1. Given the following pedigree below, use Punnett squares for each of the following possibilities: a) X-linked recessive and b) X-linked dominant in order to determine what is the mode of transmission ofthis trait. Disease allele = XA or Xa, depending on mode of transmission of the disease respectively.*Unaffected/No carrier-NormalUnaffected X chromosome = XI11* 12123a) X-linked recessive 11x12III 2 genotype (circle one): XY4XAXA235Xaya*4678b) X-linked dominant 11x12XAYXay

Name: 1. Given the following pedigree below, use Punnett squares for each o
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: 1. Given the following pedigree below, use Punnett squares for each of the following possibilities: a) X-linked recessive and b) X-linked dominant in order to determine what is the mode of transmission ofthis trait. Disease allele = XA or Xa, depend

The mode of inheritance is the pattern by which the trait or a disorder, which is passed from one…

Science

Biology

Given the following pedigree below, use Punnett squares for each of the following possibilities: a) X- linked recessive and b) X-linked dominant in order to determine what is the mode of transmission of this trait. Disease allele = XA or Xª, depending on mode of transmission of the disease respectively. Unaffected X chromosome = X *Unaffected/No carrier-Normal 11 d 1 * 1 1 2 a) X-linked recessive 9 III 2 genotype (circle one): XY 2 3 11x12 4 ΧΑΧΑ 2 3 5 Xaya *4 6 7 8 b) X-linked dominant 11x12 오 XAY Xay

Given the following pedigree below, use Punnett squares for each of the following possibilities: a) X- linked recessive and b) X-linked dominant in order to determine what is the mode of transmission of this trait. Disease allele = XA or Xª, depending on mode of transmission of the disease respectively. Unaffected X chromosome = X Unaffected/No carrier-Normal 11 d 1 1 1 2 a) X-linked recessive 9 III 2 genotype (circle one): XY 2 3 11x12 4 ΧΑΧΑ 2 3 5 Xaya *4 6 7 8 b) X-linked dominant 11x12 오 XAY Xay

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter4: Pedigree Analysis In Human Genetics

Section: Chapter Questions

Problem 10QP: A proband female with an unidentified disease seeks the advice of a genetic counselor before...

See similar textbooks

Similar questions

1.What are the possible genotypes of the parent gametes of the corn dihybrid cross described above? 2.What is the genotype of the F1 generation of the corn dihybrid cross described above? 3.What is the phenotype of the F1 generation of the corn dihybrid cross described above? 4.What are the possible maternal and paternal genotypes of the F1 gametes of the corn dihybrid cross described above?
1.What is the genotype of the F1 generation of the corn monohybrid cross described above? 2.What is the phenotype of the F1 generation of the corn monohybrid cross described above? 3.What are the possible maternal and paternal genotypes of the F1 gametes of the corn monohybrid cross described above?
1.What are the phenotypes of the parents of the corn dihybrid cross described above? 2. What are the possible genotypes of the parent gametes of the corn dihybrid cross described above? 3.What is the phenotype of the F1 generation of the corn dihybrid cross described above?
1. Pedigree #2 shows the inheritance of a X-linked trait. What is the genotype of individual 11 from this pedigree? Group of answer choices A. aa B. Aa C. AA D. XaXa E. XAXa F. XAXA 2. Refer to Pedigree #2 determine the mode of inheritance for the indicated X-linked trait before answering the question. If individuals 3 and 4 have another SON, what is the probability that the son will be AFFECTED? Group of answer choices A. 0% B. 25% C. 50% D. 75% E. 100% PLEASE PLEASE PLEASE PLEASE DO ALL
An individual heterozygous for four genes, A/a • B/b •C/c • D/d, is testcrossed with a/a • b/b • c/c • d/d, and 1000progeny are classified by the gametic contribution ofthe heterozygous parent as follows:a • B • C • D 42A • b • c • d 43A • B • C • d 140a • b • c • D 145a • B • c • D 6A • b • C • d 9A • B • c • d 305a • b • C • D 310a. Which genes are linked?b. If two pure-breeding lines had been crossed toproduce the heterozygous individual, what would theirgenotypes have been?c. Draw a linkage map of the linked genes, showing theorder and the distances in map units.d. Calculate an interference value, if appropriate
The following pedigree shows the pattern of inheritance of red green color blindness in a family. Females are shown as circles and males as squares; the squares or circles of individuals affected by the trait are filled in red. What is the chance that a son of the third-generation female indicated by the arrow will be color-blind if the father is a normal man? If the father is color-blind?
A heterozygous individual has a _______ for a trait being studied. a. pair of identical alleles b. pair of nonidentical alleles c. haploid condition, in genetic terms
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?
Examine the following pedigrees. Which is the most likely mode of inheritance of each disorder? (a) autosomal recessive (b) autosomal dominant (c) X-linked recessive (d) a, b, or c (e) a or c 10.
Individuals affected by a condition known as polydactyly have extra fingers or toes. The following pedigree shows the pattern of inheritance of this trait in one family: From the pedigree, can you tell if polydactyly comes from a dominant or recessive allele? Is the trait sex-linked? As far as you can determine, what is the genotype of each person in the pedigree with respect to the trait?
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?
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