SAE - T or F

Question 1

Short-winged purebred Drosophila females are mated to long-winged purebred Drosophila males. The wing length is determined by a single gene named W and the long-winged is dominant to the short-winged. If W is on the X chromosome, 50% of the F1 females will have long wings and the other 50% of F1 females will have short wings

Answer 1

F

Question 2

You have two purple flower pea plants in which purple is dominant over white. One plan is homozygous of the purple allele while the other is heterozygous of purple and white. If testcross is not possible, self-pollination can determine which one is homozygous and which one is heterozygous

Answer 2

T

Question 3

Pea plants that Mendel used have 7 pairs of chromosomes, 14 in total. The number of possible gamete genotype that the pea plants can produce solely via independent assortment is 2 to the power of 14.

Answer 3

F

Question 4

Molecular markers on the same chromosome will always segregate together from one generation to the next.

Answer 4

F

Question 5

Heterozygous animals of a mutant allele have no obvious phenotype (appears wild type). However, the same mutant allele is recessive lethal. Self-crossing of the heterozygous animals will result in the offspring that will all appear wild type (no phenotype).

Answer 5

T

Question 6

In the complementation test, two recessive mutant alleles of the same gene will successfully complement each other.

Answer 6

F

Question 7

If a is epistatic to b, the phenotype of the a, b double mutant will be the same as that of a.

Answer 7

T

Question 8

Autosomal recessive polycystic kidney disease is a severe genetic disorder characterized by numerous cysts in affected patients. The frequency in the general population is 1 out of 10,000 individuals. If we assume Hardy-Weinberg proportions for the genotypes, then the frequency of carriers is 0.04.

Answer 8

F

Question 9

A population has a SNP with genotypes G/G, G/A, and A/A. The frequencies are of the genotypes are 0.50, 0.40, 0.10, respectively. These are Hardy-Weinberg proportions.

Answer 9

F

Question 10

If a female lizard mates with one of its offspring, the offspring will have an inbreeding coefficient of 0.5.

Answer 10

F

Question 11

The relative fitnesses of the genotypes B/B, B/b, and b/b are 1.0, 1.0, and 0, respectively. Therefore, if the present genotype frequencies are 0.5, 0.4, and 0.1, respectively, the genotype frequencies after selection will be 0.65, 0.35, and 0.0.

Answer 11

F

Question 12

Over time in a large population, genetic drift can lead to an increase in genetic diversity

Answer 12

F

Question 13

The population mean of a quantitative trait with a broad sense heritability of 0 can be changed by natural or artificial selection provided the environmental variance is greater than 0

Answer 13

F

Question 14

When the correlation of trait values is measured for a set of twins separated at birth, the correlation can be equated to the broad sense heritability for the trait. But if the trait is influenced in the same way by the shared in-uteroenvironment, the estimate of heritability for the trait will be biased downward.

Answer 14

F

Question 15

The heritability value of a trait is a characteristic of that trait and is affected only by the genetic make-up of the population.

Answer 15

F

Question 16

The rutabega (B. napus) is a fertile allopolyploid that arose in the wild through hybridization of two different parent species: B. oleracea (2n = 18) and B. campestris (2n = 20). In a breeding experiment, you cross this rutabega plant to B. oleracea and then examine the chromosomes in the F1 progeny.

True or false: The somatic cells of the F1 progeny will contain 19 chromosomes.

Answer 16

F

Question 17

Trisomic individuals produce normal haploid gametes very rarely, with a frequency that depends on their “n”.

Answer 17

F

Question 18

The early studies that led to the discovery of Drosophila P-elements involved crossing wild-caught Drosophila with lab strains.

True or false: When wild-caught males are crossed to lab-strain females, the resulting F1 progeny do not survive.

Answer 18

F

Question 19

In C. elegans, when the mobility of a Tc1 transposon inserted into the unc-22 locus is silenced, the twitching mutant phenotype is reverted to normal.

Answer 19

F

Question 20

After mutagen treatment, a molecule of 2-aminopurine (an adenine analogue) incorporates into DNA. During replication, the 2-AP protonates and, in this form, bonds like guanine.

True or false: The mutational event caused by this will be AŸT to CŸG.

Answer 20

F

Question 21

A deletion mutation in a gene can result in the production of a larger protein product.

Answer 21

T

Question 22

Strand slippage during DNA replication can lead to either insertion or deletion mutations.

Answer 22

T

Question 23

Analysis of cDNA sequences can provide evidence that a conserved region of non-coding sequence is an exon of a functional gene.

Answer 23

T

Question 24

True or False : The activator protein CAP gets turned off when glucose levels are high because it is inhibited by glucose binding to its allosteric site.

Answer 24

False
CAP is not directly inhibited by glucose. Instead, high glucose lowers intracellular cAMP levels. Since CAP requires cAMP to bind
DNA and activate transcription, it becomes inactive indirectly in high-glucose conditions, not through glucose bindin

Question 25

True or False : The IS allele in E. coli acts dominantly and in trans.

Answer 25

True The IS (superrepressor) mutation produces a repressor protein that cannot be inactivated by lactose. It acts in trans because it can diffuse throughout the cell and repress any lac operon, and it acts dominantly over a wild-type allele by continuously repressing transcription

Question 26

True or False : Regulation of the Lac operon in response to lactose is an example of positive gene regulation.

Answer 26

False Lactose removes repression by binding to the repressor and preventing it from blocking transcription. This is an example of negative regulation, not positive activation

Question 27

True or False : Mutating an insulator protein that binds to barrier insulator sequences would lead to a general increase in gene transcription.

Answer 27

False Barrier insulators block the spread of heterochromatin. If the insulator is mutated, heterochromatin can spread into active gene regions, leading to gene silencing rather than activation

Question 28

True or False : Unlike the operator in E. coli, a silencer in eukaryotic cells can act independent of its position relative to the gene it regulates.

Answer 28

True Operators in prokaryotes must be positioned near the promoter to function. In contrast, eukaryotic silencers can repress gene expression from a distance, regardless of whether they are upstream, downstream, or within intro

Question 29

True or False : The phenotypic differences between a wild-type fly and a fly showing position-effect variegation are the result of both genetic and epigenetic changes.

Answer 29

True The gene's relocation near heterochromatin is a genetic change, while the silencing effect due to heterochromatin spreading is epigenetic

Question 30

True or False : DNA methyltransferases, histone H3K9 methyltransferases, histone acetylases, and HP-1 are all associated with gene silencing.

Answer 30

False Histone acetylases (HATs) are associated with gene activation because they open chromatin structure. The other proteins listed (DNA methyltransferases, H3K9 methyltransferases, HP-1) do promote gene silencing

Question 31

True or False : Woman affected by a disease associated with mutation of an X-chromosome gene will always pass on the disease to their sons.

Answer 31

False Not always. Whether a son inherits the disease depends on whether the mutation is dominant or recessive, and whether the mother is heterozygous or homozygous. Additionally, X-inactivation can influence gene expression, leading to variable outcome

Question 32

Alpers syndrome is a mitochondrial disease caused by mutations in the POLG gene, located on chromosome 15. This gene encodes a subunit of the DNA polymerase that replicates the mtDNA in mitochondria. TRUE or FALSE: Mutations in the POLG gene will have a maternal effect, whereby mothers affected by Alpers syndrome will pass on the disease to all their children.

Answer 32

F

Question 33

The gap gene products Giant and Krüppel work through the stripe 2 enhancer to prevent activation of eve by Bicoid and Hunchback in a spatially-dependent pattern. TRUE or FALSE: The main effect on eve stripe 2 in a giant mutant embryo would be an increase in the level of eve transcription within stripe 2.

Answer 33

F

Question 34

TRUE or FALSE: Ras mutations found in cancers increase the GTPase activity of Ras

Answer 34

F

Question 35

TRUE or FALSE: The reciprocal translocation frequently observed in Burkitt lymphoma, t(8:14), promotes cancer development in B-cells because of the enhancer activity of Myc.

Answer 35

F

Question 36

TRUE or FALSE: Although p53 is a tumor suppressor gene, cancer cells could have a wild-type allele of the p53 gene.

Answer 36

T

Question 37

TRUE or FALSE: The viral oncoproteins of DNA tumor virus originated from the human genome.

Answer 37

F

Question 38

TRUE or FALSE: ADA-SCID patients treated with gene therapy (the first disease treated with gene therapy) developed leukemia due to a vial oncogene introduced by the viral vector.

Answer 38

F

Question 39

TRUE or FALSE: Children of a patient treated with a somatic cell gene therapy will not benefit from the therapy.

Answer 39

T