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Mrrafil [7]
4 years ago
15

A dog breeder liked yellow and brown labrador retrievers. in an attempt to produce yellow and brown puppies, he mated a yellow l

abrador male and a brown labrador female. unfortunately, all the puppies produced in this cross were black. (see pp. 117–119 for a discussion of the genetic basis of coat color in labrador retrievers.)
a. explain this result.
b. how might the breeder go about producing yellow and brown labradors?
Biology
1 answer:
adell [148]4 years ago
3 0

For question a, Labrador retrievers vary in two loci, B and E. Black dogs have dominant alleles at both loci (B-E-), brown dogs have bbE-, and yellow dogs have B-ee or bbee. Since all the puppies were black, a dominant E allele from the brown parent and they must all have inherited a dominant B allele from the yellow parent. The yellow male must have been BBee, and the brown female parent must have been bbEE. The black puppies were all BbEe.


For question b, simply mating yellow with yellow will produce all yellow Labrador puppies. If at least one of the parents is homozygous EE, mating two brown Labradors will produce either all brown puppies, or ¾ brown and ¼ yellow if both parents are heterozygous Ee.

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In cats, the Dense pigment gene, D/d, codes for melanophilin, a protein involved in the transportation and deposition of pigment
SVETLANKA909090 [29]

Answer:

a. M_D_ : 9/16 Black furred individuals

b. M__dd : 3/16 gray furred individuals (1/16 MMdd + 2/16 Mmdd)

c. mmD_ : 3/16 Brow-furred individuals (1/16 mmDD + 2/16 mmDd)

d. mmdd: 1/16 gray furred individuals.

Explanation:

Available data:

  • Dense pigment gene, D, codes for melanophilin.
  • D allele dominant over d.
  • Gene M/m is responsible for the production of the pigment
  • M is dominant over m
  • If D is present, M/m is expressed
  • If dd is present, M/m is not expressed  
  • DDMM, DdMM, DDMm, DdMm --->Black fur
  • DDmm, Ddmm ---> Brown fur
  • ddMM, ddMm -----> Light gray fur

Cross: between two dihybrids

Parental)         DdMm     x     DdMm

Phenotype) Black-furred      Black-furred

Gametes) DM, Dm, dM, dm   DM, Dm, dM, dm

Punnet Square)    DM        Dm         dM          dm

             DM       DDMM    DDMm   DdMM   DdMm

             Dm       DDMm    DDmm   DdMm   Ddmm

             dM        DdMM    DdMm    ddMM   ddMm

             dm        DdMm    Ddmm    ddMm   ddmm

F1) 9/16 Black-furred individuals, D-M- (1/16 DDMM + 2/16 DDMm + 2/16    

                                                                 DdMM + 4/16 DdMm)

    3/16 Brown-furred individuals, D-mm ( 1/16 DDmm + 2/16 Ddmm)

    4/16 Gray-furred individuals, dd-- ( 1/16ddMM + 1/16 ddmm + 2/16 ddMm)

   

   

7 0
3 years ago
As an analogy to gene expression in eukaryotes, let's compare it to the process of cooking recipes from a library of cookbooks.
scoundrel [369]
Two major steps are needed to obtain a protein from a gene. Transcription makes it possible to obtain a copy of the gene present on the chromosome, in the form of a messenger RNA (mRNA). Translation is the stage during which this mRNA is read by the ribosome to obtain a protein.
DNA replication is associated with photocopy because replication will give an identical copy of the cell DNA.
Transcription is associated with book printing because in transcription the genetic information will be transferred to another platform (RNA)
Nucleus in a Library since this is where the "book printing" occurs, and where the books (DNA) are
DNA is the book, where all the information are
RNA transcript is the photocopy
Translation is the cooking since it is the protein synthesis
Proteins/polypeptide is the cooking objective, so it is the prepared dish
RNA polymerase is the Xerox machine, it will copy the information in the DNA to RNA transcript
Ribosomes is the cooks, because this is where translation occurs
Activator protein are the bookmarks because it choose wich part of DNA it has to be transcript.
6 0
3 years ago
Fully explain how cellular respiration is a chemical process whereby the bonds of food molecules and oxygen molecules are broken
Marianna [84]
Cellular respiration is a metabolic pathway that breaks down glucose and produces ATP. The stages of cellular respiration include glycolysis, pyruvate oxidation, the citric acid or Krebs cycle, and oxidative phosphorylation.

During cellular respiration, a glucose molecule is gradually broken down into carbon dioxide and water. Along the way, some ATP is produced directly in the reactions that transform glucose. Much more ATP, however, is produced later in a process called oxidative phosphorylation. Oxidative phosphorylation is powered by the movement of electrons through the electron transport chain, a series of proteins embedded in the inner membrane of the mitochondrion.
These electrons come originally from glucose and are shuttled to the electron transport chain when they gain electrons.


As electrons move down the chain, energy is released and used to pump protons out of the matrix, forming a gradient. Protons flow back into the matrix through an enzyme called ATP synthase, making ATP. At the end of the electron transport chain, oxygen accepts electrons and takes up protons to form water. Glycolysis can take place without oxygen in a process called fermentation. The other three stages of cellular respiration—pyruvate oxidation, the citric acid cycle, and oxidative phosphorylation—require oxygen in order to occur. Only oxidative phosphorylation uses oxygen directly, but the other two stages can't run without oxidative phosphorylation.). As electrons move down the chain, energy is released and used to pump protons out of the matrix, forming a gradient. Protons flow back into the matrix through an enzyme called ATP synthase, making ATP. At the end of the electron transport chain, oxygen accepts electrons and takes up protons to form water.
Glycolysis can take place without oxygen in a process called fermentation. The other three stages of cellular respiration—pyruvate oxidation, the citric acid cycle, and oxidative phosphorylation—require oxygen in order to occur. Only oxidative phosphorylation uses oxygen directly, but the other two stages can't run without oxidative phosphorylation.
6 0
3 years ago
How would a tsunami most likely affect the abundance of aquatic organisms in the intertidal zone of the ocean?
AlexFokin [52]

Answer: C). A tsunami would decrease abundance by causing physical damage to the organisms and their ecosystems.

An intertidal zone is also known as littoral zone. It is the region which is covers the foreshore and seabed. This region is exposed to the affects of tides. This region is highly vulnerable to tsunamis. The tsunami will disturb the habitat of organisms living in intertidal zone. Therefore, will result in decrease in abundance of aquatic organisms.

5 0
3 years ago
Read 2 more answers
Material through which water readily flows is termed ____.
Simora [160]
Material through which water flows is permeable. This material can be either natural or artificial. Naturally permeable material can be clean sand and gravel for example, or a sandstone and in the latter case the sandstone can have both porosity which is not synonymous with permeability as the pores must be connected to be permeable. An impermeable membrane could be used to line ditches for example to prevent leakage near mines. 
3 0
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