Bio 97 Midterm 1

Suppose you provide an actively dividing culture of E. coli with radioactive dTTPs, which can be incorporated into newly synthesized DNA. What would you expect to see if the cell replicates its DNA and divides once in the presence of the radioactive base?

-DNA in both daughter cells will be radioactive

-Radioactive thymine will pair with non-radioactive guanine in both daughter cells

-All of the bases of the DNA would be radioactive

-One of the daughter cells but not the other would have radioactive DNA

-Neither daughter cell would have radioactive DNA

DNA in both daughter cells will be radioactive

(DNA replication is semiconservative, so both copies of the genome will have radioactive thymine incorporated. This concept is mentioned in the learning objectives and covered in the reading guide for this lecture)

The polymerase chain reaction (PCR) is used to amplify fragments of DNA. During a PCR reaction, what breaks hydrogen bonds between nucleotides?

-72 degree C primer extension

-ddNTPs

-55 degree C primer annealing

-Taq DNA polymerase

-95 degree C incubation

95 degree C incubation

(The first phase of PCR is the denaturation stage, where high temperature is used to "melt" the double stranded DNA into single strands, thereby breaking the hydrogen bonds that keep the strands together)

You are setting up a dideoxysequencing reaction. What is the correct ratio of dNTPs (deoxyribonucleotides) to ddNTPs (dideoxyribonucletides)?

-Adding more dNTPs than ddNTPs

-Adding equal amounts of dNTPs and ddNTPs

-Adding more ddNTPs than dNTPs

Adding more dNTPs than ddNTPs

(In a dideoxysequencing reaction, the ddNTPs will terminate the elongation of the DNA. If you a small amount of ddNTPs, most of the time the DNA polymerase will continue elongating the new DNA with dNTPs and only occasionally incorporate a ddNTP and stop. If you have too many ddNTPs, you will only get very short pieces of DNA and not very much DNA sequence)

The sequence of the coding strand of DNA is:

5'-TTAGATGCCTGTCAAGTAACTG-3'

Assuming that is sequence for the entire mRNA transcript for this gene, how many nucleotides are the in 5' UTR?

-2

-7

-6

-4

4

(Remember that the sequence of the coding strand of DNA and of the mRNA are the same, with U's replacing T's in the mRNA. To see where translation starts, read the coding strand 5' to 3' and find the first ATG. The nucleotides 5' of the ATG will be in the 5' UTR)

Which of the following is true about promoter sequences?

-They can be read in either direction.

-They are found upstream of the transcription start site.

-They are identical for all genes in the genome.

-They are the places that the ribosome lands and tell it where to start transcribing.

They are found upstream of the transcription start site.

(Promoters are where RNA polymerase lands and direct where it needs to start transcription, are read in only one direction, and can be different from one gene to another. They are found upstream of the transcription start site)

There is a mutation in a bacterial gene's promoter that changes a nucleotide from a T to an A in a postition between the -10 and -35 elements. What does this mutation do?

-Decrease the rate of gene expression

-Increase the rate of gene expression

-Nothing

Nothing

(Mutations to the -10/-35 elements may increase or decrease gene expression. Base substitutions in between will not effect gene expression)

A part of an mRNA molecule with the following sequence is being read by a ribosome: 5' CCG-ACG 3' (mRNA). The anticodon loop of the first tRNA that will complement this mRNA is:

-5' GGC 3'

-5' UGC 3'

-3' UGC 5'

-3' GGC 5'

-5' ACG 3'

3' GGC 5'

(The first tRNA that will complement this RNA will match 5' CCG 3'. It's anticodon will be the reverse complement: 3' GGC 5')

Life as we know it depends on the genetic code: a set of codons, each made up of three bases in a DNA sequence and corresponding mRNA sequence, that specifies which of the 20 amino acids will be added to the protein during translation.

Imagine that a prokaryote-like organism has been discovered in the polar ice on Mars. Interestingly, these Martian organisms use the same DNA → RNA → protein system as life on Earth, except that

-there are only 2 bases (A and T) in the Martian DNA, and

-there are only 17 amino acids found in Martian proteins.

Based on this information, what is the minimum size of a codon for these hypothetical Martian life-forms?

-3 bases

-2 bases

-4 bases

-5 bases

5 bases

(In the most general case of x bases and y bases per codon, the total number of possible codons is equal to x^y .In the case of the hypothetical Martian life-forms, is the minimum codon length needed to specify 17 amino acids is 5 (2^5 = 32), with some redundancy (meaning that more than one codon could code for the same amino acid).For life on Earth, x = 4 and y = 3; thus the number of codons is 4^3, or 64. Because there are only 20 amino acids, there is a lot of redundancy in the code (there are several codons for each amino acid))

Bicoid is an important transcription factor for early Drosophila development. The 11th codon in its protein sequence is usually 5'-UUA-3'. A mutation has been introduced that changes the codon to 5'-UUG-3'. What is the consequence of this mutation?(NOTE: you may always use your book or the internet to find a codon table -- there is no need to memorize it!)

-The identity of the 11th amino acid in the protein will change

-Nothing, since this is a silent mutation

-A truncated 10 amino acid protein will be produced

-Nothing, since this is a nonsense mutation

-Nothing, since this is a silent mutation

(Using the codon table, you can see that both UUA and UUG encode leucine, so this is a silent mutation with no effect on the amino acid sequence of the protein Bicoid)

DNA within the cell is wrapped around nucleosomes to form chromatin. When chromatin is in the form of a 30 nm fiber, how many copies of H1 and H2A histone proteins would be found in a stretch of chromatin containing 20 nucleosomes?

-20 H1, 20 H2A

-40 H1, 20 H2A

-20 H1, 40 H2A

-40 H1, 40 H2A

-10 H1, 30 H2A

20 H1, 40 H2A

(Nucleosomes are octomers made up of 2 copies of H2A/H2B and H3/H4. In the 30 nm chromatin fibril, each nucleosome is associated with a single H1 molecule that helps the chromatin fold into the solenoid structure. Therefore 20 nucleosomes will have 20 H1 molecules and 40 H2A)

The pattern of transcription for a pair of identical twins (Tweedledee and Tweedledum) is very similar at the age of 10. However by the time they are 50, transcription of the leptin gene is much lower in Tweedledum than in Tweedledee. This could be due to what change in Tweedledum?

-A crossing over event during meiosis in their mother's egg cell

-Histone acetylation at the site of Tweedledum's leptin gene

-Spontaneous mutation in their father's sperm cell

-DNA methylation at the site of Tweedledum's leptin gene

DNA methylation at the site of Tweedledum's leptin gene

(The twins have identical genetic make ups, so changes in transcription are due to epigenetic modifications. Histone acetylation increases transcription while methylation reduces gene expression)

Housekeeping genes are genes which are expressed in virtually every cell in your body because they encode proteins required for all cells to function. For example, actin and RNA polymerase are housekeeping genes. In which type of chromatin would housekeeping genes lie?

-Constitutive heterochromatin

-Euchromatin

-Facultative heterochromatin

Euchromatin

(Genes that are actively transcribed are located in euchromatin, therefore, housekeeping genes are likely like in euchromatin)

Humans have a gene, T, that is involved in muscle formation of the tongue. Individuals homozygous for one allele can roll their tongues, while individuals homozygous for the other allele cannot. If both parents can roll their tongues, but their child cannot, what can be said about the mode of inheritance?

-Tongue rolling is dominant, and both parents were heterozygous (Tt).

-Tongue rolling is recessive, and both parents were heterozygous (Tt).

-The parents were both homozygous, but the child was heterozygous.

Tongue rolling is dominant, and both parents were heterozygous (Tt).

(If the parents can roll their tongues and the child cannot that means that tongue rolling is dominant (if we see a trait in a child and not a parent, it must be recessive). We also know the parents are both heterozygotes (Tt), otherwise, a homozygous recessive child is impossible)

Which of the following DOES NOT describe why pea plants were a good model system for Mendel?

-Pea plants have dichotomous traits with distinguishable phenotypes.

-Mendelian inheritance in pea plants follows the same principles as autosomal inheritance in humans.

-Pea plants have fewer chromosomes than humans.

-Pea plants can be artificially fertilized.

-Pea plants produce many offspring per cross, allowing for large sample sizes.

Pea plants have fewer chromosomes than humans.

(Pea plants do have fewer chromosomes than humans, but this didn't particularly make them a good model system. The other choices are all good reasons that pea plants made a good model system)

By convention, when the p-value for the difference between the observed experimental outcomes and the expected outcome is less than 5 percent (< 0.05), the experimental results are considered to be:

-not significant.

-within normal expected range.

-statistically significantly and different from the expected outcome.

-equal to the mean.

-less than one standard deviation from the mean.

statistically significantly and different from the expected outcome.

(This is a straightforward test of understanding. Please refer to pages 47-50 of the text if you weren't sure of your answer)

A scientist observes a congenital birth defect in a breed of cats and predicts that in a cross between heterozygotes, 25% of the kittens will be born with the birth defect. She surveys several litters, and finds that 44 out of 128 kittens have the defect. To see if her hypothesis is correct, she decides to use Chi-square analysis and determines that the expected number of UNAFFECTED kittens should be:

-84

-44

-104

-32

-96

96

(If 25% of the kittens are expected to be affected, 75% would be unaffected = 128 x 3⁄4 = 96)

With what form of inheritance is this pedigree consistent?

-both autosomal dominant and recessive

-autosomal dominant

-autosomal recessive

autosomal recessive

(This pedigree is only consistent with an autosomal recessive mode of inheritance. If you look at person 2's parents, neither is affected, but person 2 is. Dominant traits cannot show up in offspring without being observed in the parents)

Using the pedigree above, assume that person 1's father is a carrier. What is the probability that person 1 is a carrier?

-2/3

-1/3

-1

-0

-1/2

1

(Since the trait is autosomal recessive, you can infer that person 1's mother is a homozygote (aa), and the prompt says person 1's father is a carrier (therefore Aa).If you do the Punnett square, you can see that the possible offspring genotypes are Aa and aa, with equal frequencies. We know that person 1 isn't aa, since she doesn't show the trait, so she must be Aa, and therefore a carrier)

A mutant allele is found to have a mutation in the -35 element of the promoter region. The mutation makes the -35 element MORE SIMILAR to the -35 consensus sequence. Which term best describes this allele?

-hypermorphic

-dominant negative

-hypomorphic

-neomorphic

-amorphic

hypermorphic

(This mutation is likely to increase transcription of the gene, making it a gain-of-function mutation, specifically a hypermorphic mutation -- excessive gene action)

Crossing true-breeding strains of petunia (red × white) produces an F1 with only pink flowers. This is an example of incomplete dominance. What phenotype ratios are expected in the F2?

-all pink flowers

-1:2:1 red to pink to white flowers

-3:1 red to white flowers

1:2:1 red to pink to white flowers

(If you do a Punnett square crossing the C^R C^W flowers together, you get:1/4 C^R C^R = red flowers2/4 C^R C^W = pink flowers1/4 C^W C^W = white flowers)

Deafness is caused by recessive mutations in any one of at least five genes. Two deaf individuals have nine children, all of whom have normal hearing. Which of the following can you conclude?

-The parents have mutations in the same gene.

-The mutations are incompletely dominant to the normal allele.

-The parents have mutations in different genes.

-The mutations are codominant to the normal allele.

The parents have mutations in different genes.

(This is an example of a test for complementation. Since all of the couples offspring have the wild-type phenotype, you can conclude that the mutations that caused the deafness in the parents complement each other, and are therefore in different genes)

The chinchilla allele (cch) is an allele of the C gene. The wild-type allele (C) produces a tyrosinase enzyme that is 100% active. The chinchilla allele produces an enzyme that has less that 20% of the activity of the wild-type allele. When you cross an animal homozygous for the cch allele to an animal homozygous for the C allele, all the progeny appear wild type. This information indicates that the chinchilla allele is a ________ allele and that the wild-type allele is ________.

-Hypomorphic; haplosufficient

-Hypermorphic; haploinsufficient

-Hypomorphic; haploinsufficient

-Hypermorphic; haplosufficient

Hypomorphic; haplosufficient

(The chinchilla allele is hypomorphic, as it produces an enzyme with reduced (but not no) activity. The wild-type allele must be halposufficient, that is what enables the heterozygous progeny (Ccch) to show the wild-type phenotype)

In the 1950's, a Russian fox-fur-farmer wanted to domesticate the silver fox. In each generation, he selected the friendliest foxes and bred them together. In about 20 generations, he had successfully bred domesticated foxes, which in addition to being friendly, also developed other traits common in domesticated animals, like floppy ears and a curly tail, which he had not specifically selected for. One hypothesis (A) to explain this observation is that there was a single mutation that controlled the behavior of a special population of cells, the neural crest cells, which underlie the formation of the ear and tail cartilage and a part of the brain that controls aggressiveness. Another (B) is that "conditions of living" for domestic animals, in particular, better diets, induced these traits. These hypotheses are consistent with the ideas of:

-Hypothesis A: epistasis; hypothesis B: gene-environment interactions

-Hypothesis A: variable expressivity; hypothesis B: epistasis

-Hypothesis A: variable expressivity; hypothesis B: pleiotropy

-Hypothesis A: pleiotropy; hypothesis B: gene-environment interactions

-Hypothesis A: epistasis; hypothesis B: pleiotropy

Hypothesis A: pleiotropy; hypothesis B: gene-environment interactions

(Hypothesis A is consistent with pleiotropy - a single mutation having many phenotypic effects. Hypothesis B is consistent with a gene-environment interaction - the environmental conditions affecting the phenotype caused by a genotype. This is a real story, and hypothesis B was first suggested by Darwin to explain commonly observed features of domestic animals)

The MN blood group in humans is an autosomal trait with only two alleles, M and N. M and N are co-dominant to each other. If a man with blood type N has a child with a woman who is blood type MN, what is probably that their child has blood type M?

-0%

-25%

-50%

-100%

-75%

0%

(If M and N are co-dominant, the male has the genotype NN and the female has the genotype MN. Their offspring can either have type N or type MN blood, but cannot have the blood type M)

Parakeet feather color is controlled by two genes. Yellow pigment is synthesized by the gene product of the dominant allele Y, and blue pigment is synthesized by the gene product of the dominant allele B. B and Y are independently assorting. Parakeets that can't produce either pigment are white, those that can only produce blue pigment are blue, those that can only produce yellow pigment are yellow, and those that produce both are green. You cross a green and a yellow parakeet and get green, blue, yellow, and white birds. What at the genotypes of the parents?

-BbYy x Bbyy

-BBYY x bbYY

-BBYy x bbYy

-BbYy x bbYy

-BbYy x bbyy

BbYy x bbYy

(Based on the description, the green parent must be B_Y_ and the yellow parent must be bbY_. Since all colors of offspring are observed, both parents must be heterozygous for the dominant traits they show)

The result of the cross between the green and yellow parakeet above is 12 green, 4 blue, 13 yellow, and 3 white. Calculate the chi-square value for these data, using the description of inheritance described above.

0.33

(Using the description above, you can do a Punnett square are figure out that the predicted phenotypic ratio is 3:1:3:1 for green:blue:yellow:white. The expected numbers of offspring are 12:4:12:4. The chi-square = (12-12)2/12 + (4-4)2/4+ (13-12)2/12 + (3-4)2/4 = 0.33)

How many degrees of freedom are associated with the chi-square test above?

3

There are four categories, so df = 4-1 = 3.