Anatomy Question Please Answer them all.A. Using a punnett square.pdf

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Anatomy Question: Please Answer them all. A. Using a punnett square, show the possible genotypes for John and Andrea\'s children? (John has O blood type and Andrea has A blood type) B. Based on the blood typing results, could Daryl be John and Andrea\'s Son? How Do you know this? ( Daryl has O blood type) Solution A. Andrea :Mother has A blood type , her genotype will be either AO or AA John :Father has O blood type ,his genotype will be OO. O allele is masked by the presence of A allele or B allele. OO blood group is expressed only in presence of OO alleles. Father Father B. Daryl has O blood type.For O blood froup to be expressed in child, both parents should have O alleles that together will contribute to make OO blood type of their son.So, yes Daryl is the son of John and Andrea.MotherOOAAO A blood groupAO A blood groupOOO O blood groupOO O blood group.

Anatomy Question: Please Answer them all.
A. Using a punnett square, show the possible genotypes for John and Andrea's children? (John
has O blood type and Andrea has A blood type)
B. Based on the blood typing results, could Daryl be John and Andrea's Son? How Do you know
this? ( Daryl has O blood type)
Solution
A. Andrea :Mother has A blood type , her genotype will be either AO or AA
John :Father has O blood type ,his genotype will be OO.
O allele is masked by the presence of A allele or B allele.
OO blood group is expressed only in presence of OO alleles.
Father
Father
B. Daryl has O blood type.For O blood froup to be expressed in child, both parents should have
O alleles that together will contribute to make OO blood type of their son.So, yes Daryl is the
son of John and Andrea.MotherOOAAO A blood groupAO A blood groupOOO O blood
groupOO O blood group

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Let S = I + T*T: H rightarrow H, where T is linear and bounded. Show that S^-1: S(H) rightarrow H exists. Solution T is a bounded linear operator. Hence, T*T is also a bounded linear operator as composition of two bounded linear operator is a bounded linear operator. I is the identity operator. Hence, it is bounded linear operator. So, S = I + T*T is a bounded linear operator as sum of two bounded linear operator is a bounded linear operator. Now, S : H --> H is a bounded linear operator. The identity operator I : H --> H can be written as S-1S : H --> H as I = S-1S. Let, h H. Then I(h) = h This can be written as S-1S(h) = h Or, S-1(S(h)) = h As h H and S : H --> H is a bounded linear operator, S(h) S(H). Let S(h) = h1 S(H) So, S-1(h1) = h where h H and h1 S(H) As h is arbitrary, we can say that S-1 : S(H) --> H exists. Proved.

Let S = I + TT H rightarrow H, where T is linear and bounded. Show.pdf

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It is a nanotechnology question. In which you are able to pick one material for both questions or one material for question one and another for question two. Solution Energy is of great importance in human life because of its benefits as the main resource for human activity. The application of nano technology or nano material in the field of energy, which involves lithium-ion battery, fuel cell, light emitting diode (LED), ultra-capacitor, and solar cell (including Grätzel cell). Nanotechnology will bring significant benefits to the energy sector, especially to energy storage and solar energy. Improved materials efficiency and reduced manufacturing costs are just two of the real economic benefits that nanotechnology already brings these fields and that’s only the beginning. Battery storage capacity could be extended, solar cells could be produced cheaper, and the lifetime of solar cells or batteries for electric cars could be increased, all thanks to continued development of nanotechnology. The following nanomaterial technologies will be of particular importance: \"organic and printed electronics\", \"nano-coatings\", \"nano-composites\", \"nano-fluids\", \"nano-catalysts\", \"nanocarbons\" and \"nano-electrodes\". Nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot sensitized solar cells QDSSC) . Nanocomposites are also used for the capture of energy in silicon-based solar cells, typically in the form of nanowires Nanocomposites can also be found in antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O fi lms via sol-gel-method) In the fi eld of energy storage the advantages of nanocomposites are needed in thin, fl exible energy storage devices with nanocomposite units, e.g. supercapacitors, Li-ion batteries and hybrid devices Nanotechnology in the sectors of solar energy and energy storage Additionally, nanocomposites can be used in the process of semiconductor-assisted photocatalysis for fuel production in order to obtain solar hydrogen or methanol For example, the nanocomposites applied here are TiO2 -Gold composite nanoparticles or semiconductor- semiconductor composites Another fi eld of application lies in the wind energy sector, where lightweight nanocomposite materials are mainly used for rotor blades. nanocomposites can be found in nano-structured solar cells and PV (e.g. in polymer-inorganic PV cells, or as polymer gel electrolytes in dye-sensitized solar cells DSSC or quantum-dot sensitized solar cells QDSSC) Nanocomposites are also used for the capture of energy in silicon- based solar cells, typically in the form of nanowires . Nanocomposites can also be found in antirefl ection coatings and higher solar transmittance coatings on collector glazing (e.g. Ti-Si-O fi lms via sol-gel-method) . In the fi eld of energy storage the advantages of nanocomposites .

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I have this problem that I was given in class, but I can\'t for the life of me figure out how to do part (d). Could someone please do it for me if it\'s possible? If the problem is leaving out some information, then let me know. I am having a problem finding a way to do this part with the information given, so I will tell my teacher if this problem can\'t be answered with what is in the problem statement. 1. During the Apollo program, the U.S. landed 12 men on the Moon and returned them safely to Earth. It was an amazing feat and an exciting time to be an aerospace engineer. The last Apollo mission was accomplished in December, 1972. The descent stage (DS) separated from the command module (CM), which remained in lunar orbit with the CM pilot. The DS carried two astronauts to a landing on the Moon. When the lunar expedition ended, the ascent stage (AS) rendezvoused with the CM for the return trip to Earth. Here\'s your chance to analyze the AS trajectory using actual values. 5.322 ft/sec (acceleration due to gravity on the Moon) 5,187 lb. (Earth weightas 10,295 lb. (Earth weight including propellant); weight of AS propellant- w weight). a) What is the mass of the AS including propellant? b) What does the AS and propellant weigh on the Moon? c) What is the propellant mass? The thrust of the ascent propulsion system (T-3,617 lb. The thrust duration is called action time (tama teton = 446 sec. Of course, the mass of the AS varies as propellant is expended. d) Assuming the propellant mass flow rate is constant, what is the mass flow rate during the motor burn (from t-0 to t-taction)? Note that the AS mass, m, is a function of time such that m(t)=m() + m (t. t.) where m=-. Also let t.-0, so that m(t)=m(0)+m t . Note that mc0 Now, the total force acting on the AS in the vertical direction is T-mt)gThe vertical acceleration (up is positive) is T-m(t)g m(t) a(t) (From F ma). e) What is the speed of the AS at the end of the motor burn, that is, when t? For a first-order approximation, assume the AS trajectory is vertical and does not pitch over to rendezvous with the CM. Further assumegarconstant. dt v()-sal)dt jdb.bm(a + bt) Some useful information: =-ln (a + bt Solution Autocorrect is almost as old as personal computers. Even some of the earliest word processors had a spellcheck feature that suggested alternative words if a word you typed did not appear in its internal dictionary. Certainly, today, autocorrect has come a long way since the Cupertino effect. What’s the Cupertino effect? Well, in its early days, word processors would replace the word “cooperation” with “Cupertino” (the city in Northern California where Apple and other computercompanies are headquartered). This spelling suggestion made its way to many documents published by the United Nations, NATO, and other official bodies, so the Cupertino In the beginning autocorrect was primarily a smart spellchecker. If you typed “bfeore,” it replaced it with “before.” Then in 2007, Microsoft int.

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Implement this in Java * Method to determine if a particular element is in this set. * @param target * the element that needs to be found in this set * @return * true if the target element is in this set, false otherwise public boolean contains(int target) { } Solution public boolean contains(int target) { boolean flag=false; for(Integer i:(this)) { if(target==i.intValue()) { flag=true; break; } } return flag; }.

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If an image is represented using 400 x 300 pixels and each pixel is represented using 8 bits, what will be the exact amount of data needed in kilobytes to represent in the image? Solution there are 1200 pixels in total. Each pixel is represented requires 8 bit for its representation which 1 byte of data. Now for 1200 pixels we require 1200 bytes of data to represent a 400x300 pixel image. 1 byte = 1/1024kb 1200 byte = 1200*(1/1024) = 1.17 kb data is required to present an image..

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How are the forces of diffusion and electrical force responsible for both the resting potential and the action potential? Solution Ans). In a resting state, diffusion and electrical force balance each other to attain an equilibrium for potassium and chloride ions. Large number of protein molecules cannot move through the membrane due to their size because of their negative charge, they contribute to the relative negativity of the intracellular fluid..

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For MIMO system, (a) Please talk about the advantage and disadvantage of MIMO technique. (b) Please explain why we need Cooperative Diversity to replace MIMO in cellular network. Solution A)MIMO advantages • Capacity scales linearly with number of antennas Channel knowledge/estimation at Rx needed • MIMO offers potential for larger data rate larger spectral efficiency larger number of users improved range/coverage better interference suppression better quality of service (QoS), lower bit-error rate (BER) lower Tx power MIMO disavantages • Hardware complexity: Each antenna needs a radio-frequency (RF) unit Powerful digital signal processing (DSP) unit required • Software complexity: Most signal processing algorithms are computationally intensive Power consumption: Battery lifetime of mobile devices Thermal problems Antennas: Antenna spacing (electromagnetic mutual coupling-e.g. mobile handsets) RF interference and antenna correlation B) Most of the present researches on cooperative network in which the user nodes are equipped with a single antenna or multiple, there have been some new results which exploit the benefits of multiple antenna deployment. Cooperative MIMO technology allows a wireless network system to attain better performance gains than provided by either usual MIMO or cooperative systems. It promised significant improvement in spectral efficiency and network coverage phenomena for different next generation wireless communication systems. In wireless communication, the path towards the various techniques that gives high service quality and data rate has been through the use of the cooperative network provided by the rich scattering wireless channels. Due to their great aspects, MIMO and cooperative systems have found their way into several standards for future wireless communication systems, especially in cellular networks and wireless local area networks (LAN) in this review article we are presenting the comparative analysis..

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Explain why and give examples. \"In OO programming Polymorphism is only possible when using Interface or Inheritance\". Solution Polymorphism is the ability of an object to take on many forms. The most common use of polymorphism in OOP occurs when a parent class reference is used to refer to a child class object. Any Java object that can pass more than one IS-A test is considered to be polymorphic. In Java, all Java objects are polymorphic since any object will pass the IS-A test for their own type and for the class Object. It is important to know that the only possible way to access an object is through a reference variable. A reference variable can be of only one type. Once declared, the type of a reference variable cannot be changed. The reference variable can be reassigned to other objects provided that it is not declared final. The type of the reference variable would determine the methods that it can invoke on the object. A reference variable can refer to any object of its declared type or any subtype of its declared type. A reference variable can be declared as a class or interface type. Polymorphism using inheritance: Polymorphism using interface: class Main { }.

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Eukaryotic cells modify RNA after transcription What critical RNA processing events usually happen to pre-mRNA\'s before they sent to the cytoplasm for translation? What is a 5\' cap? What is a poly A tail? What do these end modifications do for the mRNA transcript? What is RNA splicing? What are introns? What are exons? How are intros spliced out? What is a spliceosome? What is a \"snurp\" (snRNA)? Solution After its synthesis, the eukaryotic mRNA will undergo extensive modification like capping, polyadenylation and splicing to enter into the process of translation. (B) Capping: Here the 5\' end of the mRNA is modified by the addition of 7-methylguanosine (m7G) and the main function of this cap is to protect the 5\' end of the primary RNA transcript from attack by ribonucleases and this 5’ cap will be recognized by eukaryotic initiation factors, so that it can assemble the mature mRNA with the ribosome to start the process of translation. At the 3\' end of the RNA, we will have polyadenylation signal and during transcription itself this sequence will be chopped by an enzyme and another enzyme will add about 100100100 - 200200200 adenine (A) nucleotides to the 3’ end and this will form the poly-A tail. The main function of this poly A tail is proving stability to the transcript and also helping it to get exported from the nucleus to the cytosol.RNA splicing (C )Splicing is the third big RNA processing event and the pre-mRNA will have two sequences, exons and introns. Introns are the non-coding sequences and exons are the coding sequences, here in this step the through splicing, the introns will be removed and exons will be attached together. In RNA splicing, specific parts of the pre-mRNA (introns) will be recognized and removed by a protein-and-RNA complex called the spliceosome. Mature mRNA will have only exons, but no introns. (D)The splicing signal exon/GU-intron-AG/exon will be present in nuclear mRNA precursors and 5\' and 3\' splice sites always have consensus sequences extending beyond GU and AG motifs. During splicing, the exon-intron boundaries will be recognized by snRNA and the consensus sequences within introns will get hybridized and now the proteins other snRNAs will assemble the spliceosome on the transcript, the unpaired A present at 3\' side of the introns will attacks the 5\' exon -intron boundary with the help of 2\' OH and this will give rise to lariat structure. The free 3\' OH of the upstream exon will displace the downstream junctional nucleotide, like this introns will be removed and exons will be attached together. (E)A spliceosome is a large complex formed due to the assembly of snRNAs and protein complexes, and plays an important role in splicing of pre mRNA. snRNPs (snurps) is the small nuclear ribonucleic proteins and this is a RNA-protein complexes and they will form a larger complex with the unmodified pre-mRNA and various other proteins to form the structure called spliceosome.(A)Critical RNA processing events.

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Driver: import java.util.Scanner; //A class that keeps a fleet of different types of vehicles public class FleetOfVehicles { //An array of vehicles private Vehicle[] fleet; public static final int MAX_VEHICLES = 100; public FleetOfVehicles() { fleet = new Vehicle[MAX_VEHICLES]; } public Vehicle[] getFleet() { return this.fleet; } //Adds a new vehicle to the first empty spot in the fleet array public void addVehicle(Vehicle aV) { for(int i=0;i Solution import java.util.Scanner; //A class that keeps a fleet of different types of vehicles public class FleetOfVehicles { //An array of vehicles private Vehicle[] fleet; public static final int MAX_VEHICLES = 100; public FleetOfVehicles() { fleet = new Vehicle[MAX_VEHICLES]; } public Vehicle[] getFleet() { return this.fleet; } //Adds a new vehicle to the first empty spot in the fleet array public void addVehicle(Vehicle aV) { for(int i=0;i 0) this.cylinders = cylinders; else this.cylinders = 0; } /** * @return the ownersName */ public String getOwnersName() { return ownersName; } /** * @param ownersName * the ownersName to set */ public void setOwnersName(String ownersName) { this.ownersName = ownersName; } @Override public boolean equals(Object obj) { // TODO Auto-generated method stub if (obj instanceof Vehicle) { Vehicle vehicle = (Vehicle) obj; if ((this.getManuName().equals(vehicle.getManuName())) && (this.getOwnersName().equals(vehicle.getOwnersName())) && (this.getCylinders() == vehicle.getCylinders())) return true; else return false; } else return false; } @Override public String toString() { // TODO Auto-generated method stub return getManuName() + \" \" + getCylinders() + \" \" + getOwnersName(); } } public class Truck extends Vehicle { double loadCap, towCap; public Truck() { // TODO Auto-generated constructor stub super(); setLoadCap(0); setTowCap(0); } /** * @param manuName * @param cylinders * @param ownersName * @param loadCap * @param towCap */ public Truck(String manuName, int cylinders, String ownersName, double loadCap, double towCap) { super(manuName, cylinders, ownersName); setLoadCap(loadCap); setTowCap(towCap); } /** * @return the loadCap */ public double getLoadCap() { return loadCap; } /** * @param loadCap * the loadCap to set */ public void setLoadCap(double loadCap) { if (loadCap > 0) this.loadCap = loadCap; else this.loadCap = 0; } /** * @return the towCap */ public double getTowCap() { return towCap; } /** * @param towCap * the towCap to set */ public void setTowCap(double towCap) { if (towCap > 0) this.towCap = towCap; else this.towCap = 0; } @Override public boolean equals(Object obj) { // TODO Auto-generated method stub if (obj instanceof Truck) { if (super.equals(obj)) { Truck vehicle = (Truck) obj; if (this.getLoadCap() == vehicle.getLoadCap() && this.getTowCap() == vehicle.getTowCap()) return true; else return false; } else return false; } else return false; } @Override public String toString() { // TODO Auto-generated method stub return super.toString() + \" \" + g.

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Describe a data structure that supports both removeMin() and removeMax() with O(log(n)) complexity. You need to show the implementation of insert(), removeMin(), and removeMax(), and prove their complexity. Hint: one can build on top of the heap. /*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/ I need to understand Describe a data structure that supports both removeMin() and removeMax() with O(log(n)) complexity. You need to show the implementation of insert(), removeMin(), and removeMax(), and prove their complexity. Hint: one can build on top of the heap. /*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/ I need to understand Describe a data structure that supports both removeMin() and removeMax() with O(log(n)) complexity. You need to show the implementation of insert(), removeMin(), and removeMax(), and prove their complexity. Hint: one can build on top of the heap. /*IF YOU WRITE CODE, PLEASE PROVIDE COMMENTS*/ I need to understand Solution The idea here is to use the concept of two binary heaps along with the concept of list(doubly linked list). The doubly linked list contains all input items and indexes of corresponding min and max heap nodes. The nodes of min and max heaps store addresses of nodes of doubly linked list. The root node of min heap stores the address of minimum item in doubly linked list. Similarly, root of max heap stores address of maximum item in doubly linked list. Following are the details of operations. 1) findMax(): We get the address of maximum value node from root of Max Heap. So this is a O(1) operation. 1) findMin(): We get the address of minimum value node from root of Min Heap. So this is a O(1) operation. 3) removeMin(): We get the address of minimum value node from root of Min Heap. We use this address to find the node in doubly linked list. From the doubly linked list, we get node of Max Heap. We delete node from all three. We can delete a node from doubly linked list in O(1) time. delete() operations for max and min heaps take O(Logn) time. 4) removeMax(): is similar to deleteMin() 5) Insert() We always insert at the beginning of linked list in O(1) time. Inserting the address in Max and Min Heaps take O(Logn) time. So overall complexity is O(Logn) #include #include #include // A node of doubly linked list struct LNode { int data; int minHeapIndex; int maxHeapIndex; struct LNode *next, *prev; }; // Structure for a doubly linked list struct List { struct LNode *head; }; // Structure for min heap struct MinHeap { int size; int capacity; struct LNode* *array; }; // Structure for max heap struct MaxHeap { int size; int capacity; struct LNode* *array; }; // The required data structure struct CDS { struct MinHeap* minHeap; struct MaxHeap* maxHeap; struct List* list; }; // A utility function to create a new List node struct LNode* newLNode(int data) { struct LNode* node = (struct LNode*) malloc(sizeof(struct LNode)); node->minHeapIndex = node->maxHeapIndex = -1; node->data = data; node->prev = node->next = NULL; retu.

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Columbus Incorporated just paid $4.3 per share dividend yesterday (i.e.,D0). The dividend is expected to grow at a constant rate of 6% a year. The required rate of return on the stock, r, is 11%. What is the value per share of the company\'s stock? Round your answer to two decimal places. Solution Recent Dividend, D0 = $4.30 Growth Rate, g = 6% Required Return, rs = 11% D1 = D0 * (1 + g) D1 = $4.30 * 1.06 D1 = $4.558 Value per share = D1 / (rs - g) Value per share = $4.558 / (0.11 - 0.06) Value per share = $4.558 / 0.05 Value per share = $91.16.

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A person is lost in the desert. (S)he has no water and after a couple of days loses so much water through perspiration that his/her body fluids become hypertonic. What happens to his cells? What happens to his proteins? What happens to his carbohydrates? Solution Early sign of dehydration there is no symptom but later on there is dryness of mouth and tongue, dark yellow urine,dryness of skin cramp in arm and leg. When fluid become hypertonic i.e sodium is lost from cell this result in cell shrinkage. After two days of starvation body will first utilize carbohydrate and then glycogen from liver and muscle for energy. Protein from cell / muscle will be next to be used as source of energy for body.

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2. Describe an advantage and a limitation to tracing ancestry with both types of DNA. Solution The tracing of ancestral origins using mitochondrial DNA, Y-DNA & autosomal DNA is advantageous to know about the historical evolutionary aspects of humans from thousands of years based on short tandem repeats, radioactive labeling & based on single nucleotide polymorphism. This Genetic genealogy is advantageous to unravel familial relationships but has ethical and moral limitations when examining prenatal ancestral tracing& the information about the ancestral DNA is limited to reveal about the inheritance pattern.

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arihantstoneart

Write your thought on the following: 1. petroski notes that most engineering failures are much more involved than simple technical mis-calculations and involve the failure of the design process or management culture.However, not all engineering failures involve ethical issue.The infamous collapse of the first Tacoma Narrow Brodge, and the losses of the Mars polar lander and Mars Climate Orbiter were technical and design process failures. Solution In the face of new technology, how do we balance public welfare and progress? If Moisseiff had designed a bridge similar to the ones which had already proven their stability, Tacoma Narrows Bridge would never have collapsed costing thousands of dollars and endangering many lives. It would also have been significantly more expensive. On the other hand, if engineers had never tried innovative techniques, suspension bridges may never have been built at all. At the time of their introduction, no one believed that a suspension bridge could safely accommodate trains. Roebling, however, took a gamble, pushed the limits of the current technology, and built a suspension bridge that he believed could safely support rail traffic. Luckily he was correct, and suspension bridges soon became widely accepted (Petroski. Moisseiff also took a gamble, trying to create a longer, sleeker, less expensive bridge, by pushing the limits of technology. He, however, was not as lucky, and what could have been a breakthrough in technology turned into a catastrophic failure. Every time engineers push the limits of technology they risk a similar loss like losses of mas polar lander and mass climate orbiter in which there were technical error like Mass polar lander 1. Inadequate specifications which includes what the software was supposed to do, but no mention of what it must not to do. 2. Flawed review process: Violation of basic engineering safety pracitices in the digital part of the system. Mass Climate Orbiter 1. Units communication and attention to detail failure. The primary cause of this discrepancy was that one piece of ground software supplied by Lockheed Martin produced results in a United States customary unit, contrary to its Software Interface Specification (SIS), while a second system, supplied by NASA, expected those results to be in SI units, in accordance with the SIS. Specifically, software that calculated the total impulse produced by thruster firings calculated results in pound-seconds. The trajectory calculation software then used these results - expected to be in newton-seconds - to update the predicted position of the spacecraft. sometimes even It can cause a loss of life. How much is too much? When is a possible advance worth a risk to public safety? What can the engineering profession do to make the implementation of new technology safer? Do our current peer review and building code committee processes adequately protect public safety?.

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arihantstoneart

You have two cell lines, a macrophage cell line and a CD4 T-cell line. The macrophage cell line is resistant to infection by HIV, whereas the T-cell line is fully susceptible and fully permissive. Why are macrophage cell line resistant to infection by HIV? Why are T-cell fully susceptible and fully permissive? Please explain. Also, A heterokaryon cell made from the fusion of these two cells is able to be infected by HIV. Explain why? Please explain in detail. Solution Macrophages are antigen- presenting cells so they process the viral peptides and present it to uninfected CD4 T-Cell which is the important transmission process. Also HIV-infected macrophages release soluble cytotoxic factors which induce apoptosis in CD4 T-Cells. Macrophages produce certain host restriction factors which inhibits the release of viral progeny from infected cells. So macrophages are resistant to lysis because of its genetic make up and its phenotypes (restriction factors). So when two cell lines are fused to form heterokaryon cell, the genes of macrophages are suppressed or it is recessive or the T cell genes are dominant to make the heterokaryon susceptible to infection..

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arihantstoneart

With X-linked inheritance, the X chromosome is transmitted in a different pattern by males and females. In X-linked dominant traits, males hemizygous for trait will express the trait, but females heterozygous will be carriers. For X-linked recessive, none of the sons of an affected female will show the trait. For autosomal dominant, two affected parents may produce an unaffected child. Solution Answer: A. True Most X-linked conditions are recessive. This means that in a person with two X chromosomes (most females), both copies of a gene (i.e., one on each X chromosome) must have a change or mutation whereas in a person with one X chromosome (most males), only one copy of a gene must have a mutation. A female with a mutation in one copy of a gene on the X chromosome is said to be a “carrier” for an X-linked condition. A male with a mutation in a gene on the X chromosome is typically affected with the condition. The father\'s X-chromosome goes to the daughter while the mother can give one X-chromosome to the son and other to the daughter. B. False For an X-linked dominant condition, only one copy of a gene on the X chromosome whether in a female with two X chromosomes or males with on X chromosome must have a change or mutation for an individual to be affected with the condition. For this reason, X-linked disorders are often seen with similar frequency in males and females. However, since females also have one normal X chromosome as well as an X chromosome with a mutation, the condition is often more “mild.” C.False For X-linked recessive disorders, an affected father who has a mutation in a gene on the X chromosome can transmit either the X chromosome with this mutation or a Y chromosome to his children. If the mother is not affected or a carrier, none of his sons will be affected since they can only inherit a normal X chromosome from their mother and they inherit a Y chromosome from their father. But if the female is affected, it will receive one of the affected X chromosomes, which is enough to express the trait in the son. D. True A person affected by an autosomal dominant disorder has a 50 percent chance of passing the mutated gene to each child. The chance that a child will not inherit the mutated gene is also 50 percent..

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