4. When doing epistasis problems, do the Punnett square for the epistatic gene first (if you do 2 Punnett squares (T/F). (I do not recommend doing a single large square for epistasis problems or more than 2 genes)
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Bacterial Genomics
The study of the morphological, physiological, and evolutionary aspects of the bacterial genome is referred to as bacterial genomics. This subdisciplinary field aids in understanding how genes are assembled into genomes. Further, bacterial or microbial genomics has helped researchers in understanding the pathogenicity of bacteria and other microbes.
Transformation Experiment in Bacteria
In the discovery of genetic material, the experiment conducted by Frederick Griffith on Streptococcus pneumonia proved to be a stepping stone.
Plasmids and Vectors
The DNA molecule that exists in a circular shape and is smaller in size which is capable of its replication is called Plasmids. In other words, it is called extra-chromosomal plasmid DNA. Vectors are the molecule which is capable of carrying genetic material which can be transferred into another cell and further carry out replication and expression. Plasmids can act as vectors.
Genetics Question 4
How would I do this with multiple punnet squares?
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- 3. In the test cross r/+ s/+ w/+ z/+ rswz/rswz, the following offspring are obtained: X Phenotype r st w+ z r stw z Number 359 47 4 rs w z+ rs w+z+ 98 rt st w z 92 r+ s w z+ 351 r+ st w+ z 6 r+s w+ z+ a) Construct a map of the genes showing their relative distance apart. 43 b) Is there any interference in this region? Note: the listed gene order is arbitrary. Can you deduce the proper order by inspection?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?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?(X^X^) and one male (X"). 12. The following are the gene order on the chromosomes of an individual who is heterozygous for this translocation. translocations (a.unbalanced breciprocal C. Robertsonian MN OP QR N067 89 34 5P QR 3 4 5 6 7 8 9 a. 0 LMNOP QR LMN067 89 345P QR 3456789 hmm Draw the synapsis (Prophase/Metaphase I) configuration dark spots b. What type of translocation is depicted by these chromosomes and what are the consequences of this chromosome rearrangement to the individual? this type of translocation is unbalanced and the major consequence here is that the chromosome number is not reciprocated leading to all sorts of problems with non-disjunction.5 & :56M ******* 24 DIHYBRID CROSSES DRV 0 Stv T alı A @ zladenA 9160p2-id2 bns obidalbaneoviene da II\ MOD YR 21 $59A ... Create a dihybrid cross and determine the expected phenotypic percentages of the offspring of two corn plants both of which are heterozygous for colour and texture (RrTt X RrTt). Don't forget to include clear let statements, and follow the all six steps taught on solving genetics problems. insig moni nellog: bna. zoom
- 1. Study the given alleles. Write the correct phenotype for each genotype. X – normal Gen otype xC - Color-blind Phenotype XX XY XXC xCY 2. Study the given alleles. Write the correct genotype for each phenotype. xH - Hemophiliac Phenotype X- normal Gen otype Hemophiliac female Hemophiliac male Normal female carrier of the gene Normal male Normal female 28 3. Determine the genotype and phenotype of the offspring. A color-blind mother (XCx) married a normal sighted (XY) father. Genotype: Phenotype: Genotype: Phenotype: Genotype: Genotype: Phenotype: Phenotype: a. There are b. There are c. There are d. There are % normal sons. % normal daughters. % color-blind sons. % color-blind daughters. % normal female, carrier of the disorder. or or or or e. There are or9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.2. Using the blood types listed, indicate the possible gene pairs. Make your own table and write your answers. Blood Types Possible Gene Pairs A В АВ
- Find Y AaBbCcI AaBbCcD AaB AaBbccI AaBb CcI Replace Normal Subtitle Title 1 No Spaci... Subtle Em... Select Styles 152 Editing 1 *** F1 BI 2 + EXE 5. Question 5: Consider the following genetic map from Mischievous gremlinus: A 6.0 CM B C 13.4 CM + Interference among these genes is 0.5. A parental cross is performed between AABBCC and aabbcc. The F1 generation is then testcrossed. The F2 progeny included 1,000 offspring. What are the different classes produced, and in what numbers? Assume that reciprocal classes of recombinants are obtained in equal numbers. 4 E DAWWW Create and Share Request Adobe PDF Signature Adobe Acrobat 1The Meeting Sarah stared blankly at the blue paisley wallpaper. Her husband Mike sat by her side, bending and unbending a small paper clip. “Sarah and Michael, it’s good to meet you,” welcomed the genetic counselor, as she entered the room. “I apologize for being late, but I was just meeting with another couple. Let’s see, you’d like to have a child, but you’re concerned because of your family history of cystic fibrosis.” “Yes,” Sarah replied softly. “Mike and I met at a CF support group meeting a few years ago. He had a younger brother who died of cystic fibrosis, and I had a younger sister. We saw the painful lives they had—difficulty breathing, the constant respiratory infections. Although the treatments for CF are better now, we just don’t know if we can…” she trailed off. “I can certainly understand your concern,” the genetic counselor responded sympathetically. “That’s where I hope to help, by providing as much information and advice as I can. I’m glad that you came to see me…4. What mode of inheritance is shown in the pedigree chart on the left? Justify your answer. 5. Construct pedigree charts using the inheritance of hemophilia in figure 92 (page 113). This is X-linked inheritance so you are required to label XX for females and XY for males. The gene responsible for the trait is represented by the superscript which should be specified in the legend.