Biol 103 Final

0.0(0)
studied byStudied by 0 people
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/161

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

162 Terms

1
New cards

During the process of gastrulation, 

 

a. cells change shape.

Correct!

 

b. cells divide.

Correct!

 

c. cells migrate.

Correct!

 

d. cells signal to one another via signaling molecules.

2
New cards

Cytoplasmic determinants in the egg: 

b. Can influence gene expression

Correct!

 

c. Can consist of mRNAs

3
New cards

Cortical rotation in frog eggs: 

a. Is initiated by sperm entry

4
New cards

Inductive events: 

a. Can rely on receptor-ligand interactions

Correct!

 

b. Involve cell to cell signaling

Correct!

 

c. Are important in many developmental processes

Correct!

 

d. Can initiate changes in signal transduction cascades

5
New cards

One of the roles of the normal APC protein is regulate cell division by acting as a checkpoint to control the transition from G1 to S phase. Loss-of-function mutations in the APC gene are associated with colorectal cancer. Given this information, mark all of the following answers that you think are correct:

a) APC is a tumor suppressor gene

) There are sporadic and familial forms of APC-associated colorectal cancer

6
New cards

Axolotls are a great model organism to study regeneration because, after amputation, they can grow back complex structures such as limbs. You are studying a gene, Bmp-2, and you believe it is important for limb development.  You decide to do an in-situ hybridization experiment to “catch” the expression of Bmp-2 during limb regeneration and you get the following results, with panels from left to right at successively older times in regeneration. Based on these observations, formulate a hypothesis on how Bmp-2 affects digit formation (1-2 sentences):

The Bmp-2 Gene regulates apoptosis in the digits of the axolotl. The areas where Bmp-2 is expressed are areas that undergo programmed cell death, which allow the shaping of the axolotls digits.

7
New cards

To test your hypothesis in Question 6, you ectopically overexpress Bmp-2 by implanting beads secreting Bmp-2 into the distal end of an amputated limb (D shows position of Bmp-2 secreting beads, G shows control beads).  E and H show the results of the implantation after regeneration has occurred. F and I show corresponding X-rays to visualize the bone growth in the regenerated limbs. Does this result support your hypothesis? Why? (~2 sentences)

Yes, the Bmp-2 secreting beads caused mutations in the formation of the digits, specifically, they did not form at all. If Bmp-2 is responsible for causeing cell death it would make sense that no cells could grow out of the area where Bmp-2 is over expressed.

8
New cards

In the movie ‘Waterworld’, the Arctic and Antarctic ice caps all melt and the earth is under water. People have to build dry land to survive. Kevin Costner, as a lonely mariner, comes to dry land one day. To people’s surprise, Costner has webbed feet! Based on your above hypothesis, can you postulate how he developed his webbed feet, and the potential role of Bmp-2 in the process? (1-2 sentences)

If the world had been underwater for millions of years, there is a possibility that individuals with a loss of function in their BMP-2 gene had a reproductive advantage and were able to produce offspring with this same genetic mutation. So, Costner has webbed feet from a mutation in his BMP-2 gene that stops it from killing cells in between digits.

9
New cards

The LIN-3 gene in C. elegans encodes an extracellular ligand produced by the anchor cell but influences vulval development. Which of the following approaches would allow you to determine if this extracellular ligand is sufficient for vulval development?

Correct!

 

a. Overexpress the LIN-3 gene in all cells of an otherwise normal animal.

10
New cards

What is the temporal order of these events? 

 

a. ___ Neurulation

b. ___ Fertilization

c. ___ Gastrulation

d. ___ Organogenesis

b,c,a,d

11
New cards

The process of neurulation involves: 

The folding of the neural plate to form the neural tube

Signaling from the notochord to the neural tube

12
New cards

The promoter sequence of a gene: 

Is where the RNA polymerase binds

. Is where the TATA box is

13
New cards

Homeotic genes: 

a. Are found in both animals and plants

Correct!

 

b. Encode DNA binding proteins

Correct!

 

c. Act to define specific identities along the body axis

Correct!

 

d. When mutated, cause one body part to become transformed into another.

14
New cards

You identify a new type of rose that makes only petals in its flowers.  Given the ABC model, what would you predict to be the types of alterations that would lead to this phenotype? 

d. A loss of function mutation in the C gene and ectopic expression of the B gene

15
New cards
<p><span>Recall that the chicken forelimb has three digits, d1, d2, and d3, from anterior to posterior (see figure).&nbsp; If you treat a developing chicken forelimb bud with cyclopamine (an inhibitor of SHH signaling), what would you expect to see as a</span></p>

Recall that the chicken forelimb has three digits, d1, d2, and d3, from anterior to posterior (see figure).  If you treat a developing chicken forelimb bud with cyclopamine (an inhibitor of SHH signaling), what would you expect to see as a

d. A single digit d1

16
New cards

You inject cycloheximide, an inhibitor of protein synthesis, into the anterior end of a newly fertilized Drosophila egg.  Would you predict to see the normal distribution of hunchback (hb) transcripts in the resulting blastoderm embryo?  Why or why not? (~ 2 sentences) 

the distribution of hb transcripts will be affected because cycloheximide prevents the translation of maternal Bcd mRNA into protein, which is necessary to induce zygotic hb transcription in the anterior of the embryo.

17
New cards

Questions 7-10 rely on the data presented in this figure, depicting (top) an in situ hybridization for the even skipped (eve) gene in a syncytial blastoderm fly embryo showing the seven stripes of expression along the anterior-posterior axis, (middle) a graph schematizing the eve mRNA expression pattern, and (bottom) the protein expression patterns hb (in green), kni (in red), and gi (in blue) aligned with the upper panels.

b. Catch it

18
New cards

From the information in Question 7, how do you think the expression of even skipped stripe 5 is established?  

. Through kni activation and gi repression

19
New cards

Given the information in Question 7, in an animal mutant for kni, where would you expect to see the expression of a GFP transcriptional gene fusion containing the stripe 5 enhancer sequence?

. No expression

20
New cards

You generate a GFP transcriptional gene fusion that contains the enhancer sequences for stripe 5+stripe 7 and introduce it into a wild type animal. You then carry out an immunolocalization experiment to visualize eve protein; where would you expect to see the signal? 

In stripe 5+stripe 7

21
New cards

Mark those sentences that are correct:

 

Epigenetic modifications involve mutation of the RNA in only some cells and not others.

 

Histone deacetylation is one form of regulating chromatin condensation.

 

Environmental impacts such as diet can affect the methylation status of DNA.

 

Condensed chromatin is accessible for transcription.

Epigenetic modifications involve mutation of the RNA in only some cells and not others.

 

Histone deacetylation is one form of regulating chromatin condensation.

22
New cards

Mark those sentences that are correct:

 

a. A missense mutation causes the formation of a chiasma.

 

b. Insertions of transposable elements can cause phenotypic effects through disrupting the coding region of a given gene.

 

c. Meiosis of a diploid cell results in the formation of two haploid daughter cells.

 

d. All transposable elements encode a transposase enzyme to promote their movement.

a. A missense mutation causes the formation of a chiasma.

23
New cards

You decide to knock out the function of the ced9 gene in the worm C. elegans, using an RNAi approach. Recall that the ced9 gene product is required to repress apoptosis.  You design a double stranded RNA for this experiment and introduce it into worms; which of the following molecules would be targeted by your double stranded RNA? Mark all that are correct:

b. ced9 mRNA

24
New cards

If your experiment in Question 3 is successful, what kind of phenotype would you expect to see as a result? 

b. More cells than normal undergo apoptosis

25
New cards

Refer to the information presented in Question 5. Harry and Jill get married; none of their seventeen children have sitosterolemia.  What would you conclude from this observation? 

All 17 children not having sitosterolemia can be explained by either of the following: the sitosterolemia in Harry and Jill are caused by different mutations/genes, children have a different biological father, or children have different biological parents (adopted). While, zero penetrance and inheritance of an epigenetic modification/mutaiton that suppresses the recessive condition across 17 children is highly unlikely, statistically, it is possible.

26
New cards

A black, short haired female wombat with the genotype BB hh mates with a brown, long haired male wombat of the genotype bb HH.  The coat color B gene and the hair length H gene are very tightly linked, and alleles in capital letters are dominant to alleles in small letters.

What are the genotypes of the gametes produced by the female parent?

B h

27
New cards

Based on the information you have been given in Question 8, what are the genotypes of the progeny produced by such a mating? (choose all that are correct): 

Bb Hh

28
New cards

Based on the information you have been given in Question 8, what are the phenotypes of the progeny produced by such a mating? (choose one answer): 

a. Black, long haired

29
New cards

Wiscott-Aldrich syndrome is a recessive X-linked disease resulting in eczema and spontaneous bruising. Two healthy individuals (who do not have the syndrome) have a son with Wiscott-Aldrich syndrome. Assuming A stands for the normal allele, and a for the mutated allele, what are the possible genotypes of the parents? (Mark all that are correct). 

XA Xa and XA Y

30
New cards

Given the information in Question 1, would the son with Wiscott-Aldrich syndrome have a Barr body present in his cells?

no

31
New cards
<p>A GWAS analysis, genotyping 642,461 SNPs for reduced testosterone levels in populations of European males is shown below; the probability of a significant association with reduced testosterone levels is shown on the Y axis. Everything above the grey line is considered significant. &nbsp;Based on these data, how many genomic regions are associated with the trait of reduced testosterone?</p><p></p><p>Based on the information in Question 3, is the reduction in testosterone trait recessive, semidominant or dominant?</p><p></p><p><span>Based on the information in Question 3, is the reduction in testosterone trait sex-linked?</span></p><p>&nbsp;</p>

A GWAS analysis, genotyping 642,461 SNPs for reduced testosterone levels in populations of European males is shown below; the probability of a significant association with reduced testosterone levels is shown on the Y axis. Everything above the grey line is considered significant.  Based on these data, how many genomic regions are associated with the trait of reduced testosterone?

Based on the information in Question 3, is the reduction in testosterone trait recessive, semidominant or dominant?

Based on the information in Question 3, is the reduction in testosterone trait sex-linked?

 


 

2,  

Can't tell from these data, no

32
New cards

A SNP can be detected by using a:

Gene-Chip

33
New cards

Dee and Lee are identical human female twins. Dee has two brown eyes while Lee has one brown eye and one blue eye. Give a plausible genetic explanation for why Dee and Lee have different eye coloration. (~2 sentences)

 Since the twins are female they are genetic mosaics that have one of their X-inactivated in order to balance the expression of X- linked genes. If they have different inactivated X chromosome linked to eye color it is possible that they have different eye color.

34
New cards

Will all the daughters of a female with trisomy-X also have trisomy-X? Why or why not? Briefly justify your answer (~2 sentences) 

The segregation of the mother’s three homologous X chromosomes at meiosis can give rise to normal gametes, with one homolog going to one pole and the 2 other homologs going to the other pole at meiosis 1. Then a normal meiosis 2 would give rise to 2 normal gametes and 2 aberrant gametes.

35
New cards

Recall that birds have a ZW system of sex determination, with males being ZZ and females being ZW. A recessive lethal allele that causes death of the embryo can be present on the Z chromosome in pigeons. What would be the sex ratio in the living offspring of a cross between a male that is heterozygous for the lethal allele and a normal female?

2:1 male to female

36
New cards

If you generated a homozygous loss of function mutation in the mouse testosterone receptor, what kind of phenotype would you expect to see in genotypically XY mice?

Phenotypically female

37
New cards

Developmental Biology

is the study of how organisms change in space and time, and how
different cells with different functions are produced; the study of developmental biology relies
on observation and perturbations (physical or genetic)

38
New cards

Genetics

s is the study genes and of hereditary processes

39
New cards

genome

is the sum total of all of the DNA in a cell

40
New cards

DNA

is a double stranded molecule composed of nucleotides (also called ‘bases’

41
New cards

nucleotides (also called ‘bases’

segments of DNA that typically contain information for coding a protein are called genes

42
New cards

Mutations

alter the nucleotide sequence.

43
New cards

Point mutations

(single nucleotide changes) can affect the protein(s) encoded by a gene in different ways, which are categorized as different molecular classes of mutations

44
New cards

missense

change so that the wrong amino acid is translated

45
New cards

nonsense

change to a stop codon

46
New cards

silent

(mutation that does not disrupt the coding capacity of the gene)

47
New cards

Frameshift

mutations result in a shift of the reading frame such that an RNA encodes different amino acids, and can result from single or multiple nucleotide insertions or deletions

48
New cards

Multicellular organisms consist of cells that are differentiated

to perform specialized functions

49
New cards

cells are organized into tissues

s that are groups of similar cells that work together to perform a specific function

50
New cards

different tissues are organized into organs

that are a structural unit to carry out a function.

51
New cards

Genes control development;

molecular changes in genes that control development can lead to the evolution of new traits or forms

52
New cards

Model systems

are species that are particularly suited to laboratory study;

53
New cards

Reasons the a model organism is suited for study

. Model systems are important because we can extrapolate what we learn to other systems that are not so easy to handle in a laboratory setting. This extrapolation is made possible by the fact that all living things are related and use similar processes for growth and development; there is evolutionary conservation of genes and processessize, or ease of manipulation, or short life cycle, or other qualities.

54
New cards

phylogenetic tree

Phylogenetic trees can be built on the basis of different kinds of data; the most commonly used data now are gene sequences.

55
New cards

scientific method

relies on observation, development of a hypothesis, testing of the hypothesis, and finally revision or development of new hypotheses in light of the new information

56
New cards

Development

occurs in space and time: the appropriate sequence of events occurring in the right location is necessary to form a differentiated multicellular organism. Development is ultimately controlled by genes.

57
New cards

Allele

A specific genetic sequence variant. In a population of organisms, many alleles may exist for each gene

58
New cards

Genotype

The description of the genetic constitution of an individual (eg. let60- / let60-). This is an explicit combination of alleles

59
New cards

Phenotype

The observable properties of an individual resulting from the combination of genetic and environmental factors (eg. multivulva worm)

60
New cards

Loss of function (lof)

that the gene no longer can produce a fully functional product

61
New cards

Gain of function (gof)

gain of function indicates that the gene is producing excess product or a variant of the product with higher/different activity, or producing its product in the wrong place or at the wrong time.

62
New cards

“Catch it” (or “Show it”

: shorthand for detecting where a gene product is expressed

63
New cards

“Break it”

Shorthand for eliminating gene function to examine where and when the gene product is needed.

64
New cards

“Move it”

Shorthand for expressing the gene in a new context to see what functions it is capable of producing in that new context.

65
New cards

“Rescue it”

Shorthand for replacement of the function gene which was previously lost. Usually via introduction of a new copy of a gene or gene product.

66
New cards

Genetic screen

a procedure to generate and identify a mutation in a process or phenotype of interest

67
New cards

Forward genetics

an unbiased/random process to generate mutations in genes responsible for a process or phenotype of interest

68
New cards

Reverse genetics

generating a mutation in a specific gene or genes of interest and then seeing what kind of phenotype is produced. Even if you have a phenotype in mind, when you target a specific gene, this is termed a reverse genetics approach.

69
New cards

haploid

Human gametes (sperm or egg) are haploid and possess a single copy of each of the 23 chromosomes

70
New cards

diploid

Humans have 23 pairs of chromosomes and so are diploid.

71
New cards

‘ploidy’

refers to the numbers of homologous chromosomes in the nucleus

72
New cards

alleles

Because diploid organisms have two copies of each chromosome, and thus two copies of each gene, the gene has two alleles

The two alleles of a gene can be the same (homozygous wild type or homozygous mutant) or they can differ (heterozygous, with one mutant allele and one wild type allele).

73
New cards

a mutant allele

e is referred to by a lower case letter, or by a ‘-‘ sign (if lof)

74
New cards

wild type

(most common/predominant allele in the population) is generally referred to by an upper case letter or by a ‘+’ sign.

75
New cards

dominant

—it causes a phenotypic effect when heterozygous (+/-);

76
New cards

recessive

it causes a phenotypic effect when homozygous (-/-). Most loss of function mutations are recessive

77
New cards

haploinsufficiency

‘haplo’ – single, insufficient gene product produced in the +/- situation

78
New cards

dominant mutations

can either be so due to gain of function, or due to haploinsufficiency

79
New cards

Staining

Cells or tissues can be stained with a variety of substances to enhance the visualization of subcellular components, to highlight metabolic processes, or to distinguish between living and dead cells

80
New cards

In situ hybridization (ISH)

  • Uses a labeled probe (RNA or single-stranded DNA) complementary to the target RNA.

  • The probe binds to the RNA, allowing visualization of gene expression patterns.

  • Helps determine where and when a gene is expressed in a tissue.

81
New cards

Differential gene expression

Different tissues express different genes

82
New cards

zygote

e (single cell embryo)

83
New cards

totipotent

; it has the capacity to form all cell types

84
New cards

e determined

committed to developing into particular parts of the embryo and particular adult structures.

85
New cards

Following determination, cells eventually differentiate

e into their final, specialized, forms. During the normal course of development, cells that are differentiated do not become other cell types. H

86
New cards

in response to exogenous (external or abnormal) conditions, differentiated cells can be induced to become

pluripotent (able to form most other cell types).

87
New cards

nuclear equivalence

that the nuclei of differentiated cells still retain all the information to make virtually all cell types.

88
New cards

stem cells

s (cells that are pluripotent and have the capacity to give rise to other cell types)

89
New cards

embryonic stem cells

(cells that have not yet progressed down the path of becoming differentiated

90
New cards

induced pluripotent stem cells

iPSCs can be induced to form through expressing four different transcription factors that can then reprogram the cell back to an embryonic state

91
New cards

in vitro

‘in glass’, refers to experiments done in a dish or test tube, outside of a biological context

92
New cards

organoids

iPSCs that have been reprogrammed to an embryonic state can then be induced to differentiate into specific cell types, or even into whole organs in a petri dish

93
New cards

Apoptosis

programmed cell death

94
New cards

Double mutant analyses

an be used to order genes into a hypothesized pathway

95
New cards

Cancer

Disease cause by alterations in our own genes and can be due to many different mutations

96
New cards

Proto-oncogenes

genes that are required for cell division and can become cancerous if mutated

97
New cards

Tumor suppressors

genes required to stop cell division

98
New cards

Somatic cells

Cells fo the body

99
New cards

Germ line cells

Cdells that undergo meiosis to give rise to gametes

100
New cards

Oncogenes

Mutated gene that leads to uncontrolled cell proliferation