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The complete absence of one or more teeth (tooth agenesis) is a common trait in humans—indeed, more than 20% of humans lack one or more of their third molars. However, more severe tooth agenesis, defined as the absence of six or more teeth, is less common and is frequently an inherited condition. L. Lammi and colleagues examined tooth agenesis in the Finnish family shown in the pedigree below.
Q.Are the two sets of twins in this family monozygotic or dizygotic twins? What is the basis of your answer?
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- Cystic fibrosis (CF) is an autosomal recessive trait. A three-generation pedigree is shown below for a family that carries the mutant allele for cystic fibrosis. Note that carriers are not colored in to allow you to figure out their genotypes. Normal allele = F CF mutant allele = f What is the genotype of individual #13? A) ff B) FF C) Ff D) it is impossible to tellIn mice, the presence of a dominant A allele results in the agouti pattern of fur color, whereas the aa homozygous recessive results in a solid color pattern. The presence of another dominant B allele results in black fur color, whereas the homozygous recessive bb results in brown fur color. The presence of a third dominant C allele is required for any color to be observed (brown or black), whereas the homozygous recessive cc results in lack of any color whatsoever (albino). What is the phenotype of a mouse with the following genotype? Aa bb cc albino Agouti color pattern and brown Solid color pattern and brown Agouti color pattern and black Solid color pattern and blackSuppose a geneticist is using a three-point test cross to map three linked rabbit morphology and behavioral mutations called si, sf, and H. The gene si is associated with the silky fur phenotype, and sf is associated with the short-footed phenotype. Both si and sf are recessive mutations with respect to wild type. H is a dominant mutation that confers the hyper phenotype. The geneticist first crosses true-breeding hyper rabbits to true-breeding silky fur, short-footed rabbits. Next, the geneticist backcrosses the F₁ progeny to the silky fur, short-footed parents, and obtains the results reported in the table. Phenotype hyper silky fur, short-footed short-footed silky fur, hyper silky fur short-footed, hyper silky fur, short-footed, hyper wild type Place the genes in the correct order in the chromosome. LLIIN HEL H Number 815 807 175 169 4 5 27 29 Answer Bank si sf
- Consider the following three autosomal recessive mutations in Drosophila:vestigial wings (v); wild type is long (v+)black body color (b); wildtype is gray (b+)plum eyes (p); wildtype is red (p+)A vestigal, gray, red female (homozygous for all three genes) is crossed with a long wing, black, plum male (homozygous for all three genes). The F1 female progeny are mated with triple homozygous recessive males. Here is the phenotypic data for the F2 progeny:vestigal; gray; red 580long wings; black; plum 592vestigal; black; red 45long; gray; plum 40vestigal; black; plum 89long; gray; red 94vestigal; gray; plum 3long; black; red 5A total of 1448 progeny were counted.Which one of the following values is the approximate distance between the plum eye color and black body color loci?In mice, the trait for high cholesterol is specified by a dominant allele designated HC, whereas the wild-type allele for normal cholesterol levels is designated hc. Black fur is specified by a recessive allele designated bl, whereas the wild-type allele which gives brown fur is designated BL. The genes for both of these traits are 30cM apart on the same autosome. A brown female (#1) with high cholesterol is mated to a black male (#2) with normal cholesterol. The progeny from this cross include a brown male (#3) with high cholesterol and a black female (#4) with normal cholesterol. What is the probability that the black mouse in the progeny of the first cross will also have high cholesterol?In c. elegans, genetics model organism, movement problems (unc) and small body size (sma) are encoded by two mutant alleles that are recessive to those that produce wild-type traits (unc+ and sma+). A worm homozygous for movement problems and small body is crossed with a worm homozygous for the wild-type traits. The F1 have normal movement and normal body size. The F1 are then crossed with worms that have movement problems and small body size in a testcross. The progeny of this testcross is: Normal movement, normal body size 210 Movement problems, normal body size 9 Normal movement, small body size 11 Movement problems, small body size 193 a)From the test cross results, can you tell if the two genes are on the same chromosome or not? Explain your reasoning. b)What phenotypic proportions would be expected if the genes for round eyes and white body were located on different chromosomes? (please explain hot to get to these conclusions)
- In wheat, aleurone cells form a thin layer of the seed coat that is critical to early gene expression in plant development. The color of this layer of cells is controlled by two alleles of a gene [colored aleurone (R) is dominant to colorless (r)]. A second gene is known to control the color of leaf tips [green leaf tip (G) is dominant to yellow (g)]. Two plants, each heterozygous for both characteristics, are test crossed to homozygous recessives, and their progeny are combined to produce the following totals: colored green 102 colored yellow 98 colorless green 103 colorless yellow 97 a) Use chi-square analysis to test these data for an independent assortment of the two characteristics (table provided). Please show work, how your expected values are calculated, and explain what your results indicate about the data. b) You decide to be cautious in your analysis, and decide to analyze the progeny from each of the crosses individually (instead of adding them together as shown above).…11). This pedigree illustrates a family in which some members have a completely penetrant disease caused by a dominant mutation. This mutation is linked at a distance of 10 map units from a SNP marker with three different alleles (1, 2 and 3). The SNP alleles found in each family member are indicated below each pedigree symbol. It is not yet evident whether the very young individuals labeled A and B will develop the disease. a. What is the probability that individual A will develop the disease? b. What is the probability that individual B will develop the disease? 1,3 2,2 1,2 3,2 O 1,2 3,2 A BThere 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…
- a) Which of the four modes of inheritance are consistent with the disease shown in this human pedigrees below? (List the compatible mode or modes) Give an answer for a, b and c b) If the parents in pedigree c have 2 other children, what is the probability that they will carry the disease?In the red backed vole (a small mouse-like mammal) there are two traits under investigation. The first trait is a distinct tuft of long hairs at the tip of the tail with a tufted tail being recessive to non-tufted tail. The letter T is used to represent the tail tuft gene. The second trait is a patch of red fur on the head which is known to be recessive to the more common lack of a red patch. The letter R is used to represent this trait. A female to be used in the cross was drawn from a pure breeding line that consistently has both the red patch on the head and a tufted tail. The female is to be crossed with a male from a highly variable wild population. The male used had no red head patch and a normal (untufted) tail. The results were as follows: 7 red head patch with tail tuft, 8 no red head patch with tail tuft, 8 red head patch without tail tuft, 8 no red head patch without tail tuft. What would the results be if the male was crossed with a female that was known to be…In squirrels, individuals that are heterozygous for the mutant LDL receptor gene( Fa) begin to experience heart attacks at the squirrel equivalent of the 30’s and 40’s in humans, while individuals that are homozygous for the mutant LDL receptor allele experience heart attacks much earlier. Closer examination reveals that the hepatocytes of the homozygous normal squirrels contain ONLY normal receptors. In the hepatocytes of the heterozygous squirrels, 50% of the receptors are of the mutant type, and fail to bind the LDL cholesterol, while the other 50% are normal. In individuals homozygous for the mutant LDL receptor allele, only mutant receptors are present. NAME and DEFINE the genetic phenomenon observed here.