In humans, the mic2 gene (involved in antibody production) is found on sex chromosomes and has pseudoautosomal inheritance. If a female that is a mic2a is mated to a phenotypically mic2b male then what would be resulting ratios of the offspring? (note; the mic2a and 2b alleles are codominant)
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- Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With respect to X-linked alleles, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele? c. If a female is heterozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele?Two genes control color in corn snakes as follows:O– B– snakes are brown, O– bb are orange, oo B– areblack, and oo bb are albino. An orange snake wasmated to a black snake, and a large number of F1progeny were obtained, all of which were brown.When the F1 snakes were mated to one another, theyproduced 100 brown offspring, 25 orange, 22 black,and 13 albino.a. What are the genotypes of the F1 snakes?b. What proportions of the different colors wouldhave been expected among the F2 snakes if the twoloci assort independently?In humans, color vision depends on genes encodingthree pigments. The R (red pigment) and G (green pigment) genes are close together on the X chromosome,whereas the B (blue pigment) gene is autosomal. A recessive mutation in any one of these genes can cause colorblindness. Suppose that a color-blind man married awoman with normal color vision. The four sons from thismarriage were color-blind, and the five daughters werenormal. Specify the most likely genotypes of both parents and their children, explaining your reasoning. (Apedigree drawing will probably be helpful.) (Problem 50is by Rosemary Redfield.)
- Researchers discovered recently that the sole functionof the SRY protein is to activate an autosomal genecalled Sox9 in the presumptive gonad (before it has“decided” to become a testis or an ovary).a. What would be the sex of an XY individual homozygous for nonfunctional mutant alleles of Sox9?Explain.b. Given your answer to part (a), why is SRY, ratherthan Sox9, considered the male determining factor?(Hint: What do you think would happen if you didan experiment like the one in the Fast Forward BoxTransgenic Mice Prove That SRY Is the MalenessFactor, except that you used a Sox9 transgeneinstead of SRY?)Sex determination in birds is different from that in humans. The sex chromosomes in birds are called Z and W, because males have two of the same chromosome (ZZ), whereas females have two different chromosomes (ZW). There is a Z-linked allele in some birds that causes the death of the embryo when the normal dominant allele is not present. What would be the sex ratio in the living offspring of a cross between a male heterozygous for the lethal allele and a normal female? A) What are the genotypes of the parents? Male____ Female____ B) Which gametes would each form? Male____ Female____ C) Draw your Punnett square below and determine the sex ratios of living offspring.. For several years, Hans Nachtsheim investigated an inherited anomaly of the white blood cells of rabbits. Thisanomaly, termed the Pelger anomaly, is the arrest ofthe segmentation of the nuclei of certain white cells. Thisanomaly does not appear to seriously burden the rabbits.a. When rabbits showing the Pelger anomaly were matedwith rabbits from a true-breeding normal stock,Nachtsheim counted 217 offspring showing the Pelgeranomaly and 237 normal progeny. What is the geneticbasis of the Pelger anomaly?b. When rabbits with the Pelger anomaly were matedwith each other, Nachtsheim found 223 normal progeny,439 with the Pelger anomaly, and 39 extremely abnormalprogeny. These very abnormal progeny not only haddefective white blood cells, but also showed severedeformities of the skeletal system; almost all of themdied soon after birth. In genetic terms, what do yousuppose these extremely defective rabbits represented?Why were there only 39 of them?c. What additional experimental evidence…
- Let’s suppose that a gene affecting pigmentation is found on the Xchromosome (in mammals or insects) or the Z chromosome (in birds)but not on the Y or W chromosome. It is found on an autosome inbees. This gene exists in two alleles: D (dark) is dominant to d (light).What would be the phenotypic results of crosses between true-breedingdark females and true-breeding light males, and the reciprocal crossesinvolving true-breeding light females and true-breeding dark males,in the following species? Refer back to Figures 4.1 and 4.2 for themechanism of sex determination in these species.A. BirdsB. Fruit fliesC. BeesD. Humans(This one is more complicated). Two true breeding fruit flies are allowed to mate. One fly ishomozygous dominant for body color and eye color while the other is homozygous recessive forbody color and eye color. (g+:gray body, g:black body, e+:ebony eyes, e:red), The F1 flies areallowed to mate. What are the following probabilities in the F2 generation? For this problemmake sure to show the Punnett Square for your F2 generation and all of your calculations. Youdo not need to show the Punnett Square for the F1 generation. Answer in percent and round to 2decimals.a. Two of three fruit flies are gray and red.b. The first fruit fly is gray and ebony, the second fruit fly is gray and red, and the third fruit fly is black and ebony.Mules result from a cross between a horse (2 n = 64) and a donkey(2 n = 62), have 63 chromosomes, and are almost always sterile.However, in the summer of 1985, a female mule named Krause who waspastured with a male donkey gave birth to a male foal (O. A. Ryder et al.1985. Journal of Heredity 76:379–381). Blood tests established that thefoal, appropriately named Blue Moon, was the offspring of Krause andthat Krause was indeed a mule. Both Blue Moon and Krause werefathered by the same donkey (see the accompanying pedigree). The foal,like his mother, had 63 chromosomes—half of them horse chromosomesand the other half donkey chromosomes. Analyses of genetic markersshowed that, remarkably, Blue Moon seemed to have inherited acomplete set of horse chromosomes from his mother, instead of therandom mixture of horse and donkey chromosomes that would beexpected with normal meiosis. Thus, Blue Moon and Krause were notonly mother and son, but also brother and sister.a. With the use of a diagram,…
- Given your answers to below problem, is it possible to distinguishbetween the Protenor and Lygaeus mode of sex determinationbased on the outcome of these crosses? Probelm An insect species is discovered in which the heterogametic sex isunknown. An X-linked recessive mutation for reduced wing (rw)is discovered. Contrast the F1 and F2 generations from a crossbetween a female with reduced wings and a male with normalsizedwings when(a) the female is the heterogametic sex.(b) the male is the heterogametic sex.You have a female snail that coils to the right, but you do not knowits genotype. You may assume that right coiling (D) is dominant toleft coiling (d). You also have male snails of known genotype.How would you determine the genotype of this female snail? Inyour answer, describe your expected results depending on whetherthe female is DD, Dd, or dd.In a vial of Drosophila, a research student noticedseveral female flies (but no male flies) with bag wingseach consisting of a large, liquid-filled blister insteadof the usual smooth wing blade. When bag-wingedfemales were crossed with wild-type males, 1/3 of theprogeny were bag-winged females, 1/3 were normalwinged females, and 1/3 were normal-winged males.Explain these results.