An easier way to approach this problem is to first think about the probability of NOT having any normal child and consider the sum of all probabilities.) 7/8 1/4 3/4 63/64 1/2 37/64

63/64

For an individual who has 10 gene pairs on 10 different chromosomes, how many genetically distinct gametes can form if four of the gene pairs are homozygous and the remaining six gene pairs are heterozygous? 12 32 64 6 16

64

In the common daisy, genes col-1 and col-2 control flower color. Function of both genes (col-1 and col-2) is required to have color and is conferred by the wild-type allele of each gene. At least one wild-type copy (allele) of each gene is required for flowers to be colorful instead of white. Predict the phenotypic ratio of the F2 progeny (produced from inbreeding F1) of a cross between two pure-breeding white plants (Parental), one mutant for col-1 but homozygous wild-type for col-2, and the other mutant for col-2 but homozygous wild-type for col-1. 15 white : 1 colorful 3 colorful : 1 white 9 white : 7 colorful 3 white : 1 colorful 9 colorful : 7 white 15 colorful : 1 white

9 colorful : 7 white

A certain disease is caused by a dominant mutant allele. (The wild-type allele is recessive.) However, the penetrance of the disease is 75%. Two individuals known to be heterozygotes have a child. What is the probability that the child exhibits the disease? 1/16 3/16 1/4 3/4 9/16

9/16

Compounds A-F appear to be intermediates in a biochemical pathway in Neurospora (bread mold). Conversion of one intermediate to the next is controlled by an enzyme encoded by a gene. Several mutations are discovered, and Neurospora strains 1-4 each contain a single gene mutation. Strains 1-4 are grown on minimal media supplemented with one of the compounds A-F. The table depicts the ability of each strain to grow in the presence of different compounds (+ = growth; o = no growth). The biochemical pathway that fits the data presented is _____________ , and the strain that has a mutation at the last step (rightmost) of the pathway is ___________ .

A or B → C → D → F → E; strain 3

The pairwise map distances for four linked genes are as follows: A-B = 22 m.u. B-C = 7 m.u. C-D = 9 m.u. B-D = 2 m.u. A-D = 20 m.u. A-C = 29 m.u. What is the order of these four genes? ADCB ACDB ABDC ACBD ADBC ABCD

ADBC

What level of protein structure could be influenced by a nucleotide substitution that changes the identity of a single amino acid in a polypeptide? All four levels of protein structure could be influenced by a single amino-acid change. Quaternary structure Primary structure Secondary structure Tertiary structure

All four levels of protein structure could be influenced by a single amino-acid change.

A replication bubble is subdivided into four quadrant: A-D. The origin of replication and the direction of fork movement are also shown. What regions will make the lagging strand? (Select two that apply.) B A D C

B C

Part II: The trinucleotide repeat region of the Huntington disease (HD) locus in six individuals is amplified by PCR and analyzed by gel electrophoresis as shown; the numbers to the right indicate the sizes of the PCR products in base pairs. The 5' end of one primer is located 70 nucleotides upstream of the first CAG repeat, and the 5' end of the other primer is located 60 nucleotides downstream of the last CAG repeat. Normal alleles for HD gene contain up to 35 CAG repeats, while disease-causing alleles carry 36 or more. The more repeats, the earlier the age of disease onset. The onset of disease would be earliest in the individual ["D", "B", "A", "F", "C", "E"] , and there are ["1", "3", "2", "4", "5", "6"] individual(s) most likely to be affected by Huntington disease, which is an autosomal dominant disease.

B; 4

Part II (Do Part I first): In rats, two genes control the coat pigmentation: B and P. The allele for black pigmentation (B) is dominant to the allele for cream pigmentation (b). However, in the absence of the dominant P allele, rats are albino regardless of the alleles of the other gene (B/b). When a pure-breeding, homozygous black rat was crossed to a homozygous albino, all of the F1 rats were black. When the F1 rats were crossed to each other, the F2 consisted of 161 rats: 93 black, 30 cream, and 38 albino. What is the genotype of the black F1? BBPP BBPp BBpp Bbpp BbPP bbpp BbPp bbPp bbPP

BbPp

A researcher is studying coat color in mice. Wild-type fur in mice is brown. Three pure-breeding strains of mice with white fur have been isolated: milky, blanc, and weiss. White fur is a recessive trait in each strain. These mice are crossed in pairs and the progeny phenotypes are recorded. milky x blanc = all brown progeny milky x weiss = all brown progeny blanc x weiss = all white progeny What conclusion can be drawn from these results? All three strains have mutations in different genes. Blanc and weiss have mutations in the same gene; milky has a mutation in a different gene. Milky and blanc have mutations in the same gene; weiss has a mutation in a different gene. Milky and weiss have mutations in the same gene; blanc has a mutation in a different gene. All three strains have mutations in the same gene.

Blanc and weiss have mutations in the same gene; milky has a mutation in a different gene.

Part I: In humans, the genes for red-green color blindness (C = normal, c = colorblind) and hemophilia A (H = normal, h = hemophiliac) are both X-linked and only 3 map units apart. A normal woman has a colorblind mother (with normal blood clotting) and a hemophiliac father (with normal vision). What is the genotype (with correct arrangement of alleles) of the normal woman? C H / c h C h / c H C H / C H c h / c h

C h / c H

Pick two applications of the Basic Local Alignment Search Tool (BLAST). Find nearby genes on the same chromosome Display all the open reading frames (ORF) in the query sequence Compare a nucleotide or amino acid sequence to database Find the direction of transcription Find a best match sequence in database Determine the pattern of inheritance: recessive vs. dominant mutation

Compare a nucleotide or amino acid sequence to database Find a best match sequence in database

Which of these enzymes forms phosphodiester bonds in RNA or DNA? (Select all that apply.) DNA pol III DNA pol I DNA ligase Primase RNase DNase Topoisomerase Single-strand binding protein Protease DNA helicase Telomerase

DNA pol III DNA pol I DNA ligase Primase Telomerase

How can you create overlapping fragments? Select all that apply. Digest multiple copies of the genome by sonication Digest multiple copies of the genome to completion with a restriction enzyme Digest a single copy of the genome by sonication Digest a single copy of the genome to completion with a restriction enzyme Digest multiple copies of the genome partially with a restriction enzyme Digest a single copy of the genome partially with a restriction enzyme

Digest multiple copies of the genome by sonication Digest multiple copies of the genome partially with a restriction enzyme

Which of these is an example for how the same genotype can give rise to different phenotypes in population? - There are 50 different genes that generate the developmental pathway for hearing; nonfunctional mutation at any one of these genes can result in deafness. - The human histocompatibility gene has hundreds of different alleles that generate phenotypic variation of cell surface molecules that distinguish self from foreign cells. - Even when two siblings inherit the same mutant allele that predisposes to heart disease, diet and exercise can influence the likelihood of developing the disease. - A dominant mutation in a gene important for bone growth causes dwarfism; the same gene is also critical for viability of the organism. - Interaction of multiple pairs of incompletely dominant alleles can produce continuous distribution of phenotypes such as skin color or height.

Even when two siblings inherit the same mutant allele that predisposes to heart disease, diet and exercise can influence the likelihood of developing the disease.

In the United States, most newborns undergo a screening test for up to 60 genetic disorders. A few drops of blood are taken from each baby and the levels of various blood components are determined. The screening identifies babies who might have a genetic disorder but is not diagnostic. Further testing is required if the screen shows a blood component is out of the normal range. Which of the following technologies could potentially be a low-cost replacement for the current newborn screening and would directly test most babies for the presence of many genetic disorders? RNA seq to compare the sequence reads for the FMR-1 gene responsible for the fragile X syndrome Preimplantation embryo diagnosis, which involves genotyping single embryonic cells for particular disease gene loci prior to implantation PCR amplification and sequencing of the PKU (phenylketonuria) gene Exome sequencing and comparison with known mutations in databases SNP microarray to genotype each baby for millions of common SNPs DNA fingerprinting using CODIS SSRs

Exome sequencing and comparison with known mutations in databases

You generate overlapping fragments of the human genome and obtain the following fragments from the same DNA strand. If the top fragment is at the 5' end of the complete sequence, what is the correct order of these four fragments following it? 5' ATATATAT--------------------------ATATATAT 3' F1: 5' ATGGCATT--------------------------CCCCCCCC 3' F2: 5' CCCCCCCC--------------------------GCGGCGTA 3' F3: 5' ATATATAT--------------------------TAGCATGG 3' F4: 5' CGTATTCC--------------------------GAGAGAGA 3' Correct Order: 5' Top Fragment - ["Fragment 4", "Fragment 2", "Fragment 1", "Fragment 3"] - ["Fragment 4", "Fragment 3", "Fragment 1", "Fragment 2"] - ["Fragment 2", "Fragment 3", "Fragment 1", "Fragment 4"] - ["Fragment 2", "Fragment 1", "Fragment 4", "Fragment 3"] 3'

Fragment 3, Fragment 1, Fragment 2, Fragment 4

Why do scientists think that all forms of life on earth have a common origin? Select all that apply. Genes from different organisms have similar nucleotide sequences Different organisms have related genes that carry out similar function All living organisms have the same number of chromosomes Proteins from different organisms have similar amino acid sequences Genomes of all living organisms are similar in size All living organisms have the same number of genes All living organisms use the same genetic code

Genes from different organisms have similar nucleotide sequences Different organisms have related genes that carry out similar function Proteins from different organisms have similar amino acid sequences All living organisms use the same genetic code

Red-green color blindness is controlled by an X-linked gene in humans. The allele that causes color blindness is recessive to the allele for normal vision. A man and a woman both with normal vision are married as first cousins. The mother of the woman and the father of the man are siblings and they are both colorblind. Which one of the grandparents of these first cousins must be colorblind for sure? (Assume that the mutant allele is rare and completely penetrant, and carriers have normal vision.) Grandfather Grandmother It cannot be determined Neither grandmother nor grandfather

Grandfather

Which of these is an example of simple sequence repeat (SSR)? The most common cystic fibrosis allele has a deletion of three base pairs in the coding region of the CFTR gene This is an example of InDel/DIP. A single nucleotide insertion in an exon of a gene results in a truncated protein, which is nonfunctional Individuals with fragile X syndrome have 200 or more CGG repeats in the untranslated region of the FMR-1 gene A single base substitution in the GLI3 gene makes a nonsense mutation that can cause polydactyly

Individuals with fragile X syndrome have 200 or more CGG repeats in the untranslated region of the FMR-1 gene

A protein has the amino acid sequence of N... Ala-Gly-Leu-Trp-Val-Thr ...C. A researcher identified several mutants that all produce a shortened protein with Leu in the given sequence as the terminal amino acid. The codon for Trp (tryptophan) is 5' UGG 3'. What DNA single-base change in the Trp codon would NOT cause truncation of this protein at Leu? (You may need to look up the genetic code.) A substitution of the third G with A A substitution of the second G with A Insertion of G between T and G Insertion of A between G and G Insertion of A between T and G

Insertion of G between T and G

What pieces of evidence support that a DNA sequence is part of a protein-coding gene? Select all that apply. It has one ORF in each reading frame The putative protein sequence has no ortholog It contains sequences conserved at predicted exon-intron boundaries It contains an ORF, which is 15 bp long The DNA sequence is at least 40% identical to its closest match in human, mouse, and zebrafish, and the putative protein sequence is at least 80% identical It has a start codon and a stop codon out of frame The sequence is aligned to a cDNA clone with a high match score and a low E-value

It contains sequences conserved at predicted exon-intron boundaries The DNA sequence is at least 40% identical to its closest match in human, mouse, and zebrafish, and the putative protein sequence is at least 80% identical The sequence is aligned to a cDNA clone with a high match score and a low E-value

The following picture shows ethidium bromide-stained DNA bands revealed by gel electrophoresis of the same circular DNA digested in three different tubes containing the following restriction enzymes: EcoRI only, BamHI only, or a mixture of the two enzymes. The arrow represents the direction of electrophoresis. (FYI, a restriction enzyme digests DNA by cutting both strands of a specific, short DNA sequence it recognizes.) What is the smallest DNA fragment? F A H B E D G J C I

J (The smaller the DNA molecule, the faster it migrates through the gel.)

In corn, having ligules (L) is dominant to liguleless (l), and green leaves (G) is dominant to white leaves (g). If a pure-breeding plant that is liguleless and has green leaves is crossed to a pure-breeding plant with ligules and white leaves, their F1 offspring would have the genotype of ["Ll Gg", "Ll GG", "LL Gg", "LL GG", "llgg"] and the phenotype of ["green and ligules", "white and liguleless", "green and liguleless", "white and ligules"] .

Ll Gg; green and ligules

You want to test a hydroxylating agent that adds an -OH group for mutagenicity by doing the Ames test. Hydroxylamine adds -OH to cytosine, and hydroxylated C pairs with A instead of G. There are different types of his- auxotrophs that you can choose from in order to detect reversion to his+ prototrophs in response to a compound under test. What type of his- mutants do you need to use for testing hydroxylamine in the Ames test? Hint: Consider the base pair before and after mutation as well as the type of mutation detected by the Ames test. Single-base insertion or deletion mutant Mutant with substitution of G-C (his+) base-pair with A-T (his-) Mutant with substitution of A-T (his+) base-pair with G-C (his-) Any type of point mutants

Mutant with substitution of A-T (his+) base-pair with G-C (his-)

Which of these could be caused by mutation in the coding sequence of a gene, thereby altering the protein sequence without affecting the expression pattern? (Select all that apply.) Nonfunctional protein is made (null) A protein important for embryonic limb development is expressed throughout the lifetime of an organism (with no change in protein structure) A leg-specific protein is also made in the antenna (with no change in protein structure) An incorrectly folded protein acquires a new activity that the wild-type protein does not have Missense mutation causes the protein to lose its function at temperature above threshold Lower than normal amount of protein is made (with no change in protein structure)

Nonfunctional protein is made (null) An incorrectly folded protein acquires a new activity that the wild-type protein does not have Missense mutation causes the protein to lose its function at temperature above threshold

Which of these mutations would have the most severe effect on the protein function? Deletion of one amino acid in the unstructured/flexible region of a protein Nonsense mutation occurring near the 5' end of the coding sequence Frameshift mutation occurring near the 3' end of the coding sequence Conservative missense mutation Silent mutation

Nonsense mutation occurring near the 5' end of the coding sequence

Cisplatin is a major anticancer drug that kills cancer cells by damaging the DNA and inducing apoptosis. Cisplatin is a small and simple molecule composed of one platinum atom linked to two amides and two chlorides. In its reactive form (with two H2O molecules replacing two chlorines), cisplatin covalently binds to DNA bases, particularly purines, forming DNA adducts on the same strand (intrastrand) or on the opposite strands (interstrand), and distorts the helix. The repair mechanism that can remove the bulky DNA adducts or lesions generated by cisplatin is ["Base Excision Repair (BER)", "Homologous Recombination (HR)", "Mismatch Repair (MMR)", "Non-Homologous End Joining (NHEJ)", "Nucleotide Excision Repair (NER)"] , and ["Radio waves", "X-rays", "Gamma rays", "Ultraviolet (UV)"] is the type of radiation that can be tolerated by cells resistant to cisplatin due to effective repair.

Nucleotide Excision Repair (NER); Ultraviolet (UV)

When the his- Salmonella strain used in the Ames test is exposed to substance X, no his+ revertants are seen. However, when rat liver enzymes are added to the cells along with substance X, his+ revertants do occur. Rat liver enzymes alone cannot induce reversion. How can you explain this result? Rat liver enzymes had no effect on substance X in terms of mutagenicity. Rat liver enzymes are mutagenic. Rat liver enzymes converted substance X into a mutagen. Rat liver enzymes converted a mutagenic substance X into a benign substance.

Rat liver enzymes converted substance X into a mutagen.

Part III: When you go to UCSC Genome Browser and search for the aforementioned gene, the vertebrate that has a homologous gene most similar to your human gene is ____________ , and the vertebrate that has a homologous gene least similar to your human gene is ___________ . (Hint: Analyze "Multiz Alignments of 100 Vertebrates.")

Rhesus; Zebrafish

Which polymorphism is likely to cause a rare disease? Select all that apply. SNP in the start codon of a protein-coding gene InDel within the conserved sequence of a promoter A rare SNP consistently present in the affected individuals SSR in an intron of a protein-coding gene (with no effect on splicing) SSR in a DNA region (with no known function) between genes A common SNP closely linked to the disease locus SNP in a telomeric repeat sequence

SNP in the start codon of a protein-coding gene InDel within the conserved sequence of a promoter A rare SNP consistently present in the affected individuals

Which of these restriction enzymes will generate smallest number of fragments of the human genomic DNA upon complete digestion? (^ denotes the site of cleavage.) SphI (GCATG^C) HpaII (C^CGG) SbfI (CCTGCA^GG)

SbfI (CCTGCA^GG) (The larger the length of each fragment, the fewer the number of fragments produced.)

What is the best evidence that shows each nucleotide is part of only one codon (nonoverlapping codons)? Single base substitutions change only one amino acid. Artificial messages with different base sequences give rise to different proteins. A single-base insertion followed by a single-base deletion produces the wild-type phenotype. Different point mutations that affect the same codon can recombine to give the wild-type codon. A gene's nucleotide sequence is colinear with the corresponding amino acid sequence. Different codons can specify the same amino acid.

Single base substitutions change only one amino acid.

Which of these is transcribed to RNA from DNA? (Select all that apply.) Stop codon poly-A tail Enhancer within a gene Start codon Exon Promoter 5' cap Intron Enhancer outside a gene

Stop codon Enhancer within a gene Start codon Exon Intron

Auxotrophic mutant strains of Neurospora are grown in minimal media with supplements as follows. Strains may carry more than one mutation. Growth is shown by (+) and no growth is shown by (-). Below the table is the biochemical pathway for the synthesis of arginine required for growth. Both strains a and b accumulate ornithine. Indicate the statement that is most correct regarding these two strains. Strain a has a mutation in ARG-H only. Strain a has mutations in ARG-F and ARG-H. Strain a has mutations in ARG-G and ARG-H. Strain a has a mutation in ARG-G only. Strain a has a mutation in ARG-F only. Strain b has a mutation in ARG-G. Strain b has two mutations.

Strain a has mutations in ARG-F and ARG-H.

A partial sequence of one DNA strand within a coding exon of a yeast gene is shown. Is this the template strand or the RNA-like strand? (You may need to look up the genetic code.) Hint: Translate the sequence in all possible reading frames. 5' ...GTAAGTTAACTTTCGACTAGTCCAGGGT... 3' RNA-like strand Template strand

Template strand

Griffith worked with two types of S. pneumoniae bacteria - the virulent smooth forms (S) and the non-virulent rough (R) forms. Neither the heat-killed S strain nor the live R strain produced infection when injected into laboratory mice, but a mixture of the two killed the animals. What was Griffith's conclusion that was followed up by three other scientists? Transformation did not occur, but mice died of secondary infection. The heat-killed S strain was revived through transformation. The R strain was transformed into the S strain due to genetic material from the heat-killed S. The heat did not completely kill the S strain. The R strain was transformed into the S strain due to mutation caused by heat.

The R strain was transformed into the S strain due to genetic material from the heat-killed S.

Two alleles of gene C control hair color in horses: C¹ and C². Horses homozygous for C¹ allele are red, heterozygotes are yellow, and C² homozygotes are cream. Assuming C¹ makes an enzyme that produces red pigment, and C² makes a nonfunctional enzyme, what is a plausible molecular explanation for the inheritance of hair color in horses? There is no functional red-producing enzyme made in yellow horses The amount of functional red-producing enzyme made in heterozygotes is not enough to make them red Heterozygotes make higher amount of functional red-producing enzyme than either homozygote Yellow and red horses produce equivalent amount of functional red-producing enzyme Cream horses produce the most amount of functional red-producing enzyme

The amount of functional red-producing enzyme made in heterozygotes is not enough to make them red

Part II: In some genetically engineered corn plants, a Bt gene was added to a chromosome. The Bt gene specifies a protein called Bt that is lethal to certain flying insect pests that eat the corn plants. How would the amount of Bt protein in a corn plant homozygous for the Bt gene compare to that in a hybrid (heterozygote with just one copy of the Bt gene) assuming gene expression is comparable for each transgene present? (Hint: How does the amount of gene product - protein in this case - correlate with the number of functional gene copy?) Two types of corn plants (homozygous transgenic and hybrid) would have equal amount of Bt protein expressed in cells The homozygous transgenic plant is expected to have more Bt protein than the hybrid The amount of gene product never correlates with the amount of functional gene copy The hybrid plant is expected to have more Bt protein than the homozygous transgenic

The homozygous transgenic plant is expected to have more Bt protein than the hybrid; The greater the number of functional gene copy, the greater the amount of gene product.

Cisplatin is a major anticancer drug that kills cancer cells by damaging the DNA and inducing apoptosis. ERCC1 is one of the genes involved in repair of DNA damage induced by cisplatin. A patient's response to cisplatin chemotherapy depends on the sensitivity of tumor cells to cisplatin-induced DNA damage. What is the relationship between the level of ERCC1 gene expression and susceptibility of tumor cells to cisplatin chemotherapy? Note that cells, including cancer cells, subject to extensive amounts of DNA damage can undergo apoptosis. The level of ERCC1 gene expression in tumor cells is higher in patients who respond poorly to the cisplatin treatment (cancer cells NOT killed efficiently). The level of ERCC1 gene expression in tumor cells is higher in patients who respond well to the cisplatin treatment (cancer cells killed efficiently).

The level of ERCC1 gene expression in tumor cells is higher in patients who respond poorly to the cisplatin treatment (cancer cells NOT killed efficiently).

Part I: In rats, two genes control the coat pigmentation: B and P. The allele for black pigmentation (B) is dominant to the allele for cream pigmentation (b). However, in the absence of the dominant P allele, rats are albino regardless of the alleles at the other gene (B/b). When a pure-breeding, homozygous black rat was crossed to a homozygous albino, all of the F1 rats were black. When the F1 rats were crossed to each other, the F2 consisted of 161 rats: 93 black, 30 cream, and 38 albino. How do these results suggest that the coat color is determined in rats? - The dominant allele of one gene masks the effect of a second gene - The recessive allele of one gene masks the effect of a second gene - Dominant alleles of both genes are necessary for pigmentation - A dominant allele of either gene is sufficient for pigmentation - The dominant allele of each gene determines a distinct color, and yet another color results from having dominant alleles of both genes. No color forms in absence of any dominant alleles

The recessive allele of one gene masks the effect of a second gene

A student analyzes a genomic DNA sequence of 100 nucleotides in order to determine if it is part of a gene. When she translates the sequence beginning with the first three nucleotides at the 5' end of one strand as a triplet, she finds that the tenth codon is TAA (a stop codon). What should she (a sensible genetics student) do next? Tell her professor that the sequence encodes a protein with nine amino acids Tell her professor that the sequence does not represent an exon Ignore the stop codon and keep translating in the same reading frame Translate the sequence from the opposite end (3' to 5') of the same DNA strand Translate the sequence in the five other reading frames

Translate the sequence in the five other reading frames

Two short-haired cats mate and produce six short-haired and two long-haired kittens. What does this information suggest about the inheritance of hair length? Select all that apply. Two long-haired kittens are pure-breeding The long hair allele is dominant to the short hair allele Two long-haired kittens can each produce two genetically different gamete types Two short-haired cats that mated can each produce two genetically different gamete types Two short-haired cats that mated are pure-breeding There is only one possible genotype for short-haired kittens Six short-haired kittens have the same genotype

Two long-haired kittens are pure-breeding Two short-haired cats that mated can each produce two genetically different gamete types

Which of the following is true? Splicing can occur precisely due to the similar length of all introns Usually, only one DNA strand is transcribed for a gene RNA polymerase moves in the 5'-to-3' direction on the template strand Promoter is located downstream of a gene it regulates The longer the open reading frames (ORF), the more numerous the stop codons UTRs (untranslated regions) are made of intron sequences

Usually, only one DNA strand is transcribed for a gene

The following tree shows the evolutionary history of human hemoglobin and myoglobin genes, which are homologous. The different forms of hemoglobin are adapted to most efficiently carry oxygen in different stages of development: embryonic, fetal, and adult stages. How is this functional diversification made possible? 5 functional alpha-globin genes are together on one chromosome whereas 5 functional beta-globin genes are together on another chromosome About 500 million years ago, an ancestral globin gene duplicated and over time, the two diverged, through differential gene expression pattern, to give rise to an alpha-lineage and a beta-lineage When many copies of the globin gene formed after several rounds of duplication events, independent mutations allowed some globin genes to assume new specialized but related functions Different hemoglobin forms are identical in sequence, and their diverse functions can be solely attributed to post-translational modifications (e.g. phosphorylation, glycosylation, cleavage) Different hemoglobin forms are a result of alternative splicing of a common primary transcript

When many copies of the globin gene formed after several rounds of duplication events, independent mutations allowed some globin genes to assume new specialized but related functions

Achondroplasia is a form of dwarfism in humans. It is caused by a mutant allele of the fibroblast growth factor receptor (FGFR) gene that produces an overactive protein. Having one copy of the mutant allele results in dwarfism. Having two copies of the mutant allele results in stillbirth. (Infant dies in the womb.) Which allele, the mutant allele or the wild-type allele, is dominant with respect to viability? Wild-type allele Mutant allele

Wild-type allele

What is the most likely mode of inheritance if the trait displayed in this pedigree is rare and completely penetrant? X-linked dominant Autosomal dominant Autosomal recessive X-linked recessive

X-linked recessive

The term mutation refers to only changes in DNA resulting in new phenotypes. only changes in DNA resulting in increased or decreased fitness of an organism. any changes in DNA that are repaired prior to replication. only changes in DNA resulting in novel proteins or altered amount of proteins. any heritable changes in DNA regardless of their phenotypic effects. only changes in DNA resulting in loss or gain of gene function.

any heritable changes in DNA regardless of their phenotypic effects.

Part II (Do Part I first): A wild-type (denoted by +) female mouse is crossed to a male mouse with apricot eyes (ap) and gray body (gy). The F1 mice are all wild-type for both traits. When the F1 are interbred, 200 F2 progeny are distributed as follows: Females: All wild-type = 100 Males: wild-type = 46 apricot = 6 gray = 4 apricot, gray = 44 Which of the following statements is correct? Note: RF is calculated by scoring only those progenies that can be identified as parental vs. recombinant. - ap and gy are X-linked and 10 map units apart - ap and gy are unlinked - ap and gy are autosomal and 10 map units apart - ap and gy are autosomal and 5 map units apart - ap and gy are X-linked and 5 map units apart

ap and gy are X-linked and 10 map units apart

Part III (Do Part II first): In rats, two genes control the coat pigmentation: B and P. The allele for black pigmentation (B) is dominant to the allele for cream pigmentation (b). However, in the absence of the dominant P allele, rats are albino regardless of the alleles of the other gene (B/b). When a pure-breeding, homozygous black rat was crossed to a homozygous albino, all of the F1 rats were black. When the F1 rats were crossed to each other, the F2 consisted of 161 rats: 93 black, 30 cream, and 38 albino. What is the genotype of the albino parent? bbpp bbPP BbPp Bbpp BBPP BbPP BBPp BBpp bbPp

bbpp

In the common daisy, genes col-1 and col-2 control flower color. Function of both genes (col-1 and col-2) is required to have color and is conferred by the wild-type allele of each gene. At least one wild-type copy (allele) of each gene is required for flowers to be colorful instead of white. What gene interaction is illustrated here? - complementary gene action (reciprocal recessive epistasis) - additive interaction - recessive epistasis - redundancy - dominant epistasis

complementary gene action (reciprocal recessive epistasis)

Part I: Below is a pedigree of a human genetic disease in which solid color indicates affected individuals. Assume that the disease is caused by a gene that has two different alleles A and a, and the disease allele is rare. Based on this pedigree, the most likely mode of inheritance of the disease is ____________ , and the genotype of the female 4 is _____.

dominant, Aa

If a chemical inhibitor of DNA ligase was added to bacterial cells after initiation of DNA replication, the outcome would be our genome will contain both DNA and RNA nucleotides. elongation of both leading and lagging strands would not occur at all. elongation would occur, but Okazaki fragments would not be joined together. DNA would get shorter in each replication cycle by the length of a primer. DNA would be supercoiled ahead of the replication fork and replication would not proceed to completion. replication would proceed in only one direction from the origin and would take twice longer.

elongation would occur, but Okazaki fragments would not be joined together.

The following tree shows the evolutionary history of human hemoglobin and myoglobin genes, which are homologous. The gene that would have the greatest degree of nucleotide similarity to the human beta (β) gene is the ______________ gene, and the two genes are ____________ .

human delta (δ); paralogous

Temperature of Melting (Tm) is defined as the temperature at which 50% of double-stranded DNA is changed to single-stranded DNA. Tm depends on both the length and the nucleotide sequence composition of the molecule (GC content as opposed to AT). Tm will rise with ["increase in length of DNA", "decrease in length of DNA"] and ["increase in AT content", "increase in GC content"] .

increase in length of DNA; increase in GC content

Part I: Go to UCSC Genome Browser: https://genome.ucsc.edu/cgi-bin/hgGatewayLinks to an external site. Under Position/Search term, type F8 (pick 'Homo sapiens coagulation factor VIII' on top) and Go. The F8 gene is located on the["long (q) arm", "short (p) arm"] of chromosome X, and the direction of transcription is ["away from", "toward"] the centromere (marked by two triangles).

long (q) arm; toward

Meselson and Stahl relied on equilibrium density gradient centrifugation to resolve the DNA containing 14N from the DNA containing 15N. They started off with DNA in media containing the heavy isotope, then switched to media containing the lighter isotope and allowed DNA replication to take place. Under semiconservative model of DNA replication, after THREE rounds of replication, there will be two bands of DNA, which are ________________ , and the ___________ will contain more DNA molecules.

one intermediate and one light; light band

Nondisjunction can occur at either the first or the second division of meiosis. XXX individuals would arise from nondisjunction at the ______ meiotic division in the ______. Select all that apply. (Note: Assume that only one parent experiences meiotic nondisjunction while meiosis occurs normally in the other parent.) first; father second; father first; mother second; mother

second; father first; mother second; mother

Does KpnI that cuts 5' GGTAC^C 3' make sticky ends (with what overhangs) or blunt ends? (^ denotes the site of cleavage.) blunt ends sticky ends with 3' overhangs sticky ends with 5' overhangs

sticky ends with 3' overhangs

In the process of homologous recombination, (Select all that apply.) Correct! the first step is a double-strand break in one chromatid. gene conversion always occurs whenever cells repair DNA mismatch within a heteroduplex. Correct Answer DNA strands from nonsister chromatids are able to base-pair with each other through sequence homology. Correct! a heteroduplex, consisting of one strand from each nonsister chromatid, forms in each chromatid involved. Correct! phosphodiester bonds break and reform as the Holliday junctions are resolved. enzymes are not required because the process occurs spontaneously during meiotic prophase. You Answered the entire DNA molecule becomes single-stranded before the two strands re-associate.

the first step is a double-strand break in one chromatid. DNA strands from nonsister chromatids are able to base-pair with each other through sequence homology. a heteroduplex, consisting of one strand from each nonsister chromatid, forms in each chromatid involved. phosphodiester bonds break and reform as the Holliday junctions are resolved. (Only a small part of DNA becomes single-stranded shortly after a double-strand break.)