Answer this. fun question My dear colleagues,I was also able to trace the lineage of a family of these remarkable Rabandubirds, this time a group expressing the clawed wing trait. It was quite thefascinating task!The matriarch of the family is Snippy, a normal winged female, and her mateis Chomp, a normal winged male. These weretheir offspring:The claus seemto be usedfor climbinggrastlngSpotina: Female, normal wingedandbrånchisReginald: Male, clawed wingedDuke: Male, normal wingGenie: Female, normal wingGenie mated wíth Jasper, a normal-wingedmale, and they had seven offspring. Three ofthe offspring were normal winged males, twowere normal winged females, one was a clawed-wing male, and one was aclawed-wing female. Reginald mated with Honeydew, a clawed-wing female,and they had three offspring. TWO were clawed-wing males, and one was aclawed-wing fenmale.Please let me know what you find out, it could be quite the discovery if we canfigure out how these traits are inherited!Your Colleague,Dr. Maldona

Diseases that are transmitted genetically is of five types - autosomal recessive, autosomal…

My dear colleagues, I was also able to trace the lineage of a family of these remarkable Rabandu birds, this time a group expressing the clawed wing trait. It was quite the fascinating task! The matriarch of the family is Snippy, a normal winged female, and her mate is Chomp, a normal winged male. These were their offspring: The claws seem to be for cli Spotina: Female, normal winged and br

My dear colleagues, I was also able to trace the lineage of a family of these remarkable Rabandu birds, this time a group expressing the clawed wing trait. It was quite the fascinating task! The matriarch of the family is Snippy, a normal winged female, and her mate is Chomp, a normal winged male. These were their offspring: The claws seem to be for cli Spotina: Female, normal winged and br

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Neurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe list
. An allotetraploid species has a genome composed oftwo ancestral genomes, A and B, each of which havea basic chromosome number (x) of seven. In thisspecies, the two copies of each chromosome of eachancestral genome pair only with each other duringmeiosis. Resistance to a pathogen that attacks the foliage of the plant is controlled by a dominant allele atthe F locus. The recessive alleles Faand Fbconfersensitivity to the pathogen, but the dominant resistancealleles present in the two genomes have slightly different effects. Plants with at least one FAallele areresistant to races 1 and 2 of the pathogen regardlessof the genotype in the B genome, and plants with atleast one FBallele are resistant to races 1 and 3 of thepathogen regardless of the genotype in the A genome.What proportion of the self-progeny of an FA Fa FB Fbplant will be resistant to all three races of the pathogen?
Besides the ones mentioned in this textbook, look for other examplesof variations in euploidy. Perhaps you might look in moreadvanced textbooks concerning population genetics, ecology, etc.Discuss the phenotypic consequences of these changes.
Two diploid species of closely related frogs, which we will callspecies A and species B, were analyzed with regard to the genesthat encode an enzyme called hexokinase. Species A has two distinctcopies of this gene: A1 and A2. In other words, this diploidspecies is A1A1 A2A2. Species B has three copies of the hexokinasegene, which we will call B1, B2, and B3. A diploid individualof species B would be B1B1 B2B2 B3B3. These hexokinase genesfrom the two species were subjected to DNA sequencing, and thepercentage of sequence identity was compared among these genes.The results are shown here. Percentage of DNA Sequence Identity A1 A2 B1 B2 B3A1 100 62 54 94 53A2 62 100 91 49 92B1 54 91 100 67 90B2 94 49 67 100 64B3 53 92 90 64 100…
A planet is inhabited by creatures that reproduce with the same hereditary patterns seen in humans. Three phenotypic characters are height (T = tall, t = dwarf), head appendages (A = antennae, a = no antennae), and nose morphology (S = upturned snout, s = downturned snout). Since the creatures are not “intelligent,” Earth scientists are able to do some controlled breeding experiments using various heterozygotes in testcrosses. For tall heterozygotes with antennae, the offspring are tall antennae, 46; dwarf antennae, 7; dwarf no antennae, 42; tall no antennae, 5. For heterozygotes with antennae and an upturned snout, the offspring are antennae upturned snout, 47; antennae downturned snout, 2; no antennae downturned snout, 48; no antennae upturned snout, 3. Calculate the recombination frequencies for both experiments.
On average, what proportion of the genome in the following pairs ofhumans would be exactly the same if no crossing over took place? (Forthe purposes of this question only, we will ignore the special case of theX and Y sex chromosomes and assume that all genes are located onnonsex chromosomes.)a. Father and childb. Mother and childc. Two full siblings (offspring that have the same two biological parents)d. Half siblings (offspring that have only one biological parent incommon)e. Uncle and niecef. Grandparent and grandchild
The plant blue-eyed Mary grows on Vancouver Islandand on the lower mainland of British Columbia. Thepopulations are dimorphic for purple blotches on theleaves—some plants have blotches and others don’t. NearNanaimo, one plant in nature had blotched leaves. Thisplant, which had not yet flowered, was dug up and takento a laboratory, where it was allowed to self. Seeds werecollected and grown into progeny. One randomly selected (but typical) leaf from each of the progeny is shown inthe accompanying illustration.a. Formulate a concise genetic hypothesis to explainthese results. Explain all symbols and show all genotypicclasses (and the genotype of the original plant).
A species of antelope has 20 chromosomes per set. The species isdivided by a mountain range into two separate populations, whichwe will call the eastern and western population. In a comparison ofthe karyotypes of these two populations, it was discovered that themembers of the eastern population are homozygous for a largeinversion within chromosome 14. How would this inversion affectthe interbreeding between the two populations? Could such aninversion play an important role in speciation?
. As we learned in this chapter, the white mutation ofDrosophila studied by Thomas Hunt Morgan isX-linked and recessive to wild type. When truebreeding white-eyed males carrying this mutationwere crossed with true-breeding purple-eyed females,all the F1 progeny had wild-type (red) eyes. When theF1 progeny were intercrossed, the F2 progeny emergedin the ratio 3/8 wild-type females: 1/4 white-eyedmales: 3/16 wild-type males: 1/8 purple-eyed females:1/16 purple-eyed males.a. Formulate a hypothesis to explain the inheritanceof these eye colors.b. Predict the F1 and F2 progeny if the parental crosswas reversed (that is, if the parental cross wasbetween true-breeding white-eyed females andtrue-breeding purple-eyed males).
A researcher sequences the whole exome of a patientsuffering from Usher syndrome, a rare autosomal recessive condition that is nonetheless the leading causefor simultaneous deafness and blindness. The exomesequence does not show homozygosity for any polymorphisms different from the human RefSeq.a. How could the researcher examine the data alreadygathered to try to find the disease gene, assumingthe sequence is accurate?b. If the attempt described in part (a) was unsuccessful, the researcher might contemplate sequencingthe patient’s whole genome. What are the potentialpitfalls of this strategy?
The life cycle of the haploid fungus Ascobolus is similar tothat of Neurospora. A mutational treatment producedtwo mutant strains, 1 and 2, both of which when crossedwith wild type gave unordered tetrads, all of the followingtype (fawn is a light brown color; normally, crosses produce all black ascospores):spore pair 1 black spore pair 3 fawnspore pair 2 black spore pair 4 fawna. What does this result show? Explain.The two mutant strains were crossed. Most of the unordered tetrads were of the following type:spore pair 1 fawn spore pair 3 fawnspore pair 2 fawn spore pair 4 fawnb. What does this result suggest? Explain.When large numbers of unordered tetrads were screenedunder the microscope, some rare ones that containedblack spores were found. Four cases are shown here:Case A Case B Case C Case Dspore pair 1 black black black blackspore pair 2 black fawn black abortspore pair 3 fawn fawn abort fawnspore pair 4 fawn fawn abort fawn(Note: Ascospores with extra genetic material…
If homozygous recessive (aa,tetrapter) f1 produces homozygous (aa) f2, what is the dominant and recessive trait of f2? Let’s say the dominant trait of the parents of f1 is Wild type and the recessive trait is Tetrapter.