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Lesco hrf parts manual. January 6, You may have missed. Part of manual transmission that knows your speed 5 min read. If the random spores of many meiocytes are analyzed, you would expect to find about 50 percent normal-size colonies and 50 percent small colonies if the abnormal phenotype is the result of a mutation in a single gene.
Thus, the actual results of normal-size and small-size colonies support the hypothesis that the phenotype is the result of a mutation in a single gene. The following represents an ascus with four spores. Two black guinea pigs were mated and over several years produced 29 black and 9 white offspring. Explain these results, giving the genotypes of parents and progeny.
Answer: The progeny ratio is approximately , indicating classic heterozygous-by-heterozygous mating. Ascospores are the spores that constitute the four products of meiosis. Draw an ascus from each of the following crosses: a. For a certain gene in a diploid organism, eight units of protein product are needed for normal function. Each wild-type allele produces five units.
If a mutation creates a null allele, do you think this allele will be recessive or dominant? What assumptions need to be made to answer part a? This would be an example of a haploinsufficient gene since one copy of the wild-type allele does not produce enough protein product for normal function. In the absence of knowledge about the biochemistry, we could predict a dominant inheritance pattern, as having one copy of the mutant allele is sufficient to generate the abnormal phenotype.
An important assumption would be that having five of eight units of protein product would result in an observable phenotype. It also assumes that the regulation of the single wild-type allele is not affected. Finally, if the mutant allele was leaky rather than null, there might be sufficient protein function when heterozygous with a wild-type allele.
A Neurospora colony at the edge of a plate seemed to be sparse low density in comparison with the other colonies on the plate. This colony was thought to be a possible mutant, so it was removed and crossed with a wild type of the opposite mating type.
From this cross, ascospore progeny were obtained. None of the colonies from these ascospores was sparse; all appeared to be normal. What is the simplest explanation of this result? How would you test your explanation?
Note: Neurospora is haploid. Answer: The simplest explanation is that the abnormal phenotype was not due to any genetic change. Perhaps the environment edge of plate was less favorable for growth. If all spores yield wild-type colonies, the low-density phenotype was not heritable.
From a large-scale screen of many plants of Collinsia grandiflora, a plant with three cotyledons was discovered normally, there are two cotyledons. This plant was crossed with a normal, purebreeding, wild-type plant, and seeds from this cross were planted. There were plants with two cotyledons and with three cotyledons. What can be deduced about the inheritance of three cotyledons?
Invent gene symbols as part of your explanation. Answer: Since half of the F1 progeny are mutant, it suggests that the mutation that results in three cotyledons is dominant, and the original mutant was heterozygous. In the plant Arabidopsis thaliana, a geneticist is interested in the development of trichomes small projections. A large screen turns up two mutant plants A and B that have no trichomes, and these mutants seem to be potentially useful in studying trichome development. If they were determined by single-gene mutations, then finding the normal and abnormal functions of these genes would be instructive.
Each plant is crossed with wild type; in both cases, the next generation F1 had normal trichomes. What do these results show? Include proposed genotypes of all plants in your answer. Under your explanation to part a, is it possible to confidently predict the F1 from crossing the original mutant A with the original mutant B? The data for both crosses suggest that both A and B mutant plants are homozygous for recessive alleles.
Both F2 crosses give ratios of normal to mutant progeny. You do not know if the a and b mutations are in the same or different genes. If they are in the same gene, then the F1 will all be mutant. If they are in different genes, then the F1 will all be wild type.
You have three dice: one red R , one green G , and one blue B. When all three dice are rolled at the same time, calculate the probability of the following outcomes: a. No sixes at all e. So each independent probability is multiplied together. The easiest way to approach this problem is to consider each die separately. The first die thrown can be any number.
Therefore, the probability for it is 1. The second die can be any number except the number obtained on the first die. The third die can be any number except the numbers obtained on the first two dice. If you had no other information to go on, would you think it more likely that the disease was dominant or recessive?
Give your reasons. Answer: You are told that the disease being followed in this pedigree is very rare. If the allele that results in this disease is recessive, then the father would have to be homozygous and the mother would have to be heterozygous for this allele. On the other hand, if the trait is dominant, then all that is necessary to explain the pedigree is that the father is heterozygous for the allele that causes the disease.
This is the better choice, as it is more likely given the rarity of the disease. The ability to taste the chemical phenylthiocarbamide is an autosomal dominant phenotype, and the inability to taste it is recessive. If a taster woman with a nontaster father marries a taster man. What is the probability that their first two children will be tasters of either sex? Martha has a sister who has three children, none of whom have galactosemia.
Can the problem be restated as a pedigree? If so, write one. Answer: Yes. The pedigree is given below. Can parts of the problem be restated by using Punnett squares? Answer: In order to state this problem as a Punnett square, you must first know the genotypes of John and Martha. The genotypes can be determined only through considering the pedigree. A Punnett square for their mating would be:.
The probability that Martha is carrying the g allele is based on the following chain of logic. In summary: John. Can parts of the problem be restated by using branch diagrams?
Answer: While the above information could be put into a branch diagram, it does not easily fit into one and overcomplicates the problem, just as a Punnett square would. Define all the scientific terms in the problem, and look up any other terms about which you are uncertain. Answer: The scientific words in this problem are galactosemia, autosomal, and recessive.
Galactosemia is a metabolic disorder characterized by the absence of the enzyme galactose1-phosphate uridyl transferase, which results in an accumulation of galactose.
In the vast majority of cases, galactosemia results in an enlarged liver, jaundice, vomiting, anorexia, lethargy, and very early death if galactose is not omitted from the diet initially, the child obtains galactose from milk.
Autosomal refers to genes that are on the autosomes. Recessive means that, in order for an allele to be expressed, it must be the only form of the gene present in the organism. What assumptions need to be made in answering this problem?
Another assumption that may be of value, but is not actually needed, is that all people marrying into these two families are normal and do not carry the allele for galactosemia.
Which unmentioned family members must be considered? What statistical rules might be relevant, and in what situations can they be applied? Do such situations exist in this problem? It is used to calculate the cumulative probabilities described in part 2 of this unpacked. What are two generalities about autosomal recessive diseases in human populations? Answer: Autosomal recessive disorders are assumed to be rare and to occur equally frequently in males and females.
They are also assumed to be expressed if the person is homozygous for the recessive genotype. What is the relevance of the rareness of the phenotype under study in pedigree analysis generally, and what can be inferred in this problem? Answer: Rareness leads to the assumption that people who marry into a family that is being studied do not carry the allele, which was assumed in entry 6 above.
In this family, whose genotypes are certain and whose are uncertain? All other individuals have uncertain genotypes. How does this difference affect your calculations? Is there any irrelevant information in the problem as stated?
In what way is solving this kind of problem similar to solving problems that you have already successfully solved? In what way is it different?
Answer: The problem contains a number of assumptions that have not been necessary in problem solving until now. Can you make up a short story based on the human dilemma in this problem? Answer: Many scenarios are possible in response to this question. Now try to solve the problem. If you are unable to do so, try to identify the obstacle and write a sentence or two describing your difficulty.
Then go back to the expansion questions and see if any of them relate to your difficulty. Holstein cattle are normally black and white.
All the progeny sired by Charlie were normal in appearance. However, certain pairs of his progeny, when interbred, produced red-and-white progeny at a frequency of about 25 percent. Charlie was soon removed from the stud lists of the Holstein breeders. Use symbols to explain precisely why. Answer: Charlie, his mate, or both, obviously were not homozygous for one of the alleles purebreeding because his F2 progeny were of two phenotypes.
If both parents were heterozygous, then red and white would have been expected in the F1 generation. Red and white were not observed in the F1 generation, so only one of the parents was heterozygous.
The cross is: P F1. However, if the F2 progeny came only from one mate, the farmer may have acted too quickly. Suppose that a husband and wife are both heterozygous for a recessive allele for albinism.
If they have dizygotic two-egg twins, what is the probability that both the twins will have the same phenotype for pigmentation?
Near Nanaimo, one plant in nature had blotched leaves. This plant, which had not yet flowered, was dug up and taken to a laboratory, where it was allowed to self.
Seeds were collected. One randomly selected but typical leaf from each of the progeny is shown in the accompanying illustration. Formulate a concise genetic hypothesis to explain these results. Explain all symbols and show all genotypic classes and the genotype of the original plant. How would you test your hypothesis? Be specific. Answer: The plants are approximately 3 blotched:1 unblotched. This suggests that blotched is dominant to unblotched and that the original plant, which was selfed, was a heterozygote.
Can it ever be proved that an animal is not a carrier of a recessive allele that is, not a heterozygote for a given gene? Answer: In theory, it cannot be proved that an animal is not a carrier for a recessive allele. DNA sequencing can be used to prove heterozygosity, but without sequence level information, the level of certainty is limited by sample size.
In nature, the plant Plectritis congesta is dimorphic for fruit shape; that is, individual plants bear either wingless or winged fruits, as shown in the illustration. Plants were collected from nature before flowering and were crossed or selfed with the following results:.
Interpret these results, and derive the mode of inheritance of these fruit-shaped phenotypes. Use symbols. What do you think is the nongenetic explanation for the phenotypes marked by asterisks in the table?
If that is correct, the crosses become:. The five unusual plants are most likely due either to human error in classification or to contamination. Alternatively, they could result from environmental effects on development. For example, too little water may have prevented the seedpods from becoming winged, even though they are genetically winged. How is the disorder inherited? State reasons for your answer. Give genotypes for as many individuals in the pedigree as possible.
Invent your own defined allele symbols. In all four-child progenies from parents of these genotypes, what proportion is expected to contain all unaffected children?
The disorder appears to be dominant because all affected individuals have an affected parent. Four human pedigrees are shown in the accompanying illustration. The black symbols represent an abnormal phenotype inherited in a simple Mendelian manner. For each pedigree, state whether the abnormal condition is dominant or recessive.
Try to state the logic behind your answer. For each pedigree, describe the genotypes of as many persons as possible. Pedigree 1: The best answer is recessive because two unaffected individuals had affected progeny. Also, the disorder skips generations and appears in a mating between two related individuals.
Pedigree 2: The best answer is dominant because two affected parents have an unaffected child. Also, it appears in each generation, roughly half the progeny are affected, and all affected individuals have an affected parent. Pedigree 3: The best answer is dominant, for many of the reasons stated for pedigree 2. Inbreeding, while present in the pedigree, does not allow an explanation of recessive because it cannot account for individuals in the second or third generations.
Pedigree 4: The best answer is recessive. Skip to search Skip to main content. Reporting from:. Report wrong cover image. Your name. Your email.
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