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190 Cards in this Set
- Front
- Back
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Function of Mitosis and Meiosis |
they ensure the genetic continuity of nucleated cells |
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Function of Mitosis |
necessary for the copying of the same genetic material into daughter cells which happen in somatic cells |
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Function of Meiosis |
divides the number of chromosomes exactly into half during the production of gametes in the germ line |
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Cell Structure and Location of Chromatin |
chromatin is within the nucleus/nuclear envelope |
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Gap 1 (G1) |
Growth and Metabolism, 10 hours |
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Interphase (S) |
DNA synthesis and chromosome duplication, 9 hours |
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DNA Replication |
making two identical copies of a chromosome |
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Centromere |
the point on a chromosome by which it is attached to a spindle fiber during cell division. (center of a chromosome) |
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Before Replication |
1 chromosome 1 chromatid (1 piece of DNA) |
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After Replication |
1 chromosome 2 chromatids (2 pieces of DNA) |
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Sister Chromatids |
identical copies |
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Gap 2 (G2) |
Preparation for mitosis, 4 hours |
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Mitosis (M) |
division, 1 hour. prophase. metaphase. anaphase. telophase. (cytokinesis) |
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Cell Cycle |
Gap 1, 10 hours. Interphase, 9 hours. Gap 2, 4 hours. Mitosis, 1 hour. |
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Number of chromatids in an organisms with 4 chromosomes in G1? in G2? |
G1 = 4 G2 = 8 |
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Eukaryotic Chromosomes |
- DNA carries the genes. - The molecule is linear. - Most of the time it is present in the cell in its uncoiled form associated with proteins and RNA (chromatin). - It's fragile and diffuse. - Problem fro cell division. - Solution? Package by multiple condensation. |
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A diploid cell entering mitosis |
Parent cell (2n) |
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A diploid cell entering mitosis (goal) |
two daughter cells identical to parent. two daughter cells (each 2n) |
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Centrioles |
centrosomes in animal cells consist of two centrioles. |
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Microtubules |
- form around centrosomes - build the spindle apparatus - some attach to kinetochore close to the centromeres of duplicated chromosomes during cell division. |
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Mitosis |
Interphase Prophase Metaphase Anaphase Telophase |
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Interphase |
step 1. chromosomes duplicate to produce sister chromatids |
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Prophase |
step 2. duplicated chromosomes condense. |
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Metaphase |
step 3. duplicated chromosomes migrate to the equatorial plane of the cell and the nuclear membrane breaks down. |
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Anaphase |
step 4. sister chromatids of each duplicated chromosome move to opposite poles of the cell. |
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Telophase |
step 5. chromosomes de-condense and new nuclear membranes form. |
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Cytokinesis |
step 6. membrane forms between daughter cells
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Q: At which point in the cell cycle will you likely not see chromosomes but un-condensed DNA? 1. During G1, when proteins for cell growth are being manufactured. 2. During synthesis when DNA is replicated 3. During G2, when protein microtubules for the spindle fibers are being made 4. During all of the above |
4. during all of the above.
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Every organisms has a species-specific number of chromosomes |
Each chromosome has a distinct set of genes in specific locations |
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Most animals and plants are... |
diploid. they have two homologous copies of each chromosome (one from the mother, one from the father) |
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Homologous Chromosomes... |
have the same genes in the same locations, but not necessarily the same alleles. |
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Human Karyotype |
homologous pairs of chromosomes present in a diploid organism |
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Meiosis is required for the process of... |
each parent contributing one set of homologs to the offspring. |
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Meiosis I (DNA copied in S) |
1. pair up/line up homologous chromosomes 2. separate homologs from each other 3. divide the cell |
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Meiosis !! |
1. line up individual chromosomes 2. pull chromatids apart 3. divide the cell |
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Each meiotic division is divided into 4 (5) stages |
Prophase/Prometaphase Metaphase Anaphase Telophase Prophase I of Meiosis 1 is subdivived into 5 further stage. |
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Prophase 1: Leptonema |
- begin of chromatin condensation and homology search - chromosomes, each consisting of two sister chromatids, begin to condense. |
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Prophase 1: Zygonema |
- initial alignment of chromosomes (rough pairing), synaptonemal complex begins to form - homologous chromosomes begin to pair |
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Prophase 1: Pachynema |
- synapsis completes: close pairing of homologs. Two pairs of sister chromatids occur in a tetrad. Crossing over occurs. - Homologous chromosomes are fully paired. |
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Chiasmata |
Tetrads consist of four chromatids or two homologous chromosomes, respectively (also called bivalent) |
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Prophase I: Diplonema |
- chiasmata occur between non-sister chromatids. crossing over visible. - homologous chromosomes separate, except at chiasmata. |
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Prophase 1: Diakinesis |
- chromosomes are pulled apart but remain connected through chiasmata - paired chromosomes condense further and become attached to spindle fibers. |
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Metaphase I |
paired chromosomes align on the equatorial plane in the cell. |
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Anaphase I |
homologous chromosomes disjoin and move to opposite poles of the cell. |
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Telophase I |
chromosome movement is completed and new nuclei begin to form |
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At the end of Meiosis I each daughter cell... |
- has one chromosome per homologous pair. - is HAPLOID - ( = reduction division) |
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At the end of Meiosis I each chromosome... |
- still has two chromatids - but the chromatids now have DIFFERENT information from each other |
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Prophase II |
chromosomes, each consisting of two sister chromatids, condense and become attached to spindle fibers. |
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Metaphase II |
chromosomes align on the equatorial plane in each cell |
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Anaphase II |
sister chromatids disjoin and move to opposite poles in each cell |
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Telophase II |
chromosomes de-condense and new nuclei begin to form |
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Cytokinesis |
the haploid daughter cells are separated by cytoplasmic membranes |
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When does crossing-over during cell division occur? i. before the separation of identical sister chromatids ii. after the formation of chiasmata iii. before DNA replication iv. after the formation of the synaptonemal complex |
only iv |
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Levels of Genetic Analysis |
transmission genetics. molecular genetics. population genetics. |
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Transmission Genetics |
the passage of genes from one generation to the next |
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Molecular definition of a gene |
a segment of DNA encoding a polypeptide (or an RNA chain) |
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General defintiion of a gene |
the fundamental unit of heredity, which carries information from one generation to the next |
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Alleles |
variants of genes - change their frequency in populations |
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Gregor Mendel (19th century) |
founder of the science of genetics |
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19th century misconception |
blending inheritance |
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Keys to Mendel's Success (1) |
chose a good study species - each mating makes many offspring (plenty of data) - short generation time (data can be gathered fast) - easy to arrange matings between plants |
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Keys to Mendel's Success (2) |
he looked at discontinuous traits |
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Continuous Traits |
traits which vary across an unbroken range of values |
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Discontinuous (or discrete) Traits |
traits which show two or more distinct and separate forms |
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Keys to Mendel's Success (3) |
he studied true-breeding strains. matings within true-breeding strains always produce offspring that resemble their parents. |
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Keys to Mendel's Success (4) |
he studied one trait at a time |
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Hybrid |
offspring of genetically dissimilar parents |
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Monohybrid |
hybrid between two strains that differ only at a single trait |
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Keys to Mendel's Success (5) |
he counted offspring. |
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Mendel proposed that flower color and other characters are determined by discrete inherited factors that we now call _____ |
genes |
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A plant has many genes, each one controlled a different _____ |
character (flower color, seed color, seed shape, etc.) |
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Each genes comes in different forms that determine the different versions of a character. Today we call these forms _____ |
alleles |
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Each gene is present as a _____. |
pair. The plant receives one copy from its mother and one from its father. |
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Homozygote |
an individual in which the two copies of a gene are the same allele |
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Heterozygote |
an individual in which the two copies of a gene are different alleles |
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Dominance |
a condition in which one member of an allele pair is manifested to the exclusion of the other. the manifested allele is dominant and the hidden allele is recessive. |
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Gene copies segregate equally into ____. |
gametes. |
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At fertilization, gametes unite at random to produce a _______ |
zygote. this restores paired gene copies. |
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zygote |
fertilized egg |
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Mendel's Law Of Segregation |
- the two copies of a gene separate (segregate) equally during gamete formation, so that half of the gametes carry one member of the pair, and half carry the other member. - during fertilization, gametes unite at random. that is, they fuse without regard to which allele is carried by each one. |
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Phenotype |
the observable characteristic of an organism |
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Genotype |
the specific allelic composition of an individual |
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There is no one-to-one relationship between _____ and _____ |
phenotype, genotype |
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In guinea pigs coat color is determined by a single gene with two alleles. A guinea pig from a true-breeding black strain is mated with a guinea pig from a true-breeding white strain. The F1 progeny are all black. Two of the F1 progeny are mated with each other. What proportion of the F2 progeny is expected to be white? |
1/4 |
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In guinea pigs coat color is determined by a single gene with two alleles. A guinea pig from a true-breeding black strain is mated with a guinea pig from a true-breeding white strain. The F1 progeny are all black. Two of the F1 progeny are mated with each other. What proportion of the black F2 progeny is expected to be homozygous? |
1/3 |
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Is there an easy way to identify the genotype of a specific trait? |
the testcross |
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Testcross |
pair genotype in question with homozygous recessive |
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In guinea pigs coat color is determined by a single gene with two alleles. Allele B is dominant and encodes for black coat. Allele b is recessive and encodes for white coat. if you cross a black guinea pig with a white one, what is the phenotypic ratio in the F1 generation? |
I cannot tell. |
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You analyze the offspring of a cross between two pea plants. You cross two pea plants, both of which have purple flowers. You are surprised to see that in the first generation, there are plants with both purple flowers and white flowers. What is the most likely reason for this? 1. one parent was homozygous recessive and one was heterozygous 2. a new mutation occurred 3. one parent was homozygous recessive and one was homozygous dominant 4. both parental plants were heterozygous 5. one parent was homozygous dominant and one was heterozygous |
4. both parental plants were heterozygous |
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At what phase in cell division are homologous chromosomes aligning in pairs on one plane? i. metaphase of mitosis ii. metaphase I of meiosis iii. metaphase II of meiosis a. in all three b. only i c. only ii d. only iii e. only ii and iii |
c. only ii |
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Between which chromatids does crossing-over normally occur? i. chromatids of heterologous chromosomes ii. sister chromatids iii. chromatids of homologous chromosomes a. in all three b. only i c. only ii d. only iii e. only i and iii |
d. only iii |
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At what stage are 8 pieces of DNA present in cells of an organism with 2n=8 chromosomes? a. gap 1 of cell cycle b. mitosis, metaphase c. meiosis, prophase I d. gametes |
a. gap 1 of cell cycle |
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What strategies were helpful when Gregor Mendel studied patterns of inheritances? i. study an organism with a short generation time ii. study an organism with many offspring per generation iii. focus on many traits at a time iv. look at continuous traits a. all of them b. only i and ii c. only iii and iv d. only i, ii, and iv |
b. only i and ii |
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In garden peas, flower color is determined by a single gene with two alleles. You make a cross between a purple and a white pea. In the resulting F1 generation, on average half of the offspring has purple flowers, half of the offspring has white flowers. What is the genotype of the parents? a. one parent is homozygous, one is heterozygous b. both are homozygous c. both are heterozygous d. i cannot tell |
a. one parent is homozygous, one is heterozygous |
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Law of Independent Assortment |
the alleles of genes on different chromosomes segregate (or assort) independently of each other. |
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Notation |
- capital letter indicates dominant allele. - lower case letter indicates recessive - often the letter indicates the recessive condition. |
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Product Rule of Probability |
- if two events A and B are independent, then the probability that they both occur is given by: - (probability that A occurs) X (probability that B occurs) - ex: a fair coin is flipped three times. The probability that it lands heads all three times is 1/2 X 1/2 X 1/2 = 1/8 |
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What is the probability of getting a 2 and a 6 when you roll a pair of dice? |
1/36 |
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What is the proportion of the F2 offspring have recessive (wrinkled) seed shape? (dihybrid cross: yellow, round X yellow, round) |
1/4 |
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Of the offspring with green seeds, what proportion have wrinkled seeds? (dihybrid cross: yellow, round X yellow, round) |
1/4 |
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In the dihybrid pea cross, assuming independent assortment, what proportion of the F2 offspring should be heterozygous for the color gene and homozygous wild-type (dominant) for the shape gene? |
1/8 |
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Punnet squares size _______ exponentially with number of hybrid genes. |
increases |
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You have a pea plant with the following genotype: Gg Ww Cc Dd. If you self the plant, what is the probability of obtaining the genotype Gg Ww CC Dd in the progeny? |
1/32 |
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You have a pea plant with the following genotype: Gg WW Cc Dd. If you self the plant, what is the probability of obtaining the genotype Gg WW CC dd in the progeny? |
1/32 |
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Steps in Identifying Single Genes That Affect A Trait |
1. identify a biological trait of interest 2. find mutants that affect this trait 3. cross mutant and wild type to create an F1 4. intercross the F1 to generate an F2 5. examine the F2 for Mendelian ratios |
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The X^2 test (chi square) is used in genetics to test whether observed offspring ratios match those expected from a particular hypothesis |
true |
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Chi Square Test Step 1 |
state the null hypothesis |
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Chi Square Test Step 2 |
count observed offspring numbers. |
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Chi Square Test Step 3 |
determine expected offspring numbers, if null hypothesis is true. |
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Chi Square Test Step 4 |
calculate the X^2 statistic from the observed and expected offspring numbers. |
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The critical value of the X^2 distribution depends on the number of degrees of freedom. |
True |
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Degrees Of Freedom |
number of phenotypic classes minus 1 |
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Chi Square Test Step 5 |
determine the degrees of freedom for the test |
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Chi Square Test Step 6 |
compare the calculated statistic to the appropriate 5% critical value. - if it exceeds the critical value, then the null hypothesis is rejected. - if it does not exceed the value, then the null hypothesis is not rejected. |
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Steps in Chi Square Test |
1. state the null hypothesis 2. count observed offspring numbers 3. calculate expected numbers under null hypothesis 4. calculate the chi square statistic 5. determine the number of degrees of freedom 6. compare statistic w appropriate critical value to determine whether the null hypothesis should be rejected |
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A plant has fruits that are either round or oblong. A cross between a plant with round fruit and a plant with oblong fruit yielded F1 hybrids that all had round fruit. When the F1 plants were crossed, they produced 80 F2 offspring. 68 of these had round fruit and 12 had oblong fruit. Are these data consistent with the hypothesis that fruit shape is determined by a single gene with two alleles? |
No. |
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The cross of two individuals with unknown genotypes produces 3 offspring with the same phenotype of one of the parents. What are the genotypes? P = A' x A'' F1 = 3 A' A. AA x aa B. Aa x aa C. aa x Aa D. all combinations are possible |
D. all combinations are possible |
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T/F: Human genetics relies on pedigree analysis |
True. |
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Pedigree Analysis |
the study of naturally occurring families to find informative matings |
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Pitfalls of Pedigree Analysis |
- may not find right kind of mating. - families are small; hard to test for mendelian ratios. - paternity sometimes mistaken - generation time very long |
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How to draw a pedigree |
- matings are shown by horizontal connectors - offspring are linked to parents by vertical lines, to siblings by horizontal line. - person who first comes to attention of a geneticist is called the proband |
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Autosomal Recessive key features |
- affected individuals can have unaffected parents - rare recessive traits tend to be sparsely distributed in pedigrees. - consanguineous marriages (inbreeding) increase the likelihood of a rare recessive phenotype - difficult or impossible to detect mendelian ratios in most human families (too small) |
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Autosomal |
on a chromosome which is not a sex-chromosome |
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Some autosomal recessive traits in humans |
albinism (lack of pigment) phenylketonuria sickle-cell anemia tay-sachs disease cystic fibrosis galactosemia |
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Autosomal Dominant key features |
- affected individuals always have an affected parent - disease appears in every generation - affected parents pass trait to both daughters and sons (therefore, not sex-linked) |
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Some autosomal dominant traits in humans |
achondroplasia (dwarfism) congenital night blindness huntington disease marfan syndrome neurofibromatosis widow's peak |
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x-linked recessive disorder key features |
- all male carriers affected - female offspring of affected males are carriers - male offspring of affected males are not carriers |
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T/F: Genetic counseling involves analyzing pedigrees to estimate the risk of hereditary disorders. |
True. |
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Important genetic counseling tasks |
- estimate an individual's likelihood of carrying a disease-causing gene - estimate a couple's likelihood of having one or more affected children |
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Additive rule of probability |
if two events A and B are mutually exclusive, then the probability that either of them occurs is given by: (prob that A occurs) + (prob that B occurs) |
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e+/e+ |
wildtype grey (two wildtype alleles) |
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e+/e- |
wildtype grey (one wildtype, one mutant allele) |
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e-/e- |
ebony (two mutant alleles) |
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codominance |
heterozygotes express the phenotype of both their alleles |
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incomplete dominance |
the phenotype of the heterozygote is intermediate between those of the two homozygotes, on some quantitative scale (color, size, etc.) |
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codominance (vs incomplete dominance) |
the phenotype of both allele is fully expressed in the heterozygote |
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The L(M) and L(N) alleles in the MN blood system are... A. incompletely dominant B. codominant C. recessive D. dominant E. heterozygous |
B. codominant |
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You make the following testcross: DdGgWw x ddggww. What is the proportion of the phenotype DGW in the offspring? A. 1/2 B. 1/4 C. 1/8 D. 1/16 E. 1/32 |
C. 1/8 |
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A cross yields 4 phenotype classes of offspring. You want to test whether the phenotype ratios are consistent with Mendelian genetics by using a Chi-square test. How many degrees of freedom do you have? A. 1 B. 3 C. 4 D. 5 E. 8 |
B. 3 |
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How do you decide whether different genes or different alleles cause a specific phenotype? |
Hypothesis 1: each mutation is in a separate gene. Hypothesis 2: both mutations are in the same gene |
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T/F: genes can have more than two alleles |
true |
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How do you decide whether different genes or different alleles cause a specific phenotype? Which hypothesis is correct? |
the answer is found with a test for allelism (complementation test) |
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test for allelism (complementation test) |
- cross a homozygote for the first mutation with a homozygote for the second mutation - if the offspring have the mutant phenotype, the mutations are in the same gene. - if the offspring have the wildtype phenotype, the mutation are in different genes. |
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result of test for allelism if each mutation is in the same gene... |
offspring have mutant phenotype because they are homozygous recessive at both genes. |
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result of test for allelism if each mutation is in a separate gene... |
offspring have wildtype phenotype, because they are heterozygous at both genes. |
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What causes a mutation to be dominant or recessive? |
- the role of the polypeptide structure. - genes produces different polypeptides that affect the phenotype. |
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Does a wild-type allele produce a functional polypeptide? |
Yes. |
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Does a recessive amorphic loss-of-function allele produce a functional polypeptide? |
No. (severe mutant phenotype) |
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Does a hypomorphic loss-of-function allele produce a functional polypeptide? |
It produces a partially functional polypeptide. (mild mutant phenotype) |
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T/F: Dominant-negative allele produces a polypeptide that interferes with the wild-type phenotype. |
True. (severe mutant phenotype) |
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loss-of-function mutation |
causes a gene to lose some or all of its normal function |
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gain-of-function mutation |
example: a dominant mutation of the antennapedia gene causes flies to grow legs where their antennae would normally be |
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T/F: each phenotype is affected by only one gene. |
False, every phenotype is affected by many genes. |
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T/F: cellular chemistry is controlled by pathways, each involved only one gene. |
False, the pathways involve several genes. |
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Drosophila eye color is an example of ________ gene interaction. |
epistatic. |
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Gene _______ occurs when two or more genes affect the same phenotype. |
interaction. |
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T/F: genes cane only be within the same pathway. |
False, genes cane be in different pathways. |
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epistasis |
the alleles at one gene mask the effect of the alleles at another |
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How can we tell whether two genes interact within the same pathway? |
- Usually, we don't know this information. - to detect interaction between two genes within the same pathway, we must perform a dihybrid cross. |
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if two interacting genes are not within the same pathway, the dihybrid cross generally creates four offspring phenotypes in a 9:3:3:1 ratio affecting the _________ character. |
same. |
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when each of the four genotypes has a distinct phenotype, this shows that the genes __________ acting within the same pathway. A. are. B. are not. |
b. are not. |
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if two genes do interact within the same pathway, there are ______ that four phenotypes in the offspring of the dihybrid cross. |
fewer. (9:7 ratio) |
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different pathway interaction ratio |
9:3:3:1 |
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same pathway ratio |
9:7 |
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T/F: mutations in either gene disrupt the pathway, causing the mutant phenotype. |
true. |
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strategy for studying gene interations |
1. find two mutants affecting the same phenotype 2. do a test for allelism to find out if two genes are involved. 3. if so, perform a dihybrid cross and examine the offspring ratios to infer the pathways governed by the two genes. |
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recessive epistasis |
- 9:3:4 phenotypic ratio - the overriding allele is epistatic. the masked gene is called hypostatic. |
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T/F: the epistatic gene is downstream of the hypostatic gene. |
false, it is upstream of the hypostatic gene. |
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T/F: if the upstream gene is "knocked out", it does not matter what alleles are present downstream. |
true. |
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A geneticist is studying flower-color mutations in petunias. He finds two recessive mutants that have white flowers, rather than the wild-type purple. He crosses a homozygote for the first mutation with a homozygote for the second mutation. All of the offspring have white flowers. Which of the following is the best way to write the type of genotype of these offspring? A. Aa B. aa C. a1a1 D. a1a2 E. Aa1Bb1 |
D. a1a2 |
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Wild type blue-eyed mary has blue flowers. Two genes control the pathway that makes the blue pigment: the product of gene W turns a white precursor into magenta pigment. The product of gene M turns the magenta pigment into blue pigment. Each gene has a recessive loss-of-function allele: w and m, respectively. A double-heterozygote (WwMm) is self-pollinated. What will be the color of the offspring with the genotype wwmm? A. white B. magenta C. blue |
A. white. |
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Wild type blue-eyed mary has blue flowers. Two genes control the pathway that makes the blue pigment: the product of gene W turns a white precursor into magenta pigment. The product of gene M turns the magenta pigment into blue pigment. Each gene has a recessive loss-of-function allele: w and m, respectively. A double-heterozygote (WwMm) is self-pollinated. What proportion of offspring will be magenta? A. 1/16 B. 3/16 C. 1/4 D. 7/16 E. 9/16 |
B. 3/16 |
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Wild type blue-eyed mary has blue flowers. Two genes control the pathway that makes the blue pigment: the product of gene W turns a white precursor into magenta pigment. The product of gene M turns the magenta pigment into blue pigment. Each gene has a recessive loss-of-function allele: w and m, respectively. A double-heterozygote (WwMm) is self-pollinated. What proportion of the offspring will be white? A. 1/16 B. 3/16 C. 1/4 D. 7/16 E. 9/16 |
C. 1/4 |
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The epistatic gene is ________ of the hypostatic gene. |
upstream. |
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pleiotropic mutation |
one that affects multiple phenotypes |
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penetrance |
the proportion of individuals with a genotype who show the phenotype associated with that genotype |
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incomplete penetrance |
occurs when some individuals fail to show the phenotype associated with a trait. |
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expressivity |
the degree to which a particular genotype is expressed in the phenotype. (expression of a phenotype can vary among individuals with the same genotype) |
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humany karyotype |
- 22 pairs of autosomes - 1 pair of sex chromosomes |
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hemizygous |
having or characterized by one or more genes (as in a genetic deficiency or in an X chromosome paired with a Y chromosome) that have no allelic counterparts |
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euchromatic |
active DNA with many genes |
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heterchromatic |
inactive DNA with almost no genes |
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PAR |
psuedoautosomal region |
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SRY |
sex-determining region Y |
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MSY |
male-specific region of the Y |
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female sexual characteristics |
1. in the absence of a Y chromosome, no TDF is produced. 2. the lack of TDF allows the cortex of the embryonic gonads to develop into ovaries. 3. in the absence of testosterone, the embryo develops female characteristics. |
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male sexual characteristics |
1. the testis-determining factor (TDF) is produced by a gene on the Y chromosome. 2. TDF induces the medulla of the embryonic gonads to develop into testes. 3. the testes produce testosterone, a hormone that initiates development of male sexual characteristics. |
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x-inactivation |
- in each of cell of a female's body, one of the two X chromosomes is turned off, so that almost none of its genes are expressed. - therefore, both males and females express only one copy of most X chromosome genes. |
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x-chromosome inactivation |
an inactivated mammalian x-chromosome is a highly condensed heterochromatic structure, visible under the microscope as a barr body |
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random x inactivation: the Lyon hypothesis |
states that in embryonic development, X-chromosomes are randomly inactivaed |
