Which statement about gills is falseThey contain the environmenta.pdf

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Which statement about gills is false? They contain the environmental medium. The respiratory membrane has a large surface area. They are surrounded by the environmental medium. Solution Which statement about gills is false? They contain the environmental medium.---------answer The respiratory membrane has a large surface area. They are surrounded by the environmental medium. ====================================================== EXPLANATION GILLS Gills significantly increment the surface area for exchange of gases. They happen in an assortment of creature gatherings including arthropods , annelids, fish, and creatures of land and water. Gills normally are convoluted outgrowths containing veins secured by a thin epithelial layer. Commonly gills are sorted out into a progression of plates and might be interior (as in crabs and fish) or outer to the body (as in a few creatures of land and water). Gills are extremely proficient at expelling oxygen from water: there is just 1/20 the measure of oxygen present in water as in a similar volume of air. Water streams over gills in one course while blood streams the other way through gill vessels. This countercurrent stream augments oxygen exchange..

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Corals are marine invertebrates in class Anthozoa of phylum Cnidaria typically living in compact colonies of many identical individual "polyps". Corals are gastrovascular marine organisms. Each one of these animals is known as a coral "polyp". Coral Polyps are tiny, primitive marine organisms. A single polyp has a tube-shaped body with a mouth which is surrounded by tentacles. The polyp of hard corals produces a stony skeleton of calcium carbonate which form the base. Often the skeleton forms a cup-like structure in which the polyp lives. Coral polyps in colonies make up the cora reefs.

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In 2012, the percent of American adults who owned cell phones and used their cell phone to send or receive text messages was at an all-time high of 80%. Assume that 80% refers to the population parameter. More recently in 2015, a polling firm contacts a SRS of 1200 people chosen from the population of cell phone owners to confirm the percent who use their phone to text. find mean, shape, standard deviation. Solution.

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I need help with this one method in java. Here are the guidelines. Only public Employee[] findAllBySubstring(String find). EmployeeManager EmployeeManager - employees : Employee[] - employeeMax : final int = 10 -currentEmployees : int <> EmployeeManager + addEmployee( type : int, fn : String, ln : String, m : char, g : char, en : int, ft : boolean, amount : double) + removeEmployee( index : int) + listAll() + listHourly() + listSalary() + listCommision() + resetWeek() + calculatePayout() : double + getIndex( empNum : int ) : int + annualRaises() + holidayBonuses() : double + increaseHours( index : int, amount : double) + increaseSales( index : int, amount : double) + findAllBySubstring(find : String) : Employee[] - RabinKarp(name : String, find : String) : int - stringHash(s : String) : int - charNumericValue(c : char) : int - RabinKarpHashes(s : String, hashes : int[], pos : int, length : int) : int - linearSearchRecursive(nameHashes : int[], findHash : int, pos : int) : int public EmployeeManager() Constructor, creates the Employee array, sets currentEmployees to 0. public void addEmployee(int, String, String, char, char, int, Boolean, double) Takes an int representing the type of Employee to be added (1 – Hourly, 2 – Salary, 3 – Commission) as well as the required data to create that Employee. If one of these values is not passed output the line, “Invalid Employee Type, None Added”, and exit the method. If an Employee with the given Employee Number already exists do not add the Employee and output the line, “Duplicate Not Added”, and exit the method. If the array is at maximum capacity do not add the new Employee, and output the line, \"Cannot add more Employees\". public void removeEmployee(int) Removes an Employee located at the given index from the Employee array. public void listAll() Lists all the current Employees. Outputs there are none if there are none. public void listHourly() Lists all the current HourlyEmployees. Outputs there are none if there are none. public void listSalary() Lists all the current SalaryEmployees. Outputs there are none if there are none. public void listCommission() Lists all the current CommissionEmployees. Outputs there are none if there are none. public void resetWeek() Resets the week for all Employees. public double calculatePayout() Returns the total weekly payout for all Employees. public int getIndex(int) Given an Employee Number, returns the index of that Employee in the array, if the Employee doesn’t exist retuns -1. public void annualRaises() Applies annual raise to all current Employees. public double holidayBonuses() Outputs and returns the total holiday bonus of all Employees. public void increaseHours(int, double) Increase the hours worked of the Employee at the given index by the given double amount. public void increaseSales(int, double) Increase the sales of the Employee at the given index by the given double amount. public Employee[] findAllBySubstring(String find) This method will return.

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Explain how TWO (2) different structural features can be used to discriminate between the three types of muscles. (2 marks) Solution Three types of muscles found in the human body are: 1)Skeletal muscle 2)Smooth 3)Cardiac FUNCTION OF MUSCLES: 1)Regulate organs 2)Help us to move and maintain posture FUNCTIONS OF SKELETAL MUSCLE TISSUE FUNCTIONS OF CARDIAC MUSCLE Pumping of blood through the heart: Alternate contraction and relaxation of cardiac muscle pumps FUNCTIONS OF SMOOTH MUSCLE TISSUE Smooth muscle also contributes to moving fluids through the body and to the elimination of indigestible matter from the gastrointestinal system.

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Explain two (2) alternative ways in which plants can obtain nutrients when they live in nutrient poor soils. Solution When plants live in nutrient poor soils they develop specific adaptations to obtain nutrients. For example, carnivorous plants such as Nepenthes, Dionea, Urticularia, that are growing in nitrogen poor soils developed modified leaves into pitchers, fly traps and urticles that can trap insects and enzymes produced in these structures digest them. Some plants like legumes develop symbiotic association with nitrogen fixing bacteria like Rhizobium and form root nodules. The plants provide shelter to bacteria and bacteria fix the atmosphric nitrogen and supply to the plants. Mangrove plants like Avicennia growing in oxygen poor marshy soils develop aerial roots that obtain oxygen. Some plants establish mutual relationship with Mycorrhiza (fungus) to obtain nutrients like nitrogen and phosphorous. Therefore, plants develop adaptations for obtaining nutrients and survival in nutrient poor soil environments..

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Couldn\'t find the right \"subject\" that it belongs to... There are several big picture meta-theoretical themes used in this class—as presented in the syllabus and textbook, and as explained by the instructor. Find the theme that is NOT part of the assumptions from the meta-theory used in this class. Question 4 options: Development is promoted by person-centered contexts and interactions. People carry the past forward into a dynamic present. Genetics are at the root of behavioral development. Development is influenced by an active changing context and an active developing person. All people face fundamental developmental tasks, although there are many healthy pathways through them. Solution reference Yardley, L., Morrison, L., Bradbury, K., & Muller, I. (2015). The Person-Based Approach to Intervention Development: Application to Digital Health-Related Behavior Change Interventions. Journal of Medical Internet Research, 17(1), e30. http://doi.org/10.2196/jmir.4055 topic : The Person-Based Approach to Intervention Development...............Application to Digital Health-Related Behavior Change Intervention. The basic point of the individual based approach is to ground the improvement of conduct change intercessions in a significant comprehension of the viewpoint and psychosocial setting of the general population who will utilize them, increased through iterative top to bottom subjective research. There is broad accord in the eHealth inquire about group that evoking and tending to the necessities and point of view of the expected intercession client is a crucial piece of good mediation improvement to guarantee (at any rate) that intercessions are usable and locks in. This is a basic issue for eHealth in the event that it is to satisfy its potential and defeat the issues of low take-up and adherence . It is troublesome notwithstanding for master mediation designers to completely foresee the needs and needs of clients , thus intercession engineers as of now routinely evoke the perspectives of target clients in an assortment of ways , yet there is shockingly little level headed discussion and itemized direction concerning how best to do this . The individual based approach gives a procedure that empowers designers to increase imperative bits of knowledge into how diverse individuals experience and actualize mediations, and a system to help engineers distinguish the key attributes that will make an intercession more important, alluring, and valuable to the individuals who draw in with it. The individual based approach was produced by our exploration group through down to earth involvement of making and assessing various fruitful wellbeing related intercessions, including general wellbeing mediations (eg, to oversee weight and stress, advance physical action and hand cleanliness) and disease administration intercessions (to help clients adapt to dazedness, back torment, weakness, respiratory conditions, hypertension, diabetes, growth, stroke, and nu.

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Consider the following C code snippet: // C code void setArray(int num) { int i; int array[10]; for (i=0; i < 10; i=i+1) { array[i]=compare(num, i); } } int compare(int a, int b) { if (sub(a, b) >= 0) return 1; else return 0; } int sub(int a, int b) { return ab; } Implement the C code snippet above in MIPS assembly language. Use $s0 to hold the variable i. Be sure to handle the stack pointer appropriately. The array is stored on the stack of the setArray function (i.e. when setArray function is called it allocates space for the array on the stack.). Solution I have used the Test function i.e ain function to get this module of the fuction definition working. Thw below is the MIPS code for the complete program:- .file 1 \"\" .section .mdebug.abi32 .previous .gnu_attribute 4, 1 .abicalls .text .align 2 .globl _Z3subii $LFB0 = . .set nomips16 .ent _Z3subii .type _Z3subii, @function _Z3subii: .frame $fp,8,$31 # vars= 0, regs= 1/0, args= 0, gp= 0 .mask 0x40000000,-4 .fmask 0x00000000,0 .set noreorder .set nomacro addiu $sp,$sp,-8 $LCFI0: sw $fp,4($sp) $LCFI1: move $fp,$sp movz $31,$31,$0 $LCFI2: sw $4,8($fp) sw $5,12($fp) lw $3,8($fp) lw $2,12($fp) nop subu $2,$3,$2 move $sp,$fp lw $fp,4($sp) addiu $sp,$sp,8 j $31 nop .set macro .set reorder .end _Z3subii $LFE0: .size _Z3subii, .-_Z3subii .align 2 .globl _Z7compareii $LFB1 = . .set nomips16 .ent _Z7compareii .type _Z7compareii, @function _Z7compareii: .frame $fp,32,$31 # vars= 0, regs= 2/0, args= 16, gp= 8 .mask 0xc0000000,-4 .fmask 0x00000000,0 .set noreorder .cpload $25 .set nomacro addiu $sp,$sp,-32 $LCFI3: sw $31,28($sp) $LCFI4: sw $fp,24($sp) movz $31,$31,$0 $LCFI5: move $fp,$sp $LCFI6: .cprestore 16 sw $4,32($fp) sw $5,36($fp) lw $4,32($fp) lw $5,36($fp) lw $2,%got(_Z3subii)($28) nop move $25,$2 jalr $25 nop lw $28,16($fp) nor $2,$0,$2 srl $2,$2,31 andi $2,$2,0x00ff beq $2,$0,$L4 nop li $2,1 # 0x1 b $L5 nop $L4: move $2,$0 $L5: move $sp,$fp lw $31,28($sp) lw $fp,24($sp) addiu $sp,$sp,32 j $31 nop .set macro .set reorder .end _Z7compareii $LFE1: .size _Z7compareii, .-_Z7compareii .align 2 .globl _Z8setArrayi $LFB2 = . .set nomips16 .ent _Z8setArrayi .type _Z8setArrayi, @function _Z8setArrayi: .frame $fp,88,$31 # vars= 48, regs= 3/0, args= 16, gp= 8 .mask 0xc0010000,-4 .fmask 0x00000000,0 .set noreorder .cpload $25 .set nomacro addiu $sp,$sp,-88 $LCFI7: sw $31,84($sp) $LCFI8: sw $fp,80($sp) movz $31,$31,$0 $LCFI9: sw $16,76($sp) $LCFI10: move $fp,$sp $LCFI11: .cprestore 16 sw $4,88($fp) sw $0,24($fp) b $L8 nop $L9: lw $16,24($fp) lw $4,88($fp) lw $5,24($fp) lw $2,%got(_Z7compareii)($28) nop move $25,$2 jalr $25 nop lw $28,16($fp) move $3,$2 sll $2,$16,2 addiu $4,$fp,24 addu $2,$4,$2 sw $3,4($2) lw $2,24($fp) nop addiu $2,$2,1 sw $2,24($fp) $L8: lw $2,24($fp) nop slt $2,$2,10 andi $2,$2,0x00ff bne $2,$0,$L9 nop move $sp,$fp lw $31,84($sp) lw $fp,80($sp) lw $16,76($sp) addiu $sp,$sp,88 j $31 nop .set macro .set reorder .end _Z8setArrayi $LFE2: .size _Z8setArrayi, .-_Z8setArrayi .align 2 .globl ma.

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Describe a mechanism that the body uses to prevent a mutation from becoming an abnormal cell. Solution The body has several mechanisms in place to ensure that mutations do not turn lethal. The first of those steps is the proof reading activity of the DNA polymerase which reads the DNA strand and corrects the sequence as it synthesizes the new DNA fragment. Thus, the daughter cells contain the original non-mutated sequence. Secondly, organisms are mostly diploid i.e. they contain two sets of chromosomes which includes two forms of the same gene. The mutation would be lethal or cause loss of function only when the other copy of the gene is also mutated (recessive mutation). Also, mutant cells are usually cleared by the immune system which can detect and produce specific antibodies against these cells..

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classify domian of life differences Solution Classify the domain of life differences: the domain of life is mainly classified based on the following differences 1. Ribosomal RNAs (rRNA) inside the cell 2. The structure of cell membrane with lipid structure 3. Cell sensitivity to antibiotics Three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae, kingdom Animalia)\" Explanation: The mitochondria, histone (nucleosomes) are not evolved in archae domain & prokaryote domain but evolved in the “kingdom of eukaryotes”, as an evolutionary adaptation difference. This is a major differential aspect among the three domains of life in their classification according to endo-symbiotic theory. The aspects of cell wall are constructed differently in all three domains of life such as bacteria, archaic species, and eukaryotes. The archae has S -layer in their cell wall but do not possess peptidoglygan layer as bacteria do. Lipopolysaccharide, peptidoglycan layers are absent in eukaryotic cell walls and their mechanism of synthesis is absent in eukaryotes. Plant cells are eukaryotic cells and contain cell wall but animal cells do not contain cell wall. Bacteria are prokaryotes & do not possess histone proteins but eukaryotes contain histone proteins associated with DNA. Any morphological characteristics of group of species are completely different to that of three domains of life such as \"prokaryota\", \"archae\" and \"eukaryota (kingdom plantae, kingdom Animalia)\". These morphological features are completely different to the ancestral species and leading to new phylum evolvement. Internal structural features, embryological development, nucleus structure, genome nature (RNA or DNA), type of cellular respiration & energy synthesis. The major domian of life differences with morphological relations in the evolution of vertebrate animal life (at the phylum level & domain level) from invertebrates (from four -limb species): The major features of embryonic vertebrate zygote formation & followed by eight-cell stage of embryo finally blastula stage and gastrulation. This stage is absent in the archae & prokaryotes species. There are total 2 types of groups of species that are divided based on presence of morphological innovation that is \"presence of vertebral column\" in animal domain & prokaryote. Those living species which do not have vertebra\" are referred as \"invertebrates\" and those which possess vertebrae\" are referred as “chordates\" or \"vertebrates\"..

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Compare and contrast the development of a WBS in traditional project management versus development in an agile environment. Solution An Agile based WBS is organized around units of end user functionality - commonly referred to as user stories. Each user story represents the work to deliver that increment of functionality. In Agile based WBS, Features are decomposed into Epics (large user stories) and Epics into User Stories. User stories are decomposed to represent functionality that can be completed within one iteration - typically one to four weeks depending on the team/project. Each leaf level WBS element represent the working software to be produced through analysis, design, coding and testing of the requirement in the user story. Because each user story is atomic, stories can be re-prioritized, added and or removed without changing the scope, provided that the total size/effort of stories within the scope of the project does not change. Additionally, iterative development facilitates multiple releases (deployments) within a project. So, user stories are typically scheduled within an iteration and a release. Note, that until an iteration has started, the stories within that iteration can be changed, as part of iteration planning. In case of traditional project management WBS is organized around the primary products of the project (or planned outcomes) instead of the work needed to produce the products (planned actions). Since the planned outcomes are the desired ends of the project, they form a relatively stable set of categories in which the costs of the planned actions needed to achieve them can be collected. A well-designed WBS makes it easy to assign each project activity to one and only one terminal element of the WBS. In addition to its function in cost accounting, the WBS also helps map requirements from one level of system specification to another, for example a requirements cross reference matrix mapping functional requirements to high level or low level design documents..

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B. You are an evolutionary biologist in a heated argument with a creation scientist who makes the statement—\"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for chance. 2. Now play the other side and respond to the supporting statement as you think a creation scientist would respond.? Solution You are an evolutionary biologist in a heated argument with a creation scientist who makes the statement— \"Mathematically, it is inconceivable that anything as complex as a protein, let alone a living cell or a human, could spring up by chance\". 1. Make a supporting statement for chance. Chance: It plays a part in evolution (for example, random mutations in a gene pool can give rise to few new traits), but evolution does not depend ONLY on chance in creating new organisms, proteins or any other entities. This is quite opposite to natural selection. Natural selection is the principal which is known for the mechanism of evolution. By chance some mutation caused in the gene pool and it gave to new acceptable traits. If they found adjustable to the environment, they continues as a new species. The \"undesirable\" (nonadaptive) ones will be thus elimated from that gene pool. As long as the forces of selection stay constant, natural selection pushes the evolution in only one direction and produce new desired structures. Thus chance plays a part in evolution. Nature plays a major role in this. Fossil studies support this theory/ process. Where we can see many intermediate or failed species in between major species. Archaeopteryx, a link between reptiles and birds, extinctinted soon is a good example of intermediate species. Haldane and Oparin proposed a theory of the origin of life on Earth through the transformation, during the gradual chemical evolution of particles that contain carbon in the primordial soup shows an example of Evolution. 2. Now play the other side and respond to the supporting statement as you think a creation scientist would respond? As a creation scientist i do not agree with science and evolution part, all this is created by God and thus it follows biblical roots. All that it is in the nature is creation of God, all of a sudden it appeared on the earth. According to us GOD created all species like birds,animals etc., later he wanted to create a genius brained animal, so he created humans. And God creation according to us all this is due to CHOICE not because of CHANCE.Thus it involved logical and systematic process..

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Assume Hashtable is a simple array of size 8, with indices 0..7. Numeric keys are mapped by a Hashfunction that gives the mod(8,n) value for any key “n” yielding the Hashtable index for that key (0..7). A Hashtable entry is null unless a key exists with that index as its hashed index; if so, the Hashtable entry points to the first node of a linked list of keys with that hash index. The last node on this linked list has a null reference for the next referenced node. Assume the occurrence of a linked list node is represented by the object “Node” and its “Data” and “NextRef” attributes. Create pseudocode for the \"Remove\" operation. Solution Removal of a Node from a Singly Linked List To complete deletion of firstNode in the list we have to change Head pointing to NextRef of firstNode. Pseudocode: firstNode = Head Head = firstNode->NextRef free firstNode Complexity: Time Complexity: O(1) Space Complexity: O(1) Sourcecode: int delNodeData(int num) { struct Node *prev_ptr, *node; node=Head; while(cur_ptr != NULL) { if(cur_ptr->Data == num) { if(cur_ptr==Head) { Head=cur_ptr->NextRef; free(node); return 0; } else { prev_ptr->NextRef=node->NextRef; free(node); return 0; } } else { prev_ptr=node; node=node->NextRef; } } printf(\"\ Element %d is not found in the List\", num); return 1; } Deleting Last Node in the Singly Linked List Traverse to Last Node in the List using two pointers namely prevNode and node. Once node reaches the last Node in the list point NextRef in prevNode to NULL and free the node. Pseudocode: node = head forever: if node->NextRef == NULL break prevNode = node node = node->NextRef prevNode->NextRef = NULL free Node Complexity: Time Complexity: O(n) Space Complexity: O(1) Deleting Node from position \'p\' in the List To delete a Node at the position \'p\' we have to first traverse the list until we reach the position \'p\'. For this case have to maintain two pointers namely prevNode and Node. Since Singly Linked Lists are uni-directional we have to maintain the information about previous Node in prevNode. Once we reach the position \'p\' we have to modify prevNode NextRef pointing to Node NextRef and free Node. Pseudocode: curNode = head curPos = 1 forever: if curPos == P || Node == NULL break prevNode = node node = node->NextRef curPos++ if noode != NULL: prevNode->NextRef = node->NextRef free curNode Complexity: Time Complexity: O(n) worst case Space Complexity: O(3) Sourcecode: int delNodeLoc(int loc) { struct Node *prev_ptr, *node; int i; node=Head; if(loc > (length()) || loc <= 0) { printf(\"\ Deletion of Node at given location is not possible\ \"); } else { // If the location is starting of the list if (loc == 1) { Head=node->NextRef; free(node); return 0; } else { for(i=1;iNextRef; } prev_ptr->NextRef=node->NextRef; free(node); } } return 1; }.

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Amon the following, which has the lowest levels of dissolved iron? a. Atlantic Ocean b. Indian Ocean c. Pacific Ocean d. The temperate regions of all oceans Solution The correct option is Pacific ocean because of the low dust sedimentation and iron is readily soluble in low oxygen regions and the continental margins. It was believed that iron remained in the localized areas but didn\'t contribute to the overall iron content of the ocean. As it is explained by Nielsen, as soon as the iron-rich fluids hit the seawater with high oxygen content, the iron gets dumped out and is not contributed to the overall amount of the dissolved iron..

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A box contains 10 red balls and 40 black balls. Two balls are drawn .pdf

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As late as 1992, the United States was running budget deficits of nearly $300 billion. During the remainder of the 1990\'s, deficits declined and became surpluses. As the new century began, these surpluses again turned into deficits. •Explain the decline in deficits and subsequent surpluses in the late 1990\'s. •Explain the return to deficit spending since the turn of the century. •Consider the causes of the deficits and surpluses and provide your own insight as to whether these surpluses or deficits have a \"positive\" or \"negative\" effect on our economy Solution As the 1990s came to an end, virtually all financial-market observers and politicians rejoiced that budget deficits were gone and that the U.S. had entered a period of fiscal stability. Politicians on both sides of the aisle viewed the new world of budget surpluses as an opportunity to save Social Security, to cut Federal debt, and to lower taxes. Budget deficits or budget surpluses are neither good nor bad, they are fiscal policy options that can stabilize or de-stabilize an economy. The U.S. economy, under Reagan, Bush, and early Clinton, experienced higher not lower rates of economic growth at a time when budget deficits were at record levels and tax rates were on the decline. Over the past 20 years a period of record budget deficits the stock market increased from 950 to over 11,000, providing ample evidence that budget deficits are not necessarily bad for an economy.On the other hand, budget surpluses are not necessarily good. As the budget surpluses began to expand dramatically in late 1999 and into 2000, the stock market appeared to be reversing the record up-trend of the 1980s and \'90s. One possible reason for the reversal: the growing budget surplus was sapping the savings of the private sector. Consumers were able to maintain their spending only through expanding debt to record level. In the late 90s, the confluence of strong economic growth, decelerating federal government spending, and substantial increases in tax rates (revenues) had the unexpected effect of producing enormous budget surpluses. As unexpected budget surpluses burst on the scene in the late 1990s, most mainstream economists weren’t sensitive to the contractionary impact that surpluses can have on the economy. The reason was simple that none of their economic models ever included the forecast of a substantial surplus. Ronald Reagan’s was the vice president of the USA for eight years and knew firsthand how difficult it was to bring the federal budget deficit down from 6 percent of GDP in 1983 to 2.8 percent in Reagan’s last budget. Bush knew that the many budget deals of the 1980s had been hard fought and always involved higher taxes as part of the deficit reduction, Reagan signed into law 11 major tax increases that raised taxes by $133 billion in 1988 or 2.7 percent of GDP.Bush administration economists believed that an easier monetary policy was the key to stimulating growth. But Federal Reserve Board.

As late as 1992, the United States was running budget deficits of ne.pdf

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A bacteriophage population is introduced to a bacterial colony that is in stationary phase with rapidly depleting nutrients. Appraise the following scenarios and select which one would favor long-term survival of the bacteriophage. A. The bacteriophages infect the remaining viable cells and integrate their genetic material into the host cell genome thereby preserving their genome. B. The bacteriophages do not infect the bacterial cells since they are dying; they remain inert outside of the cells. C. The bacteriophages infect the bacterial cells and go through all stages of development apart from release, thus keeping the progeny phage from entering an inhospitable environment. Solution Answer : C. The bacteriophages infect the bacterial cells and go through all stages of development apart from release, thus keeping the progeny phage from entering an inhospitable environment..

A bacteriophage population is introduced to a bacterial colony that .pdf

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Xavier and Yolanda both have classes that end at noon and they agree to meet every day after class. They arrive at the coffee shop independently. Xavier\'s arrival time is X and Yolanda\'s arrival time is Y, where X and Y are measured in minutes after noon. The individual density functions are given. (Xavier arrives sometime after noon and is more likely to arrive promptly than late. Yolanda always arrives by 12:10 PM and is more likely to arrive late than promptly.) After Yolanda arrives, she\'ll wait for up to 20 minutes for Xavier, but he won\'t wait for her. Find the probability that they meet. (Round your answer to three decimal places.) f1(x) = {e^-x if x gt o and o if x Solution.

Xavier and Yolanda both have classes that end at noon and they agree .pdf

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1. Low platelet count is a recessively inherited trait. Reevaluation of the genetic screen indicates that Nikoleta is homozygous dominant normal for a trait that causes low platelet count and her husband is homozygous recessive. If the F1 offspring has normal platelets and beta-thalassemia (15pts). a. What are the parental gametes? (2pts) b. Based on the F1 progeny phenotype given and the parental gametes, write the two possible genotypes. (2 pts) c. What is the gamete frequency if the two genotypes in (b) were crossed with each other? (4pts)? (4pts) d. What is the phenotype frequency for the cross in (c)? SHOW ALL WORK. (7 pts) Solution Both the conditions are recessive. Nikoleta: Homozygous dominant for trait that causes low platelet, and homozygous recessive for beta thalessemia. Genotype is LLbb Husband: Homozygous recessive for low platelet count, and heterozygous dominant for beta- thalessemia. Genotype is llBb The possible gametes produced by these parents: Nikoleta: Lb Husband: lB, lb Possible genotype of the F1 offspring: LlBb or Llbb Since the F1 offspring are given to be affected with beta-talassemai, the F1 offspring genotype should be Llbb. Possible gametes are LB, lb, lB, and lb (if genotype is LlBb) / Lb, lb (if genotype is Llbb). Therefore, the frequency of the gametes is LB = 1, lb = 2, lB =1, and Lb = 2 The following is the phenotype frequency: LLBb, LlBb, LlBb (Normal platelet count and no beta thallasemia) = 3 LLbb, Llbb, Llbb (Normal platelet count and beta thallasemia) = 3 llBb (Low platelet count and no beta thallasemia) = 1 llbb (Low platelet count and beta thallasemia) = 1 Thus phenotypic ratio stands as: Normal platelet count and no beta thallasemia : Normal platelet count and beta thallasemia : Low platelet count and no beta thallasemia : Low platelet count and beta thallasemia = 3 : 3 : 1 : 1.

1. Low platelet count is a recessively inherited trait. Reevaluation.pdf

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