the genes are 23 millimeters apart. there are 23 other genes between the two genes of interest. | 23 % of the offspring exhibit recombination between the two genes. 23% of the offspring do not survive. 77% of the offspring exhibit recombination between the two genes.

the genes are 23 millimeters apart. there are 23 other genes between the two genes of interest. | 23 % of the offspring exhibit recombination between the two genes. 23% of the offspring do not survive. 77% of the offspring exhibit recombination between the two genes.

Biology 2e

2nd Edition

ISBN:9781947172517

Author:Matthew Douglas, Jung Choi, Mary Ann Clark

Publisher:Matthew Douglas, Jung Choi, Mary Ann Clark

Chapter13: Modern Understandings Of Inheritance

Section: Chapter Questions

Problem 2VCQ: Figure 13.4 Which of the following statements is true? Recombination of the body color and red/...

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Individuals of genotype AaBb were mated to individuals of genotype aabb. One thousand offspring were counted, with the following results: 474 Aabb, 480 aaBb, 20 AaBb, and 26 aabb. What type of cross is it? Are these loci linked? What are the two parental classes and the two recombinant classes of offspring? What is the percentage of recombination between these two loci? How many map units apart are they?
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
A proband female with an unidentified disease seeks the advice of a genetic counselor before starting a family. Based on the following data, the counselor constructs a pedigree encompassing three generations: (1) The maternal grandfather of the proband has the disease. (2) The mother of the proband is unaffected and is the youngest of five children, the three oldest being male. (3) The proband has an affected older sister, but the youngest siblings are unaffected twins (boy and girl). (4) All the individuals who have the disease have been revealed. Duplicate the counselors feat
Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?
Which recombination frequency corresponds to independent assortment and the absence of linkage? 0 0.25 0.50 0.75
Still referring to Problem 1, what will be the possible genotypes of offspring from the following matings? With what frequency will each genotype show up? a. AABB aaBB b. AaBB AABb c. AaBb aabb d. AaBb AaBb
Sunflowers with flowers 10 cm in diameter are crossed with a plant that has 20-cm flowers. The F1 plants have flowers 15 cm in diameter. In the F2 generation, 4 flowers are 10 cm in diameter and 4 are 20 cm in diameter. Between these are 5 phenotypic classes with diameters intermediate to those at the extremes. a. Assuming that the alleles that contribute to flower diameter act additively, how many genes control flower size in this strain of sunflowers? b. How much does each additive allele contribute to flower diameter? c. What size flower makes up the largest phenotypic class?
Which recombination frequency corresponds to perfect linkage and violates the law of independent assortment? 0 0.25 0.50 0.75
Pedigree Analysis Is a Basic Method in Human Genetics Using the pedigree provided, answer the following questions. a. Is the proband male or female? b. Is the grandfather of the proband affected? c. How many siblings does the proband have, and where is he or she in the birth order?
The young woman shown at right has albinismvery pale skin, white hair, and pale blue eyes. This phenotype is due to the absence of melanin, which imparts color to the skin, hair, and eyes. It typically is caused by a recessive allele. In the following situations, what are the probable genotypes of the father, the mother, and their children? a. Both parents have normal phenotypes; some of their children are albino and others are not. b. Both parents and all their children are albino. c. The mother is not albino, the father is albino, and one of their four children is albino.
Why dont genetic markers on the Y chromosome undergo recombination? Why is this lack of recombination a necessity for these markers to be used in tracing migrations?
As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progeny