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The essence of meiosis is that




Homework II;Name;All questions to be answered are in bold type. Please highlight the correct answer;using the highlight tool in MSWORD. YOU must return the answered exam as a;MSWORD document, no other format is accepted. YOU must put your name in the file;name when you submit the answered homework: an example: SmithJ_HWII.doc !;Each correct answer is worth 3 points. Students always have problems with questions;18 onward. Due date: February 18, 2014. CHANGEDUSE THIS DATE.;1.;The essence of meiosis is that;a.;b.;c.;d.;e.;gametes receive two copies of each pair of homolog chromosomes.;diploid cells are formed.;each gamete receives a pair of homologous chromosomes.;gametes are formed with 2n chromosomes.;each gamete receives one member of each pair of homologous;chromosomes, and haploid gametes are formed.;2.;There are _____ molecules of DNA in a developing human ovum at the;beginning of prophase I;a.;46;b.;23;c.;92;d.;half as many(as compared to somatic cells);e.;twice as many (as compared to mature sperm);3;In comparing mitosis and meiosis, which of the following statements is;true?;a.;Meiosis II is more like mitosis than is meiosis I.;b.;Both processes result in four cells;c.;Pairing of homologues occurs in both;d.;Chromatids are present only in mitosis;e.;Meiosis I resembles mitosis;4.;A gene locus is;a.;recessive gene;b.;the location of an allele on a gene;c.;a sex chromosome;d.;the location of an allele on a chromosome;e.;a dominant gene;5.;The most accurate description of an organism with genotype AaBb is;a.;heterozygous recessive;b.;homozygous;c.;homozygous dominant;d.;homozygous recessive;e.;heterozygous dominant;6.;For Mendels explanation of inheritance to be correct;a.;the genes for the traits he studied have to be located on the same;chromosome.;b.;c.;d.;e.;genes cannot be transmitted independently of each other.;the combination of gametes at fertilization has to be due to chance.;only diploid organisms demonstrate inheritance patterns.;none of these apply;7.;8.;9.;Heritable units of information about traits are called;A homozygous recessive is written as ______ for the gene labeled B;An X-linked carrier is a;a.;homozygous dominant female;b.;heterozygous male;c.;homozygous recessive female;d.;homozygous female;e.;heterozygous female;10.;is;The sex chromosome composition of a person with Kleinfelters syndrome;a.;b.;c.;d.;e.;11.;XXX;XO;XXY;XYY;none of these;Gel electrophoresis separates the DNA fragments according to;a.;their length;b.;their mass;c.;their speed of travel through the gel;d.;the number of nucleotides in the fragment;e.;all of these;12.;The polymerase chain reaction (PCR) uses ________ and a heat resistant;DNA polymerase to rapidly increase the number of molecules of a DNA fragment.;13.;can reveal the order of nucleotide bases in a fragment of DNA.;14.;An X-linked recessive gene (d) causes deafness. A normal woman whose;father was deaf marries a deaf man. This is not a multiple choice.;a.;What are the possible genotypes for the mother of the deaf man?;b.;What are the possible genotypes for the father of the deaf man?;c.;What are the chances that the first son with be deaf?;d.;What are the chances that the first daughter will be deaf?;15.;a.;b.;c.;d.;16.;There are three alleles controlling ABO blood types, A and B are;codominant genes. AB produces AB blood type. The third allele O is;recessive to the other two alleles. Indicate which of these parents could;produce the given child?;Parents;Child;Yes or No;A X AB;O;AxO;O;A x AB;O;AB x AB;B;A genetic locus that is analyzed in many forensic and paternity testing;laboratories is the human leukocyte antigen locus known as HLA-DQ alpha.;There are four major alleles at this locus known as 1, 2, 3, and 4. How many;different genotypes are possible for these four alleles? HINT: Construct a;Punnett square. NO PARTIAL CREDIT;a.;8;b.;10;c.;12;d.;16;e.;24;17.;In humans, brown eyes (B) are dominant over blue (b). A brown-eyed man;marries a blue-eyed woman and they have three children, two are brown-eyed and;one is blue-eyed. (stolen from;;Draw the Punnett square that illustrates this marriage?;What is the mans genotype?;What are the genotypes of the children?;Below is a table of frequencies of alleles versus ancestry for the locus CSF1PO.;There are 18 alleles starting with 6 and ending with 15.2, if I counted corrected.;We are not going to worry about the details of allele naming. The four columns;are named allele, CSF1P0 302 Cau, CSF1PO 258 African Amer and CSF1PO;140 Hispanic. The number refers to the number of individuals used to calculate;allele frequencies. The total of all frequencies in a column must equal one. You;will notice that not all the alleles listed are found in each column. This probably;means that that allele, like 5 is extremely rare and not enough individuals were;sampled to detect its presence. Reading across a row starting at allele 11, we;find.30132,.24903 and.2986 as frequencies for Cau, AA and His populations.;Working backwards we can calculate then number of individuals having that;allele. For 11 CAU, we have 0.30132 x 302 people = 91 individuals rounded up.;For AA, we 0.24903 x 258 = 64 people, and for HIS, we have 0.29286 x 140 =;41 people.;18. Calculate the number of individuals that have the 13 allele in all three groups?;19.;What is frequency of all alleles that are not 10 in Cau and Hispanics? (For;example, with AA, the frequency is 0.25681, so the remaining is 1 -.25681 =.;74309);Ok. Moving on to calculate allele frequencies when gametes combine. That is what is;the probability (which is the same as frequency) that an 11 and 12 will be found in any;individual in a population. Let us look at two cases: Homozygote and heterozygote.;Remembering our laws we can only make a homozygote one way. In this case we;simply take that frequency for that allele and square it. Lets put this in terms of an;example. In the Cau populate, one is the frequency of individuals that would be;homozygote for 10. The frequency for 10 is 0.21689. The chance of both occurring in;the same person would be the chance to find it two timeswhich is 0.21689 x 0.21689 =;0.470 sufficiently rounded.;For heterozyogote (for example: CAU CSF1PO 11, 12 above) for we use 2 times the;frequency of 11 times frequency of 12. The two comes in because we can first pick 11;first then 12 and vice versa. Basically there are two ways to get the same resulr. So;reading the chart we have one allele frequency is.30132, and then the other allele;frequency is.36093. (sometimes, the first allele is assigned the variable p and the;second, the variable q);The heterozygote frequency is two times p times q which is 2pq = 2 x.30132 x.36093;=.21751;So let us relate this to gametes recombining due to SEX, In this case a 11, 12 crossing;with another 11, 12 for individuals mating randomly in the population.. Lets illustrate this;with a Punnet Square.;11 (.30132);11(.30132) 11,11 (.30132 x.30132);12(.36093) 11, 12 (.30132 x.36093);12(.36093);11,12(.30132 x.36093);12,12 (.36093 x.36093);If you add these values in the Punnet Square, you get this relationship;p2 + 2pq + q2 = 1. This is the Hardy-Weinberg law. When we calculate genotype;frequencies in forensics in this class, we assume Hardy-Weinberg equilibrium. This;states that from generation to generation the genotype frequencies do not change.;Apply what we learned excluding using Hardy-Weinbery equation to the following;questions;20.;What is the locus frequency for 8, 8 (usually written as just 8) for;AA?;21.;What is the locus frequency for 14, 8 for Caucasians?;22.;What is the locus frequency for 10.3 for Hispanics?;23.;What is the complementary sequence for sequences a. and b. shown;below.;NOTE: Please remember DNA has polaritythe sequence needs to be written in 5;(read five prime) to 3(read three prime) direction. For example;The complement of 5-AGCT-3 is not TCGA, because as written this means 5-TCGA-3;but 3-TCGA-5. Convention has it that we write the complementary sequence from 5 to;3, so you reverse 3-TCGA-5 to 5-AGCT-3, which is the correct answer.;a. 5-AGGTTAG-3;b. 3-AGGTTAC-5


Paper#18030 | Written in 18-Jul-2015

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