Final Review for Genetics PDF

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List of items to know for final exam...

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Dear Students, To help you with your preparation for the final exam, we have prepared this list that includes the topics we discussed this semester and tells you how to prepare. This list is a bunch of statements: if you can do them, then you are ready for the test!

Genetics Final Chapter 2: 1. Understand how observations from Mendel’s one factor crosses support his law of segregation. ● ●

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Single-factor cross: involves crossing two variant of the same characteristic; only a single characteristic is observed For all seven characteristics studied ○ The F1 generation showed only one of the two parental traits ○ The F2 generation showed an approximately 3:1 ratio of the two parental traits His third idea was the Law of Segregation. ○ Mendelian unit factors are now called genes ○ Alleles are different versions of the same gene ○ An individual with two identical alleles is termed homozygous ○ An individual with two different alleles, is termed heterozygous ○ Genotype refers to the specific allelic composition of an individual ○ Phenotype refers to the observable traits of an individual The two copies of a gene segregate (or separate) from each other during the process that gives rise to gametes. ○ Therefore, each gamete carries one copy of a given gene. In other words, a gamete carries a single allele of a given gene.

2. Understand how observations from Mendel’s two factor crosses support his law of segregation. ●

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Mendel also performed two-factor crosses • ○ Crossing individual plants that differ in two characters ○ For example • Character 1 = Seed texture (round versus wrinkled) • Character 2 = Seed color (yellow versus green) There are two possible patterns of inheritance for these characters • ○ Linked assortment • ○ Independent assortment The F2 generation contains seeds with novel combinations (that is: not found in the parentals) ○ Round and Green • Wrinkled and Yellow ■ These are called nonparentals Their occurrence contradicts the linked assortment model Then the predicted phenotypic ratio in the F2 generation would be 9:3:3:1

3. Based on genotype of two parents for two different traits, predict the probability that an offspring will exhibit a particular phenotype Chapter 3: 1.

Understand the major steps in mitosis, meiosis I and meiosis II ●

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Mitosis and cytokinesis ultimately produce two daughter cells having the same number and complement of chromosomes as the mother cell.The two daughter cells are genetically identical to each other. Mitosis is Subdivided into Five Phases ○ Prophase-Nuclear envelope dissociates into small vesicles, chromatids condense into more compact structures, and centrosomes begin to separate ○ Prometaphase- Centrosomes move to opposite ends of the cell, forming spindle poles. Spindle fibers interact with the sister chromatids; spindle apparatus forms. Kinetochore microtubules grow from the two poles ○ Metaphase- Pairs of sister chromatids align themselves along a plane called the metaphase plate. Each pair of chromatids is attached to microtubules ○ Anaphase- The connection holding the sister chromatids together is broken and moves toward opposite ends. Each chromatid, now an individual chromosome, is linked to only one pole. ○ Telophase-The chromosomes reach their respective poles and decondense. Nuclear membrane reforms to form two separate nuclei ○ Cytokinesis- In animals= Formation of a cleavage furrow ○ In plants= Formation of a cell plate

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Before the 5 main phases of Mitosis comes Interphase(G1,S,G2, The Cell Cycle) ○ During the G1 phase, a cell prepares to divide ○ S phase, where chromosomes are replicated.The two copies of a replicated chromosome are termed chromatids. They are joined at the centromere to form a pair of sister chromatids or a dyad. ○ During the G2 phase, the cell accumulates the materials that are necessary for nuclear and cell division then advances into the M phase of the cycle where mitosis occurs.

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Meiosis I ○ Sexual reproduction is a common way for eukaryotic organisms to produce offspring • Parents (diploid=2n) make gametes with half the amount of genetic material (haploid= 1n) ○ Prophase 1 is further subdivided into five stages known as: ○ Leptotene- Replicated chromosomes condense. ○ Zygotene- Synapsis begins ○ Pachytene- A bivalent has formed and crossing over has occurred ○ Diplotene- Synaptonemal complex dissociates. ○ Diakinesis- End of Prophase I. Crossing over has occurred

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Goes though the same metaphase, anaphase, telophase and cytokinesis but during ANAPHASE they get pulled apart as Dyads. Meiosis II ○ Goes though the same prophase, metaphase, anaphase, telophase and cytokinesis. ○ The only difference it that PROPHASE has the result of the crossing over chromosomes from Meiosis I ○ The only difference it that ANAPHASE is pulled toward opposite poles as monads like Mitosis Mitosis versus Meiosis ○ Mitosis produces two diploid daughter cells ○ Meiosis produces four haploid daughter cells ○ Mitosis produces daughter cells that are genetically identical ○ Meiosis produces daughter cells that are not genetically identical

Understand how Mendel’s laws related to the meiotic behavior of chromosomes ●

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Relationship between Mendel’s Laws and Chromosome Transmission ○ Mendel’s law of segregation can be explained by the homologous pairing and segregation of chromosomes during meiosis ○ Homologous chromosomes segregate from each other in meiosis 1. This leads to the segregation of the alleles into separate gametes. Mendel’s law of independent assortment can be explained by the relative behavior of different (nonhomologous) chromosomes during meiosis ○ Random Alignment can occur.(Look at Slide 84) ○ Both Blue, Both Red or one Red and Blue The Chromosome Theory of Inheritance( Also Important) ○ The chromosome theory of inheritance describes how the transmission of chromosomes account for Mendelian patterns of inheritance ○ Chromosomes contain the genetic material ○ Chromosomes are replicated and passed from parent to offspring ○ The nuclei of most eukaryotic cells contain chromosomes that are found in homologous pairs, they are diploid ○ During meiosis, each homolog segregates into one of the two daughter nuclei; during the formation of gametes, different types of (nonhomologous) chromosomes segregate independently ○ Each parent contributes one set of chromosomes to its offspring

Chapter 4: 1. Understand incomplete penetrance and variable expressivity, and the distinction between them.

● Incomplete penetrance: individuals with the genotype may not always express the phenotype ○ BRCA1 gene: 80% exhibits that phenotype (penetrances) ○ Incomplete penetrance, which means some individuals in a population who should show a trait based on their genotype don't actually show the trait. But at the level

of the individual who doesn't show the trait, even though his or her genotype is one where they are expected to show the trait, we could say that it's non-penetrant for that individual. ● Variable expressivity: Individuals with the phenotypes don’t express in the same way ○ Such as Marfan syndromes ○ Although some genetic disorders exhibit little variation, most have signs and symptoms that differ among affected individuals. 2. Be able to identify a non-penetrant individual on a pedigree, or identify a pedigree of a dominant trait that exhibits incomplete penetrance

● ● Answer: III-10 because III-11 is married to individual III-10 and has a child with brachydactyly type D. The runs in the family of individual III-10 but is not expressed in individual III-10 and is expressed in her child. 3. Be able to identify whether two different recessive alleles exhibit complementation and be able to interpret this result (are they alleles of the same gene or different genes)

● Complementation: Figure 4.17. Each recessive allele is complemented by a wild type allele. This phenomenon indicates that the recessive alleles are in different genes. ( two strains exhibiting the same recessive trait will produce offspring that show the dominant( wild-type) trait.

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Be able to identify epistasis between alleles of different genes

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Epistasis is an interaction where one genotype at one locus affects the phenotypic character of the genotype on another Locus. So epistasis can inhibit the phenotypic character of the trait or modify the character of the trait.One gene masks or interferes with the expression of another ● Impacts trait expression

5. Given an epistatic relationship between two genes, and genotypes of two parents, be able to determine the probability that an offspring will display a specific phenotype.

An example of epistasis is pigmentation in mice. The wild-type coat color, agouti (AA), is dominant to solid-colored fur (aa). However, a separate gene (C) is necessary for pigment production. A mouse with a recessive c allele at this locus is unable to produce pigment and is albino regardless of the allele present at locus A. Therefore, the genotypes AAcc, Aacc, and aacc all produce the same albino phenotype. A cross between heterozygotes for both genes (AaCc x AaCc) would generate offspring with a phenotypic ratio of 9 agouti:3 solid color:4 albino (Figure 1). In this case, the C gene is epistatic to the A gene. Link: https://courses.lumenlearning.com/wm-biology1/chapter/reading-epistasis-2/

Chapter 5 Be able to: 1. Define what maternal inheritance is. ●

When the phenotype of the mother is the phenotype of the offspring regardless of genotype.

2. Explain how dosage compensation works for unequal numbers of X chromosomes in mammals.

● Dosage compensation: species with sex chromosomes, one of the sex chromosomes is altered so that males and females have similar levels of gene expression. It’s purpose is to offset differences in the number of active sex chromosomes. ○ In mammals: Barr body- highly condensed x-chromosomes. ○ Lyon hypothesis: inactivation of a single x chromosomes in females

○ ○ X-inactivation-Calico Cats: 5% to 75% white with large orange and black patches (or sometimes cream and grey patches); however, the calico cat can have any three colors in its pattern. Predominantly females because they’re coloring is related to the X-chromosomes 3. Explain the connection between X chromosome inactivation, sex determination, and the calico cat pattern of fur color.

4. Explain maternal inheritance of mitochondria in humans. ●

Mitochondrial inheritance is non-Mendelian, as Mendelian inheritance presumes that half the genetic material of a fertilized (zygote) derives from each parent.

5. Explain how mutations in the mitochondrial genome can cause disease in humans, and how the transmission of this will look in a pedigree. ●

Pathogenic mtDNA mutations invariably result in a general defect in mitochondrial respiration that leads to a reduced ability to produce cellular ATP. The clinical phenotypes of mitochondrial disorders are however extremely heterogeneous.

Chapter 6 1.

Understand physical and genetic linkage. ● ● ●

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Physical Linkage- Is when two genes are located on the same chromosome. Genetic Linkage- The tendency of alleles that are located close together on a chromosome to be inherited together during meiosis. Linked genes do not assort independently. (Meaning they don’t follow Mendel’s Law of Independent Assortment.

Be able to identify parental and recombinant allelic combinations. ● ●

If crossing over does not occur, the products are parental gametes. If crossing over occurs, the products are recombinant gametes.

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Understand the relationship between genetic linkage and recombination rate. ●

Recombination rates of 50% can be achieved by genes that reside on different chromosomes.

4. Be able to calculate the recombination rate/map distance between two different genes based on a two or three factor mapping cross.

Chapter 8:

1. Know major forms of structural variation duplication, deletion, inversion, translation, as well as their common effects on viability or fertility. 2.

Know the meaning of ploidy terms (euploid, polyploid, haploid, aneuploid) ● ● ● ●

Euploid- increase in the number of complete sets of chromosomes. It occurs frequently in plants, rarely in animals can lead to formation of new species Haploid- (1n) Organisms or cells that have one chromosome set. Haploid is the normal state for gamete cells Polyploid- Organisms that have two or more copies of the entire chromosome set. Aneuploid- gene imbalance loss or gain of one or more chromosomes producing a chromosome number that is not an exact multiple of the haploid number. Leads to an abnormal condition.

3. Understand why only some aneuploidies are viable in humans (sex chromosomes, 13,18,21) ●

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People with Turner syndrome typically have one X chromosome instead of the usual two sex chromosomes. Turner syndrome is the only full monosomy that is seen in humans — all other cases of full monosomy are lethal and the individual will not survive development.

Understand biological prevalence and significance of polyploidy in plants and animals

Chapter 9 Be able to: 1. Correctly identify the three parts of a nucleotide. ● ● ●

Phosphate Sugar Base

2. Correctly identify the drawn structure of ATP, dATP, AMP, and dAMP.

dATP

ATP

AMP

dAMP

3. Be able to determine whether a base is a purine or a pyrimidine. ● ● ●

Purines in DNA are adenine and guanine, the same as in RNA. Pyrimidines in DNA are cytosine and thymine; in RNA, they are cytosine and uracil. You Purines are larger than pyrimidines because they have a two-ring structure while pyrimidines only have a single ring.

4. Identify the differences between RNA and DNA

5. Identify the 5’ and 3’ ends of a nucleic acid strand. ●

The 5' end is identified by the presence of the phosphate group and the 3' end is identified as ending in the pentose sugar (ribose in RNA and deoxyribose in DNA).

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6. Deduce the sequence of a nucleic acid strand from its complement.

Chapter 11 Be able to: 1. Describe the process of DNA replication ●

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DNA replication is the process by which genetic material is copied. The original DNA strands are used as templates for the synthesis of new strands ○ Relies on the complementarity of DNA strands (AT/GC rule) ○ Two new strands = daughter strands ○ Two original strands = parental strands The structure of DNA is a double helix with an antiparallel alignment

2. Define and describe the role of the origin and replication forks. ●

Bacterial DNA Replication ○ Two replication forks form at the origin of replication ■ Origin of replication: where DNA synthesis begins ○ Each bacterial chromosome only has one origin of replication ○ Bidirectionality, the two replication forks meet at the opposite side of the bacterial chromosome ○ The origin of replication in E.coli is termed OriC (Origin of Chromosomal Replication). There is a specific DNA sequence within the OriC where the replication of DNA beings ■ DnaA boxes- sites for binding DnaA protein ■ AT-rich regions-sites where the DNA strands separate ■ GATC methylation sites- sites that help regulate DNA replication ○ The specific sequences of bases in a functional region of DNA is very important for that function

3. List the rules followed by DNA polymerase. Rule #1: DNA polymerases cannot initiate DNA synthesis by linking two individual nucleotides ●

Problem is overcome by the RNA primers synthesized by primase

Rule #2: DNA polymerases can attach nucleotides only in the 5’ to 3’ direction, but the two strands are antiparallel and go in opposite directions ●

Problem is overcome by synthesizing the new strands both toward, and away from, the replication fork

4. label the leading and lagging strands at a replication fork.

5. Explain why telomeres and telomerase are so important in eukaryotic chromosomes. ●

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Telomere ○ refers to the complex of telomeric DNA sequences and bound proteins ○ Sequences consist of moderately repetitive tandem arrays ○ Tend to shorten in actively dividing cells ○ Cells become senescent when telomeres are short Telomerase ○ Is not expressed in most cells of our bodies but in many cancers, it is more active ○ Allows cancer cells to keep dividing since their chromosomes don’t continually get shorter ○ Contains protein and RNA ○ Enzymatic action of telomerase ■ Step 1: binding ■ Step 2: polymerization ■ Step 3: translocation

Chapter 12 Be able to: 1. Describe the central dogma of genetics and molecular biology. DNA replication: makes DNA copies that are transmitted from cell to cell and from parent to offspring. Chromosomal DNA: stores information in units called genes Transcription: Produces an RNA copy of a gene Messenger RNA: a temporary copy of a gene that contains information to makes a polypeptide

Translation: produces a polypeptide using the information in mRNA Polypeptide: becomes part of a functional protein that contributes to an organism’s traits 2. List the three stages of transcription ●

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Initiation ○ The promoter functions as a recognition site for transcription factors. The transcription factors enable RNA polymerase to bind to the promoter. Following binding, the DNA is denatured into a bubble known as the open complex Elongation ○ RNA polymerase slides along the DNA in an open complex to synthesize RNA Termination ○ A terminator is reached that causes RNA polymerase and the RNA transcript to dissociate from the DNA

3. Identify the functions of sigma factor and General Transcription factors in prokaryotic and eukaryotic transcription. ●

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Sigma factor ○ Transcription factor that recognizes bacterial promoter sequences and facilitates the binding of RNA polymerase to the promoter General transcription factors ○ Needed along RNA polymerase II to initiate transcription of protein-encoding genes

4. List the three RNA polymerases in eukaryotes and the genes they transcribe (mostly, which one does mRNAs?) ● ●

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Rna polymerase I ○ Transcribes all rRNA genes (except for the 5S rRNA) RNA polymerase II ○ Transcribes all protein-encoding (structural) genes (that is, it synthesizes all mRNAs) ○ Transcribes some snRNA genes needed for splicing RNA polymerase II ○ Transcribes all tRNA genes ○ Transcribes the 5S rRNA gene ○ Transcribes microRNA genes

5. Explain what the core promoter is. ● ● ●

It typically consists of the TATA box and transcriptional start site; TATA box not always present Important in determining the precise start point for transcription Produces a low level of transcription called basal transcription

6. Describe chopping, splicing, capping and tailing in RNA processing. ●

Chopping ○ The cleavage of a large RNA transcript into smaller pieces. One of more of the

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smaller pieces becomes a functional RNA molecule ○ Processing occurs in both prokaryotic and eukaryotic rRNAs and tRNAs. Splicing ○ Splicing involves both cleavage and joining of RNA molecules. The RNA is cleaved at two sites, which allows an internal segment of RNA, known as an intron, to be removed. After the intron is removed, the two ends of the RNA molecules are joined together ○ Splicing is common among eukaryotic pre-mRNAs, and it also occurs occ...