2. What are the genotypes (in symbols) of the individuals you crossed to obtain F,?..F,?F;: Male parent: XbTYFemale parent: X8tXBtF2: Male parent: XBtYFemale parent: X8TXbtBristles: X® - morula,Xb - hookTarsi: X" - wild type,X' - jointed3. Differentiate between autosomal genetic linkage and sex linkage.CriteriaAutosomal GeneticSex LinkageLinkage

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Description: 2. What are the genotypes (in symbols) of the individuals you crossed to obtain F,?..F,?F;: Male parent: XbTYFemale parent: X8tXBtF2: Male parent: XBtYFemale parent: X8TXbtBristles: X® - morula,Xb - hookTarsi: X" - wild type,X' - jointed3. Different

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genotyp ...F2? F,: Male parent: XbTY Female parent: X®'X8t F2: Male parent: X®tY Female parent: X®TXbt Bristles: X - morula, X - hook Tarsi: XT - wild type, X'- jointed

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 2TYK: In peas, the allele Le produces tall plants and the allele le produces dwarf plants. The Le allele...

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1) By decomposing the cross, MmNn X Mmnn, into single gene outcomes, the possibility of obtaining a double recessive outcome for the ‘N’ gene is: a) 50%! b) 100! c) 0%! d) 25%! 2) A 16-square Punnett is time-consuming to draw out. Dr. Thompson can easily solve this problem by decomposing her dihybrid crosses into separate monohybrid Punnett squares, which would also be quicker than using a forked line diagram. If so, how many individual monohybrid Punnett squares will she need to draw? a) Four Punnett squares. b) Two Punnet squares. c) One Punnett square. d) Sixteen Punnett squares since there are 16 possible outcomes.
1.If the AaBBCc individual will be test-crossed, what is the probability that the resulting offspring will show all the recessive traits? a.three-fourth b.zero c.nine-sixteenth d.one-sixteenth 2. Who conclusively demonstrated that the substance responsible for the transformation of non-virulent Streptococcus pneumoniae to virulent S. pneumoniae is the DNA? a.Frederick Griffith b.Hershey and Chase c.Watson and Crick d.Avery, MacLeod, and McCarthy
4) A testcross is used to determine which of the following? a) The phenotype of an individual with a dominant genotype b) The phenotype of an individual with a recessive genotype c)The genotype of an individual with a dominant phenotype d) The genotype of an individual with a recessive phenotype
2.In dogs, there is a hereditary deafness caused by a recessive allele, d. A kennel owner has a male dog that she wants to use for breeding purposes if possible. The dog can hear, but the owner is unsure of the genotype. She does a testcross (crosses it to a homozygous recessive dog), and two of the five offspring are deaf. This means that the male dog A.has the genotype DD B.has the genotype Dd C.is still of unknown genotype since there were offspring of both deaf and hearing phenotypes. D.has the genotype dd
2. Be able to compare and contrast (discuss similarities and differences in) in your own words the terms genotype and phenotype, wild-type and mutant phenotypes. Step by step.
1. a) If you were at risk of a fatal disease such as Huntington’s disease and only the 95% accurate preliminary genetic test was available, would you take the test? After all, 95% of the time the test would tell you whether you would eventually get Huntington’s disease or whether you did not have to worry about getting this disease. b) The linkage distance between the DNA sequence used in the test and the actual Huntington’ disease locus was 5%. In a large sample of the Venezuelan family, what percentage of people would inherit the DNA sequence but not the Huntington’s allele (i.e. be a false positive test for Huntington’s)? What percentage of that large Venezuelan family would not inherit the DNA sequence but would inherit the Huntington’s allele (i.e., be a false negative test for Huntington’s)?
11. Shown below are the offspring of the testcross qr/q +r+ X q r/q r. Which groups are in the recombinant class? q+r and qr+ q+r+ and qr+ qr and qr+ q+r+ and qr
Which recombination frequency corresponds to perfect linkage and violates the law of independent assortment? 0 0.25 0.50 0.75
In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant W form (WW or Ww), this gene blocks the formation of fur color and all the offspring are white; ww individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross AaWw Aa Ww?
Two pairs of traits are segregating in a cross. Two parents produce 156 progeny that fall into 4 phenotypes. The numbers of offspring in the 4 phenotypes are 89, 31, 28, and 8. What are the genotypes of the two parents?
In peas, the allele Le produces tall plants and the allele le produces dwarf plants. The Le allele is dominant to le. If a tall plant is crossed with a dwarf, the offspring are distributed about equally between tall and dwarf plants. What are the genotypes of the parents?
A testcross is a way to determine ________. a. phenotype b. genotype c. dominance