A testcross is used to determine which of the following? The phenotype of an individual with a dominant genotype The phenotype of an individual with a recessive genotype O The genotype of an individual with a dominant phenotype O The genotype of an individual with a recessive phenotype
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Q: You are given two data sets that provide counts of F1 and F2 offspring with given genders and…
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- O e. Parent 2: Parent 1: Parent 2: ¡Ai QUESTION 9 Bi QUESTION 8 Let's assume that, in dragons, red scales (B) are dominant to green scales (b), and long tongues (S) are dominant to short tongues (s). The genes that determine these characteristics assort independently. A homozygous red, long-tongued dragon is crossed with a homozygous green, short-tongued dragon. If an F1 dragon is crossed to a homozygous green and homozygous short-tongued dragon, what phenotypes and proportions are expected in the offspring? O a. 100% green and long-tongued O b. ½ green and short-tongued & ½ red and long-tongued O c. ½ red and short-tongued & ½ green and long-tongued O d.% red and long-tongued, % red and short-tongued, ½ green and long-tongued, % green and short-tongued O e. 9/16 red, long-tongued, 3/16 green, long-tongued, 3/16 red, short-tongued, 1/16 green, short-tongued Save and Submit to save and submit. Click Save All Answers to save all answers. 000 MacBook Air100% - Normal teRI 52 Search the menus (Alt+/) 4 2. that a parent The pattern of on to their Heredity (to Inherit) Аny that can be passed or Trait from parent to offspring. the of a trait Allele This allele will be or shown, no Dominant Trait matter what if it is given to the offspring, This allele will be ONLY if there are given to the offspring; both parents Recessive Trait must have and give a copy of the recessive trait for the offspring.to have the trait. Word Bank Illustration Bank characteristic • passes different versions • shown • expressed • traits inherited two copiesIncomplete dominance Codominance Dosage Dominant negative LOF ||| ||| An allele (or gene) that produces a new functional product. A mutation whose gene product produces a misshaped protein in the recessive condition and whose recessive allele will also cause the heterozygous protein product to misfold. Wild type or the phenotype of the typical form of a species as it occurs in nature. A mutation that results in a LOF. An allele (or gene) that no longer produces ✓ Open functional product.
- Could the trait presented in the pedigree shown be caused by an X-linked recessive allele? Why or why not? 11 ||| IV ■ 5 6 Yes, with individual 2 of generation I being heterozygous. No, the offspring of 7 and 8 contradict an X-linked recessive inheritance. No, the offspring of 3 and 4 of generation II contradict an X-linked recessive inheritance. No, the offspring of 1 and 2 of generation I contradict an X-linked recessive inheritance.Classical Mendelian Genetics, Incomplete Dominance, Codominance, and Multiple Alleles 1. Complete the table given below regarding the phenotype and genotype ratios in completely dominant traits. R and r represent the dominant and recessive allele, respectively. Type of Cross rrx rr RR x rr Rrx rr Rrx Rr Genotype Ratio Phenotype Ratio RR x Rr RR X RR *How would the genotype ratios be affected if the mode o inheritance in incomplete dominance? Codominance? 2. In dogs, barer trait is controlled by a dominant gene D and the silent trait by the recessive gene d. Normal tail is dependent on a dominant gene M and the screw m. Give the probable genotypes of the parents in the following crosses: Phenotype of parents Phenotypes of progeny Genotypes of parents Barker Barker Silent Silent normal screw normal normal a. silent normal x silent normal b. barker normal x silent normal 0 0 6 2 7 2 8 3 c. barker normal x silent screw 4 5 5 3 d. barker screw x silent normal 6 0 0 0 e. barker screw x…PEDIGREES: Problem 7 This pedigree shows the inheritance of atype of X-linked color blindness. It is a recessive trait. Carriers have NOT been half-shaded in this pedigree. • QUESTIONS • • • I ODO0 I. Work out who is definitely a carrier & drag the correct shape onto them. 1 2 3 4 II 1 3 2. Is it possible to work out the geno- type of individual II-I? Explain III your reasoning. 1 2 3 IV 1
- Dominant negative Incomplete dominance Epistasis Recessive lethal allele III ||| E A condition where one gene has the ability to override the expression of another gene no matter what the relationship is between the other gene's alleles. A condition when a new mutation is able to suppress or revert an earlier mutation allowing wildtype function to reappear. A condition where a recessive allele influences the shape of a protein dimer product in the heterozygous condition so that it neither resembles the homozygous dominant nor the homozygous recessive conditions leading to a LOF in the heterozygous state and the recessive state. A condition where two recessive alleles will be fatal to an offspring although it will not affect add-ons Help в I U A Calibri 12 三 三1 |:三 6. Consider a guinea pig with a homozygous genotype and a white fur color phenotype. a. What is the probability this parent will produce a gamete with the dominant allele? b. What is the probability this parent will produce a gamete with the recessive allele? C. If 31 sperm cells are collected from this guinea pig, how many would you expect to have the recessive allele (as determined by sequencing the gene)? !!!1 pts The pedigree below shows the expression of Huntington's disease in a family. Huntington's disease is a fatal genetic disorder that causes the progressive breakdown of nerve cells in the brain. It deteriorates a person's physical and mental abilities usually during their prime working years and has no cure. Huntington's is inherited as a dominant allele (H). Examine the pedigree below and determine the genotype for individual "2" DODO0O 2. O HH O Hh О h O either HH or Hh
- More Crosses with Pea Plants: The Principle of Independent Assortment Determine the possible genotypes of the following parents by analyzing the phenotypes of their children. In this case, we will assume that brown eyes (B) is dominant to blue (b) and that right-handedness (R) is dominant to left-handedness (r). a. Parents: brown eyes, right-handed brown eyes, right-handed Offspring: 3/4 brown eyes, right-handed 1/4 blue eyes, right-handed b. Parents: brown eyes, right-handed blue eyes, right-handed Offspring: 6/16 blue eyes, right-handed 2/16 blue eyes, left-handed 6/16 brown eyes, right-handed 2/16 brown eyes, left-handed c. Parents: brown eyes, right-handed blue eyes, left-handed Offspring: 1/4 brown eyes, right-handed 1/4 brown eyes, left-handed 1/4 blue eyes, right-handed 1/4 blue eyes, left-handedA 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?