BIOL 3000 Genetics Mastering + DSM questions

euchromatin chromosome regions V.S heterochromatin chromosome regions

*Chromatin exists in different states of compaction and relaxation that regulate access to DNA by regulatory proteins. __________________________________ -Euchromatin: --less chromatin condensation --many expressed genes --transcriptional proteins and enzymes are more easily able to gain access to DNA. __________________________________ -Heterochromatin: -- condensed chromatin -- few expressed genes -- are relatively inaccessible to transcriptional proteins either transiently (in the case of facultative heterochromatin) or nearly always (in the case of constitutive heterochromatin).

*The ABO blood group assorts independently of the Rhesus (Rh) blood group and the MN blood group. Three alleles, IA, IB, and i, occur at the ABO locus. Two alleles, R, a dominant allele producing Rh+, and r, a recessive allele for Rh−, are found at the Rh locus, and codominant alleles M and N occur at the MN locus. Each gene is autosomal -A child with blood types A, Rh−, and M is born to a woman who has blood types O, Rh−, and MN and a man who has blood types A, Rh+, and M. --What is the mother's blood type? --What is the father's blood type?

--The mother's blood type is: ii rr MN --The father's blood type is: I^A_ Rr MM

-For the cross aabb×AaBb , what is the expected genotype ratio? Express your answer as four numbers separated by colons (example 3:3:1:1). -What is the expected phenotype ratio?

-AaBb:Aabb:aaBb:aabb = 1:1:1:1 -A_B_ : aaB_ : A_bb : aabb = 1:1:1:1

-For the cross Bb×Bb, what is the expected genotype ratio? Express your answers as three numbers separated by colons. -What is the expected phenotype ratio? Express your answers as two numbers separated by colon.

-BB : Bb : bb = 1:2:1 B:b=3:1

____________ are capable of remodeling chromatin by adding acetyl groups to various lysine residues in histones that comprise the nucleosome. Following this modification, the lysine residue no longer ________________________.

-Histone acetyltransferases (HATs) -has a positive charge.

In a diploid species of plant, the genes for plant height and fruit shape are syntenic and separated by 18 m.u. Allele D produces tall plants and is dominant to d for short plants, and allele R produces round fruit and is dominant to r for oval fruit. -A plant with the genotype DRdr produces gametes. Identify gamete genotypes, label parental and recombinant gametes, and give the frequency of each gamete genotype. -Give the same information for a plant with the genotype DrdR

-Parental: 41% DR, 41%dr; recombinant: 9% Dr, 9%dR -Parental: 41% Dr, 41%dR; recombinant: 9% DR, 9%dr.

*Genetic linkage mapping for a large number of families identifies 4% recombination between the genes for Rh blood type and elliptocytosis. At the Rh locus, alleles R and r control Rh+ and Rh- blood types. Allele E producing elliptocytosis is dominant to the wild-type recessive allele e. Tom and Terri each have elliptocytosis, and each is Rh+. Tom's mother has elliptocytosis and is Rh- while his father is healthy and has Rh+. Terri's father is Rh+ and has elliptocytosis; Terri's mother is Rh- and is healthy. -Part A: Based on the information provided, determine the genotypes of Tom and Terri. -Part B: What is the probability that the first child of Tom and Terri will be Rh- and have elliptocytosis? Enter your answer to four decimal places (example 0.2365). -Part C: What is the probability that a child of Tom and Terri who is Rh+ will have elliptocytosis? Enter your answer to three decimal places

-Part A: --Tom: rE/Re --Terri: RE/re -Part B: 0.2404 -Part C: 0.679

** BLANK OPTIONS: -half (the gametes) - none of (the gametes) -all (the gametes) -75% of (the gametes) -25% of (the gametes) -anaphase II -prophase I -metaphase I -prophase II ________________________________________ Mendel′s law of segregation states that D and d segregate during meiosis such that ________________ the gametes Mendel ′ s law of segregation states that D and d segregate during meiosis such that _______________ the gametes will contain D and half the gametes will contain d . This is demonstrated by the pairing of homologous chromosomes during ________________________________ through_____________________ in meiosis. _________ the gametes the gametes will be derived from the sister chromatids carrying D at one pole, and __________ the gametes will be derived from sister chromatids carrying d at the other pole.

-half (the gametes) -half (the gametes) -prophase I -metaphase I -half (the gametes) -half (the gametes)

Some heterozygotes have a phenotype that is intermediate between the phenotypes of the two homozygotes This situation is the result of___

-incomplete dominance * Neither allele is completely dominant to the other.

The genes for miniature wings (m) and garnet eyes (g) are approximately 8 map units apart on chromosome 1 in Drosophila. Phenotypically wild-type females (m + g / mg +) were mated to miniature-winged males with garnet eyes. PART A: Which of the following phenotypic classes reflect offspring that were generated as a result of a crossover event? Select all that apply. -miniature wings, garnet eyes -miniature wings -wild type -garnet eyes

-miniature wings, garnet eyes -wild type *When a single genetic exchange occurs between two nonsister chromatids during the tetrad stage, two noncrossover (parental) and two crossover (recombinant) gametes are produced. Therefore, a female with the m + g / mg + genotype would produce two parental gametes which were not involved in genetic exchange -- m + g and mg + -- and two recombinant gametes from the single genetic exchange between the m and g genes -- m + g + and mg. When crossed with a male with miniature wings and garnet eyes (mg / mg), the phenotypes observed in the resulting offspring would reflect the genotypes of the female gametes.

Stern observed all of the following results EXCEPT _______ in his experiment. -one of the recombinant phenotypes was associated with an X chromosome of normal length -offspring with red, round eyes resulted from fertilization of eggs containing recombinant X chromosomes -the number of car, B+ male offspring was roughly equal to the number of car+, B male offspring -the number of males was roughly equal to the number of females

-offspring with red, round eyes resulted from fertilization of eggs containing recombinant X chromosomes

Histones in general have a net _____________ charge that allow them to bind to DNA. Acetylation of histones, _______ their positive charge and _________ the histone-DNA interaction.

-positive - decreases -weakens

-Mendel studied the seed color trait which is controlled by what gene?

-the sgr gene

All of the following events occur during normal meiosis except _______. -sister chromatids separate from one another during meiosis II -one diploid cell produces four haploid cells -homologous chromosomes separate from one another during meiosis I -two haploid gametes fuse to form a diploid cell

-two haploid gametes fuse to form a diploid cell BECAUSE: Fusion of haploid gametes occurs after meiosis.

In order to reject the null hypothesis, you look for a chi-square value that exceeds the value at ________, meaning that there is less than a 5% chance of obtaining results this disparate from the expected values by ____________. Presumably, something is affecting your genes, causing them to be inherited in a non-Mendelian manner. For a dihybrid cross, there are four possible offspring.

0.05 sheer random chance

An organism of the genotype a+b+c+/abc was testcrossed to a triply recessive organism, abc/abc. The phenotypic classes of the progeny are as follows: 28 a+b+c+ 27 abc 10 a+bc 12 ab+c+ 9 a+b+c 11 abc+ 1 a+bc+ 2 ab+c What is the coefficient of coincidence?

0.68 The expected number of double recombinants is equal to the product of the two single crossovers. (0.2)(0.22) = 0.044, or 4.4%. The observed number of double crossovers is 3/100 = 0.03, or 3%. Thus, c = 3/4.4 = 0.68. So, I = 1-c, or 0.32. This means that the observed number of double recombinants is 32% lower than expected. **To calculate c: divide the observed double recombinants by the expected.

MATCHING: match the terms with the definitions (will all be used once) ______________________________________________________________________ terms: -RFLP analysis -dot plot -P value -Manhattan plot -LOD score curves ______________________________________________________________________ definitions: 1) A ____________ is a profile of chromosomes scattered with dots and bars that represent where linkage disequilibrium has been detected between a SNP haplotype and a potential contributing gene. 2) A ____________ is used to determine the potential significance of a gene influencing the appearance of a trait, so the P value will contribute data toward the development of the Manhattan plot. 3) A ___________ is a graph showing the percent alignment between two sequences. If you are looking for regions of variability between alleles, a dot plot would illustrate the percentage of conserved DNA sequence between individuals or between species. It does not represent linkage disequilibrium, though. 4) _________ is a pattern of bands on a gel representing the presence of a specific haplotype that is part of a specific restriction enzyme cut site. While it can contribute toward the understanding of how haplotypes are linked, it is not illustrating the physical location along the chromosomes. 5) _____________ are visual representations of the statistic that argues in favor of genetic linkage. A LOD score of greater than 3 is indicative of linkage, so that would lead to the presence of a bar or dot on the chromosome profile. Thus, multiple LOD score curves are required to generate the data needed to create the type of chromosome linkage map described here.

1) A Manhattan plot is a profile of chromosomes scattered with dots and bars that represent where linkage disequilibrium has been detected between a SNP haplotype and a potential contributing gene. 2) A P value is used to determine the potential significance of a gene influencing the appearance of a trait, so the P value will contribute data toward the development of the Manhattan plot. 3) A dot plot is a graph showing the percent alignment between two sequences. If you are looking for regions of variability between alleles, a dot plot would illustrate the percentage of conserved DNA sequence between individuals or between species. It does not represent linkage disequilibrium, though. 4) RFLP analysis is a pattern of bands on a gel representing the presence of a specific haplotype that is part of a specific restriction enzyme cut site. While it can contribute toward the understanding of how haplotypes are linked, it is not illustrating the physical location along the chromosomes. 5) LOD score curves are visual representations of the statistic that argues in favor of genetic linkage. A LOD score of greater than 3 is indicative of linkage, so that would lead to the presence of a bar or dot on the chromosome profile. Thus, multiple LOD score curves are required to generate the data needed to create the type of chromosome linkage map described here.

MATCHING ___________________________________________________________________________ TERMS: have to know what they stand for -VNTRs -SNPs -RELPs ___________________________________________________________________________ DEFINITIONS: 1) _________, aka________________ , are DNA sequence variants in which one base pair is substituted for another. In this case, the difference between allele A and B is a single nucleotide mutation from a C to a G. 2)________ ,aka _____________________ , are genetic markers consisting of short sequences of DNA repeated end-to-end in a non-coding region of a chromosome. 3)___________ aka_______________________, are related to SNPs, but are analyzed using restriction enzymes rather than PCR sequencing.

1) SNPs, or single nucleotide polymorphisms, are DNA sequence variants in which one base pair is substituted for another. In this case, the difference between allele A and B is a single nucleotide mutation from a C to a G. 2) VNTRs, or variable number tandem repeats, are genetic markers consisting of short sequences of DNA repeated end-to-end in a non-coding region of a chromosome. 3) RFLPs, or restriction fragment length polymorphisms, are related to SNPs, but are analyzed using restriction enzymes rather than PCR sequencing.

In a large metropolitan hospital, cells from newborn babies are collected and examined microscopically over a 5-year period. Among approximately 7500 newborn males, six have one Barr body in the nuclei of their somatic cells. All other newborn males have no Barr bodies. Among 7500 female infants, four have two Barr bodies in each nucleus, two have no Barr bodies, and the rest have one. -What is the cause of the unusual number of Barr bodies in a small number of male and female infants? Select the six correct statements.

1. The unusual females with two Barr bodies result from a normal egg fusing with a sperm containing two X chromosomes (nondisjunction during meiosis II in the father). 2. The unusual females with no Barr bodies result from a normal egg fusing with a sperm containing no sex chromosomes (nondisjunction during meiosis I or II in the father). 3. The unusual males with two X chromosomes result from fusion of a normal sperm containing a Y chromosome with an egg containing two X chromosomes (nondisjunction during meiosis I or II in the mother). 4. The unusual males with two X chromosomes result from a normal egg fusing with a sperm containing an X and a Y chromosome (nondisjunction during meiosis I in the father). 5. The unusual females with no Barr bodies result from fusion of a normal sperm containing an X chromosome with an egg that has no sex chromosomes (nondisjunction during meiosis I or II in the mother). 6. The unusual females with two Barr bodies result from fusion of a normal sperm containing an X chromosome with an egg containing two X chromosomes (nondisjunction during meiosis I or II in the mother).

What is the probability of rolling one six-sided die and obtaining a 1 or a 2?

1/3

What is the probability of rolling two six-sided dice and obtaining two 4's?

1/36

A pea plant is heterozygous for two genes; one controlling height, one controlling color. The genotype is written PpTt. Based on the Law of Independent Assortment, approximately what proportion of the pollen produced by this plant should have the genotype PT?

1/4 *There are four different possible genotypes. Independent Assortment assumes that all are equally likely. Therefore, each of the four should occur with roughly equal frequency (1/4).

Suppose you have one plant with the genotype TTPpYy, and another plant that is recessive for all three traits. If you were to cross these two plants, what proportion of the offspring are expected to be dominant for all three traits?

1/4 *To calculate the proportion of offspring that will be dominant, you can calculate the probability for each gene and multiply the numbers together. For TT x tt, all offspring will be dominant (Tt). For both Pp x pp and Yy x yy, half of the offspring are dominant and half are recessive. So the chances of being dominant for all three is 1 x ½ x ½ = ¼. *** If the first gene were also a heterozygote instead of being TT, you would multiply ½ x ½ x ½ = 1/8. This is the proportion predicted by a trihybrid cross for three heterozygous genes, not two. If this were a cross between two dihybrids, you would expect to see the 9/16, 3/16, or 1/16 ratios. However, one parent is recessive, so that makes this an example of a test-cross for three genes simultaneously.

In one of Morgan's test-crosses, he crossed pr+ vg+ / pr vg females with recessive males. Assuming you collect 100 offspring, which of the following results would indicate genetic linkage between these two genes? 12 red eye, vestigial wing 25 red eye, full wing 25 red eye, vestigial wing 12 red eye, full wing 25 purple eye, vestigial wing 12 purple eye, vestigial wing

12 red eye, vestigial wing *In a cross between pr+ vg+ / pr vg females and recessive males, collecting only 12 offspring with red eye, vestigial wing is indicative of genetic linkage. If the genes assort independently, you would expect to observe equal numbers of all four possible genotypes (25 of each). The red eye, full wing and the purple eye, vestigial wing represent the two parental genotypes, whereas the red eye, vestigial wing or the purple eye, full wing represent recombinant genotypes. So, 25 red eye, full wing is what you expect if there is no linkage. If the genes are linked, you would have expected to observe >>25 of each of these two parentals. 12 red eye, full wing is a parental genotype. If there is linkage, you would expect the number of recombinant gametes to decrease, not the parental genotypes. Thus, you would predict to observe >>25 of this genotype if the genes are linked. 25 purple eye, vestigial wing also is expected with no linkage because this is a parental genotype. If the genes are linked, you would have expected to observe >>25 of this parental genotype. 12 purple eye, vestigial wing is a parental genotype. So if there is linkage, you would expect the number of recombinant gametes to decrease, not the parental genotypes. Thus, you would predict to observe >>25 of this genotype if the genes are linked.

Parental chromosomes have an arrangement with DE on one chromosome and de on the other. Assuming the genes exhibit incomplete linkage and the percentage of the parental gamete DE is 38%, what is the expected frequency of the De gamete?

12% **Assuming the genes exhibit incomplete linkage and the percentage of the parental gamete DE is 38%, the expected frequency of the DE gamete is 12%. If the frequency of DE is 38%, that means the frequency of de is also 38%. Thus, the frequency of parental gametes is 76%. The frequency of the two classes of recombinant gametes is 100-76 = 24%. However, there are two possible classes, so 24/2 = 12% for just the De gamete.

How many possible gametes would you expect to see in an individual who is heterozygous for four genes?

16 *To determine the number of allelic combinations for 4 genes, the equation is 2n, where n = number of genes. Thus, for a heterozygote at 4 genes, the number of possible combinations is 24 = 16. The multiplication rule applies here, but if you used the addition rule by mistake, you may have added 2 + 2 + 2 + 2 = 8 instead of multiplying the probabilities. The possible allele combinations in offspring if this individual is self-crossed would be calculated using the equation 3n, where n = 4. Thus, a self-crossed individual heterozygous at 4 loci could produce 81 possible genotypes in the offspring. You would expect a Punnett square for a trihybrid cross to have 64 squares, while a Punnett square for a four gene cross would have 16 rows and 16 columns, or 256 squares.

For a dihybrid cross, suppose two true-breeding parents are crossed and their offspring (F1) are then test-crossed. What phenotypic ratios do you expect to observe in the F2 generation?

1:1:1:1 *For a dihybrid cross, suppose two true-breeding parents are crossed and their offspring (F1) are then test-crossed. The phenotypic ratios are 1:1:1:1. ***If an individual that is heterozygous for two genes is self-crossed, you would predict a 9:3:3:1 ratio. However, in a test-cross, an individual with the dominant phenotype is crossed to a recessive individual to establish whether they are heterozygous or homozygous. In this case, the F1 offspring are heterozygotes (AaBb) being crossed to a recessive individual (aabb). There are four possible genotypes for their offspring - AaBb, Aabb, aaBb, or aabb. Because of independent assortment, these four possible combinations occur with equal likelihood.

For a monohybrid cross, suppose two true-breeding parents are crossed and their offspring (F1) are self-crossed. What genotypic and phenotypic ratios do you expect to observe in the F2 generation?

1:2:1 genotypic and 3:1 phenotypic

If a heterozygote is self-crossed, what proportion of their progeny that express the dominant phenotype could be self-crossed to produce a 3:1 phenotypic ratio?

2/3 * If a heterozygote is self-crossed, 2/3 of the dominant progeny could be self-crossed to produce a 3:1 phenotypic ratio. When a heterozygote is self-crossed, the offspring will be 1/4 homozygous dominant, 1/2 heterozygotes, and 1/4 homozygous recessive. Self-crossing a homozygous plant would not give a 3:1 ratio, so only the heterozygotes would be able to produce this ratio. However, the question asks what proportion of the dominant offspring, so you could exclude the homozygous recessive offspring from your calculations. Of all offspring, ½ are predicted to be heterozygous. Of only the dominant individuals, 2/3 would be heterozygotes.

If there are 13 nucleosome core particles connected by 12 linker regions, how much DNA is present?

2546 base pairs **(13 x 146) + (12 x 54) = 2546 bp

The parental cross between a rabbit with the Himalayan phenotype (genotype chch) and an albino rabbit (genotype cc) results in F1 rabbits that all have the Himalayan phenotype (genotype chc). -P: c^h c^h (Himalayan) X cc (albino) -F1: c^hc If the resulting F1 rabbits are crossed, what proportion of the F2 offspring will have the Himalayan phenotype and what causes the hypomorphic ch allele to be unstable and result in the distinctive fur pattern?

3/4; temperature

The genes for miniature wings (m) and garnet eyes (g) are approximately 8 map units apart on chromosome 1 in Drosophila. Phenotypically wild-type females (m + g / mg +) were mated to miniature-winged males with garnet eyes. PART B: If 800 offspring were produced from the cross, in what numbers would you expect the following phenotypes? __wild type : __ miniature wings : __ garnet eyes : __ miniature wings, garnet eyes

32:368:368:32 *If the genes for miniature wings (m) and garnet eyes (g) are linked with 8 map units between them, then the two recombinant classes must add up to 8% of the total (4% each) and the two parental classes must add up to 92% of the total (46% each). As a result, the following distribution would be expected: wild type: 4% x 800 = 32 miniature wings: 46% x 800 = 368 garnet eyes: 46% x 800 = 368 miniature wings, garnet eyes: 4% x 800 = 32

A patient has a disease-causing allele, D, and the genotype A1B1D / A2B2d. You are trying to decide if this haplotype can be used to predict the inheritance of the allele, D, in this patient's children. First, you need to decide how often crossing over occurs between these genetic markers to decide if it will occur. The distance between A and B is 20 cM, and between B and D is 15 cM. What is the probability of an affected individual inheriting the parental haplotype with the disease-causing allele (A1B1D)?

34% The parental haplotype A2B1D occurs when there is no crossing over between genes A, B, and D. So, if the probability of crossing over between A and B is 0.2, the probability of no crossover is 0.80. If the probability between B and D is 0.15, the probability of no crossover is 0.85. So, (0.5)(0.80)(0.85) = 0.34.

A plant with the genotype AaCcTt is self-crossed. What proportion of the offspring would have at least one recessive trait?

37/64 *If a plant that is heterozygous for three traits is self-crossed, 37/64 offspring would have at least one recessive trait. For each gene, a heterozygous self-cross would give you a probability of ¼ recessive, or ¾ dominant. So, how many possible combinations would have at least one recessive trait? For that calculation, you can look at the possible combinations. You could have all three recessive (aacctt), and the probability of that genotype is ¼ x ¼ x ¼ = 1/64. You could have two out of three recessive (aaccT-, aaC-tt, A-cctt), so the probability for each of those combinations is ¼ x ¼x ¾ = 3/64. You could also have only one out of three recessive (aaC-T-, A-ccT-, A-C-tt), which is ¼ x ¾ x ¾ = 9/64. By asking "at least one", you would use the sum rule to add all of these probabilities. So, 1/64 + 3/64 + 3/64 + 3/64 + 9/64 + 9/64 + 9/64 = 37/64.

Our closest primate relative, the chimpanzee, has a diploid number of 2n=48. For each of the following stages of M phase, identify the number of chromosomes present in each cell. -End of meiotic anaphase II.

48 chromosomes

Our closest primate relative, the chimpanzee, has a diploid number of 2n=48. For each of the following stages of M phase, identify the number of chromosomes present in each cell. -Early mitotic prophase.

48 chromosomes, 96 chromatids.

Our closest primate relative, the chimpanzee, has a diploid number of 2n=48. For each of the following stages of M phase, identify the number of chromosomes present in each cell. -Early prophase I.

48 chromosomes, 96 chromatids.

Our closest primate relative, the chimpanzee, has a diploid number of 2n=48. For each of the following stages of M phase, identify the number of chromosomes present in each cell. -Meiotic metaphase I.

48 chromosomes, 96 chromatids.

Our closest primate relative, the chimpanzee, has a diploid number of 2n=48. For each of the following stages of M phase, identify the number of chromosomes present in each cell. -Mitotic metaphase.

48 chromosomes, 96 chromatids.

Our closest primate relative, the chimpanzee, has a diploid number of 2n=48. For each of the following stages of M phase, identify the number of chromosomes present in each cell. -End of mitotic telophase.

48 chromosomes: 46 autosomes, 2 sex chromosomes,each chromosome will be composed of a single chromatid.

An individual with the genotype AABbccDd is crossed to an individual with the genotype aaBbCCdd. How many possible genotypes do you expect to see in the F1 progeny?

6 *In the cross AABbccDd x aaBbCCdd, you will see 6 possible genotypes in the offspring. In this case, you can look at each gene independently. First, a cross between AA x aa will only generate one genotype (Aa). For the cross between Bb x Bb, you can have one of three possible genotypes (BB, Bb, or bb). A cross CC x cc will also only generate heterozygotes (Cc). Finally, the cross between a heterozygote Dd and a homozygote dd will result in Dd or dd offspring. Because these genes assort independently, you can multiply the number of possible genotypes for gene A (1) by the possible genotypes for gene B (3), gene C (1), and gene D (2). 1 x 3 x 1 x 2 = 6. Because the genes can assort in various combinations, you must use the multiplication rule, not the addition rule, which would have given you 1 + 3 + 1 + 2 = 7. If the parents were heterozygous for all four genes, you could have as many as 3 x 3 x 3 x 3 = 81 possible offspring.

Cystic fibrosis (CF) exhibits a recessive inheritance pattern, so affected individuals have a genotype of cc. Suppose two healthy parents have one child with CF and one child without CF. What are the chances that their unaffected child is a carrier (heterozygote)?

67%

Suppose you have a plant with the genotype RrTtPp. How many possible gametes could this individual generate?

8 * For each trait, a gamete can inherit one of two alleles (dominant or recessive) and there are three traits (each designated by a different letter of the alphabet). Therefore, there are 23 possibilities or 8 possible gametic combinations possible. For each gene individually, there are 2 possible alleles - R or r, T or t, P or p. There are a total of 3 genes here, but each gene has two possible options. You cannot simply multiply the number of genes (3) by the number of alleles (2), because there are multiple combinations possible here. The equation is 23, where 3 is the number of genes. If you reversed these values and calculated 32, you may have calculated 9. If you used the equation 22n, that would give you 64. This represents the number of possible offspring or combinations if you were to cross this individual (the number of squares in the Punnett square for a trihybrid cross), but does not tell you the types of gametes that can be produced.

If an individual that is heterozygous for two genes is self-crossed, you would predict a __________ ratio.

9:3:3:1

Linkage disequilibrium describes a nonrandom relationship between alleles of closely linked genes, which means that specific alleles are linked to a specific haplotype. What type of visual graph can be used to illustrate locations of chromosomes where linkage disequilibrium has been detected?

A Manhattan plot

What the type of molecule on the surface of red blood cells that is responsible for the determination of ABO blood type?

ABO blood type is determined by the terminal sugar on a glycolipid molecule

When a true-breeding plant with yellow peas was crossed to a true-breeding plant with green peas, what results did Mendel observe?

All of the plants generated yellow peas. * When a true-breeding plant with yellow peas was crossed to a true-breeding plant with green peas, all of the plants generated yellow peas. In a cross between two true-breeding plants (YY and yy), the offspring would all be heterozygotes (Yy) and display the dominant phenotype.

FLIP CARD OVER FOR QUESTIONS..... the answers are on the bottom of this card. ________________________________________ II-1 * Likely experienced crossover during gamete formation, resulting in a recombinant gamete passed to offspring III-6. This daughter is affected, but she does not have the P1 gene marker.

Allelic phase in Family A has been linked to the P1 genetic marker. Which individual in this pedigree likely experienced a crossover event during meiosis, leading to a recombinant gamete in which P1 is not linked to the disease allele?

Review Considering the Mendelian traits tall (D) versus dwarf (d) and violet (W) versus white (w), consider the crosses below and determine the genotypes of the parental plants by analyzing the phenotypes of the offspring. Parental Plants: tall, violet x tall, white Offspring: 1/2 tall, white 1/2 tall, violet

DDWw x Ddww DdWw x DDww DDWw x DDww

Review Considering the Mendelian traits tall (D) versus dwarf (d) and violet (W) versus white (w), consider the crosses below and determine the genotypes of the parental plants by analyzing the phenotypes of the offspring. Parental Plants: tall, violet x tall, white Offspring: 9/16 tall, violet 3/16 tall, white 3/16 dwarf, violet 1/16 dwarf, white

DdWw x DdWw * The only way to achieve a 9:3:3:1 ratio for a dihybrid cross is if both parents are heterozygous for both traits being studied. Recall that Mendel observed such a ratio only after he self-fertilized the F1 generation he had obtained after crossing plants that were true breeding for two traits -- he crossed plants that were heterozygous for both traits to obtain this ratio.

T or F? The 9:3:3:1 ratio exhibited in the F2 generation of a dihybrid cross is a genotypic ratio

False

T or F? In order to create the possibility of generating a trisomy, nondisjunction must occur during meiosis II.

False. *Nondisjunction during either meiosis I or meiosis II creates gametes that will generate trisomies if fertilized

*A researcher interested in studying a human gene on chromosome 21 and another gene on the X chromosome uses FISH probes to locate each gene. The chromosome 21 probe produces green fluorescent color, and the X chromosome probe produces red fluorescent color. -If the subject studied is female, how many green and red spots will be detected? Explain your answer.

Females contain two copies each of chromosome 21 and the X chromosome, thus there will be two green spots and two red spots in cells from a female.

Which of the traits that Mendel studied is controlled by the gene bHLH?

Flower color * The bHLH gene produces a transcription factor that activates the genes required to synthesize the pigment anthocyanin. If the gene is mutated and the transcription is not activated, the flowers lacking this pigment will be white. Thus, the dominant trait is purple, and the recessive trait is white. Seed color is controlled by the gene Sgr. Seed shape is controlled by the gene Sbe1. Stem length is controlled by the gene Le.

The histone protein _____ plays a key role in stabilizing the 30-nm solenoid structure.

H1 ***The long N-terminal and C-terminal ends of the H1 protein attach to adjacent nucleosome core particles pulling them into an orderly solenoid array.

HATs stands for what?

Histone acetyltransferases

What are the basic proteins that interact with negatively charged DNA?

Histones

regions of the genome with increased frequency of recombination events

Hotspots

__________ indicates that less than 100% of individuals with a mutant genotype have the mutant phenotype.

Incomplete penetrance

Rank the following levels of chromatin compaction in eukaryotes from the least compact to the most compact. METAPHASE CHROMOSOME LOOP DOMAINS NAKED DNA SOLENOID CHROMATID NUCLEOSOME

LEAST COMPACT naked DNA nucleosome solenoid loop domains chromatid metaphase-chromosome MOST COMPACT **Chromatin compaction is required for the nucleus to accommodate genomes that are often more than 1,000 times longer than the nuclear diameter. As the 2nm wide DNA is organized into chromatin, each level of compaction has a chromatin structure with a signature diameter (2nm DNA, 11nm nucleosome, 30nm solenoid, 300nm loop domains, 700nm chromatid, and 1400nm metaphase chromosome).

Compair and contrast: codominance and incomplete dominance

Like incomplete dominance, codominance produces a distinct phenotype for each of the three genotypes in a monohybrid cross. Unlike incomplete dominance, however, codominance results in both alleles fully expressing their phenotype in the heterozygote.

*A researcher interested in studying a human gene on chromosome 21 and another gene on the X chromosome uses FISH probes to locate each gene. The chromosome 21 probe produces green fluorescent color, and the X chromosome probe produces red fluorescent color. -If the subject studied is male, how many green and red spots will be detected? Explain your answer.

Males contain two copies of chromosome 21 but only one copy of the X chromosome, thus there will be two green spots and one red spot in cells from a male.

The following statement is an accurate description of which of Mendel's Laws? During gamete formation, two alleles for a trait will segregate with equal probability into the gametes.

Mendel's Law of Segregation (his first law) --his second law is independent assortment

FLIP CARD OVER FOR QUESTIONS..... the answers are on the bottom of this card. ________________________________________ Part A: c-w-s Part B: c+w+s/cws+ Part C: 0.134 Part D: 0.252 Part E: 0.45

Part A: Determine the order of the genes on the chromosome. PartB: Identify the alleles that are present on each of the homologous chromosomes in the trihybrid rabbits. Part C: Calculate the recombination frequency between c and w pair of genes. Part D: Calculate the recombination frequency between w and s pair of genes. Part E: Determine the interference value for this cross.

FLIP CARD OVER FOR QUESTIONS..... the answers are on the bottom of this card. ________________________________________ Part A: Dominant trait Part B: Yes, NPS appears to segregate with blood type A in this pedigree, indicating genetic linkage between these traits. Part C: The frequency of recombinant progeny is less than 50%. Part D: I−1 is in / in , I−2 is IAN / in II−2 is in / in, II−4 is I A N / in, II−6 is IAN / in, II−7 is in / in and II−9 is IAN / in. Part E: III−6 has NPS because he inherited the recombinant chromosome i N from his mother, whereas III−8 inherited the nonrecombinant i n from her mother. Part F:III−6 is i N / i n andIII−8 is i n/ i n. Part G: The genotypes of III−11 and III−12 cannot be unambiguously determined.

Part A: Is nail-patella syndrome a dominant or a recessive condition? Part B: Does this family give evidence of genetic linkage between nail-patella syndrome and ABO blood group? Part C: why or why not? Part D: Using N and n to represent alleles at the nail-patella locus and I A , IB , and i to represent ABO alleles, write the genotypes of I−1 and I−2 as well as their five children in generation II. Part E:Why does III−6 have nail-patella syndrome and III−8 not have the disorder? Part F: Give genotypes for these two individuals. Part G: Give genotypes for III−11 and III−12 individuals.

According to the blending theory of heredity, a cross between a red flower and a white flower would produce what types of offspring?

Pink flowers

Mendel used several types of crosses that have since become a standard approach to genetic analysis. Suppose you have identified a trait that you suspect is uniquely inherited through pollen only, so it is passed only from the male plants to the offspring. Which type of cross would allow you to test this hypothesis?

Reciprocal cross *If you suspect a trait is being passed through pollen and not eggs, a reciprocal cross would allow you to directly test this hypothesis. Plants with the same phenotypes would be crossed, but the sexes of the donating parents would be switched.

Allele A has the sequence 5'-ATCCGA-3', while allele B has the sequence 5'-ATCGGA-3'. You have designed a PCR-based test to screen patients for the presence of either allele. Which type of common genetic marker is seen in these variants?

SNP

-Mendel studied the seed shape trait which is controlled by what gene?

Sbe1

What causes the histones loosen their grip on the negatively charged DNA?

The strength of histone's interaction with negatively charged DNA is modulated by epigenetic modifications. One of these modifications is acetylation. Acetylation adds an acetyl group to the positively charged amino group present on the side chain of the amino acid lysine effectively changing the net charge of the protein by neutralizing the positive charge. When the positive charge is reduced

True-breeding parental plants (P generation) are tall with terminal flowers (TTaa) and short with axial flowers (ttAA). If the F1 generation were crossed, what proportion of the F2 generation are expected to be tall with axial flowers?

True-breeding parental plants (P generation) are tall with terminal flowers (TTaa) and short with axial flowers (ttAA). If the F1 generation were crossed, the proportion would be --------> 9/16.

__________indicates that the severity of a mutant phenotype differs among individuals with the same mutant genotype.

Variable expressivity

A Manhattan plot

a profile of chromosomes scattered with dots and bars that represent where linkage disequilibrium has been detected between a SNP haplotype and a potential contributing gene.

An allele is never intrinsically "dominant" or "recessive." Instead, these terms describe ______ between two alleles.

a relationship *This relationship is evaluated by examining a heterozygote: The allele that determines the phenotype of the heterozygote is dominant to the other (recessive) allele. Some heterozygotes have a phenotype that is intermediate between the phenotypes of the two homozygotes.

the following descriptions is true of pedigree for what trait? Each person with the trait must have at least one parent who also has the trait.

an autosomal dominant trait *A recessive trait often skips generations, but a dominant trait cannot be inherited if neither parent exhibits the trait. Males and females are affected with equal frequency, and the trait can be inherited from either parent. Since an individual with a dominant trait can be heterozygous, not all of their children may have the trait. If both parents are heterozygotes, there is a ¼ chance that their child would be recessive. If only one parent is dominant, you cannot assume that half of the children will be affected. First, it's possible the dominant parent is homozygous, in which case all children would be dominant (heterozygous). Additionally, it is not necessarily true that half of the children will be dominant. It is more appropriate to say that each child has a 50% chance of being dominant if their parent is a heterozygote. So if they only have 2 children, the fact that both are recessive is not enough information to make that conclusion. The total data set is simply too small. For autosomal dominant traits, they can be inherited from either parent. If a trait is X-linked dominant, it can exhibit a unique pattern because affected males can pass it to their daughters, but not sons. Mothers would be equally likely to pass it to daughters or sons, so sex linkage is not observed for dominant traits.

During which stage of M phase do these two alleles segregate from one another? (Assuming no crossovers between homologs). -prophase I -metaphase I -anaphase I -telophase I

anaphase I

Recall Mendel's law of segregation for autosomal alleles D and d and consider the behavior of homologous chromosomes in meiosis. Suppose a crossover occurs between the homologous chromosomes. At what stage or stages of M phase could alleles D and d segregate? -metaphase I -telophase I or II -anaphase I or II -prophase

anaphase I or II

The Online Mendelian Index in Man (OMIM) is a database of human genes and the phenotypes associated with mutations in those genes. The majority of heredity conditions in OMIM are the result of which type of gene mutation?

autosomal *The second most common type of mutation as X-linked

Alleles of the Sbe1 gene, which controls the pea shape, can be detected by DNA analysis. In heterozygous plants, two distinct bands of DNA are visible. What is the relationship between the alleles of the heterozygote at the molecular level?

codominant

In a trihybrid chromosome d+e+f+ / def, which of the following represents a double crossover? de+f de+f+ d+e+f d+e+f+ def

de+f ** In a trihybrid chromosome d+e+f+ / def, chromosome de+f represents a double crossover. In this case, one crossover occurred between d and e, and one between e and f. The chromosome de+f+ represents one crossover between d and e. The chromosome d+e+f represents one crossover between e and f. The chromosome d+e+f+ represents no crossovers (parental chromosome). The chromosome def also represents no crossovers (parental chromosome).

For a _________ cross, you would expect to see 9 possible genotypes, and a 9:3:3:1 phenotypic ratio in the offspring.

dihybrid

The 9:3:3:1 ratio requires that the four gamete classes produced occur with _____ frequency. This requires ________

equal frequency independent assortment.

Light G bands that appear along chromosomes contain ...

euchromatic regions.

The results of the Stern experiment supported the general idea that _______.

genetic recombination is a result of physical exchange between homologous chromosomes *Whenever recombinant phenotypes occurred, the cytological markers indicated that a physical exchange between the X chromosomes had also occurred.

Dark G bands that appear along chromosomes contain ...

heterochromatic regions.

repetitive DNA telomeric DNA Barr body and centromeric DNA are all examples of _______.

heterochromatin *Repetitive DNA is often associated with heterochromatin. -At the centromere, these heterochromatic repetitive sequences facilitate binding of spindle fibers during segregation of homologous chromosomes and sister chromatids. -Telomeric repetitive sequences are involved in the maintenance of the chromosome's structural integrity. In other cases, entire chromosomes can be heterochromatic as occurs during X-inactivation in mammalian females where the inactive chromosome forms a highly condensed Barr body.

The discernible difference in length between the two X chromosomes of the female fruit fly was important in Stern's experiments because _______.

it allowed cytological detection of physical exchange between the chromosomes *The differences in structure between the two chromosomes allowed Stern to track the inheritance of recombinant and nonrecombinant X chromosomes.

Ordinarily, homologs separate during meiosis I. Failure of this separation is one of the ways in which ____________ can occur.

nondisjunction

A _________ cross would also be helpful to support your hypothesis, provided you are doing multiple crosses with each parental combination.

replicate cross

A ______ cross would make it difficult to evaluate whether the pollen contained the gene, since both gametes would be provided by the same individual.

self-cross

A ________ cross is useful if you are trying to identify an individual as being true-breeding, so it would not be able to predict a pollen-only inheritance pattern.

test cross

-Mendel studied the stem length trait which is controlled by what gene?

the gene Le

A ______________ cross or a heterozygote cross would allow you to establish whether a trait is dominant or recessive, but neither would establish whether it was unique to the pollen unless you also include reciprocal crosses.

true-breeding cross

Assume that the DNA associated with a nucleosome core particle plus the DNA in the linker adds up to 200 bp. Approximately how many base pairs are found in the linker region?

~54 *Approximately 146 bp of DNA is wrapped around the nucleosome core particle. Therefore 200 bp - 146 bp = 54 bp of linker DNA. This is the amount of DNA present in the "string."

Since an individual with a dominant trait can be heterozygous, not all of their children may have the trait. If both parents are heterozygotes, there is a ________ chance that their child would be recessive.

¼

A plant has with the genotype RrTt is self-crossed. What proportion of the offspring would be true-breeding?

¼ ** In a self-cross of a plant that is heterozygous for two traits, ¼ of the offspring will be true-breeding. True-breeding could mean one of four possible genotypes: RRTT, RRtt, rrTT, or rrtt. In a heterozygote cross, ¼ of the offspring are either homozygous dominant or homozygous recessive. The chances of each of the possible true-breeding genotypes is ¼ x ¼ = 1/16. Since there are four possible combinations and the probability of each combination is 1/16, you would add the probabilities to get a total of 1/16 + 1/16 + 1/16 + 1/16 = 4/16, or ¼. ¾ of the offspring would be dominant for one trait, but that is not true-breeding. 9/16 of the offspring would be dominant for both traits. There is a ½ chance of each heterozygote producing a gamete containing the dominant or the recessive allele, but you must calculate the probability of all possible combinations of being dominant and/or recessive in order to account for all true-breeding combinations.