Introduction:
During pre-embryonic stage, the zygote starts multiplying into more than hundreds of cells. These cells establish into primary germ layers, and the conceptus attaches to the uterine wall. This process is divided into cleavage, implantation, and embryogenesis.

Answer to Problem 1TYR
Correct answer:
The conceptus reaches the uterus about 72 hours after ovulation. It consists of more than 16 cells and resembles a mulberry. Hence, it is called a morula. Therefore, option b is correct.
Explanation of Solution
Justify reasons for the correct statement:
The morula is smaller than the zygote. Small blastomeres are produced by cleavage. Cleavage also initiates the production of large amounts of cells, by which the embryonic tissues are formed. Meanwhile, the morula produces identical twins and divides into two cell masses.
Option (b) is given as, “morula”.
In the uterine cavity, the morula lies freely for 4 to 5 days and starts multiplying into more than 100 cells.
Hence, option (b) is correct.
Justify reasons for the incorrect statements:
Option (a) is given as, “zygote”.
Zygote multiplies into hundreds of cells. These cells are established into three primary germ layers, and the conceptus is attached to the uterine wall firmly. Hence, it is a wrong answer.
Option (c) is given as, “blastomere”.
After fertilization, the first cleavage produces two daughter cells or blastomeres. These cells proliferate at shorter time intervals. Hence, the number of blastomeres is increased. Hence, it is a wrong answer.
Option (d) is given as, “blastocyst”.
During embryonic development, a hollow thin structure consists of a group of cells (inner cell masses) called blastocyst. The embryo arises from the blastocyst. Hence, it is a wrong answer.
Option (e) is given as, “embryo”.
Fertilized egg is called as an embryo. Hence, it is a wrong answer.
Hence, options (a), (c), (d), and (e) are incorrect.
Morula is a spheroidal stage, and it consists of more than 16 blastomeres.
Want to see more full solutions like this?
Chapter 29 Solutions
A&P UNITY AND FUNCTION ACCESS
- An 1100 pound equine patient was given 20 mg/kg sucralfate 3 times a day, 2.8 mg/kg famotidine twice a day, and 10mg/kg doxycycline twice a day. Sucralfate comes as a 1 gm tablet, famotidine as 20 mg tablets, and doxycycline as 100mg tablets. All are in bottles of 100 tablets.How many total mg are needed for the patient and how many tablets of each would be needed to provide each dose?How many bottles of each would be needed to have available if this patient were to be on this drug regimen for 5 days?arrow_forwardThe patient needs a solution of 2.5% dextrose in Lactated Ringer’s solution to run at 75 ml/hr for at least the next 12hours. LRS comes in fluid bags of 500 ml, 1 Liter, 3 Liters and 5 Liters. How can a 2.5% solution be made by adding50% dextrose to the LRS?arrow_forward“Gretchen” was a 68-pound canine who came to the VMTH as small animal surgery patient. She receivedacepromazine, 0.2 mg/kg from a 10 mg/ml solution and oxymorphone, 0.08 mg/kg from a 1 mg/ml solution before surgery.What are the mechanisms of action of acepromazine and oxymorphone? Why would they be given together?How many mg provide each dose and how many ml of each of these solutions were given?arrow_forward
- After surgery, “Gretchen” was put on carprofen, 1 mg/pound bid (twice a day). The tablets come in 25, 75 and 100 mgsizes. Which size tablet would be appropriate?What is the mechanism of action of carprofen?An outpatient prescription was written for her so she would have enough for 10 days. How many tablets did she need?What information needs to be on her out-patient prescription?arrow_forwardJoden Koepp olor in chickens is due to incomplete dominance. BB = Black chicken, WW = White BLOOD TYPES Arhite chicken is In humans, Rh positive blood is dominant (R) over Rh negative blood (r). A man with type 0, Rh positive blood (whose mother had Rh negative blood), marries a woman with type AB, Rh negative blood. Several children were born. is? R R Genotypes Phenotypes RRR RR Rr Rr 4/16 RR R RR RK Rr Rr 4/16 rr 3/4 Rh posi 1/4 Rh negu 1/2 Rr rr rr rrrr 88 888 75 e genotype of the man? the genotype of the woman? The mother of the man had type AB blood.arrow_forwardPlease indentify the unknown organismarrow_forward
- 5G JA ATTC 3 3 CTIA A1G5 5 GAAT I I3 3 CTIA AA5 Fig. 5-3: The Eco restriction site (left) would be cleaved at the locations indicated by the arrows. However, a SNP in the position shown in gray (right) would prevent cleavage at this site by EcoRI One of the SNPs in B. rapa is found within the Park14 locus and can be detected by RFLP analysis. The CT polymorphism is found in the intron of the Bra013780 gene found on Chromosome 1. The Park14 allele with the "C" in the SNP has two EcoRI sites and thus is cleaved twice by EcoRI If there is a "T" in that SNP, one of the EcoRI sites is altered, so the Park14 allele with the T in the SNP has only one EcoRI site (Fig. 5-3). Park14 allele with SNP(C) Park14 allele with SNPT) 839 EcoRI 1101 EcoRI 839 EcoRI Fig. 5.4: Schematic restriction maps of the two different Park14 alleles (1316 bp long) of B. rapa. Where on these maps is the CT SNP located? 90 The primers used to amplify the DNA at the Park14 locus (see Fig. 5 and Table 3 of Slankster et…arrow_forwardFrom your previous experiment, you found that this enhancer activates stripe 2 of eve expression. When you sequence this enhancer you find several binding sites for the gap gene, Giant. To test how Giant interacts with eve, you decide to remove all of the Giant binding sites from the eve enhancer. What results do you expect to see with respect to eve expression?arrow_forwardWhat experiment could you do to see if the maternal gene, bicoid, is sufficient to form anterior fates?arrow_forward
- Concepts of BiologyBiologyISBN:9781938168116Author:Samantha Fowler, Rebecca Roush, James WisePublisher:OpenStax CollegeBiology (MindTap Course List)BiologyISBN:9781337392938Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. BergPublisher:Cengage Learning
- Medical Terminology for Health Professions, Spira...Health & NutritionISBN:9781305634350Author:Ann Ehrlich, Carol L. Schroeder, Laura Ehrlich, Katrina A. SchroederPublisher:Cengage Learning



