Discuss about the Biology for Hungarian Press.
The function is to make lactose and the breaking of sugars into simple proteins such as milk. B/b has been describes as follows: B = the capacity of the genotype to make production of the lactase enzyme and be able to digest the final result of the milk. Small letter b is used to represent the genotype inability to digest the lactose in milk. AA Means that the individual is in a position to biologically make the lactase and hence digest milk into the results of being fine for the purpose of assimilation in the body digestive system. When the person or the character under consideration has got the both genotypes versions he/she in a position to produce the complex sugars in the aspect of D/d and E/e. E/e means that the individual is in a position to make the lactase and also be able to digest the simple substance of protein just in an easy way.
The probability of two 1’s is 0.25, (b) a 1 and 2 is 0.3.
A 1 on the first die and a 2 on the second dice are 0.5.
The probability that you will be derived from values that sum up the probability of getting one particular value is 1/6. ... 7, 1+6, 2+5, 3+4, 4+3, 5+2, 6+1, 6/36 = 17% or 0.17, to 2 or values that sum up to 11 is 0.2. Finally the values that sum up to anything except 6 are 0.4 as the probability. The genotypes that could be seen in the offspring of A/A: B/b: D/d: E/e and A/a: b/b: D/d: e/e are lactase enzyme and be able to digest the final result of the milk. “Peterson argued genotype x contains the content that is defined as the recipe for the structure of the body” (Richard, 2012, p.578).
There are four possible genotypes in the sample: B/b has been describes as follows: B = the capacity of the genotype to make production of the. Small letter b is incompatible the genotype that has resistance to digest the lactose in milk. AA biologically make the lactase and hence digest milk into the results of being fine for the purpose of assimilation in the body digestive system. The presence of both genotypes versions shows him/her in a position to produce the complex sugars in the aspect of D/d and E/e.
- B) There are six phenotypes as given: E/e is the genotype that portrays the individual is in a position to make the lactase and also be able to digest the simple substance of protein just in an easy way. The description of e/e means that people in these criteria of gene modification cannot at any given moment be in a position to conduct a digestion of milk in a proper method because the individuals cannot make the lactase and hence known as the individual who are lactose intolerant. D /d phenotype shows the dominance phenotype that prevails and controls the rhesus factor in the individual.
- b) The types of complete phenotypes
L which is Lactose intolerance is an example of phenotype that is utilized in the digestive system in metabolism hence it assists in the digestion of sugar and lactose into milk. The sample above has got a composition of gene that assists in the making of enzymes and thus lactase produced is used to digest the simple sugars.
The phenotype H: height of human being and the color of the eye pupil are led and determined by the phenotype composition. Dwarfs height is represented by phenotype h since Genes are not developed in a large extent in the centralization of height of a person. “Mathew said the behavior can also be deemed to be dominance phenotype” (Thompson, 2012, p.244).
Collies is a phenotype example engaged in breeding in order to produce a productive and super breed of animal in the context of sheep. In case the individual have never seen a sheep before he will be in a position to portray behavior that resembles that of herding. Running around the crib collecting the junk and pillow cases is an example.
The chance that the offspring from A/a: B/b: D/D: E/e would be 70 % have the genotype which is thick and it is represented by a higher probability of 70% chance that the offspring will contain the genotype from the following calculations. A/a: B/b: E/e = 3 genotypes inclusive parents and also known biologically as carriers. Total genes = 4 the phenotype is represented by DD and thus the computation is ¾ times 100% = 70%.
The chance that the offspring from these parents would produce the phenotype is A_ bb D e e is 50% since the gene A and D are carriers and hence the potentially dominance genes in this aspect. Therefore the two genes and the phenotypes are two in number giving sum of four hence 2/4 times 100/% =50%.
In the case of Arabiodapsis thaliana, the plant has got content of five chromosomes in total each having a composition of producing two different gametes by the parent. These are autosomes members hence they are fully homogenous and the criteria is known as the homozygous of the locus. Alleles which are different in structure and this affects the gametes production limiting to two in number.
There are ten different genotypes which prevail in the case of the F1 , the genotypes are replicated by the gamete production which makes the double of the number of available genes.
Suppose that F1 plant from the above part is self-pollinated the genotypes that are possible to be in the F2 are four. The gamete genotypes carried by the mode of self- crossed pollination and their host adds up to two. The plant F2 already has got two different genotypes in the structure hence the total genotypes are 4.
In the case of two brown mice, 9 brown pups and 3 white pups in the progeny I would use the linear extrapolation cross to determine the type of genotype found in the structure of brown offspring. The reason behind the selection of this cross that I have adopted is because the organisms in this context cannot portray a state of haploid nature of the mice.
The possible phenotypic ratios that are found in the cross above are 3:1 and 2: 1. The relative likelihood of each and every results above is that the ratio of 3;1 obtained from the 9 brown pups and 3 white pups in the progeny is high compared to the 2 brown mice and one white mice.
the chances that their first child of John and Martha will have galactosemia is low and the representation is 0.2 % since the genetic disorder is autosomal recessive inheritance. This will not occur because the in the case of Martha the sisters has got already given birth and the three children has no galactosemia: The likelihood is nil or negative.
- c) In the case of the first child being infected with the genetic disorder the likelihood of their third child being affected is high. “Erick said the autosomal recessive disease will multiply by three hence a 70% chance of being infected with the disorder” (Willy, 2014, p.637).
- The phenotype that F1 could have is known as heterozygous phenotype of the pure breeding mouse with short haired round ears mouse.
The percentage of the F2 that would be phenotypically resembled by the parents in the original cross is 60% given that the F2 have got a higher score in number in the self -crossing of F1.
Chances that the man’s first great grandchild will have wooly hair are 80% because of the co dominance or the genes in the people of Scandinavian.
Suppose that the man have four great grandchildren in total, the chances that exactly one will be in the danger of wooly hair is 20%. “Peterson said a higher likelihood of the children will be having the trait that is autosomal” (Gabriel, 2013, p.444).
There is inheritance of the dusky and clipped traits are supposedly obtained from the long transparent wings where the phenotypes are the observable traits consisting of features. “Thomas said an elaboration of this point is seen in the pea plant where the appearance is either color green or yellow depending on the environment” (Luke, 2009, p.123). Adaptation factor to the environment is the reason behind the scenario.
- Genotypes of the parents W and w for the long transparent wings and the Dd for the dusky wings.
- a) The allele symbols show the genotypic structure hence the representation of the information from the pedigree provided: AA and bb. This individual is at risk of contracting the diseasefrom the history of the family. The history provided has been utilized as a diagnostic tool and thus helps to guide the decisions arrived of genetic. There are diseases that are common in the family tree and powerful screening tool which is updated on each check by patients. Probability for the next generation to have the weakness in their genes and hence suffer is high.
- b) The person marked with the question mark hence low chances of 70 % being unaffected by the trait.
- ( a) The predicted phenotype frequencies of males and females are drone male with extremely short wings almost in capable of flying and the black eyes will be double the number in terms of fertilization because of the co dominance of diploid nature of the eggs.
The phenotype frequencies I would expect incase the eggs were 80% females is haplodiploids hence the ants will be more than three which have white eyes.
The male whom the woman might have a child affected by OTD is the husband’s mother who is a carrier for OTD.
In the above pairing where the woman’s father has ornithine transcarbamylase deficiency the proportion of sons and daughters that I expect to have OTD is 1 daughter and 2 sons.
Thomas, P. (2012). Elaboration of this point is seen in the pea plant where the appearance is either color green or yellow depending on the environment. Australia: Melbourne press.
Peterson, N. (2013). Genotype x contains the content that is defined as the recipe for the structure of the body. Sydeny :Times press.
Mathew, L. (2012). The behavior can also be deemed to be dominance phenotype. Melbourne: Hungarian press.
Erick, A. (2014). The autosomal recessive disease will multiply by three hence a 70% chance of being infected with the disorder. Current press: New York.