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1
Q

Haploid vs Diploid

A

-Diploid cells contain two complete sets (2n) of chromosomes.Haploid cells have half the number of chromosomes (n) as diploid.
-Diploid cells reproduce by mitosis making daughter cells that are exact replicas. Haploid cells are a result of the process of meiosis.

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2
Q

Sexual vs Asexual

A

In sexual reproduction, an organism combines the genetic information from each of its parents and is genetically unique. In asexual reproduction, one parent copies itself to form a genetically identical offspring.

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3
Q

Phases of Meiosis

A

Prophase 1
Metaphase 1
Anaphase 1
Telophase 1
Prophase 2
Metaphase 2
Anaphase 2
Telophase 2

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4
Q

Prophase 1

A

Starting cells are diploid. Homologous chromosomes pair up and exchange fragments (crossing over)

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5
Q

Metaphase 1

A

Homologous pairs line up at the metaphase plate.

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6
Q

Anaphase 1

A

Homologs separate to opposite ends of the cell, sister chromatids stay together.

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7
Q

Telophase 1

A

Newly formed cells are haploid. Each chromosome has two non identical sister chromatids.

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8
Q

Prophase 2

A

chromosomes condense and the nuclear envelope breaks down, if needed. The centrosomes move apart, the spindle forms between them, and the spindle microtubules begin to capture chromosomes.

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9
Q

Metaphase 2

A

he chromosomes line up individually along the metaphase plate.

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10
Q

Anaphase 2

A

the sister chromatids separate and are pulled towards opposite poles of the cell.

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11
Q

Telophase 2

A

nuclear membranes form around each set of chromosomes, and the chromosomes decondense. Cytokinesis splits the chromosome sets into new cells, forming the final products of meiosis: four haploid cells in which each chromosome has just one chromatid.

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12
Q

How many gamete types are produced from an
individual with the genotype EeHh?

A

2n (n= number of heterozygous loci)
n=2; 4
EH, eh, Eh, eH

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13
Q

A diploid cell is 2n=6
How many chromosomes present at the end of meiosis 1? Meiosis 2?

A

Meiosis 1: Haploid
Meiosis 2: Haploid (n)

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14
Q

Mitosis vs Meiosis

A

Mitosis is a process where a single cell divides into two identical daughter cells (cell division). Produces diploid. Meiosis is a process where a single cell divides twice to produce four cells containing half the original amount of genetic information.Produces haploid.

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15
Q

In pea plants, round seeds is dominant to wrinkled seeds.
R – round; r - wrinkled
If an individual is heterozygous, how many unique gamete types can be produced
by that individual? Give the genotype of the gamete(s).

A

2 unique gamete types
(R and r)
50% of each type

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16
Q

A man is phenotypically normal but has a
mother with phenylketonuria, an autosomal
recessive metabolic disorder. What is the
likelihood that a gamete chosen at random
from the man will carry the phenylketonuria
allele?

A

1/2

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17
Q

A man and woman are both of normal pigmentation, but
both have one parent who is albino. Albinism is a
recessive trait. What is the probability that their first
child will be an albino?

A

1/4

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18
Q

In tigers, a recessive allele causes a white tiger (absence of
fur pigmentation). If one phenotypically normal tiger that is
heterozygous is mated to another that is phenotypically white,
what percentage of their offspring is expected to be white?

A

50%

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19
Q

n sheep, lustrous fleece results from an allele that
is dominant over an allele for normal fleece. A ewe
with lustrous fleece is mated with a ram with
normal fleece. The ewe then gives birth to a single
lamb with normal fleece. From this single
offspring, is it possible to determine the genotypes
of the two parents? If so, what are their
genotypes? If not, why not?

A

t is possible to determine the genotypes of the two
parent.
The lustrous ewe must have at least one dominant
allele: A_
Because she gives birth to a normal-fleeced lamb (aa),
then the ewe must also carry a recessive allele: Aa
The ram has normal fleece and so must be
homozygous recessive: aa
Aa X aa →aa (1 Aa : 1 aa)

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20
Q

In watermelons, bitter fruit (F) is dominant over sweet fruit (f), and
yellow spots (S) are dominant over no spots (s). The genes for
these two characteristic assort independently. A homozygous plant
that has bitter fruit and yellow spots is crossed with a homozygous
plant that has sweet fruit and no spots. The F1 are mated to
produce the F2.
Write out the P generation mating:
Write out the F1 generation mating:
Predict ratio.

A

FFSS x ffss

FfSs x FfSs

9 Bitter, Spotted
3 Bitter, No spots
3 Sweet, Spotted
1 Sweet, No spots

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21
Q

wo true-breeding stocks of pea plants are crossed. One parent has
red, axial flowers and the other has white, terminal flowers; all F1
individuals have red, axial flowers. Among the F2 offspring, what is
the probability of plants with red axial flowers?
a. 9/16 b. 1/16 c. 3/16 d. 1/8 e. 1/4
If 1,000 F2 offspring resulted from the cross, approximately how
many of them would you expect to have red, terminal flowers?
a. 65 b. 190 c. 250 d. 565 e. 750

A

. 9/16

190

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22
Q

Principle of Segregation

A

describes how pairs of gene variants are separated into reproductive cells.

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23
Q

The Principle of Independent Assortment

A

The Principle of Independent Assortment describes how different genes independently separate from one another when reproductive cells develop.

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24
Q

In Guinea pigs, curly hair is recessive to straight hair.
Guinea pigs can also have a condition called bowlegged, where their legs curve
noticeably outward. Bowleggedness is a dominant lethal allele in that the
homozygous dominant genotype is lethal before birth.
Show the cross between a curly haired, bowlegged guinea pig and a
heterozygous straight haired pig that is also bowlegged.
What proportion of their offspring would you expect to be normal with curly hair?

A

hhBb x HhBb
1/2 Hh (straight)
1/2 hh (curly)
2/3 Bb (bowlegged)
1/3 bb (normal)
Normal, curly = 1/3 * 1/2 = 1/6

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25
In beets, white-colored root is recessive to red- colored root. A heterozygous plant will produce a pink-colored root. Two pink-colored plants are mated. What phenotypic ratio will be seen among the offspring?
1 white: 2 pink: 1 red
26
Also in beets, tall plants are dominant to dwarf plants. When the two pink-colored plants above were mated, a mix of tall and dwarf plants were found. The ratio of tall to dwarf was a 3:1 ratio. What proportion of the offspring were both pink-colored and dwarf?
1/8
27
If a woman with type AB blood marries a man with type O blood, which of the following blood types could their children possibly have?
a. A and B
28
A cross between a black cat and a tan cat produces a tabby pattern (black & tan fur together). What percent of kittens would have tan fur if a tabby cat is crossed with a black cat?
A1A2 x A1A1 A1-black; A2-tan 50% A1A2 (tabby) 50% A1A1 (black) The answer is 0%
29
Body color for horses is influenced by several genes, one of which has multiple alleles. Two of these alleles, chestnut (dark brown) and cremello (pale cream), display incomplete dominance. A horse heterozygous for these two alleles is palomino (golden body color with flaxen mane and tail). Is it possible to produce a herd of pure-breeding palomino horses?
A palomino horse is heterozygous. There is no “palomino” allele and so we cannot have a homozygous palomino genotype.A mating between two palominos will yield a phenotypic ratio of 1 chestnut: 2 palomino: 1 cremello.
30
In chickens, the form of the comb can vary and is controlled by two genes designated as P and R. The following interactions produce various phenotypes: If at least one dominant allele is found only at the P locus, a pea-type comb is produced. If at least one dominant allele is found only at the R locus, a rose-type comb results. At least one dominant allele at both loci produces a walnut-type comb Homozygous recessive genotype at both loci produces a single-type comb. Which of the following represents a mating between a pure-breeding pea-type chicken and a pure-breeding rose-type comb chicken.
PPrrX ppRR
31
Duroc Jersey pigs are typically red, but a sandy variation is also seen. When two different varieties of true-breeding sandy pigs were crossed to each other, they produced F1 offspring that were red. When these F1 offspring were crossed to each other, they produced red, sandy and white pigs in a 9:6:1 ratio. Explain this pattern of inheritance by assigning alleles and giving genotypes of the two true-breeding pigs, the F1 pigs and their F2 progeny.
9:6:1 A_B_ 9 red A_bb 3 sandy aaB_ 3 sandy aabb 1 white F1 must be AaBb to give a modified 9:3:3:1 ratio So P must be AAbb X aaBB Having either dominant allele (but not both) gives sandy color.
32
A pure-breeding black lab and a pure-breeding golden lab produce a litter of 6 black puppies. These puppies are mated and produce offspring in the following phenotypes and ratio: 9 black: 4 golden: 3 brown What are the genotypes of the pure-breeding dogs? What are the genotypes of the F1 generation?
BBEE X bbee P generation BbEe F1 generation BbEe X BbEe B-E- black 9/16 B-ee golden 3/16 bbE- brown 3/16 bbee golden 1/16 9/16 black 4/16 golden 3/16 brown
33
In humans, red-green colorblindness is an X- linked (sex-linked) recessive trait. A woman who is heterozygous for the colorblindness allele marries a colorblind man. What proportion of their children will be sons with colorblindness?
25%
34
A woman who is not color-blind (and is heterozygous at this locus) and does not have albinism marries a color- blind man without albinism. Both had albino fathers. What are the genotypes of the two individuals?
AaXCXc x AaXcY
35
In birds, sex is determined by a ZWchromosome scheme. Males are ZZand females are ZW. A recessive lethal allele that causes death of the embryo is sometimes present on the Zchromosome in pigeons. What would be the sex ratio in the offspring of a cross between a male that is heterozygous for the lethal allele and a normal female?
2:1 male to female
36
Draw a DNA nucleotide. Include base pair rules.
sugar, phosphate and base group. denine (A) and thymine (T) pair together, and cytosine (C) and guanine (G) pair together
37
DNA Polymerase
-Synthesizes new DNA in a 5’ to 3’ direction. -Synthesizes new DNA antiparallel to the template DNA. -Requires a nucleic acid primer (3’ OH) to initiate synthesis. -Most can proofread.
38
What binds at the origin to unwind dsDNA?
Initiator Proteins
39
Leading Strand
continuous synthesis and synthesis directly follows the opening of the fork.
40
What do DNA Polymerases require that RNA Polymerases do not?
a 3’ OH. So – primase lays down an RNA primer which provides the 3’ OH for DNA polymerase.
41
What chemical group is associated with the 3’ and 5’ ends of DNA?
phosphate groups
42
Okazaki fragments
short sequences of DNA nucleotides which are synthesized discontinuously and later linked together by the enzyme DNA ligase to create the lagging strand during DNA replication.
43
Ligase
The DNA ligase is responsible for sealing any breaks in the sugar-phosphate backbone of the strand
44
Telomerase
Extends the un replicated end to preserve the length of telomeres
45
How are DNA Polymerase mistakes minimized?
1. Proofreading - done in DNA synthesis 2. Mismatch repair- done immediately after synthesis.
46
Prokaryotic Polymerases
I – Primer removal and replacement II – Repair III – Main synthesis enzyme
47
Prokaryotic Polymerase Duties
5’-3’ Synthesis activity(Pol I, II and III) 3’-5’ Exonuclease activity – proofreading(Pol I, II and III) 5’-3’ Exonuclease activity – removal of primers(Pol I only)
48
How do DNA Polymerases proofread?
a 3’→5’ exonuclease.
49
Primer Removal
5'-3' exonuclease
50
Which DNA pol removes and replaces primers?
DNA Polymerase I
51
Why is DNA synthesis “discontinuous” on the lagging strand?
new RNA primers must be added as opening of the replication fork continues to expose new template. This produces a series of disconnected Okazaki fragments.
52
DNA Pol function
polymerizes deoxyribonucleotide monomers into DNA.
53
RNA Primer
must be added to DNA Template strand to supply 3' end for DNA polymerase.
54
How is primer added?
Primase (type of RNA pol)
55
Synthesis of leading strand(bacteria)
1. RNA primer added 2. DNA polymerase reads, adding deoxyribonucleotides.
56
Synthesis of lagging strand (bacteria)
-synthesized moving away from the fork. 1. primer added 2. first okazaki fragment synthesized 3. second okazaki fragment synthesized 4. primer replaced 5. nick closed
57
How are okazaki fragments synthesized?
In bacteria, DNA pol 3 dissociates from the 3' end of an okazaki fragment when the polymerase encounters primer for next segment. Pol 1 attaches to 3' end and moves in a 5'-3' direction removing RNA primer and replacing the ribonucloetides it with deoxyribonucleotides.
58
Lagging strand synthesis: differences in organisms.
1. Eukaryote okazaki fragments are shorter than bacteria. 2. Eukaryote lagging strand synthesis has a specific DNA pol. 3. Removal of primers is different.
59
DNA Polymerase III
In bacteria DNA synthesis, extends the leading strand, or extends okazaki fragments.
60
DNA Pol I
5'-3' synthesis 3'-5' proofreading 5'-3' primer removal
61
DNA Pol II
5'-3'synthesis 3'-5' proofreading
62
DNA Pol III
3'-5' proofreading 5'-3' synthesis
63
Transcription
Transcription is the process of copying a segment of DNA into RNA.
64
Translation
translation is the process in which ribosomes in the cytoplasm or endoplasmic reticulum synthesize proteins after the process of transcription of DNA to RNA in the cell's nucleus
65
Where does prokaryotic transcription and translation occur?
The cytoplasm
66
Where does eukaryotic transcription and translation occur?
transcription occurs in the nucleus, translation occurs in the cytoplasm.
67
Promoters
regions of DNA that promote the start, sites where transcription should begin.
68
tRNA
(transfer) – carries amino acids to the ribosome -this type participates in the translation process but is not translated.
69
mRNA
(messenger) – encodes a polypeptide -this is the only type of RNA that gets translated.
70
rRNA
(ribosomal) – part of ribosome structure -this type is also not translated.
71
Coding vs Template strand
Coding: 1. The coding strand determines the correct nucleotide sequence of mRNA. 2. They do not take part in the transcription process. 3. The coding strand is directed in the 3’ to 5’ direction. Template: 1. The template strand acts as a base for mRNA transcription. 2. Here, they take part in the transcription. They help in the formation of mRNA. 3. The template strand is directed in the 5’ to 3’ direction.
72
The Initiation of Translation in bacteria
1. Small subunit and initiator tRNA bind to the 5’ end of the mRNA. 2. Scanning begins to find the correct AUG start codon(euks). 3. Large subunit binds to the small subunit.
73
Elongation of Translation
1. Charged tRNA enters at the A site of the ribosome. 2. A peptide bond is created between the amino acids in the A and P sites 3. The bond is broken between tRNA-a.a. in the P site 4. The ribosome translocates toward the 3’ end of the mRNA
74
How does translation termination occur?
When the translocating ribosome reaches one of the stop codons (UAA, UAG, UGA), a protein called release factor recognizes the stop codon and fills the A site
75
Silent mutation
change in nucleotide sequence that does not change amino acid specified by codon.
76
Missense mutation
change in nucleotide sequence that changes amino acid specified by codon.
77
Nonsense mutation
change in nucleotide that results in early stop code.
78
Frameshift mutation
addition or deletion of a nucleotide.
79
Positive regulation of the lac operon: cAMP present vs not present
present: cAMP binds to CAP and the cAMP-CAP complex binds to DNA at the CAP site. RNA polymerase binds the promoter efficiently. Transcription occurs frequently. not: CAP does not bind to DNA. RNA polymerase binds the promoter inefficiently. Transcription occurs rarely.
80
cAMP & glucose relationship
The amount of cAMP and the rate of transcription of the lac operon are inversely related to the concentration of glucose.
81
Full expression of lac operon requires:
“relieving” of negative regulation: -the release of the lac repressor -occurs in the presence of lactose and the “application” of positive regulation: -the binding of CAP-cAMP -occurs in the absence of glucose
82
Glucose inhibits transcription of the lac operon by _____. a. strengthening the binding of the repressor to the operator b. reducing levels of intracellular cAMP c. inhibiting RNA polymerase from opening the strands of DNA to initiate transcription. d. weakening the binding of repressor to the operator
B
83
What happens to the expression of the lacI gene if lactose is not available in the cell? a. The lacI gene turns off. b. The lacI gene turns on. c. The lacI gene increases its rate of transcription. d. There is no change in lacI gene expression.
D
84
lacI
Prevents transcription of lacZ and lacY when lactose is absent
85
Forms of control in bacteria
Transcriptional (saves most energy for the cell), Translational (allows more rapid changes) , Post Translation (most rapid response)
86
Inducer
small molecule that triggers transcription of a specific gene.
87
catabolite activator protein (CAP)
a bacterial protein that activates the transcription of operons involved in the use of sugars other than glucose. *must be bound to cAMP
88
core promoter
the specific sequence where RNA Pol binds.
89
enhancer
regulatory sequence far from the promoter, activates transcription.
90
silencer
DNA sequence that will stop transcription if a repressor is bound.

Bio Final (1 decks)

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