In mice, the A allele causes agouti fur, the AY allele causes yellow fur and is dominant to the A allele. The homozygous AY AY genotype is lethal. Why is the AYAY homozygous condition lethal? The yellow pigment produced by A allele is required for survival. The AY allele lacks an upstream merc gene which is required for normal embryonic development. The AY allele highly transcribes the merc gene which is lethal in a homozygous AYAY individual. The AY allele contains a null mutation in the A gene, which is required for normal embryonic development. None of the other answers are correct.

In mice, the A allele causes agouti fur, the AY allele causes yellow fur and is dominant to the A allele. The homozygous AY AY genotype is lethal. Why is the AYAY homozygous condition lethal? The yellow pigment produced by A allele is required for survival. The AY allele lacks an upstream merc gene which is required for normal embryonic development. The AY allele highly transcribes the merc gene which is lethal in a homozygous AYAY individual. The AY allele contains a null mutation in the A gene, which is required for normal embryonic development. None of the other answers are correct.

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 1QP: The gene controlling ABO blood type and the gene underlying nail-patella syndrome are said to show...

See similar textbooks

Similar questions

In mice, the A allele causes agouti fur, the AY allele causes yellow fur and is dominant to the A allele. The homozygous AY AY genotype is lethal. Why is the AYAY homozygous condition lethal? The AY allele highly transcribes the merc gene which is lethal in a homozygous AYAY individual. None of the other answers are correct. The AY allele lacks an upstream merc gene which is required for normal embryonic development. The AY allele contains a null mutation in the A gene, which is required for normal embryonic development. The yellow pigment produced by A allele is required for survival.
Suppose that you are studying the role of Protein B, which you believe plays a role in regulating PCD/Apoptosis in mice. You create two lines of mutant mice. One (bb) is homozygous for a loss-of-function allele of gene B. The other (Bb) is heterozygous, with one wild-type allele and one loss-of function allele. Initially you pay particular attention to two phenotypes of the resulting mice:(i) The morphology of their paws (see picture) (ii) The size of their brains & shape of their skulls. The bb mice have unusually large brains and unusual protrusions from their skulls. Based on these data, does it appear that Protein B, when present and active, favors or inhibits PCD/Apoptosis?Briefly explain your reasoning. The answer should address both the paw and brain/skull data.
Colorblindness and hemophilia are both X-linked traits in humans. Explain how a female who has a defective color vision gene on one X chromosome and a defective blood clotting gene causing hemophilia on the other X chromosome can be neither a hemophiliac nor colorblind? Please discuss the effect of Gene dosage compensation in your answer and in your answer describe the molecular process by which this occurs.
There are two genetic disorders that result from mutation in imprinted genes: Prader-Willi syndrome, Angelman syndrome. Angelman syndrome results from deletion of UBE3A, which is a gene imprinted such that only the maternal copy is expressed. In the pedigree above, individual I-1 is heterozygous for a deletion of UBE3A and does not have Angelman syndrome. Individual I-2 is homozygous wild type for UBE3A. Which individuals in the pedigree are at risk for exhibiting Angelman syndrome, if any? (Who could potentially have the syndrome, based on what alleles it is possible for them to inherit and express?) Question 8 options: Only I-1 could have been at risk. If he does not have the syndrome, no one in the pedigree could. Only III-1 is at risk I-1, II-2, and III-1 are all at risk Only II-2 is at risk No one in the pedigree is at risk Both II-2 and III-1 are at…
In mice, coat color is due to the inheritance of three genes located on three different chromosomes. The A gene which determine if a yellow band is present on the hair ( A_= yellow band, aa= no yellow band ) and the B gene which determines if the hair is brown ( bb ) or Black ( B_). The A and B genes interact to produce the following coat colors: Agouti: A_B_ Cinnamon: A_bb black: aaB_ brown aabb The third gene is the C gene, which determines if the mice has any hair pigment at all. If the mouse has at least one C allele then the coat color will be determined by the A and B genes as described above. However, if the mouse is cc in genotype it will be albino, regardless of the mouse’s A and B genotype. 1. In mice coat color (a) the C gene is dominant to the A and B genes (b) the cc genotype is epistatic to the A and B genes (c) the C gene is codominant to the A and B genes (d) The A gene is X-linked (e) none of the above
In mice, coat color is due to the inheritance of three genes located on three different chromosomes. The A gene which determine if a yellow band is present on the hair ( A_= yellow band, aa= no yellow band ) and the B gene which determines if the hair is brown ( bb ) or Black ( B_). The A and B genes interact to produce the following coat colors: Agouti: A_B_ Cinnamon: A_bb black: aaB_ brown aabb The third gene is the C gene, which determines if the mice has any hair pigment at all. If the mouse has at least one C allele then the coat color will be determined by the A and B genes as described above. However, if the mouse is cc in genotype it will be albino, regardless of the mouse’s A and B genotype. 3. If a AabbCc mouse mates with a aaBbcc mouse, what is the probability that a mouse will be born an albino? (a) ½ (b) ¼ (c) 1/16 (d) 3/16 (e) none of the above
In mice, coat color is due to the inheritance of three genes located on three different chromosomes. The A gene which determine if a yellow band is present on the hair ( A_= yellow band, aa= no yellow band ) and the B gene which determines if the hair is brown ( bb ) or Black ( B_). The A and B genes interact to produce the following coat colors: Agouti: A_B_ Cinnamon: A_bb black: aaB_ brown aabb The third gene is the C gene, which determines if the mice has any hair pigment at all. If the mouse has at least one C allele then the coat color will be determined by the A and B genes as described above. However, if the mouse is cc in genotype it will be albino, regardless of the mouse’s A and B genotype. 1.If a AabbCc mouse mates with a aaBbcc mouse, what is the probability that the mouse will be born brown? (a) ½ (b) ¼ (c) 1/8 (d) 1/16 (e) none of the above
A recessive mutation pd causes purple eyes in Drosophila, in contrast to the wildtype red eyes. A dominant suppressor called Su can restore the color of pd/pd fly eyes to red. If you cross a pd/+ ; Su/+ fly to a pd/pd ; +/+ fly, what proportion of the offspring will have purple eyes? A recessive mutation pd causes purple eyes in Drosophila, in contrast to the wildtype red eyes. A dominant suppressor called Su can restore the color of pd/pd fly eyes to red. If you cross a pd/+ ; Su/+ fly to a pd/pd ; +/+ fly, what proportion of the offspring will have purple eyes? a. 1/8 b. 1/16 c.1/2 d. 3/16 e. 1/3 f. 3/4 g. 15/16 h. 2/3 i. 1/4
The expression of antigen A or antigen B in red blood cells requires the help of an H antigen. A recessive mutation (h) that prevents the synthesis of the H antigen also prevents the expression of A and B antigens. This is called the Bombay effect. There is no ill effect in an individual with this mutation, but complications with blood transfusions or parental disputes may arise. a. Individuals with the Bombay genotype (hh) produce anti-H antigen. How can this be a problem during blood transfusion?
In Labrador retrievers, two genes work together to determine coat colour.The first gene affects the colour of the dark pigment, eumelanin. A genotype containing at least one dominant allele of this pigment gene results in black pigmentation (B_). A homozygous recessive genotype of this gene results in chocolate (brown) pigmentation. The second gene affects whether the pigment eumelanin is present in the fur. If a dog has one dominant allele for this expression gene (E_), fur colouration is determined by the pigment gene. If a dog is homozygous recessive for the extension trait, the dog will have a yellow coat. Description Genotype(s) Black coat B_E_ Chocolate coat bbE_ Yellow coat _ _ ee What percentage of chocolate labs would be produced from the mating of a yellow lab who is homozygous recessive for both genes to a black lab who is heterozygous for both coat colour and expression? Select one: a. 50% b. 75% c. 33% d. 25%
. Neurofibromas are tumors of the skin that can arisewhen a skin cell that is originally NF1+/ NF1− losesthe NF1+ allele. This wild-type allele encodes a functional protein (called a tumor suppressor), while theNF1− allele encodes a nonfunctional protein.A patient of genotype NF1+ / NF1− has 20 independent tumors in different areas of the skin. Samplesare taken of normal, noncancerous cells from thispatient, as well as of cells from each of the 20 tumors.Extracts of these samples are analyzed by a techniquecalled gel electrophoresis that can detect variantforms of four different proteins (A, B, C, and D) allencoded by genes that lie on the same autosome asNF1. Each protein has a slow (S) and a fast (F) formthat are encoded by different alleles (for example, ASand AF). In the extract of normal tissue, slow and fastvariants of all four proteins are found. In the extractsof the tumors, 12 had only the fast variants of proteinsA and D but both the fast and slow variants of proteins B and…
Which of the following statements is/are true and which is/are false? The expression of lethal genes in heterozygotes is called penetrance. Expressivity is the balanced genetic output from a hemizygous condition.