Pigmentation gene (dark allele) found in a dark butterfly Pigmentation gene (light allele) found in a light butterfly Transcription and translation Highly functional pigmentation enzyme Poorly functional pig entation enzyme (a) Molecular level Pigment molecule Wing cells Lots of pigment made Little pigment made (b) Cellular level Dark butterfly Light butterfly (c) Organism level

Pigmentation gene (dark allele) found in a dark butterfly Pigmentation gene (light allele) found in a light butterfly Transcription and translation Highly functional pigmentation enzyme Poorly functional pig entation enzyme (a) Molecular level Pigment molecule Wing cells Lots of pigment made Little pigment made (b) Cellular level Dark butterfly Light butterfly (c) Organism level

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 27QP: Variations in Phenotype Expression A genetic disorder characterized by falling asleep in genetics...

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A+ 70°F, a cell cultue is able to graw and divide farever, but after a new mutation is found in the čeus, and the cells are at 99°F, mey Stop dividing after a few generahons.what * 16 ikery disiupted by the mutation for the cells to stop dividing after a few gener ution when Compared to the 70°F when they diide and qrow indefinitely! grown they aure 95°F grown at
Neanderthal Hair Color The MC1R gene regulates pigmentation in humans (Sections 14.1 and 15.1 revisited), so loss-of-function mutations in this gene affect hair and skin color. A person with two mutated alleles for this gene makes more of the reddish melanin than the brownish melanin, resulting in red hair and pale skin. DNA extracted from two Neanderthal fossils contains a mutated MC1R allele that has not yet been found in humans. To see how the Neanderthal mutation affects the function of the MC1R gene. Carles Lalueza-Fox and her team introduced the allele into cultured monkey cells (FIGURE 26.16). FIGURE 26.16 MC1R activity in monkey cells transgenic for an unmutated MC1R gene, the Neanderthal MC1R allele, or the gene for green fluorescent protein (GFP). GFP is not related to MC1R. 1. How did MCR1 activity in monkey cells with the mutant allele differ from that in cells with the normal allele?
It was clear from the early days of studying devel-opment that certain “morphogenetic” substances werepresent in the egg and segregated asymmetrically intocells of the developing embryo. One such investigationin ascidian (sea squirt) embryos examined endodermalalkaline phosphatase, which could be visualized by a his-tochemical stain. Treatment of embryos with cytochalasinB stopped cell division, but did not block expression ofalkaline phosphatase at the appropriate time. Treatmentwith actinomycin D, which blocks transcription, did notinterfere with expression of alkaline phosphatase. Treat-ment with puromycin, which blocks translation, elim-inated expression of alkaline phosphatase. What is thelikely nature of the morphogenetic substance that givesrise to alkaline phosphatase?
Bong Question #1: The diagram below depicts the regulatory regions for two (made-up) genes, which contain cis-regulatory sequences X, Y, and Z and bind to transcriptional regulatory proteins: zelo led diogot bolgate SMARTY – a transcriptional ACTIVATOR protein, which is present in all neuronal cells and binds to cis-regulatory sequence, X1oq & vino 19vewod.152 moldong sai mut tum BRAWNY-a transcriptional ACTIVATOR protein, which is present in all muscle cells and binds to cis-resgulatory sequence, Yolgulum di ko malo na SNARKY - a transcriptional REPRESSOR protein, which is present in peripheral neurons only and binds to cis-regulatory sequence Z 100 bio se i da se lotimo broup gniwollt od 19 bolgate ons zegg or we de base do no me to stir noitesup od went of sistemos seu anoitesup 15wens horle 10oldog woy ni gnius stoted 1910 ni tatayot ovizasovo got no rade od lliw anioq azia oo ingene Aroom or b X y Jeol VELY gan 100 Tonnodige ΤΑΤΑ, 229nibrow dong H .aodto diw atse meldong mov.no o…
I. The retinoic acid receptor (RAR) is a transcription factor that is similar to steroid hormone receptors. Thesubstance (ligand) that binds to this receptor is retinoicacid. One of the genes whose transcription is activatedby retinoic acid binding to the receptor is myoD. Thediagram that follows shows a schematic view of theRAR proteins produced by genes into which one oftwo different 12-base double-stranded oligonucleotides had been inserted in the ORF. The insertion site(a–m) associated with each mutant protein is indicatedwith the appropriate letter on the polypeptide map.For constructs encoding proteins a–e, oligonucleotide 1(5′ TTAATTAATTAA 3′ read off either strand) wasinserted into the RAR gene. For constructs encoding proteins f–m, oligonucleotide 2 (5′ CCGGCCGGCCGG 3′)was inserted into the gene.NH2 f g h i j k l m COOHa b c d eThe wild-type RAR protein can both bind DNA and activate transcription weakly in the absence of retinoic acid(RA) and strongly in RA’s presence. Each…
Would you simply answer the question for me, i saw many answers for it but still not sure. The flower color of genetically identical hemp plants (clones) is purple when the flowers develop under cold environment (30 °C). Describe in detail a molecular mechanism that could explain how these two groups of genetically identical plants differ in their gene expressions to produce different flower colors?
Gene Vellow Pigment ,in some plants White By-product ENZYME Rabbits that have the active form of this gene produce white-colored fat. Rabbits who have the defective form of this gene produce yellow- colored fat. However, there is an environmental complication of this trait. (See the following questions.) 1. If a rabbit that has the active form of the gene is fed plants with the pigment, what color is its fat? 2. If a rabbit with the defective gene is fed plants with the pigment, what çolor is its fat?
genetics question please explain We know that multiple genes (some with many different alleles) contribute to skin color, but genes alone donot account for the diversity of pigmentation we see among humans; environmental factors play a role. b) Propose an explanation for how one of these factors could alter the expression of skin color genes.
Biologists have long been interested in the effects of radiation on cells. In one experiment, researchers examined the effect of radium on mitosis of chick embryo cells growing in culture. A population of experimental cells was examined under the microscope for the number of cells in telophase (as a measure of mitosis occurring) before, during, and after exposure to radium. The results are shown in the Figure. What is the effect of radium exposure on mitosis? Source: R. G. Canti and M. Donaldson. 1926. The effect of radium on mitosis in vitro. Proceedings of the Royal Society of London, Series B, Containing Papers of a Biological Character 100:413419.
The molecular mechanism of SCE is similar to homologousrecombination between homologs except that the two segmentsof DNA are sister chromatids instead of homologous chromatids.If branch migration occurs during SCE, will a heteroduplex beformed? Explain why or why not. Can gene conversion occurduring sister chromatid exchange?
Like Hurler syndrome, Fabry disease involves an abnormal accumulationof substances within lysosomes. However, the lysosomesof individuals with Fabry disease show an abnormal accumulationof lipids. The defective enzyme is α-galactosidase A, which is alysosomal enzyme that functions in lipid metabolism. The enzymaticdefect causes cell damage, especially to the kidneys, heart,and eyes. The gene that encodes α-galactosidase A is found on theX chromosome. Let’s suppose a phenotypically unaffected coupleproduces two sons with Fabry disease and one phenotypicallyunaffected daughter. What is the probability that the daughter willhave an affected son?
Chromosome 3 in orangutansdiffers from chromosome 3 in humansby two inversion events that occurredin the human lineage (Figure Q4–2).Draw the intermediate chromosomethat resulted from the first inversionand explicitly indicate the segmentsincluded in each inversion.