How do the microsporangium and microspore in seed plants differ from.pdf

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How do the microsporangium and microspore in seed plants differ from primitive heterosporous plants? Solution Microsporangium and Microspores in seed plants: Microspores are the first cells to form into the male gametophyte and give rise to pollen grains. The megaspores are formed within the developing seed. Microsporangia develop in pairs toward the bases of the scales. Each of the microsporocytes in the microsporangia undergoes meiosis, producing four haploid microspores. These then develop into pollen grains and each grain consists of four cells and a pair of external air sacs. Primitive Heterosporous plants: The small female gametophyte formed on germination of the megaspore produces one or few archegonia in a very short time. The microspore also develops rapidly, and spermatozoids are soon liberated from the male gametophyte..

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2. Sketch the life cycle of a conifer. Include all relevant terms from p.10 of this chapter. One characteristic of Pinophytes (and many other non-flowering plants) worthy of note is their cones (which are actually strobili, the same structure observed in vascular non-seed plants like club mosses and horsetails). Non-flowering seed plants typically produce both male and female cones (the plants may either be monoecious or dioecious, depending on the lineage). Both male and female cones are formed of many scales (leaf-like structures) surrounding a central axis (the stem-like structure the scales emerge from); microsporangia or megasporangia are found at the base of the scales. The male cones contain numerous microsporangia that produce haploid microspores (there are typically hundreds of microspores produced per microsporangium; a microsporangium typically ends up looking like a large hollow sac containing hundreds of tiny free-floating microspores). These microspores grow into small (four-celled) microgametophytes; when the microgametophytes are mature they are called pollen grains. Female cones have ovules in them; each ovule contains a megasporangium surrounded by integument. The ovules of the female cones are typically found at the base of the cone's scales (which are typically much larger and stronger than the scales of male cones). The megasporangium produces megaspores that grow into megagametophytes. Each megagametophyte houses at least one archegonium, which develops an egg that is fertilized by the sperm from a pollen grain. The fertilized egg becomes a zygote, which eventually develops into a embryo (that is diploid) surrounded by the tissue of the meqagametophyte develops into a embryo (that is diploid) surrounded by the tissue of the megagametophyte (which develops into haploid nutritive tissue) and the megasporangium (which develops into the seed coat). A mature seed is composed of a seed coat surrounding nutritive tissue and the embryo; gymnosperm seeds do not have endosperm, and are not contained in a fruit. The seeds are released into the environment, where they can germinate and grow into a sporophyte plant..

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Homework question with 2 parts A. Compare and contrast the geologic and atmospheric characteristics of the following two moons: Miranda and Triton. B. In what respects is Pluto more like a moon than a Jovian or terrestrial planet? Solution A. Geologically triton made of solid surface and an evidance of geological activity as IOs and its surface young with sparse impart craters. Spectroscopic studies reveasl that Triton consists of thin nitrogen and methane atmosphere.Mirinda atmosphere composes of half of ice and half of rocky materials (silicate rock) at its center. The geological activity caused by tidal heating. B. Size of Pluto is very less(1/6 th mass of our moons). Pluto is more like a moon than jovian planets.Like terristrial planets, Pluto does not consists of rocky (solid) surface and like jovian planets, pluto does not consists of gases and liquids composition ..

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Client Business Risk The risk that the client will fail to achieve its obiectives. Auditor\'s primary concern is: The risk of material misstatements in the financial statements due to client business risk. Management is the primary source for identifying client business risks. Explain and provide examples Solution Auditors primary concern is \" The risk of material misstatements in the financial statements due to clients business risk.\" In such a situation, Management is the primary source for identifying client business risks because management is the ultimate responsible persons in the concern for day to day affairs of the business. Examples: Risks of material misstatement can arise from a variety of sources, including external factors, such as conditions in the company\'s industry and environment, and company-specific factors, such as the nature of the company, its activities, and internal control over financial reporting. For example, external or company-specific factors can affect the judgments involved in determining accounting estimates or create pressures to manipulate the financial statements to achieve certain financial targets. Also, risks of material misstatement may relate to, e.g., personnel who lack the necessary financial reporting competencies, information systems that fail to accurately capture business transactions, or financial reporting processes that are not adequately aligned with the requirements in the applicable financial reporting framework..

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create a new interface called DropoutStackADT for representing a dropout stack. It should contain exactly the same methods as StackADT, and have documentation for the file and each method in proper javadoc. Your file will be different from StackADT since a dropout stack acts differently than a normal stack. When documenting methods that should support throwing an exception, make a note in its javadocdescription. Implement a drop-out stack using a linked structure. You use the LinearNode or LinearDoubleNode classes. Follow the DropoutStackADT interface. See Base_A11Q3.java for a starting place. STACKADT LinearDoubleNode /** * Represents a node in a linked list. * * @author Java Foundations * @version 4.0 */ public class LinearDoubleNode { private LinearDoubleNode next,prev; private T element; /** * Creates an empty node. */ public LinearDoubleNode() { next = null; element = null; prev = null; } /** * Creates a node storing the specified element. * @param elem element to be stored */ public LinearDoubleNode(T elem) { next = null; element = elem; prev = null; } /** * Returns the node that follows this one. * @return reference to next node */ public LinearDoubleNode getNext() { return next; } /** * Sets the node that follows this one. * @param node node to follow this one */ public void setNext(LinearDoubleNode node) { next = node; } /** * Returns the element stored in this node. * @return element stored at the node */ public T getElement() { return element; } /** * Sets the element stored in this node. * @param elem element to be stored at this node */ public void setElement(T elem) { element = elem; } public LinearDoubleNode getPrev(){ return prev; } public void setPrev(LinearDoubleNode node){ prev = node; } } Base File DROP-OUT STACK TESTING The size of the stack is: 4 The stack contains: The size of the stack is 4 The stack contains: Solution public class DropOutStack implements StackADT { private final int DEFAULT_CAPACITY = 10; // refers to the item in the array being indexed private int cIndex; // holds the count of the stack private int count; private T[] stack; public DropOutStack() { cIndex = count = 0; stack = (T[]) (new Object[DEFAULT_CAPACITY]); } public DropOutStack(int initialCapacity) { cIndex = count = 0; stack = (T[]) (new Object[initialCapacity]); } public T peek() throws EmptyCollectionException { if (isEmpty()) throw new EmptyCollectionException(\"Drop-out Stack\"); return stack[(cIndex + stack.length - 1)%stack.length]; } public T pop() throws EmptyCollectionException { if (isEmpty()) throw new EmptyCollectionException(\"Drop-out Stack\"); cIndex = (cIndex + stack.length - 1)%stack.length; T result = stack[cIndex]; stack[cIndex] = null; count--; return result; } public void push(T element) { cIndex = cIndex % stack.length; stack[cIndex] = element; cIndex++; if (count != stack.length) count++; } public int size() { return count; } public boolean isEmpty() { return size() == 0; } public String toString() { String str = \"\"; i.

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Translate the following into English. Let C denote an arbitrary collection of individuals (we use different brackets to aid in pairing them up.) Vk eNrzhe N |(ICI > n) S C Cr (ISI > k) ((Vz, ye S(Fiz. yl)V (tr.ue S. ,r y(F(z.yn Solution It states that for all k innatural numbers there exist a natural number n such that cardinality of C 8s greater than or equal to n . This implies that there exist a subset S of C where cardinality of S is greater than equal to k and for all x,y in S define F(x,y) or for all x,y in S x is not same as y and is nit in the family of yF(x,y).

Translate the following into English. Let C denote an arbitrary coll.pdf

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The force on an object is F = - 21j. For the vector v = 2 i - 2j, find: The component of F parallel to v: The component of F perpendicular to v: The work, W, done by force F through displacement v: Solution Given that F=-21j V=2i-2j |v|={2^2+2^2} |v|=(4+4) |v|=8 |v|=22 The unit vector along v is n Then n= v/|v| n=(2i-2j)/22 n=(i-j)/2 |F|=21 a) the component F parallel to v is F_n F_n=|F|*n=-21*((i-j)/2) F_n=-21*n b) the component F perpendicular to v Let this component be (F_n)`,then F_n+(F_n)`=F F_n)`=F-F_n (F_n)`=-21j+{(21/2)*n} =-21j+{(21/2)*((i-j)/2)} =-21j+{(21/2)*(i-j)} =-21j+((21i-21j)/2) ={(-21j*2)+(21i-21j)}/2 =(-42j+21i-21j)/2 =(21i-63j)/2 C) the work done by F through displacement v W=F.v W=(-21j).(2i-2j) W=(-21)*(-2) W=42.

The force on an object is F = - 21j. For the vector v = 2 i - 2j, fin.pdf

f3apparelsonline

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