7. In a population of 1000 oats, there is a single Mendelian locus.pdf
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7. In a population of 1000 oats, there is a single Mendelian locus with two alleles O1 and O2 that are codominant. Initially, 360 individuals have genotype O1O1, 480 have genotype O1O2, and 160 have genotype O2O2. What are the initial allele frequencies in this generation (lets call it Generation 1)? Frequency of O1 = and frequency of O2 = . 8. In the same population of oats, not all individuals survive to reproduce equally. Specifically, only half of the O1O1 individuals survive, all of the heterozygotes survive, and none of the O2O2 individuals survive to adulthood. What will the genotype frequencies be in this same generation, Generation 1, but after the selection event? Give you answer to the nearest hundredth (two places past the decimal). Frequency of O1O1 = , frequency of O1O2 = , and frequency of O2O2 = ..
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A geneticist discovers an obese mouse in his laboratory colony. He br.pdf
A geneticist discovers an obese mouse in his laboratory colony. He breeds this obese mouse
with a normal mouse. All the F1 mice from this cross are normal in size. When he interbreeds
two F1 mice, eight of the F2 mice are normal in size and two are obese. The geneticist then
intercrosses two of his obese mice, and he finds that all of the progeny from this cross are obese.
These results lead the geneticist to conclude that obesity in mice results from a recessive allele.
A second geneticist at a different university also discovers an obese mouse in her laboratory
colony. She carries out the same crosses as the first geneticist and obtains the same results. She
also concludes that obesity in mice results from a recessive allele. One day, the two geneticists
meet at a genetics conference, learn of each other\'s experiments, and decide to exchange mice.
They both find that, when they cross two obese mice from the different laboratories, all the
offspring are normal. However, when they cross two obese mice from the same laboratory all the
offspring are obese. Explain their results by drawing crossing diagrams and assigning genotypes
and phenotypes to each parent and offspring.
Solution
The first scientist has detected an obese allele ---> recessive
His identified alleles ---> o1 (recessive) & the normal allele ---> O1
The series crosses; the obese allele has produce recessive
Cross-1: possible genotype
Phenotypes
Obese recessive (o1o1) × Normal (O1O1) --->All normal (O1o1) (F1-generation)
Cross-2: possible genotypes: The following cross is between F1 offspring
Normal (Oo1) × normal (O1o1) ---> total eight normal genotypes includes O1O1 and O1o1 &
two obese with genotypes o1o1
Cross 3: genotypes:
Homozygous recessive obese (o1o1) × Obese (o1o1) ---> all are obese with genotypes “o1o1”
The second geneticist has depicted that the obese mouse in colony --> obese allele as o2
(recessive) and the normal allele as O2 (dominant).
It has found that the cross between mice in different laboratories generated normal mice even
though gene loci different alleles are both recessive therefore, it can be concluded that the obese
mice of different laboratories associated with “obese genes” mainly independent of each other
The prospective genotypes of the obese mice are:
Parents crosses: phenotype Obese mouse -1 (o1o1O2O2) × phenotype Obese mouse 2
(O1O1o2o2) --> All are normal mice with genotypes O1o1O2o2 in F1 generation.
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Foundations of Biological Sciences I Evolutionary Agents - 1 .docx
Foundations of Biological Sciences I Evolutionary Agents - 1
A quick recap….
There are several terms that need to be clarified so that you can more easily follow the exercise. A gene is a
piece of DNA that directs the expression of a particular characteristic (trait). Genes are located on
chromosomes, and the location where a particular gene is found is referred to as the locus (plural: loci) of that
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organism is a description of the way a trait is displayed in the structure, behavior, or physiology of the organism.
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by uppercase letters (e.g., “A”). The alleles that are masked are called recessive alleles. The recessive condition
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expression A). When both recessives are present in the genotype (aa), the organism is said to be homozygous
recessive for the trait, and the organisms will show the recessive phenotype (trait expression a). In the case of
complete dominance, the dominant allele completely masks the recessive allele, and an organism with a
heterozygous genotype (Aa) will show the dominant phenotype (trait expression A).
Evolutionary Agents
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time. The mechanisms of evolution include selection (which can cause change over time & adaptation), and
forces that provide variation and cause change over time (but not adaptation). Factors that change gene
frequencies over time are referred to as evolutionary agents.
A powerful way to detect the presence of evolutionary agents is the use of the Hardy–Weinberg model. This
model can be applied to traits that are influenced by several loci; the simplest case is for a trait that is regulated
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A geneticist discovers an obese mouse in his laboratory colony. He br.pdf
A geneticist discovers an obese mouse in his laboratory colony. He breeds this obese mouse
with a normal mouse. All the F1 mice from this cross are normal in size. When he interbreeds
two F1 mice, eight of the F2 mice are normal in size and two are obese. The geneticist then
intercrosses two of his obese mice, and he finds that all of the progeny from this cross are obese.
These results lead the geneticist to conclude that obesity in mice results from a recessive allele.
A second geneticist at a different university also discovers an obese mouse in her laboratory
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also concludes that obesity in mice results from a recessive allele. One day, the two geneticists
meet at a genetics conference, learn of each other\'s experiments, and decide to exchange mice.
They both find that, when they cross two obese mice from the different laboratories, all the
offspring are normal. However, when they cross two obese mice from the same laboratory all the
offspring are obese. Explain their results by drawing crossing diagrams and assigning genotypes
and phenotypes to each parent and offspring.
Solution
The first scientist has detected an obese allele ---> recessive
His identified alleles ---> o1 (recessive) & the normal allele ---> O1
The series crosses; the obese allele has produce recessive
Cross-1: possible genotype
Phenotypes
Obese recessive (o1o1) × Normal (O1O1) --->All normal (O1o1) (F1-generation)
Cross-2: possible genotypes: The following cross is between F1 offspring
Normal (Oo1) × normal (O1o1) ---> total eight normal genotypes includes O1O1 and O1o1 &
two obese with genotypes o1o1
Cross 3: genotypes:
Homozygous recessive obese (o1o1) × Obese (o1o1) ---> all are obese with genotypes “o1o1”
The second geneticist has depicted that the obese mouse in colony --> obese allele as o2
(recessive) and the normal allele as O2 (dominant).
It has found that the cross between mice in different laboratories generated normal mice even
though gene loci different alleles are both recessive therefore, it can be concluded that the obese
mice of different laboratories associated with “obese genes” mainly independent of each other
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Parents crosses: phenotype Obese mouse -1 (o1o1O2O2) × phenotype Obese mouse 2
(O1O1o2o2) --> All are normal mice with genotypes O1o1O2o2 in F1 generation.
Biodiversity Jewett
BioDiversity & Animal Diversity lecture covering from Campbell & Reece "Biology" 8th edition (International). For ACS Biology 10, Sofia Bulgaria. March 2010
Powerpoint tips for scientists
Powerpoint tips for researchers, scientists, and science students.
More info at https://sciencetell.wordpress.com/
Foundations of Biological Sciences I Evolutionary Agents - 1 .docx
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A quick recap….
There are several terms that need to be clarified so that you can more easily follow the exercise. A gene is a
piece of DNA that directs the expression of a particular characteristic (trait). Genes are located on
chromosomes, and the location where a particular gene is found is referred to as the locus (plural: loci) of that
gene. An allele is a gene for which there is an alternative expression, which can lead to the alterative form of a
trait. For example, a diploid organism carries the allele “A” on one homologous chromosome, and the allele “A”
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carry two different alleles. For example on one chromosome it could carry the allele “A” and on the other it
could carry the allele “a”. The genotype of such an organisms is then Aa, and it is described as heterozygous for
this chromosomal locus.
The genotype of an organism is the listing of the two alleles for each trait that it possesses. The phenotype of an
organism is a description of the way a trait is displayed in the structure, behavior, or physiology of the organism.
Some alleles are dominant to others and mask the presence of other alleles. The dominant condition is indicated
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is indicated by lowercase letters (e.g., “a”). When both dominants are present in the genotype (AA), the organism
is said to be homozygous dominant for the trait, and the organisms will show the dominant phenotype (trait
expression A). When both recessives are present in the genotype (aa), the organism is said to be homozygous
recessive for the trait, and the organisms will show the recessive phenotype (trait expression a). In the case of
complete dominance, the dominant allele completely masks the recessive allele, and an organism with a
heterozygous genotype (Aa) will show the dominant phenotype (trait expression A).
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Evolution is a process resulting in changes in gene frequencies (= the genetic make-up) of a population over
time. The mechanisms of evolution include selection (which can cause change over time & adaptation), and
forces that provide variation and cause change over time (but not adaptation). Factors that change gene
frequencies over time are referred to as evolutionary agents.
A powerful way to detect the presence of evolutionary agents is the use of the Hardy–Weinberg model. This
model can be applied to traits that are influenced by several loci; the simplest case is for a trait that is regulated
by one locus with two alleles.
With the Hardy–Weinberg model, the frequency of genotypes in the population can be predicted from the
probability of encounters between gametes bearing the different alleles. With alleles R .
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5.2 Par�s a Mosc�. En su programa de estudios de verano en el extr.pdf
5.2
Pars a Mosc. En su programa de estudios de verano en el extranjero en Europa, se queda dos
semanas ms para viajar de Pars a Mosc. Te vas de Pars con 2 coma 200 euros (euro o EUR) en tu
rionera. Al querer cambiar todo esto por rublos rusos (R o RUB), obtiene las siguientes
cotizaciones:
Tasa al contado
USD es igual a 1,00 EUR (o $/euro) 1,1187
Tasa al contado RUBequals=1.00 USD (o R/$) 61.96
a. Cul es el tipo de cambio del rublo ruso al euro?
b. Cuntos rublos rusos obtendrs por tus euros?.
5. Cyanobacteria live mostly in aquatic habitats but they can also b.pdf
5. Cyanobacteria live mostly in aquatic habitats but they can also be found on landin deserts of
all places! They lie dormant for much of the year, as it is so dry, but when it rains they become
active metabolically. Although rain is good for them, it is potentially harmful for us, because
these desert cyanobacteria produce harmful toxins that can be kicked up in dust storms. One
might guess that the concentration of these harmful toxins is highest in soil just after it rains. To
test this, imagine running an experiment in which you have four treatment groups: 1.)
intact/unwatered; 2.) disturbed/unwatered; 3.) intact/watered; and 4.) disturbed/wateredwhere
intact means that the surface layer of desert soil is left in place and disturbed means that 0.5cm
from the surface has been scraped off, and watered versus unwatered means what it sounds like.
You have a sample size of 5 in each treatment group. After ten days of watering or not watering,
you measure the average concentration of the toxin in each treatment group and compare the four
averages. The null hypothesis is that the mean toxin concentrations are equal across the four
treatment groups. You perform a Kruskal-Wallis H test, which is a null-hypothesis significance
test. The critical value for this experiment at the 0.05 level is H = 7.4; at the 0.01 level it is H =
9.8. For your data, the observed H = 10.6. First, in light of these results, what decision should
you make, and second, how would you write that result up in a single sentence for the Results
section of a scientific paper?.
5. The Federal Reserves organization1) There are (5712) Feder.pdf
5. The Federal Reserve's organization
1) There are (5/7/12) Federal Reserve regional banks.
2) The Federal Reserve's role as a lender of last resort involves lending to which of the following
financially troubled institutions?
- U.S. banks that cannot borrow elsewhere
- U.S. state governments when they run short on tax revenues
- Governments in developing countries during currency crises
3) The Federal Reserve's primary tool for changing the money supply is ( the discount rate/the
reserve requirement/open market operations) . In order to decrease the number of dollars in the
U.S. economy (the money supply), the Federal Reserve will (buy/sell) government bonds..
5. Mrs. Toguchi wants to determine if politicians tend to have highe.pdf
5. Mrs. Toguchi wants to determine if politicians tend to have higher salaries, on average, than
first responders (specifically emergency medical technicians, EMTs). Independent random
samples from each of these two professions were selected and their salaries were recorded. The
data are summarized below.
Construct and interpret a 99% confidence interval for the difference in the true mean salaries for
politicians and EMTs..
41. a and b 41. Introductory Statistics Course Withdrawals. A univer.pdf
41. a and b 41. Introductory Statistics Course Withdrawals. A university found that 20% of its
students withdraw without completing the introductory statistics course. Assume that 20 students
registered for the course. a. Compute the probability that 2 or fewer will withdraw. b. Compute
the probability that exactly 4 will withdraw. c. Compute the probability that more than 3 will
withdraw. d. Compute the expected number of withdrawals..
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41) La justificacin legal como nico criterio tico es adecuada debido a cul de los siguientes?
R. Algunas conductas poco ticas son legales.
B. La ley siempre se basa en definiciones precisas.
C. La ley encarna muchos de los principios morales de un pas.
D. La ley tarda en desarrollarse en reas emergentes.
E. El comportamiento legal es siempre un comportamiento tico..
40. What is the common challenge in the world of financea. It gen.pdf
40. What is the common challenge in the world of finance?
a. It generally operates with a high degree of uncertainty
b. It causes investor anxiety due to a volatile stock market
c. It is difficult to abide bt generally accepted accounting principles (GAAP)
d. It must comply with banking regulations that are subject to change
It might be A, but I am not sure..
4.a. Calculate Target�s current ratio as of January 29, 2022 (show cur.pdf
4.a. Calculate Targets current ratio as of January 29, 2022 (show current assets, current liabilities
and the calculated current ratio) See accompanying Notes to Consolidated Financial Statements..
5. Earth HistoryPaleogeography - World. Check and double-click the .pdf
5. Earth History/Paleogeography - World. Check and double-click the Problem 5 placemark to
advance time to approximately 270-280 Ma. A major continent-continent collision is occurring
to form what supercontinent around this time? Review the continental configurations in Chapter
11 of your textbook and their corresponding ages.
a. Rodinia
b. Pangaea
c. Laurentia
d. Gondwana.
5. Design and implement a method contains 2 for BinarySearchTree, fu.pdf
5. Design and implement a method contains 2 for BinarySearchTree, functionall equivalent to
the contains method, that does not use recursion.
show the driver code and do this in java
this is the binarysearchtree
import java.util.*; // Iterator, Comparator
public class BinarySearchTree implements BSTInterface
{
protected BSTNode root; // reference to the root of this BST
protected Comparator comp; // used for all comparisons
protected boolean found; // used by remove
public BinarySearchTree()
// Precondition: T implements Comparable
// Creates an empty BST object - uses the natural order of elements.
{
root = null;
comp = new Comparator()
{
public int compare(T element1, T element2)
{
return ((Comparable)element1).compareTo(element2);
}
};
}
public BinarySearchTree(Comparator comp)
// Creates an empty BST object - uses Comparator comp for order
// of elements.
{
root = null;
this.comp = comp;
}
public boolean isFull()
// Returns false; this link-based BST is never full.
{
return false;
}
public boolean isEmpty()
// Returns true if this BST is empty; otherwise, returns false.
{
return (root == null);
}
public T min()
// If this BST is empty, returns null;
// otherwise returns the smallest element of the tree.
{
if (isEmpty())
return null;
else
{
BSTNode node = root;
while (node.getLeft() != null)
node = node.getLeft();
return node.getInfo();
}
}
public T max()
// If this BST is empty, returns null;
// otherwise returns the largest element of the tree.
{
if (isEmpty())
return null;
else
{
BSTNode node = root;
while (node.getRight() != null)
node = node.getRight();
return node.getInfo();
}
}
private int recSize(BSTNode node)
// Returns the number of elements in subtree rooted at node.
{
if (node == null)
return 0;
else
return 1 + recSize(node.getLeft()) + recSize(node.getRight());
}
public int size()
// Returns the number of elements in this BST.
{
return recSize(root);
}
public int size2()
// Returns the number of elements in this BST.
{
int count = 0;
if (root != null)
{
LinkedStack> nodeStack = new LinkedStack>();
BSTNode currNode;
nodeStack.push(root);
while (!nodeStack.isEmpty())
{
currNode = nodeStack.top();
nodeStack.pop();
count++;
if (currNode.getLeft() != null)
nodeStack.push(currNode.getLeft());
if (currNode.getRight() != null)
nodeStack.push(currNode.getRight());
}
}
return count;
}
private boolean recContains(T target, BSTNode node)
// Returns true if the subtree rooted at node contains info i such that
// comp.compare(target, i) == 0; otherwise, returns false.
{
if (node == null)
return false; // target is not found
else if (comp.compare(target, node.getInfo()) < 0)
return recContains(target, node.getLeft()); // Search left subtree
else if (comp.compare(target, node.getInfo()) > 0)
return recContains(target, node.getRight()); // Search right subtree
else
return true; // target is found
}
public boolean contains (T target)
// Returns true if this BST contains a node with info i such that
// comp.compare(target, i) == 0; otherwise, re.
7. Para comprender mejor la prehistoria de la isla hawaiana de Lana.pdf
7. Para comprender mejor la prehistoria de la isla hawaiana de Lana'i, los antroplogos Maria
Sweeney, Melinda Allen y Boyd Dixon utilizaron la datacin por radiocarbono del carbn
encontrado en una antigua vivienda, el Monumento Histrico Nacional de Kaunolu Village, el
monumento arqueolgico ms grande complejo en la isla. En una de sus muestras, encontraron que
quedaba aproximadamente el 94% del carbono 14 original. Utilizando el hecho de que el
carbono 14 se descompone en un 1,202 % cada 100 aos, determine la edad aproximada de esta
muestra.
a. En su documento de Word, pegue la tabla de Excel que hizo para obtener su respuesta.
b. Usando logaritmos, estime la edad de la muestra. Muestra al menos 2 decimales en tu
respuesta..
7. Lawn Spray Inc. desarrolla y produce equipos de pulverizaci�n.pdf
7.
Lawn Spray Inc. desarrolla y produce equipos de pulverizacin para el mantenimiento del csped y
usos industriales. El 31 de enero del ao en curso, Lawn Spray Inc. readquiri 18,000 acciones de
su capital social
Las acciones en circulacin cuando una corporacin ha emitido solo una clase de acciones.
a $19 por accin. El 14 de junio se vendieron 13,500 de las acciones readquiridas a $27 por accin
y el 23 de noviembre se vendieron 3,200 de las acciones readquiridas a $20.
Requerido:
Capital aportado a una corporacin por los accionistas y otros.
CATLOGO DE CUENTAS rociador de csped inc. Libro mayor ACTIVOS 110 Dinero 120
Cuentas por cobrar 131 Documentos por cobrar 132 Intereses por cobrar 141 Inventario de
mercanca 145 Material de oficina 151 Seguro prepagado 181 Tierra 193 Equipo 194 Depreciacin
Acumulada-Equipos PASIVO 210 Cuentas por pagar 221 Pagar 226 Los intereses a pagar 231
Dividendos en Efectivo por Pagar 241 Salarios por Pagar 261 Pagar hipotecario EQUIDAD 311
Acciones comunes 312 Capital pagado en exceso de la par-acciones ordinarias 315 Autocartera
321 Acciones preferentes 322 Capital pagado en exceso de acciones preferentes a la par 331
Capital pagado por venta de acciones propias 340 Ganancias retenidas 350 Dividendos en
acciones distribuibles 351 Dividendos en efectivo 352 Dividendos en acciones 390 Resumen de
ingresos.
701 () � Kenneth_Laudon,_J... a a+ as Part One Ogriesions, Mrugemer.pdf
7:01 () Kenneth_Laudon,_J... a a+: as Part One Ogriesions, Mrugemere, snd the Netacked
Enterprise.
71 of all students at a college still need to take another math class.pdf
71% of all students at a college still need to take another math class. If 5 students are randomly
selected, find the probability that a. Exactly 2 of them need to take another math class. b. At most
2 of them need to take another math class. c. At least 3 of them need to take another math class.
d. Between 2 and 4 (including 2 and 4) of them need to take another math class. Round all
answers to 4 decimal places..
7.4.3 4.3 Consider a set of independent data observations x1,�.pdf
7.4.3 4.3 Consider a set of independent data observations x1,,xn that have an exponential
distribution with an unknown parameter . Show that the method of moments and maximum
likelihood estimation both produce the point estimate =x1.
7.16 LAB Smallest and largest numbers in a listWrite a program th.pdf
7.16 LAB: Smallest and largest numbers in a list
Write a program that reads a list of integers into a list as long as the integers are greater than
zero, then outputs the smallest and largest integers in the list.
Ex: If the input is:
the output is:
You can assume that the list of integers will have at least 2 values.
I keep getting an error "Traceback (most recent call last): File "main.py", line 8, in user_input =
int(input()) ValueError: invalid literal for int() with base 10: '10 5 3 21 2 -6'".
7. El balance general de Tucker Company reflejaba activos por $12,.pdf
7.
El balance general de Tucker Company reflejaba activos por $12,000, pasivos por $4,000 y
capital comn.
acciones de $3,000 al 31 de diciembre de 2014. Si las ganancias retenidas al 31 de diciembre de
2015 el saldo
hoja es de $8,000 y Tucker pag un dividendo de $2,000 durante 2015, entonces el monto de la
utilidad neta
para 2015 fue cul de los siguientes?
a.$1,000
b. $3,000
c.$5,000
d. Ninguna de las anteriores.
4.3 Expresi�n positiva y cort�s Revise las siguientes declaracio.pdf
4.3 Expresin positiva y corts
Revise las siguientes declaraciones para que sean ms positivas y corteses.
a) La construccin de su edificio est paralizada porque el contratista no puede verter los cimientos
hasta que el suelo ya no est empapado.
B) No se puede emitir un pasaporte hasta que se complete una solicitud y se incluya una foto
reciente.
4.4 Lenguaje libre de prejuicios
Reduzca el sesgo de las siguientes declaraciones.
En 18 o ms estados, un empleado tiene derecho a ver su registro de empleado.
4.5 Lenguaje sencillo y palabra familiar
Use un lenguaje sencillo y palabras familiares para las siguientes declaraciones.
a) El escritor trat de ofuscar el tema con datos extraos y superfluos.
4.6 seleccionar las palabras ms precisas y vigorosas
a) Si te encuentras (teniendo, haciendo, haciendo malabarismos) con muchas tareas, busca
formas de reducir tu participacin.
b) El excelente informe de Rana contiene (muchos, muchos, un almacn de) datos tiles..
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