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
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- 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…In a cross of Lymnaea, the snail contributing the eggs was dextralbut of unknown genotype. Both the genotype and the phenotypeof the other snail are unknown. All F1 offspring exhibited dextralcoiling. Ten of the F1 snails were allowed to undergo self-fertilization.One-half produced only dextrally coiled offspring, whereasthe other half produced only sinistrally coiled offspring. Whatwere the genotypes of the original parents?Hello Heliodors! Trait B Heliodors are either red (R), yellow (Y) or an intermediate phenotype, orange. What is the genotype of a yellow heliodor? СС Сс cc C'CY CRCY CRCR
- In a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…corn id purple smooth Purple wrinkled yellow smooth yellow wrinkled W20-AB1-D-A13 486 152 162 54 W21-BB1-A-19 445 152 154 94 W20-BB1-D-19 501 153 167 88 W20-AB1-C-16 489 153 94 56 W20-BB1-H-11 480 153 94 56 MONOHYBRID CROSS1. Present the data as monohybrids (Xx, XX, xx) for all five corn cobs andfor the plant. Present the colour and texture phenotypes with their associatedgenotypes in separate tables. Your first table should include the corn ID, the number of purple kernels, thenumber of yellow kernels, and the phenotypic ratio between the two kernelcolours. Present the texture results in the same format to a separate table.2. What is the dominant phenotype for kernel colour in your assigned cornplant? What is the dominant phenotype for kernel texture in your assigned cornplant?3. What are the probable phenotypes for the previous generation of cornplants with regard to kernel coloration?4. What are the probable phenotypes of the previous generation of cornplants with…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.
- Species I is diploid (2n=8) with chromosomes AABBCCDD. Related species II is diploid (2n=8) with chromosomes MMNNOOPP. Individuals with the ff. sets of chromosomes represent what types of chromosome mutations (be very specific)? Give the level of ploidy (in equation form) and resultant chromosome number for each plant. AABBCCCCDD (c) MMNNOOOPPP (e) AABBCDD AABBDD (d) AABBCCDDMNOP38 sday, November 5 DOLL Safari File Edit View History Bookmarks Window Help A canton.open.suny.edu B5 Take Te.. S https:.. DCultu. S Lace. S Drago Wom. Does... S Crop. Credi. QUESTION 1 A pea plant that is heterozygous for the flower color gene makes gametes. What is the probability that a specific gamete contains the recessive allele for flower color? Oa. 75 percent Ob.25 percent Oc. 50 percent Od. 100 percentAn autosomal recessive disease affects /200 dogs and the population is in Hardy-weinlerg. a) what propornon of he poulanon are cauners for the disease? b) A carrrer dog and an unknown dog are randomy sevectedfrom the popuaron and thay have one offipring, what is the fosioi'ing the oříspring is a carrier for he disease?
- Many modern people have some Neanderthal DNAin their genome, but the Neanderthal alleles are notuniformly distributed across their genome. A person ofEuropean or Asian ancestry typically has 1.5 to 2 percent Neanderthal DNA in their autosomes. However, researchers have never found any Neanderthal alleles on the Y chromosome of a modern human. By one hypothesis, Neanderthal Y alleles disappeared because theywere incompatible with H. sapiens genes. Explain howreduced fitness in hybrids arising from genetic incompatibility could have, over time, led to the elimination of theNeanderthal Y alleles from the H. sapiens gene pool.9:31 Dh O N ll 97% 20220331_200252.jpg Phoblims 84 In Drovphila a chros was made buetween females ell enprening the 3X- linkage, recuesive thaits, Slut bristus (sc), Scale body (5) and vermillion eye (v) and wild type males. On the FI, ell femalis cuere wild type whil all mals enprunid ell the matonttraits. The Chos was labrid to the F2 generation Ond 1000 offspring were countid arith resalt shoued beloo: Phonatype Se S V Set st yt. Set s v- Sc st v+ Sc st v Sct s vt. s vt. 314 280 156 46 30 10 14 Sc sct st v- To00 Uing the nommelature, detumine genatyps of Pond Fl patnt. D Osder f genes ond mapdistonce. 9 Aru there moto or fuver double crou thon onfectid D aleulate the Caeffeicint of Coincidence. 62 Femole hutrazygous for ehony le/e), searlit (st"/st) end spineles sst/ss) were itest chomud ord the following F praginy wrere olitainid:- Preginy Phenolypt Wild type e st ss - et st ss.- 67 8. 68 347 Spiniless - e st sst Scarlt fineles - e st so Ebony Ebony seorlet- ct set ss Ehgrny…. 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?