A black Labrador Retriever is mated with a brown lab, but their offspring consists of black, brown AND golden labs. The two genes involved in producing this phenotype are a gene that controls pigment color and a gene that controls if pigment is deposited. This is an example of: O Incomplete Dominance Pleiotropy Epistasis O Multiple Alleles
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- In individuals affected by cystic fibrosis, salt crystals may appear afterperspiration dries up. In addition, the disease causes respiratory disorderswhich can be both debilitating and lethal. It occurs in individuals homozygousfor the recessive gene. Two normal parents had a daughter with thesymptoms of this disease, and a normal son who marries a normal womanwith an afflicted A test (salt concentration in perspiration of heterozygotes ishigher than normal) disclosed that both are indeed carriers of the gene. If thefirst child born to the mating in (b) was defective, what is the probability thatthe 2nd child would also be defective?Express answer in fraction formThe mutant FMR-1 allele that causes fragile X syndrome is considered to be X-linked dominant withincomplete penetrance and variable expressivity.Why do most females heterozygous for one mutantand one normal allele have at least some symptomsof the disease?Albino rabbits (lacking pigment) are homozygous forthe recessive c allele (C allows pigment formation).Rabbits homozygous for the recessive b allele makebrown pigment, while those with at least one copy ofB make black pigment. True-breeding brown rabbitswere crossed to albinos, which were also BB. F1 rabbits, which were all black, were crossed to the doublerecessive (bb cc). The progeny obtained were 34black, 66 brown, and 100 albino.a. What phenotypic proportions would have beenexpected if the b and c loci were unlinked?b. How far apart are the two loci?
- The gene for the production of eye colour in this species of fruit fly can be expressed aseither normal red-eyes or as brown-eyes. The allele for the normal red eyes is dominant tothat of brown, and is transmitted in normal Mendelian fashion.A gene involved with body colour in this species of fruit fly is located on the non-homologousportion of the X chromosome, and has two alleles, grey and yellow, where the grey allele isdominant to yellow.The two genes are NOT linked.a) Produce a key to clearly show the nature of the alleles associated with the eye colour inthis species of Drosophila, and in each case justify your choice of letters and / or style ofpresentation to best depict the genetics involved.The maternal-effect mutation bicoid (bcd) is recessive. Inthe absence of the bicoid protein product, embryogenesis isnot completed. Consider a cross between a female heterozygousfor the bicoid mutation (bcd+/ bcd-) and a homozygousmale(bcd-/ bcd-). Predict the outcome (normal vs. failed embryogenesis) inthe F1 and F2 generations of the cross described.A wild-type fruit fly (heterozygous for gray body color andred eyes) is mated with a black fruit fly with purple eyes. Theoffspring are wild-type, 721; black purple, 751; gray purple, 49;black red, 45. What is the recombination frequency betweenthese genes for body color and eye color? Using informationfrom problem 3, what fruit flies (genotypes and phenotypes)would you mate to determine the order of the body color, wingsize, and eye color genes on the chromosome?
- Mutations in DNA that result in altered proteins can causehereditary diseases. Pedigree studies and genetic testing mayclarify the risk of disease. At the chromosome level, nondisjunctionduring meiosis can result in gametes with too few or too manychromosomes, most of which produce inviable offspring.Imprinting refers to inactivation of alleles depending on whichparent the alleles come from; offspring in whom imprinting occursappear haploid for the affected gene even though they are diploid. During spermatogenesis, is there any difference in outcome between first- and second-division nondisjunction?In rabbits, fur colour can be either grey, black or blotchy. Blotchy is present as grey and black spotspresent in no specific pattern. Males are almost always either grey or black, while females can begrey, black or blotchy. However, a male individual is found that shows the blotchy pattern, and youexpect a non-disjunction event might be to blame. Explain the inheritance of coat colour in rabbits,and also provide an explanation for the blotchy male.A sex-influenced trait is dominant in males of a certain speciesof animal and causes bushy tails. The same trait is recessive infemales. Fur color is not sex-influenced.Yellow fur is dominantto white fur. A true-breeding female with a bushy tail and yellowfur was crossed to a white male without a bushy tail (i.e., a normaltail). The F1 females were then crossed to white males withoutbushy tails. The following results were obtained: A. Conduct a chi square test to determine if these two genes arelinked.B. If the genes are linked, calculate the map distance betweenthem. Explain which data you used in your calculation.
- In 1952, an article in the British Medical Journalreported interesting differences in the behavior ofblood plasma obtained from several people who suffered from X-linked recessive hemophilia. Whenmixed together, the cell-free blood plasma from certain combinations of individuals could form clots inthe test tube. For example, the following table showswhether clots could form (+) or not (−) in variouscombinations of plasma from four people withhemophilia:1 and 1 − 2 and 3 +1 and 2 − 2 and 4 +1 and 3 + 3 and 3 −1 and 4 + 3 and 4 −2 and 2 − 4 and 4 −What do these data tell you about the inheritance ofhemophilia in these individuals? Do these data allowyou to exclude any models for the biochemical pathway governing blood clotting?In sweet pea plant, an allele for purple flowers (P) is dominant when paired with a recessive allele for red flowers (p). An allele for long pollen grains (L) is dominant when paired with a recessive allele for round pollen grain (l). Bateson and Punnett crossed a plant having purple flowers/long pollen grains with one having white/flowers/round pollen grains. All F1 offspring had purple flowers and long pollen grains. Among the F2 generation, the researchers observed the following phenotypes: 296 purple flowers/long pollen grains 19 purple flowers/round pollen grains 27 red flowers/long pollen grains 85 red flowers/round pollen grains What is the best explanation for these results?Manx cats have no tail. This is caused by a dominantmutation M, which is embryo lethal when homozygous. An unlinked gene B controls the color of the fur onthe cat's tail. The dominant B allele produces black fur on tails, while the recessive allele b produceswhite fur on tails. A Manx cat that is B/B is crossed to a white-tailed cat. Half of the F1 progeny hadthe Manx phenotype and the other half had normal, black tails.The F1 cats with the Manx phenotype were crossed to each other, producing F2 cats. What is the ratioof the following phenotypes in the F2