Exam III end of book questions

name three essential structural elements of a functional eukaryotic chromosome and describe their function

(1) Centromere: Serves as the point of attachment for the spindle fibers (microtubules). (2) Telomeres or the natural ends of the linear eukaryotic chromosome: Serve to stabilize the ends of the chromosome; may have a role in limiting cell division. (3) Origins of replication: Serve as the starting place for DNA synthesis.

In the snail Cepaea nemoralis, an autosomal allele causing a banded shell (BB) is recessive to the allele for an unbanded shell (BO). Genes at a different locus determine the background color of the shell; here, yellow (CY) is recessive to brown (CBw). A banded, yellow snail is crossed with a homozygous brown, unbanded snail. The F1 are then crossed with banded, yellow snails (a testcross). a. What will the results of the testcross be if the loci that control banding and color are linked with no crossing over?

(a) P generation - BB -----> CYxBO -------> CBW ---------> BB CYBO CBW F1 generation BB ------------> CY ------------> BO CBWF or F2 generation: BB ------------> CYxBB -------> CY --------> BO CBWBB CYBB -------->CYBB ---------> CY --------------> BB CYBO CBW which has 50% banded yellow and 50% unbanded brown. (b) P generation is BBBBCYCYx BOBOCBWCBW F1 generation BOBBCBWCYF or F2 generation BOBBCBWCY x BBBBCYCYBO CBWBOCYBBC BWBBCYBBCY BOBBCBWCY Unbanded brown - BOBBCYCY Unbanded yellow - BBBBCBWCY Banded brown - BBBBCYCY . Banded yellow = 25% unbanded brown = 25% unbanded yellow = 25% banded brown = 25% banded yellow (c) P generation - BB ------> CYxBO --------> CBW ------------> BB CYBO CBW F1 generation - BB --------> CY ----------> BO CBWF or F2 generation: BB ----------> CYxBB --------> CY ------------> BO CBWBB CYGenes are linked and are 20 mu apart - the F1 double heterozygote should form recombinant gametes 20% of the time (10% BBCBW and 10%BOCY), and nonrecombinant gametes 80% of the time (40% BBCY and 40% BOCBW).BB --------> CYBO -------->CBWBB -------> CBWBO ---------> CY --------> BB CYBB CYBB CYBB CY with 40% banded yellow , 40% unbanded brown , 10% banded brown and 10% banded yellow

What two processes unique to meiosis are responsible for genetic variation?

-crossing over of homologous pairs (zygotene stage of prophase I) -random distribution of homologous chromosomes (anaphase I)

In cats, curled ears result from an allele (Cu) that is dominant over an allele (cu) for normal ears. Black color results from an independently assorting allele (G) that is dominant over an allele for gray (g). A gray cat homozygous for curled ears is mated with a homozygous black cat with normal ears. All the F1 cats are black and have curled ears. An F1 cat mates with a stray cat that is gray and possesses normal ears. What phenotypes and proportions of progeny are expected from this cross? Is it: 1) 9/16 black cats, curled ears; 3/16 black cats, normal ears; 3/16 gray cats, curled ears; and 1/16 gray cats, normal ears 2) All gray cats; normal ears 3) All black cats with curled ears 4) 1/4 black cats, curled ears; 1/4 black cats, normal ears; 1/4 gray cats, curled ears and 1/4 gray cats, normal ears 5) 9/16 gray cats, normal ears; 3/16 gray cats, curled ears; 3/16 black cats, normal ears; and 1/16 black cats, curled ears

1) 9/16 black cats, curled ears; 3/16 black cats, normal ears; 3/16 gray cats, curled ears; and 1/16 gray cats, normal ears Given that, the gray cat homozygous for curled ears (gg CuCu) is mated with a homozygous black cat with normal ears (GG cucu). The genotypes of the offspring resulting from this cross are as follows. gg CuCu* GG cucu = Gg Cucu (100% black and curled ears). ----------> F1 generation. Cross between F1 offspring will have the progeny with the following genotypes and phenotypes. Gg Cucu* Gg Cucu = GG CuCu (1/16, black cats, curled ears), GG Cucu (2/16, black cats, curled ears), Gg CuCu (2/16, black cats, curled ears), Gg CuCu (4/16, black cats, curled ears), GG cucu (1/16, black cats, normal ears), gG cucu (2/16, black cats, normal ears), gg Cucu (2/16, gray cats, curled ears), gg CuCu (1/16,gray cats, curled ears), gg cucu (1/16, gray cats, normal ears).

list the stages of interphase and the major events that take place in each stage also add in check points

1. G1: In this phase, the cell grows and synthesizes proteins necessary for cell division. During G1, the G1/S checkpoint takes place. Once the cell passes the checkpoint, it is ready to divide. 2. S phase: During S phase, DNA replication takes place 3. G2: Additional biochemical reactions take place that prepare the cell for mitosis. A major checkpoint in G2 is the G2/M checkpoint. Once the cell has passed this checkpoint, it enters mitosis G0 phase: at the G2/S checkpoint. Cells may exit the active cell cycle and enter into a nondividing phase.

similarities and difference between mitosis and meosis

1. Mitosis is a single cell division that produces two GENETICALLY IDENTICAL progeny cells. Meiosis has two cell divisions that usually result in four progeny cells that are not genetically identical 2. Chromosome number of progeny cells and the original cell remain the same. Daughter cells are haploid and have half the chromosomal complement of the original cell as a result of the separation of homologous pairs during anaphase I 3. Daughter cells and the original cell are genetically identical. No separation of homologous chromosomes or crossing over takes place. In Meiosis, crossing over in prophase I and separation of homologous chromosomes in anaphase I produce daughter cells that are genetically different from each other and from the original cell. 4. Homologous chromosomes do not synapse in Mitosis. In Meiosis, prophase I has synapsis. 5. In metaphase, individual chromosomes line up on the metaphase plate. In metaphase I, homologous chromosomes line up on metaphase and metaphase II has sister chromatids 6. In anaphase, sister chromatids separate. Anaphase 1, homologous chromosomes separate, but sister chromatids separate in Anaphase II

what is the difference between autopolyploidy and allopolyploidy? How does each arise?

10. * What is the difference between autopolyploidy and allopolyploidy? How does each arise? In autopolyploidy, all sets of chromosomes are derived from a single species. In allopolyploidy, the sets of chromosomes are derived from two or more different species. Autopolyploidy may arise through nondisjunction in an early 2n embryo or through meiotic nondisjunction that produces a gamete with extra sets of chromosomes. Allopolyploidy is usually preceded by hybridization between two different species, followed by chromosome doubling.

what are the major results of meiosis

4 unique daughter cells that are haploid (contain half the number of chromosomes of the original diploid cell)

a green-nose fly ormally has six chrosomes: two metacentric and four acrocentric. A geneticist exmines the chrosomes of an odd-looking green-nose fly and discoveres that it only has 5 chrosomes: three of them are metacentric and two are acrocentric. Explain how this changes in chromosome number might have taken place.

A Robertsonian translocation between two of the acrocentric chromosomes would result in a new metacentric chromosome and a very small chromosome that may have been lost.

What is the difference between a genetic map and a physical map?

A genetic map gives the order of genes and relative distance between them based on recombination frequencies observed in genetic crosses. A physical map locates genes on the actual chromosome or DNA sequence, and thus represents the physical distance between genes.

How is the chi-square goodness-of-fit test used to analyze genetic crosses? What does the probability associated with a chi-square value indicate about the results of a cross?

A statistical method used to evaluate the role of chance in causing deviations between the observed and expected numbers of offspring produced in a genetic cross. The probability value obtained from the chi-squared table refers to the probability that random chance produced the deviations of the observed numbers from the expected numbers.

How can inversions in which no genetic information is lost or gained cause phenotypic effects?

Although inversions do not result in loss or duplication of chromosomal material, inversions can have phenotypic consequences if the inversion disrupts a gene at one of its breakpoints or if a gene near a breakpoint is altered in its expression because of a change in its chromosomal environment, such as relocation to a heterochromatic region. Such effects on gene expression are called position effects

what features are exhibited by a pedigree of recessive trait? what features are exhibited if the trait is dominant?

Assuming autosomal inheritance and complete penetrance, recessive inheritance is indicated by the presence of affected offspring from two unaffected parents. If the trait is rare, frequently the recessive allele will be passed for a number of generations without the trait appearing in the pedigree. For rare recessive traits, the trait often appears as a result of mating between two close relatives. Finally, if the trait is recessive, all of the offspring of two affected parents will be affected. Assuming autosomal inheritance and complete penetrance, at least one of the parents of affected children should be affected (unless a new mutation has led to creation of a new dominant allele). The trait should not skip generations within a lineage. Two affected parents who are heterozygotes can have unaffected offspring. Unaffected parents do not transmit the trait to their offspring.

What is a Barr body? How is it related to the Lyon hypothesis

Barr bodies are darkly staining bodies in the nuclei of female mammalian cells. Mary Lyon correctly hypothesized that Barr bodies are inactivated (condensed) X chromosomes. By inactivating all X chromosomes beyond one, female cells achieve dosage compensation for X-linked genes.

Bob has XXY chromosomes (Klinefelter syndrome) and is color blind. His mother and father have normal color vision, but his maternal grandfather is color blind. Assume that Bob's chromosome abnormality arose from nondisjunction in meiosis. In which parent and in which meiotic division did nondisjunction take place? Explain your answer

Because Bob must have inherited the Y chromosome from his father, and his father has normal color vision, a nondisjunction event in the paternal lineage cannot account for Bob's genotype. Bob's mother must be heterozygous X+Xc because she has normal color vision, and she must have inherited a color-blind X chromosome from her color-blind father. For Bob to inherit two color-blind X chromosomes from his mother, the egg must have arisen from a nondisjunction in meiosis II. In meiosis I, the homologous X chromosomes separate, and so one cell has the X+ chromosome and the other has Xc. The failure of sister chromatids to separate in meiosis II would then result in an egg with two copies of Xc.

list the different types of chromosome mutations and define each one

Chromosome rearrangements: Deletion: Loss of a part of a chromosome. Duplication: Addition of an extra copy of a part of a chromosome. Inversion: A part of a chromosome is reversed in orientation. Translocation: A part of one chromosome becomes incorporated into a different (nonhomologous) chromosome. Aneuploidy: Loss or gain of one or more chromosomes, causing the chromosome number to deviate from 2n or the normal euploid complement. Polyploidy: Gain of entire sets of chromosomes, causing the chromosome number to change from 2n to 3n (triploid), 4n (tetraploid), and so on.

What is a complementation test and what is it used for?

Complementation tests are used to determine whether different recessive mutations affect the same gene or locus (are allelic) or whether they affect different genes. The two mutations are introduced into the same individual by crossing homozygotes for each of the mutants. If the progeny show a mutant phenotype, then the mutations are allelic (in the same gene). If the progeny show a wild-type (dominant) phenotype, then the mutations are in different genes and are said to complement each other because each of the mutant parents can supply a functional copy (or dominant allele) of the gene mutated in the other parent.

What effect does crossing over have on linkage?

Crossing over generates recombination between genes located on the same chromosome, and thus reduces the amount of linkage.

Why is the frequency of recombinant gametes always half the frequency of crossing over?

Crossing over occurs at the four-strand stage, when two homologous chromosomes, each consisting of a pair of sister chromatids, are paired. Each crossover involves just two of the four strands and generates two recombinant gametes. The remaining two strands that were not involved in the crossover generate two non recombinant gametes. Therefore, the frequency of recombinant gametes is always half the frequency of crossovers.

Explain how to determine, using the numbers of progeny from a three-point cross, which of three linked loci is the middle locus.

Double crossovers always result in switching the middle gene with respect to the two nonrecombinant chromosomes. Hence, one can compare the two double crossover phenotypes with the two nonrecombinant phenotypes and see which gene is reversed. In the diagram below, we see that the coupling relationship of the middle gene is flipped in the double crossovers with respect to the genes on either side. Therefore, whichever gene on the double crossover can be altered to make the double crossover resemble a nonrecombinant chromosome is the middle gene. If we take either of the double crossover products l M r or L m R, changing the M gene will make it resemble a non recombinant.

the amount of DNA per cell of a particular species is measured in cells found at various stages of meiosis and the following amounts are obtained ____3.7 pg . _____7.3 pg . _____14.6 pg

G1 and Metaphase II : 7.3pg Following Telophase and cytokinesis: 3.7pg Prophase I, G2, Anaphase I: 14.6 pg

What is gene interaction? What is the difference between an epistatic gene and a hypostatic gene?

Gene interaction is the determination of a single trait or phenotype by genes at more than one locus; the effect of one gene on a trait depends on the effects of a different gene located elsewhere in the genome. One type of gene interaction is epistasis. The alleles at the epistatic gene mask or repress the effects of alleles at another gene. The gene whose alleles are masked or repressed is called the hypostatic gene.

What is the difference between genes in coupling configuration and genes in repulsion? What effect does the arrangement of linked genes (whether they are in coupling configuration or in repulsion) have on the results of a cross?

Genes in coupling have two wild-type alleles on one homologous chromosome and the two mutant alleles on the other homologous chromosome. Genes in repulsion have a wild-type allele of one gene together with the mutant allele of the second gene on one homologous chromosome, and vice versa on the other homologous chromosome. The two arrangements have opposite effects on the results of a cross. For genes in coupling configuration, most of the progeny will be either wild type for both genes, or mutant for both genes, with relatively few that are wild type for one gene and mutant for the other. For genes in repulsion, most of the progeny will be mutant for only one gene and wild-type for the other, with relatively few recombinants that are wild-type for both or mutant for both.

What is the difference between genotype and phenotype?

Genotype refers to the genes or the set of alleles found within an individual. Phenotype refers to the manifestation of a particular character or trait.

a summer squash plant that produces disc-shaped fruit is crossed wit a summer-swash plant that produces long fruit all the F1 have discshaped fruit what the F1 are intercossed F2 progeeney are produced in the following ratio 9/16 disc-shaped fruit, 6/16 spherical, 1/16 long.Give the genotypes of F2 prog.

Genotypic ratios: 1/16 will be homozygous dominant for both traits (WWDD) 2/16 will be homozygous dominant for color and heterozygous for shape (WWDd) 2/16 will be heterozygous for color and homozygous dominant for shape (WwDD) 1/16 will be homozygous dominant for color and homozygous recessive for shape (WWdd) 4/16 will be heterozygous for both traits (WwDd) 2/16 will be heteozygous for color and homozygous recessive for shape (Wwdd) 1/16 will be homozygous recessive for color and homozygous dominant for shape (wwDD) 2/16 will be homozygous recessive for color and heterozygous for shape (wwDd) 1/16 will be homozygous recessive for both traits (wwdd) This is a 1:2:2:1:4:2:1:2:1 genotypic ratio! Yuck! Phenotypic ratios: 9/16 will have white, disk-shaped fruit 3/16 will have white, sphere-shaped fruit 3/16 will have yellow, disk-shaped fruit 1/16 will have yellow, sphere-shaped fruit This is a 9:3:3:1 phenotypic ratio.

outline the processes of spermatogenesis and oogenesis in animals

In animals, spermatogenesis occurs in the testes. Primordial diploid germ cells divide mitotically to produce diploid spermatogonia that can either divide repeatedly by mitosis or enter meiosis. A spermatogonium that has entered prophase I of meiosis is called a primary spermatocyte and is diploid. Upon completion of meiosis I, two haploid cells, called secondary spermatocytes, are produced. Upon completing meiosis II, the secondary spermatocytes produce a total of four haploid spermatids. Female animals produce eggs through the process of oogenesis. Similar to what takes place in spermatogenesis, primordial diploid cells divide mitotically to produce diploid oogonia that can divide repeatedly by mitosis, or enter meiosis. An oogonium that has entered prophase I is called a primary oocyte and is diploid. Upon completion of meiosis I, the cell divides, but unequally. One of the newly produced haploid cells receives most of the cytoplasm and is called the secondary oocyte. The other haploid cell receives only a small portion of the cytoplasm and is called the first polar body. Ultimately, the secondary oocyte will complete meiosis II and produce two haploid cells. One cell, the ovum, will receive most of the cytoplasm from the secondary oocyte. The smaller haploid cell is called the second polar body. Typically, the polar bodies disintegrate, and only the ovum is capable of being fertilized.

What is meant by genetic sex determination?

In organisms that follow this system, there is no cytogenetically recognizable difference in the chromosome contents of males and females. Instead of a sex chromosome that differs between males and females, alleles at one or more loci determine the sex of the individual.

How does sex determination in the XX-XY system differ from sex determination in the ZZ-ZW system?

In the XX-XY system, males are heterogametic and produce gametes with either an X chromosome or a Y chromosome. In the ZZ-ZW system, females are heterogametic and produce gametes with either a Z or a W chromosome.

how do incoplete dominance and codominance differ?

Incomplete dominance means the phenotype of the heterozygote is intermediate to the phenotypes of the homozygotes. Codominance refers to situations in which both alleles are expressed and both phenotypes are manifested simultaneously.

how do translocation in which no genetic information is lost of gained produce phenotypic effects?

Like inversions, translocations can produce phenotypic effects if the translocation breakpoint disrupts a gene or if a gene near the breakpoint is altered in its expression because of relocation to a different chromosomal environment (a position effect)

what characteristics are exhibited by an X-linked trait?

Males show the phenotypes of all X-linked traits, regardless of whether the X-linked allele is normally recessive or dominant. Males inherit X-linked traits from their mothers, pass X-linked traits to their daughters, and through their daughters, to their daughters' descendants but not to their sons or their sons' descendants.

what are the stages of meiosis and what major evets take place in each stage

Meiosis 1: separation of homologous chromosomes Prophase 1: The chromosomes condense and homologous pairs of chromosomes undergo synapsis. While the chromosomes are synapsed, crossing over occurs. The nuclear membrane disintegrates and the meiotic spindle begins to form. Metaphase 1: The homologous pairs of chromosomes line up on the equatorial plane on the metaphase plate Anaphase 1: Homologous chromosomes separate and move on opposite poles of the cel. Each chromosome possesses two sister chromatids Telophase 1: The separated homologous chromosomes reach the spindle poles and are at opposite ends of the cell ~~Followed by cytokinesis: results in the division of the cytoplasm and the production of two haploid cells. These cells may skip directly into meiosis II or enter interkinesis, where nuclear envelope reforms and spindle fibers break down Meiosis II: Separation of sister chromatids Prophase II: Chromosomes condense, nuclear envelope breaks down, and spindle fibers form Metaphase II: Chromosomes line up at the equatorial plane of the metaphase plate Anaphase II: The centromeres split, which results in the separation of sister chromatids Telophase II: The daughter chromosomes arrive at the poles of the spindle. The nuclear envelope reforms, and the spindle fibers break down. Following meiosis II, cytokinesis takes place.

Why was Mendel's approach to the study of heredity so successful?

Mendel was successful for several reasons. He chose to work with a plant, Pisum sativum, that was easy to cultivate, grew relatively rapidly, and produced many offspring whose phenotype was easy to determine, which allowed Mendel to detect mathematical ratios of progeny phenotypes. The seven characteristics that he chose to study exhibited only a few distinct phenotypes and did not show a range of variation. Finally, by looking at each trait separately and counting the numbers of the different phenotypes, Mendel adopted a reductionist experimental approach and applied the scientific method. From his observations, he proposed hypotheses that he was then able to test empirically.

How are Mendel's principles different from the concept of blending inheritance discussed in Chapter 1?

Mendel's principles assert that the genetic factors or alleles are discrete units that remain separate in an individual organism with a trait encoded by the dominant allele being the only one observed if two different alleles are present. According to Mendel's principles, if an individual contains two different alleles, then the individual's gametes could contain either of these two alleles (but not both). Blending inheritance proposes that offspring are the result of blended genetic material from the parent and the genetic factors are not discrete units. Once blended, the combined genetic material could not be separated from each other in future generations.

four major types of aneuplodiy

Nullisomy: having no copies of a chromosome Monosomy: having only one copy of a chromosome Trisomy: having three copies of a chromosome Tetrasomy: having four copies of a chromosome

In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics 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 intercrossed to produce the F2. a. What are the phenotypic ratios in the F2? b. If an F1 plant is backcrossed with the bitter, yellow-spotted parent, what phenotypes and proportions are expected in the offspring? c. If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring?

P - BBSS x bbss F1 - BbSs x BbSs a. 9 (bitter w/ spots): 3 (bitter w/ no spots) : 3 (sweet w/ spots) : 1 (sweet w/ no spots) b. 100% bitter with spots . BBSS - 1/4; BbSs - 1/4; BBSs - 1/4; BbSS- 1/4 c. BbSs - Bitter w/ spots - 1/4 ; Bbss - bitter w/ no spots - 1/4 ; bbSs - sweet w/ spots - 1/4 ; bbss - sweet with no spots - 1/4

what is the difference between primary down syndrome and familial down syndrome? How does each type arise?

Primary Down syndrome is caused by spontaneous, random nondisjunction of chromosome 21, leading to Trisomy 21. Familial DS arises from Robertsonian translocation between chromosome 21 and 14. Translocation carriers do not have DS, but their children have an increased incidence of it. If the translocated chromosome segregates with a normal 21, the gametes will have two copies of chromosome 21 and result in DS

list the stages of mitosis and the major events that take place in each stage

Prophase: The chromosomes condense and become visible, the centrosomes move apart, and microtubule fibers form from the centrosomes Prometaphase: nucleoli disappear and the nuclear envelope begins to disintegrate, allowing for the cytoplasm and nucleoplasm to join. Sister chromatids begin to attach to microtubule from the opposite centrosomes Metaphase: The spindle microtubules are clearly visible and the chromosomes arrange themselves on the equatorial plane of the cell Anaphase: The sister chromatids separate at the centromeres after the breakdown of cohesin protein, and the newly formed daughter chromosomes move to the opposite poles of the cell Telophase: the nuclear envelope reforms around each set of daughter chromosomes. Nucleoli reappear. Spindle microtubules disintegrate

what does the term recombination mean? What are the two causes of recombination?

Recombination means that meiosis generates gametes with different allelic combinations than the original gametes the organism inherited. If the organism was created by the fusion of an egg bearing AB and a sperm bearing ab, recombination generates gametes that are Ab and aB. Recombination may be caused by loci on different chromosomes that assort independently or by a physical crossing over between two loci on the same chromosome, with breakage and exchange of strands of homologous chromosomes paired in meiotic prophase I.

Red-green color blindness in humans is due to an X-linked recessive gene. Both John and Cathy have normal color vision. After 10 years of marriage to John, Cathy gave birth to a color-blind daughter. John filed for divorce, claiming that he is not the father of the child. Is John justified in his claim of nonpaternity? Explain why. If Cathy had given birth to a color-blind son, would John be justified in claiming nonpaternity?

Since color blindness is a recessive trait, the color-blind daughter must be homozygous recessive. Because the color blindness is X-linked, John has grounds for suspicion. Normally, their daughter would have inherited John's X chromosome. Because John is not color blind, he could not have transmitted ac color-blind X chromosome to his daughter. A few remote alternative possibilities could be considered: 1) the daughter is XO, having inherited a recessive color-blind allele from her mother & no sex chromosome from her father. In that case, the daughter would have Turner syndrome. 2) John has Klinefelter's (XXY) & is heterozygous for the color-blind allele. 3) daughter is heterozygous, but the X w/ the normal vision allele was inactivated in both eyes. 4) A new X-linked color-blind mutation is possible, albeit even less likely. We do not consider autosomal color blindness bc the problem states this is an X-linked trait. If Cathy had a color-blind son, then John would have no grounds for suspicion. The son would have inherited John's Y chromosome and the color-blind X chromosome from Cathy.

What is the concept of dominance?

The concept of dominance states that when two different alleles are present in a genotype, only the dominant allele is expressed in the phenotype. Incomplete dominance occurs when different alleles are expressed in a heterozygous individual, and the resulting phenotype is intermediate to the phenotypes of the two homozygotes.

Why do extra copies of genes sometimes cause drastic phenotypic effects?

The expression of some genes is balanced with the expression of other genes; the ratios of their gene products, usually proteins, must be maintained within a narrow range for proper cell function. Extra copies of one of these genes cause that gene to be expressed at proportionately higher levels, thereby upsetting the balance of gene products

What are the addition and multiplication rules of probability and when should they be used?

The multiplication rule allows for predicting the probability of two or more independent events occurring together. According to the multiplication rule, the probability of two independent events occurring together is the product of their probabilities of occurring independently. The addition rule allows for predicting the likelihood of a single event that can happen in two or more ways. It states that the probability of a single mutually exclusive event can be determined by adding the probabilities of the two or more different ways in which this single event could take place. The multiplication rule allows us to predict how alleles from each parent can combine to produce offspring, while the addition rule is useful in predicting phenotypic ratios once the probability of each type of progeny can be determined.

in which phase of mitosis and meiosis are the principals of segregation and independent assortment at work?

The principle of independent assortment states that alleles at different loci segregate independently of one another. The principle of independent assortment is an extension of the principle of segregation: the principle of segregation states that the two alleles at a locus separate; according to the principle of independent assortment, when these two alleles separate, their separation is independent of the separation of alleles at other loci.

What is the principle of segregation? Why is it important?

The principle of segregation states that an organism possesses two alleles for any particular characteristic. These alleles separate during the formation of gametes. In other words, one allele goes into each gamete. The principle of segregation is important because it explains how the genotypic ratios in the haploid gametes are produced

assume that long earlobes in humans are an autosomal dominate trait that exhibits 30% penetrance. A person who is heterozygous for long earlobes mates with a person who is homozygous fro normal earlobes. What is the probability that thier child will have long earlobes?

To have long ear lobes, the child must inherit the dominant allele and also express it. The probability of inheriting the dominant allele is 50%; the probability of expressing it is 30%. The combined probability of both is 0.5(0.3) = 0.15, or 15%

Explain why tortoiseshell cats are almost always female and why they have a patchy distribution of orange and black fur.

Tortoiseshell cats have 2 different allele of an X-linked gene: X+ (non-orange, or black) and Xo (orange). The patchy distribution results from X-inactivation during early embryo development. Each cell of the early embryo randomly inactivates one of the two X chromosomes, and the inactivation is maintained in all of the daughter cells. So each patch of black fur arises from a single embryonic cell that inactivated the X+. Normal male cats have only one X chromosome, so they cannot have patches of black and orange fur.

E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings 853 virescent-white seedlings and 260 yellow seedlings a) Give the genotypes for the green, virescent-white, yellow seedlings b) Explain how color is determined in these seedlings (which genotype gives which phenotype) *Show/explain how you got to your result, i.e., do not just write a number or phenotype with no explanation as to why you say so. Show the relevant branches of branch diagrams.

Total number of seedlings = 3583 + 853 + 260 = 4696 The frequency of green seedlings = 3583 / 4696 = 0.76 The frequency of virescent white seedlings = 853 / 4696 = 0.18 The frequency of yellow seedlings = 260 / 4696 = 0.05 Of these frequencies, green seedlings appear to be dominant. Only three phenotypes are observed and this might be due to incomplete dominance at one locus. Therefore, the genotypes would be GG or Gg for green, YY o Yy for virescent - white and yy for yellow seedlings. These three phenotypes should segregate according to dihybrid ratio of 9:3:3:1 Instead of this ratio, the phenotypes segregated as 12 :3:1 GGYY or Gg yy appears green (9+3 = 12) ggYY appeasr virescent - white (3) ggyy appears yellow (1) The epistatic interaction of G over Y results in this ratio. Since, G is dominant and mask the effect of allele Y, the GGYY genotype also appears green. When the 'g' is recessive Y is expressed and appears as virescent - white. Both recessive alleles g and y appear as yellow.

what characteristics of an organism would make it suitable for studies of the principles of inheritance? Can you name several organisms that have these characteristics?

Useful characteristics • Are easy to grow and maintain • Grow rapidly, producing many generations in a short period • Produce large numbers of offspring • Have distinctive phenotypes that are easy to recognize Examples of organisms that meet these criteria • Neurospora, a fungus • Saccharomyces cerevisiae, a yeast • Arabidopsis, a plant • Caenorhabditis elegans, a nematode • Drosophilia melanogaster, a fruit fly

What is the chromosome theory of heredity? Why was it important?

Walter Sutton's chromosome theory of inheritance states that genes are located on the chromosomes. The independent segregation of pairs of homologous chromosomes in meiosis provides the biological basis for Mendel's two principles of heredity.

Alkaptonuria is a metabolic disorder in which affected persons produce black urine. Alkaptonuria results from a disease allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, but her brother has alkaptonuria. Sally's father has alkaptonuria, and her mother has normal metabolism. If Sally's parents have another child, what is the probability that the child will have alkaptonuria?

a) Sally: Aa Brother: aa Mom: Aa Dad: aa b) 50% chance c) 50% chance

Species I is diploid (2n = 8) with chromosomes AABBCCDD; related species II is diploid (2n = 8) with chromosomes MMNNOOPP. What types of chromosome mutations do individual organisms with the following sets of chromosomes have?

a. AAABBCCDD [tandem duplication] (there is an extra copy of A right next to the other A's) b. MMNNOOOOPP [tetrasomic] c. AABBCDD [deletion] d. AAABBBCCCDDD [autotriploid] e. AAABBCCDDD [double trisomic] f. AABBDD [nullisomic] g. AABBCCDDMMNNOOPP [allotetraploid] (created as a result of both chromosome sets of each parents being present in the gametes) h. AABBCCDDMNOP [allotriploid] (2n from one species and 1n from another species)

Bill and Betty have had two children with Down syndrome. Bill's brother has Down syndrome and his sister has two children with Down syndrome. On the basis of these observations, indicate which of the following statements are most likely correct and which are most likely incorrect. a. Bill has 47 chromosomes. b. Betty has 47 chromosomes. c. Bill and Betty's children each have 47 chromosomes. d. Bill's sister has 45 chromosomes. e. Bill has 46 chromosomes. [ f. Betty has 45 chromosomes. [ g. Bill's brother has 45 chromosomes.[f

a. Bill has 47 chromosomes. [false] b. Betty has 47 chromosomes. [false] c. Bill and Betty's children each have 47 chromosomes. [false] d. Bill's sister has 45 chromosomes. [true] e. Bill has 46 chromosomes. [false] f. Betty has 45 chromosomes. [false] g. Bill's brother has 45 chromosomes.[false] he high incidence of Down syndrome in Bill's family and among Bill's Relatives is consistent with familial Down syndrome, caused by a Robertsonian(creates a chromosome with two long arms and a chromosome with two short arms-(which is usually lost) translocation involving chromosome 21. Bill and his sister, who are unaffected, are phenotypically normal carriers of the translocation and have 4 chromosomes. Their children and Bill's brother, who have Down syndrome, have 46 chromosomes, but one of these chromosomes is the translocation that has an extra copy of the long arm of chromosome 21. From the info given, there is no reason to suspect that BIll's wife Betty has any chromosomal abnormalities. Therefore, statement (d) is most likely correcty.

species I has 2n=16 chrosomes. How many chromosomes will be found per cell in each of the following mutants in this species?

a. Monosomic [15] (missing a single chromosome) b. Autotriploid [24] (each haploid chromosome set occurs in groups of three homologous) c. Autotetraploid [32] (occurs in groups of four so you might think that it's 164=64 but REMBER that this means the cell will be 4n. So what do we need to multiply both sides with to get 4n on one side? Two, so 22n=16*2 which equals 4n=32). d. Trisomic [17] (one chromosome in addition to the usual diploid number: 2n+1) e. Double monosomic [14] (missing two chromosomes) f. Nullisomic [14] (lack of one of the normal chromsomal pairs) g. Autopentaploid [40] (haploid set occurs in groups of five so a haploid set would be half of 16 (1n) which would be 8. Penta is five so eight times five is 40) h. Tetrasomic [18] (Presence of four copies of a chromosome instead of the normal 2)

In chickens, comb shape is determined by genes at two loci (R, r and P, p). A walnut comb is produced when at least one dominant gene R is present at one locus and at least one dominant gene P is present at a second locus (genotype R_ P_). A rose comb is produced when at least one dominant gene is present at the first locus and two recessive genes are present at the second locus (genotype R_ pp). A pea comb is produced when two recessive genes are present at the first locus and at least one dominant gene is present at the second (genotype rr P_). If two recessive genes are present at the first and the second locus (rr pp), a single comb is produced. Give genotypes for comb shape of the parents in the following cross: Rose crossed with single produces 31 rose offspring. a. RR PP × rr pp b. RR pp × rr pp c. Rr PP × rr pp d. Rr Pp × rr pp e. Rr pp × rr pp

b) Rr pp × rr Pp cross will produce 2 walnut, 1 single, and 1 pea offspring. The Rose plant genotype will be Rrpp The pea plant genotype will be rrPp Rr pp × rr Pp, this will give rise to RrPp, rrPp, rrpp combinations in 2:1:1 ratio.

Explain why autopolyploids are usually sterile, whereas allopolyploids are often fertile.

because the alignment of homologous chromosomes in prophase I accompanied with segregation in anaphase I affected by unevenly distributed gametes, autopolyploids are enable to produce offspring with correct chromosomes allopolyploids however contain diploid chromosomal sets from each parent, segregation and lining up is possible.

why are the two cells produced by the cell cycle generally identical?

because they split from the same cell

Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he gets ½ white kittens and ½ black kittens. When the black kittens are interbred, all of the kittens they produce are black. On the basis of these results, would you conclude that white or black coat color in cats is a recessive trait? Explain your reasoning.

black is recesive Bb * bb

Hairlessness in American rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?

cross it with a homozygous recessive

what are the genetically important results of the cell cycle and mitosis

daughter cells have copies of genetic material identical to the parents thus making their genomes identical

The following two genotypes are crossed: Aa Bb Cc dd Ee and Aa bb Cc Dd Ee. What will the proportion of the following genotypes be among the progeny of this cross? A) Aa Bb Cc Dd Ee B) Aa bb Cc dd ee C) Aa bb cc dd ee D) AA DD CC DD EE

do it

which types of chrosome mutations which types of chromosome mutations increase the amount of genetic material in a particular chromosome? increase the amount of genetic material in all chromosomes? decrease the amount of genetic material in a particular chromosome? change the position of DNA sequences in a single chromosome without changing the amount of genetic material? move DNA from one chromosome to a nonhomologous chromosome? translocation

duplication polyploidy deletion inversion translocation

a certain species has three pairs of chromosomes: an acrocentric pair, a metacentric pair, and a submetacentric pair. Draw a cell of this species and it would appear in metaphase of mitosis

fjksladf

the fruit gly drosophila melogaster has four paits of chrosomes wheras the house fly has six pairs of chrososomes. In which species would you experct to see more genetic variation amout the progeny of a cross?

fly

all of the following cells shown in various staes from from same plant what is the diploid number give the names of each stage give number and number of DNA

found on desktop

List three fundamental events that must take place in cell reproduction.

gebetic information copied copies must seperate cell must divied

Coat color in cats is determined by genes at several different loci. At one locus on the X chromosome, one allele (X+) encodes black fur; another allele (XO) encodes orange fur. Females can be black (X+X+), orange (XOXO) or a mixture of orange and black called tortoiseshell (X+XO). Males are either black (X+Y) or orange (XOY). Bill has a female tortoiseshell cat named Patches. One night Patches escapes from Bill's house, spends the night out, and mates with a stray male. Patches later gives birth to the following kittens: one orange male, one black male, two tortoiseshell females, and one orange female. Give the genotypes of Patches, her kittens, and the stray male with which Patches mated.

he correct answer is option c) X0Y. In this case, Bill has a female cat. Thus, the stray cat has to be a male. So, all the other options with no Y chromosome are eliminated. The genotypes of the offsprings will be: Orange male: X0Y Black male: X+Y Tortoiseshell females: X+X0 The cross that takes place is as follows: (Patches) -> X+X0 X0Y <- (Stray Cat) \/ X+X0 , X0 X0 , X0Y, X+Y (Kittens: Tortoiseshell females, Orange Female, Orange male and Black male respectively).

a cell in G1 of interphase has 12 chromosomes how many chromosomes and DNA molecules will be found per cell when this original progresses to the following stages? a) G2 b) metaphase c) prophase d) anaphase I e) anaphase II f) prophase II g) after cytokinesis following mitosis h) after cytokinesis following meiosis II

heh

What is incomplete penetrance and what causes it?

incomplete penetrance occurs when an individual with a particular genotype does not express the expected phenotype. Environmental factors, as well as the effects of other genes, may alter the phenotypic expression of a particular genotype.1

what characteristics are exhibited by a cytoplasmicaly inherited trait?

inherited traits are encoded by genes in the cytoplasm. Because the cytoplasm usually is inherited from a single (most often the female) parent, reciprocal crosses do not show the same results. Cytoplasmically inherited traits often show great variability because different egg cells (female gametes) may have differing proportions of cytoplasmic alleles from random sorting of mitochondria (or plastids in plants).

The Talmund, an acient book of jewish civil and religous laws, states that if a woman bears two sons who die of bleeding after circumcision, any additional sons that she has should not be circumcised,

its x linkedyes

How is sex determined in humans?

loci of genes

Sketch and identify four different types of chromosomes based on the position of the centromere.

look on slides

a cell in prophase II of meiosis has 12 chromosomes. How many chromosomes would be present in a cell from the same organism if it were in prophase of mitosis? Prophase I of meiosis?

same about of chrosomes

In silkmoths (Bombyx mori), red eyes (re) and white-banded wing (wb) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (re+and wb+); these two genes are on the same chromosome. A moth homozygous for red eyes and white-banded wings is crossed with a moth homozygous for the wildtype traits. The F1 have normal eyes and normal wings. The F1 are crossed with moths that have red eyes and white-banded wings in a testcross. The progeny of this testcross are: wild-type eyes, wild-type wings 418 red eyes, wild-type wings 19 wild-type eyes, white-banded wings 16 red eyes, white-banded wings 426 a. What phenotypic proportions would be expected if the genes for red eyes and for white-banded wings were located on different chromosomes?

see photo

J. A. Moore investigated the inheritance of spotting patterns in leopard frogs (J. A. Moore, 1943, Journal of Heredity 34:3-7). The pipiens phenotype had the normal spots that give leopard frogs their name. In contrast, the burnsi phenotype lacked spots on its back. Moore carried out the following crosses, producing the progeny indicated. U se a chi-square test to evaluate the fit of the observed numbers of progeny to the number expected from the cross burnsi × burnsi that produced 39 burnsi, 6 pipiens. Assume that the burnsi allele is dominant to the pipiens allele and that the burnsi parents are heterozygous. The determination cannot be made with the information given.

see screen shots

a chrosome has the follow segments where * is centromere: AB*CDEFG

see screen shots

Red-green color blindness is an X-linked recessive trait in humans. Polydactyly (extra fingers and toes) is an autosomal dominant trait. Martha has normal fingers and toes and normal color vision. Her mother is normal in all respects, but her father is color blind and polydactylous. Bill is color blind and polydactylous. His mother has normal color vision and normal fingers and toes. If Bill and Martha marry, what types and proportions of children can they produce?

ya

a cell has eight chromosomes in G1 of interphase, draw a picture of the cell with it's chrosomes at the following stages. Indicate how many DNA molecules are present metaphase of mitosis anaphase of mitosis anaphase II of mitosis

ye

in cucumbers, dull fruit (D) is dominant over glossy fruit (d) orange fruit (R) is dominant oer cream fruit (r) and bitter cotyledons (B) are dominant over non-biter cotyledons (b). The three characteristics are encoded by genes located on different pairs of chrososomes, A plant homozygous for dull orange fruit and bitter cotydeons is crossed with a plant that has glossy cream fruit and non bitter cotyledons. The F1 are intercross to product f2 give phenotypes and proportions

ye

In cucumbers, orange fruit color (R) is dominant to cream fruit color (r). A cucumber plant homozygous for orange fruits is crossed with a plant with cream fruit. The F1 plants are intercrossed to produce an F2 generation. What proportion of the F2 plants will have cream fruit color? give parents F1 F2

yes

fill out chart of number chromosomes per cell and number of DNA molecules that begins with four chromosomes

yes

look at horse screenshot

yes

When a Chinese hamster with white spots is crossed with another hamster that has no spots, approximately 1/2 of the offspring have white spots and 1/2 have no spots. When two hamsters with white spots are crossed, 2/3 of the offspring possess white spots and 1/3 have no spots. How might you go about producing Chinese hamsters that breed true for white spotting?

you cant because it's lethal