7. Complete the following problem. Assume that a diploid cell heterozygous for the MET15/met1540; LYS2/lys240, and MatA/Mata (i.e. the BY4743 strain) was sporulated. For one tetrad, you obtained the following results. a. Assuming only independent assortment, draw out the one independent assortment that led to this result. tetrad YPD SD-met SD-lys mating type tester 1a + + 1b + + - 1c + + + 1d + + + b. Use the branch method to identify all possible gamete genotypes that could have arisen from diploid strain.

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?

How many Barr bodies are present in the interphase cells of the ff. individuals? A) Klinefelters where 2n = 48 (assume no autosomal aneuploidy in this individual) (b) metafemales with chromosome number of 48 (no autosomal aneuploidy in this individual). Show solutions.

3 Other (mutant) Gene: Clot, Please use "cl" abbreviation C A homozygous wt organism has been crossed to a homozygous mutant for your gene and for withered wings (whd). The resulting heterozygote is test-crossed. Predict the test cross results.

Give typed explanation

A female patient 19 years old, whose symptoms areanemia and internal bleeding due to a massive buildupof leukemic white blood cells, is diagnosed withchronic myelogenous leukemia (CML). Karyotypeanalysis shows that the leukemic cells of this patientare heterozygous for a reciprocal translocation involving chromosomes 9 and 22. However, none of thenormal, nonleukemic cells of this patient contain thetranslocation. Which of the following statements istrue and which is false?a. The translocation results in the inactivation (loss offunction) of a tumor-suppressor gene.b. The translocation results in the inactivation (loss offunction) of an oncogene.c. There is a 50% chance that any child of this patientwill have CML.d. This patient is a somatic mosaic in terms of thekaryotype.e. DNA extracted from leukemic cells of this patient,if taken up by normal mouse tissue culture cells,could potentially transform the mouse cells intocells capable of causing tumors.f. The normal function of the…

1. The figure below represents a karyotype performed on a somatic cell of an unknown organism. Answer the following questions using the figure. a. Circle (and label) 3 different sets of homologous chromosomes? b. This individual is heterozygote for gene A (Aa), Homozygote recessive for gene B (bb) and gene homozygote dominant for gene c (CC). Also, gene B and gene C are linked. Using the homologous chromosomes you circled in part a), provide a representation of the location of the A, B and C, genes and alleles on the homologous chromosomes and sister chromatids in such a way that is consistent with the scenario described above in part b). Choose the chromosomes of your liking to show an appropriate representation. TMIID IL MI

Equalizing the Expression of X Chromosome Genes in Males and Females Individuals with an XXY genotype are sterile males. If one X is inactivated early in embryogenesis, the genotype of the individual effectively becomes XY. Why will this individual not develop as a normal male?

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?