For two homologous sequences (S1 and S2) with a JC69 distance (D), compute the expected proportion of sites that are identical between the two sequences. (hint: transition probability p_ii given D and the JC69 model) D = 0.47 p_ii
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![For two homologous sequences (S1 and S2) with a JC69 distance (D), compute the expected proportion of
sites that are identical between the two sequences. (hint: transition probability p_ii given D and the JC69
model) D = 0.47 p_ii](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd92f3dca-de6f-4e94-845c-b356136462a2%2F81beb3c6-accf-4ee4-b3ad-add6005fb585%2Fcu6yybh_processed.png&w=3840&q=75)
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- Please answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?Short hair (S) in rabbits is dominant over long hair (s). The following crosses are carried out, producing the progeny shown. Give all possible genotypes of the parents in each cross.A Neurospora cross was made between a strain that carried the mating-type allele A and the mutant allele arg-1and another strain that carried the mating-type allele aand the wild-type allele for arg-1 (+). Four hundred linear octads were isolated, and they fell into the sevenclasses given in the table below. (For simplicity, they areshown as tetrads.)a. Deduce the linkage arrangement of the mating-typelocus and the arg-1 locus. Include the centromere orcentromeres on any map that you draw. Label all intervalsin map units.b. Diagram the meiotic divisions that led to class 6. Labelclearly
- Consider three genes L, U, and W, for which the count of F2 phenotypes after a 3-point cross is as follows: Phenotype F2 count: L U w 19 L u W 1 l u W 21 L U W 33 l U W 274 l u w 41 l U w 2 L u w 259 Which of the following statements about genes L, U, and W are TRUE? (may be more than one correct ans) A. L, U, and W are each on a different chromosome B. Only U and L are on the same chromosome C. Only U and W are on the same chromosome D. Only W and L are on the same chromosome E. L, U, and W are all on the same chromosomeIn a three point cross problem for genes Q, L, and T, you have identified the parentals and double recombinants. Using that data (shown below), what is the order of the genes? Type: Alleles parental q L t parental Q l T double cross over q L T double cross over Q l t Group of answer choices L Q T L T Q T L Q the correct answer is not availableOn Neurospora chromosome 4, the leu3 gene is just to theleft of the centromere and always segregates at the firstdivision, whereas the cys2 gene is to the right of the centromere and shows a second-division segregation frequency of 16 percent. In a cross between a leu3 strain anda cys2 strain, calculate the predicted frequencies of thefollowing seven classes of linear tetrads where l = leu3 andc = cys2. (Ignore double and other multiple crossovers.)
- Consider two maize plants:a. Genotype C/c m ; Ac/Ac+, where cm is an unstable allele caused by a Ds insertionb. Genotype C/c m, where cm is an unstable allele caused by Ac insertionWhat phenotypes would be produced and in what proportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a is crossed with the plant in part b? Assume that Ac and c are unlinked, that the chromosome-breakage frequency is negligible, and that mutant c /C is Ac+.Recombination frequencies between four genetically-linked loci in corn are shown in the following table: Loci Recombination Frequency (%) L and Q 20 Q and R 50 R and L 30 Q and W 13 L and W 7 What is the order of the genes on the chromosome? (note: The same answer can be represented forward or backwards. e.g. A B C D = D C B A) LQWR RQWL LRQW QRLW RLWQA double heterozygote has the repulsion configuration Ab/aB of two linked genes that have a frequency recombination of 0.2. If a randomly chosen gamete carries A, what is the probability that it also carries B?
- A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What if the couple wanted prenatal testing so that a normal fetus could be aborted?
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