Biology homework please help Iniciar reunión - ZoomEAKS 6a/b/b1 Mutations xG In humans, the allele fo xOLSdv6s9c6ISZpFULpxeSkekPHNA3ptICMhWbM4XLEpCG-SGCH9/formResponseThis question requires at least one response per rowBrown eyes are dominant to blue eyes. Identify the phenotype for the followinggenotype BB, Bb, bb and determine if the genotype is heterozygous orhomozygous.b.BBBbBbBbBbbbbbbbBBbrown eyesB.- dominant brown eye alleleBbbrown eyesb- recessive blue eye alleleblue eyes - Iniciar reunión - ZoomE AKS 6a/b/b1 Mutations XG In humans, the allele fo xDLSdv6s9c6ISZpFULpxeSkekPHNA3ptICMhWbM4XLEpCG-SGCH9/formResponseBbbbbbbbB -dominant brown eye alleleBBbrown eyesb recessive blue eye alleleBbbrown eyesbbblue eyesBlue eyesBrown eyesHeterozygousHomozygousBBBbbbIn humans, the allele for unattached earlobes (L) is dominant to the allele for

Human cells carry two copies of each chromosomes and they have two versions of each gene. An allele…

Answered: Blue eyes Brown eyes Heterozygous…

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Blue eyes Brown eyes Heterozygous Homozygous BB Bb bb O O O

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 8QP: Analysis of Autosomal Recessive and Dominant Traits In the following pedigree, assume that the...

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The phenotype of individuals heterozygous for ________ alleles comprises both homozygous phenotypes. a. epistatic b. codominant c. pleiotropic d. hybrid
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?
Analysis of Autosomal Recessive and Dominant Traits Huntington disease is a rare, fatal disease that usually develops in the fourth or fifth decade of life. It is caused by a single autosomal dominant allele. A phenotypically normal man in his twenties who has a 2-year-old son of his own learns that his father has developed Huntington disease. What is the probability that he himself will develop the disease? What is the chance that his young son will eventually develop the disease?
While sitting at home during Movement Control Order (MCO) because of pandemic covid19, observe two different traits of a couple in your family (eg. your mom & dad or your sister & her husband or your brother & his wife, etc). Draw a genetic cross that involves cross of the parents with the chosen 2 pairs of their contracting traits. Imagine that the cross obeys the Mendelian Laws, show the cross and gametes production for each generation (P, F1 and F2). By Using a Punnet square as symbolic representation of the results for the cross, determine the phenotypes, genotypes, phenotypic ratio and genotypic ratio of F2 generation in the family.
5. In the fruit fly Drosophila melanogaster, very dark(ebony) body color is determined by the e allele. Thee+ allele produces the normal wild-type, honeycolored body. In heterozygotes for the two alleles (butnot in e+e+ homozygotes), a dark marking called thetrident can be seen on the thorax, but otherwise thebody is honey-colored. The e+ snd e alleles are thusconsidered to be incompletely dominant.a. When female e+e+ flies are crossed to male e+eflies, what is the probability that progeny will havethe dark trident marking?b. Animals with the trident marking mate amongthemselves. Of 300 progeny, how many would beexpected to have a trident, how many ebony bodies, and how many honey-colored bodies?
Female fruit flies homozygous for the X-linked white-eye alleleare crossed to males with red eyes. On very rare occasions, an offspringof such a cross is a male with red eyes. Assuming these rareoffspring are not due to a new mutation in one of the mother’s Xchromosomes that converted the white-eye allele into a red-eyeallele, explain how a red-eyed male arises.
The XG locus on the human X chromosome has twoalleles, XG+ and XG. The XG+ allele causes the presence of the Xg surface antigen on red blood cells,while the recessive XG allele does not allow antigento appear. The XG locus is 10 m.u. from the STSlocus. The STS+ allele produces normal activity ofthe enzyme steroid sulfatase, while the recessive STSallele results in the lack of steroid sulfatase activityand the disease ichthyosis (scaly skin). A man withichthyosis and no Xg antigen has a normal daughterwith Xg antigen. This daughter is expecting a child.a. If the child is a son, what is the probability he willlack Xg antigen and have ichthyosis?b. What is the probability that a son would have boththe antigen and ichthyosis?c. If the child is a son with ichthyosis, what is theprobability he will have Xg antigen?
. Among adults with Turner syndrome, it has beenfound that a very high proportion are genetic mosaics.These are of two types: In some individuals, themajority of cells are XO, but a minority of cellsare XX. In other Turner individuals, the majorityof cells are XO, but a minority of cells are XY.Explain how these two patterns of somatic mosaicscould arise.
Determining Gene Order 1. We want to map the distance between genes E (tall height), B (blue flower color), and G (thick stalk). Each geen has a recessive allele (e= short, b= white and g= thin). (The actual configuration of the Trihybrid will be deduces as a part of the process) A. Your testcross genotypes will be (use slash notation). Female: ______ Male:______ Based on these results, are the genes linked or not? _______
A man is brachydactylous (very short fingers; rare autosomal dominant), and his wife is not. Both can taste thechemical phenylthiocarbamide (autosomal dominant;common allele), but their mothers could not.a. Give the genotypes of the couple.If the genes assort independently and the couple hasfour children, what is the probability ofb. all of them being brachydactylous?c. none being brachydactylous?d. all of them being tasters?e. all of them being nontasters?f. all of them being brachydactylous tasters?g. none being brachydactylous tasters?h. at least one being a brachydactylous taster?
6. In certain plant species such as tomatoes and petunias, a highly polymorphic incompatibility gene Swith more than 100 known alleles prevents self-fertilization and promotes outbreeding. In this form ofincompatibility, a plant cannot accept sperm carrying an allele identical to either of its own incompatibility alleles. If, for example, pollen carryingsperm with allele S1of the incompatibility genelands onto the stigma (a female organ) of a plantthat also carries the S1allele, the sperm cannot fertilize any eggs in that plant. (This phenomenonoccurs because the pollen grain on the stigma cannotgrow a pollen tube to allow the sperm to unite withthe egg.)For the following crosses, indicate whether anyprogeny would be produced, and if so, list all possiblegenotypes of these progeny. a. ♂ S1S2× ♀ S1S2b. ♂ S1S2× ♀ S2S3c. ♂ S1S2× ♀ S3S4d. Explain how this mechanism of incompatibilitywould prevent plant self-fertilization.
Fast forward 10 years, and you are working as a dermatologist specializing on hands. You notice almost all your patients who have nail patella syndrome (N: disease; n: wild-type) are also O blood type (genotype at the ABO gene: OO). Seems like a weird coincidence, until you remember that the genes causing these two traits are linked and 10 cM apart. If someone who is NBnO has children with someone who is NBnO, what is the probability they would have a kid without nail patella syndrome who is also B-blood type (genotype: BB). a. What is heritability, and what does it measure? b. Would you do a GWAS on a trait with heritability of 0? Why or why not? c. The effect sizes for the SNPs linked to performance on IQ tests are very very small. Why does that make it unlikely that we can genetically engineer humans with super high IQ?