Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 23.3, Problem 3COMQ
To map the distance between molecular markers via testcrosses, the markers must be
a. polymorphic.
b. monomorphic.
c. fluorescently labeled.
d. on different chromosomes.
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Examination of a karyotype reveals 23 pairs of chromosomes, with each homologous pair illustrating similar size and centromere location. Which of the following can be concluded from this karyotype?a. The individual has Down syndrome.b. The individual is a female.c. The individual does not have cystic fibrosis.d. The individual does not have sickle-cell anemia.
Chromosome translocations include:
A. Alterations in which the genetic material remains the same but rearranged
B. Alterations in which the total amount of genetics information increases
C. Alterations in which the total amount of genetics information decreases
D. Variations in the chromosome number
You breed two monster parents (green, hairless) GGhh and ggHH (white, hairy) to produce F1 offspring (green, hairy). Genes G and H are on the same chromosome.
a. What are the dominant phenotypes?
b. What are ALL of the potential gamete genotypes that could be created by an F1? Clearly indicate the recombinant genotypes.
c. Explain how recombinants are created.
d. What are the potential offspring genotypes from the cross of a F1 monster with a homozygous recessive monster.
e. From this test cross you get 152 green, hairless; 23 green, hairy; 148 white, hairy, and 27 white, hairless. Calculate the chi squared to test the independent assortment hypothesis and determine if these genes assort independently.
Chapter 23 Solutions
Genetics: Analysis and Principles
Ch. 23.1 - Prob. 1COMQCh. 23.2 - Prob. 1COMQCh. 23.3 - A molecular marker is a _____ found at a specific...Ch. 23.3 - 2. Which of the following is an example of a...Ch. 23.3 - To map the distance between molecular markers via...Ch. 23.4 - 1. What is a contig?
a. A fragment of DNA that...Ch. 23.4 - A vector that can carry a large fragment of...Ch. 23.4 - 3. Chromosomal walking is a method of _____ in...Ch. 23.5 - Prob. 1COMQCh. 23.5 - Prob. 2COMQ
Ch. 23.5 - 3. A prokaryotic genome is about 4 million bp in...Ch. 23.6 - Metagenomics is aimed at a. determining the...Ch. 23 - 1. A person with a rare genetic disease has a...Ch. 23 - For each of the following, decide if it could be...Ch. 23 - Which of the following statements about molecular...Ch. 23 - 1. Is each of the following a method used in...Ch. 23 - Prob. 2EQCh. 23 - Prob. 3EQCh. 23 - The cells from a persons malignant tumor were...Ch. 23 - 5. Figure 23.2 describes the technique of FISH....Ch. 23 - Explain how DNA probes with different fluorescence...Ch. 23 - 7. A researcher is interested in a gene found on...Ch. 23 - Prob. 8EQCh. 23 - Prob. 9EQCh. 23 - Prob. 10EQCh. 23 - Prob. 11EQCh. 23 - Prob. 12EQCh. 23 - In the Human Genome Project, researchers have...Ch. 23 - 14. Take a look at question 3 in More Genetic...Ch. 23 - 15. Place the following stages of a physical...Ch. 23 - 16. What is an STS? How are STSs generated...Ch. 23 - 17. Four cosmid clones, which we will call cosmids...Ch. 23 - A human gene, which we will call geneX, is located...Ch. 23 - 19. Describe how you would clone a gene by...Ch. 23 - 20. A bacterium has a genome size of 4.4 Mb. If a...Ch. 23 - 21. Discuss the advantages of next-generation...Ch. 23 - Prob. 22EQCh. 23 - Prob. 23EQCh. 23 - What is a molecular marker? Give two examples....Ch. 23 - Which goals of the Human Genome Project do you...
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- How many Barr bodies will be visible in cells from a person with the karyotype depicted below? a. None. b. One. Two. $$ ку K ) K Kral 55 }} )) se (5 11 }} C. d. Three. e. Four. 21 }< SF 31 (rarrow_forwardno need to explain just need n=to know if it a.b.c and d. 1.) In a species of dog, a mutant gene that causes deafness is found on the Y chromosome. What type of offspring will be produced between a cross of male dog whose father is deaf, and female dog whose father is not deaf? a.) 25% normal female dogs, 25 % deaf female dogs, 25 % normal male dogs, 25 % deaf male dogs b.) 50% normal female dogs, 50 % deaf male dogs c.) 50% deaf female dogs, 50 % deaf male dogs d.) 50% normal female dogs, 25 % deaf male dogs, 25% normal male dogs 2.) Mendel crossed true breeding plants that had yellow pods with true breeding plants that had green pods. Yellow pod colour (Y) is dominant over green pod colour (y). Predict the genotypes and phenotypes of F1 generation. a.) yellow pods with genotype yy b.) yellow pods with genotype Yy c.) green pods with genotype Yy d.) green pods with genotype Yyarrow_forwardA geneticist determines that the order of four genes on a specific chromosome is CABD because the recombination frequencies were 35% for C-D, 24% for A-D, 23% for B-D, and 10% for A-B. What has the geneticist constructed? A. a physical map B. a karyotype C. a cytogenic map D. a linkage maparrow_forward
- Which of the following statements is/are true of the variegated eye color phenotype in Figure 12-23?a. The white gene is active in the white sectors and inactive in the red sectors.b. The white gene is inactive in the white sectors and active in the red sectors.c. The white gene is inactive in the white sectors due to the spread of heterochromatin.d. b and c are both true.arrow_forwardFor single crossovers, the frequency of recombinant gametes is half the frequency of crossing over because a. a testcross between a homozygote and a heterozygote produces ½ heterozygous and ½ homozygous progeny.b. the frequency of recombination is always 50%.c. each crossover takes place between only two of the four chromatids of a homologous pair.d. crossovers take place in about 50% of meioses.arrow_forwardIn the lab, you discover two yeast haploid mutants that cannot produce arginine. You cross them together and the resulting diploid produces arginine. If producing arginine is the wild type phenotype, what can you definitively conclude? A.The haploid strains have identical mutations in the same genes. B.The haploid strains have identical mutations in different genes. C.The haploid strains have mutations in different genes. D.The haploid strains have mutations in the same gene. E.The haploid strains must belong to the complementation group encoding the first enzyme in the biosynthetic pathway.arrow_forward
- Which of the following statements about molecular markersare true?A. All molecular markers are segments of DNA that carryspecific genes.B. A molecular marker is a segment of DNA that is found at aspecific location in a genome.C. We can follow the transmission of a molecular marker byanalyzing the phenotype (i.e., the physical characteristics)of offspring.D. We can follow the transmission of molecular markers usingmolecular techniques such as gel electrophoresis.E. An STS is a molecular marker.arrow_forwardConsider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forwardGive typing answer with explanation and conclusion to all parts The chromosomes drawn below are found in a human skin cell. The lines represent the chromosomes themselves. The letters indicate the location of a particular version of a gene. a) T/F An individual’s offspring are more likely to inherit a D1 and R together than a D1 and T9 together. b) T/F About 50% of the egg cells from this individual will have both T9 and BA5. c) T/F Because r is recessive, the frequency at which the r version of the gene occurs in a population is lower than the frequency at which the R version occurs in the population. d) T/F If the protein produced by the BA gene is involved in the formation of ovaries, then the BA gene is likely located on a sex chromosome. e) T/F Because the genes shown here are found in skin cells, these genes code for proteins required for skin cell function.arrow_forward
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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license