Module 2 structure Flashcards by kayla w

Briefly describe the structure of DNA.

An antiparallel, double helix with a phosphodiester bond–linked sugar phosphate backbone oriented such that the bases of the nucleotide monomers are oriented to the middle of the double helix
(perpendicular to the axis of the helix) and the sugar-phosphate backbone is on the outside.
The helix is stabilised by hydrogen bonding between opposing bases from each strand of the helix, with
specific base pairing between adenine and thymine, and guanine and cytosine bases.

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State how the structure of DNA would allow for the semiconservative replication of DNA.

Because of the specific base-pairing (A-T, G-C) each strand can serve as a template for the synthesis of a new strand of DNA that is complementary to the template.
The newly synthesised strand would form half of a new double helix, which also has one parent strand, thus conserving half of the original
starting material in the new replicated product.

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The enzyme DNA pol III

adds nucleotides to the 3’ end of growing strands.

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What direction is DNA replicated?

5’ to 3’ direction.

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What direction do the two strands of the DNA double helix run in?

an antiparallel direction.

The simultaneous synthesis of both strands presents a problem.

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What is the leading strand?

New 5’ to 3’ strand being formed along the template by DNA polymerase in the progressing replication fork.

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The lagging strand is created as a series of

short segments, called Okazaki fragments,

Formed in the 5’ to 3’ direction away from the replication fork.

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DNA ligase

An enzyme that joins the sugar-phosphate backbones of the fragments.

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What does PCR stand for, and what is its purpose?

PCR: Polymerase chain reaction
Purpose: to amplify (make many copies) of a specific (target) region of DNA.

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List the components required for PCR.

Template DNA, primers (forward and reverse)
dNTPs (deoxyribonucleotide triphosphates – A,T,G and C),
DNA polymerase.

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What determines which region of DNA is synthesized by PCR?

The primers.

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PCR involves three steps; denaturing, annealing and extension. Briefly explain what occurs during the extension step.

Starting from the 3’ end of the primer, DNA polymerase adds in complementary bases/nucleotides to extend/synthesise the new strand in the 5’ to 3’ direction.

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Draw Anaphase and what events take place?

The 2 sister chromatids of each pair separate, and are now
referred to as chromosomes. They move towards opposite
ends of the cell as their kinetochore microtubules shorten.

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Draw Interphase (G2) and what events take place?

Chromosomes replicate during S phase but cannot be seen

as they have not yet condensed.

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Draw Metaphase and what events take place?

The centrosomes are now at opposite ends of the cells.
The chromosomes are located on an imaginary plane between the spindle’s two poles.
The kinetochores of the
sister chromatids are attached to kinetochore microtubules
from opposite poles.

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Draw the stages of meiosis

Prophase l
Metaphase l 
Anaphase l 
Prophase II 
Anaphase II 
Telophase II/cytokinesis

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Describe the number and appearance of the chromosomes in a horse sex cell at the end of
meiosis I.

You would expect to see 32 independent replicated chromosomes at each pole of the mother cell (which may be undergoing cytokinesis to produce two cells).

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Name TWO sources of variation in the genetic make-up of horse gametes and where they occur in the cell cycle.

Large number of genetically different gametes produced by horses (e.g., horses produce 2^32 genetically different types of gametes through the segregation of chromosomes at anaphase 1 of meiosis)
Crossing over occurs during Prophase I and vastly increases gamete variety

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Mules are a hybrid of a male donkey and a female horse. Mules are sterile and have 63
chromosomes in their somatic cells.

What is the diploid number of chromosomes in a
donkey somatic cell?

Horse gamete has 32 (unreplicated) chromosomes

Mule somatic cells have 63 chromosomes, therefore donkey gamete has 31 chromosomes

Diploid number of chromosomes in donkey = 62

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What is the clinical name for the genetic abnormality 3 chromosomes at 21?

Trisomy 21 or Down Syndrome

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How many Barr bodies would you expect to see in interphase nuclei in cells from a down syndrome?

None (only has 1 X chromosome)

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Down syndrome occurs sporadically in humans at a rate of about 1 in 700 births. Occasionally it can occur repeatedly over several generations in the same family.

What is the cause of this familial version of the abnormality?

Familial Down Syndrome is caused by translocation of an extra chromosome 21 onto one of the acrocentric chromosomes, usually chromosome 14. Carriers of this translocation (i.e. 14, 14+21,
21) are phenotypically normal, but produce a variety of gametes (eggs or sperm) which can give rise to Down syndrome, normal, or more carrier zygotes (offspring produced), as well as increased zygote lethality.

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What are non-coding gene regions?

Region of a gene that is not translated into a protein.

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What is (are) the function(s) of the non-coding gene regions?

They are involved in regulating gene expression.

regulated at the transcriptional or translational level.

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What are possible consequences of a DNA change (mutation) in non-coding gene regions?

The gene product is not expressed as it should be, i.e. the protein is not produced as normal. (no protein at all produced, too little protein or too much protein produced).

The DNA of genes contains sequences that: (a) regulate the transcription of RNA from that gene, and (b) determine which nucleotides are incorporated into the RNA. Describe the role of these DNA sequences in the initiation and elongation phases of transcription.

a) Promoter sequence to which transcription factors and RNA polymerase binds. b) The template strand of the DNA is read by RNA polymerase, which inserts complementary nucleotides into the growing RNA.

The process of transcription - Template - Location - Product

- DNA - Nucleus - Primary Transcript (pre-mRNA)

The process of transcription | - Molecules involved

RNA nucleotides DNA template strand RNA polymerase transcription factors

The process of transcription | - Control – start and stop

Transcription factors locate promoter region with TATA box, polyadenylation sequence.

The process of Translation - Template - Location - Product

- mRNA - Cytoplasm - Protein

The process of Translation | - Molecules involved

``` Amino acids tRNA mRNA rRNA ribosomes enzymes release factors ```

The process of Translation | - Control – start and stop

Initiation factors initiation sequence (AUG) stop codons release factors

Codon recognition: | elongation process

1st The anticodon of an incoming tRNA base-pairs with the complementary mRNA codon in the A site.

Peptide bond formation: | elongation process

between the amino group of the new amino acid in the A site and the growing polypeptide in the P site. This step removes the polypeptide from the tRNA in the P site and attaches it to the amino acid on the tRNA in the A site.

What recognises the stop codon?

Release factor

Provide an explanation as to how valine can be encoded by different sequences. A short segment of DNA and the polypeptide it encodes are shown below: 5’ – ATGGTTGTGTGTTGA : METHIONINE-VALINE-VALINE-CYSTEINE-STOP

More than one codon codes for some amino acids

If the DNA sequence was mutated to 5’ – ATGGTTGTTTGTTGA, what polypeptide would it now encode? A short segment of DNA and the polypeptide it encodes are shown below: 5’ – ATGGTTGTGTGTTGA : METHIONINE-VALINE-VALINE-CYSTEINE-STOP

Same (GTT and GTG both code for valine)

Co-dominance

Both alleles are fully expressed in heterozygotes.

Homozygote

True-breeding variety. | Genotype with same alleles.

Heterozygote

Genotype with two different alleles.

Phenotype

The physical characteristics of an individual.

Test cross

Cross with recessive homozygote to determine genotype of unknown.

Dihybrid cross

Cross between hybrids that are | heterozygous for two genes.

Incomplete dominance

Heterozygotes intermediate between phenotypes of homozygotes.

Genotype

The genetic make-up, or set of alleles, of an organism.

Law of segregation:

The two alleles for each gene separate during gamete formation.

Law of independent assortment:

Alleles of genes on non-homologous chromosomes assort independently during gamete formation.

List 6 factors that may alter the basic phenotypic inheritance patterns that were observed by Mendel.

``` Incomplete dominance, Co-dominance, Polygenic traits, Environment, Linked genes, Sex-linked genes. ```

determine the probability of obtaining the indicated genotype in an offspring: AAbb AAbb x AaBb

1⁄4 | 0.25

determine the probability of obtaining the indicated genotype in an offspring: aaBB AaBB x AaBb

1/8 | 0.125

determine the probability of obtaining the indicated genotype in an offspring: aaBB AaBB x AaBb

1/8 | 0.125

determine the probability of obtaining the indicated genotype in an offspring: AaBbCc AABbcc x aabbCC

1⁄2 | 0.5

determine the probability of obtaining the indicated genotype in an offspring: aabbcc AaBbCc x AaBbcc

1/32 | 0.03125

A species of flowering plant has one gene (R) that controls flower colour and another gene (T) that determines plant height. Both genes show complete dominance. A geneticist performs a cross between a dihybrid red-flowered, tall plant line with a pure-breeding white-flowered, short plant line of the same species. The progeny from this cross comprise: 591 red-flowered, tall plants; 387 red-flowered, short plants; 406 white-flowered, tall plants; 616 white-flowered short plants. Write out the genotypes associated with these parent and progeny phenotypes using standard genetic symbols.

Parents: RrTt (red, tall) rrtt (white, short) ``` Progeny: RrTt (red, tall); Rrtt (red, short); rrTt (white, tall); rrtt (white, short) ```

Test cross

A species of flowering plant has one gene (R) that controls flower colour and another gene (T) that determines plant height. Both genes show complete dominance. A geneticist performs a cross between a dihybrid red-flowered, tall plant line with a pure-breeding white-flowered, short plant line of the same species. The progeny from this cross comprise: 591 red-flowered, tall plants; 387 red-flowered, short plants; 406 white-flowered, tall plants; 616 white-flowered short plants. Do genes R and T behave as predicted under Mendel’s Second Law and what is the genetic explanation for this?

No, they do not behave as predicted under Mendel’s Second Law. Alleles at genes R and T are not assorting independently, because progeny show a substantial numerical bias towards parental phenotypes/genotypes. Therefore, genes R and T must be linked (i.e., on the same chromosome).

A species of flowering plant has one gene (R) that controls flower colour and another gene (T) that determines plant height. Both genes show complete dominance. A geneticist performs a cross between a dihybrid red-flowered, tall plant line with a pure-breeding white-flowered, short plant line of the same species. The progeny from this cross comprise: 591 red-flowered, tall plants; 387 red-flowered, short plants; 406 white-flowered, tall plants; 616 white-flowered short plants. Calculate the distance between R and T in map units.

``` Recombinant phenotypes (or genotypes) = red, short (387) + white, tall (406) = 793. ``` Total genotypes = 591+387+406+616 = 2000 Recombination frequency = 793/2000 = 0.3965 Distance between R and T = 100 x 0.3965 = 39.65 centimorgans (map units)

A researcher performed a cross between two pure-bred pea plants: one tall with purple flowers, and the other short with white flowers. The F1 plants produced by this cross were all tall with purple flowers. When the researcher then allowed these F1 plants to self-fertilise, the next generation (F2) consisted of: 182 purple-flowered tall plants; 58 white-flowered tall plants; 61 purple-flowered short plants; and 19 white-flowered short plants. Why do all the F1 plants have the same phenotype?

They all have the same genotype/they are heterozygotes or One allele is completely dominant over the other (1⁄2 mark) or Purple and tall are dominant (1⁄2 mark)

Two genes each with two alleles or two genes and four alleles.

Parents: PPTT (purple/tall) and pptt (white/short) F1: PpTt F2: P-T- (purple/tall), P-tt (purple/short), ppT- (white/tall), pptt (white/short)

The phenotype ratio is 9:3:3:1 a dihybrid cross. The genes are assorting independently (are not linked) following Mendel’s 2nd Law.

A European population has a high frequency of a mutant allele that causes cystic fibrosis in individuals homozygous for the mutant. What type of inheritance does this genetic disease show?

Autosomal recessive

A European population has a high frequency of a mutant allele that causes cystic fibrosis in individuals homozygous for the mutant. If both parents of a child are carriers, what is the probability that the child will have cystic fibrosis?

0.25 (because Aa x Aa will give 25% AA; 50% Aa and 25% aa)

A European population has a high frequency of a mutant allele that causes cystic fibrosis in individuals homozygous for the mutant. If one in 1600 children is born with cystic fibrosis, and assuming random mating, what proportion of the population are carriers for the disease?

``` p^2 = 1/1600 therefore p = 1/40 = 0.025 p+q = 1 therefore q = 0.975 Carriers = 2pq = 2 x 0.025 x 0.975 = 0.04875 (4.88% of the population) ```

Autosomal dominant disorders - skip generation? - affects offspring (M/F)?

do not skip a generation affect 50% of the offspring of affected individuals, regardless of sex.

autosomal recessive disorders - skip generation? - affects offspring (M/F)?

do often skip a generation. Offspring of an affected individual may be unaffected carriers, although if they mate with another carrier, 25% of offspring, regardless of sex, will be affected.

X-linked dominant disorders, - skip generation? - affects offspring (M/F)?

heterozygous females produce 50% affected children, regardless of sex of the child, Affected males can only pass their X chromosome to their daughters and not to their sons.

X-linked recessive disorders - skip generation? - affects offspring (M/F)?

female offspring of an affected individual can be carriers but male offspring cannot.

Colouration of beach mice and inland mice appear to “match” the colour of the habitat that they live in. Beach mice are white and live on white sand dunes while inland mice are brown and live on darker soil. Mice that “match” the colour of their habitat are less likely to be caught by predators than mice that do not “match” the colour of the habitat. Are the changes in coat colour likely to be the result of genetic drift or selection? Explain your choice.

selection, rather than genetic drift. Camouflaged mice are less likely to be caught by predators than mice that stand out against their habitat. As camouflaged mice are more likely to escape predation long enough to produce offspring, camouflaged mice are selected for and non-camouflaged mice are selected against.