If Tom Harris had type AB blood and Mary had type O, could the type O child they brought home be their child? * No, the baby would have to have inherited a gene for O from both parents and Tom has only a gene for A or B. Yes, both parents had the recessive hidden gene for O blood in their genotype which they passed on to the baby. No; the baby would have to have Type AB blood that it inherited from the father since A and B are dominate to O. Yes; the baby inherited type O blood from the mother Mary Harris which is dominant to the genes for both A and B.

Answers

Answer 1
Answer:

The study of blood is called hematology. The component of blood is RBC, WBC, platelets, and plasma.

The correct answer to the question is option B.

What is a blood group?

  • A blood type is a classification of blood, based on the presence and absence of antibodies and inherited antigenic substances on the surface of red blood cells.
  • These antigens may be proteins, carbohydrates, glycoproteins, or glycolipids, depending on the blood group system.

The genes A and B are dominant over O therefore the O blood group can not be seen in offspring.

Hence, the correct answer to the question is option B is Yes, both parents had the recessive hidden gene for O blood in their genotype which they passed on to the baby. No; the baby would have to have Type AB blood that it inherited from the father since A and B are dominant to O.

For more information about the blood group, refer to the link:-

brainly.com/question/787658

Answer 2
Answer:

Explanation:

No, the baby would have to have inherited a gene for O from both parents and Tom has only a gene for A or B.


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Which organisms are most likely to survive in a population in which disruptive selection is occurring?

Answers

The answer would be : Organisms that have extreme traits.

Hope this helps !

Photon
"Organisms that have extreme traits" most likely to survive in a population in which disruptive selection is occurring

Hope this helps!

PLEASE HELP. I need the answer now

Answers

Explanation:

i didnt pay attention last year soooo sorry

Have this done correctly for brainiest​

Answers

LABELING

From Top to bottom

1) Ocular lens (Eye Piece)

__________________________________________________________

First Row Of the Labeling

2) Arms ( Right Box)

3) Revolving Nose Piece (left Box)

__________________________________________________________

Second Row Of the Labeling

4) Stage Clip (Right Box)

5) Objective Lenses (Left box)

__________________________________________________________

The Knobs in the Right

6) Course Focus Knob (the big Knob)

7) Fine Focus Knob (the small knob)

8) Mechanical Stage adjustment knobs (Small circular knobs in front of the course focus and fine focus knob)

__________________________________________________________

The Left Labeling below the objective lens

9) Stage

10) Condensor Lens

11) lamp or Illuminator

__________________________________________________________

Right Bottom 2 Labeling

12) Brightness adjustment

13) Base

__________________________________________________________

Functions

A) OCULAR LENS

B) MECHANICAL STAGE KNOBS

C) FOCUS KNOBS

D) MECHANICAL STAGE

E) STAGE CLIP

F) CONDENSOR LENS

G) LOW POWER OBJECTIVE LENS

H) BASE AND ARM

I) DIAPHRAGM

J) COARSE FOCUS KNOB

K) ILUMINATOR OR LAMP

L) HIGH POWER OBJECTIVE LENS

You want to make 50 ml of 1X tricaine solution in order to euthanize some fish. How much of a 20X tricaine stock solution will you need to dilute in order to make your solution?a. 1 ml b. 5 ml c. 0.4 ml d. 10 ml e. 2.5 ml

Answers

Answer: e. 2.5 ml

Explanation:

According to the dilution law,

C_1V_1=C_2V_2

where,

C_1 = concentration of stock solution = 20X

V_1 = volume of stock solution = ?

C_2 = concentration of required solution= 1X

V_2 = volume of  required solution= 50 ml

20* V_1=1* 50ml

V_2=2.5ml

Thus 2.5 ml much of a 20X tricaine stock solution is needed to dilute in order to make your solution.

Which situation probably involves autoantibodies?

Answers

Answer:

Biological markers

Homeostasis

House keeping

Explanation:

Autoantibodies are also reffered to as natural antibodies. This antibody react with self molecules in the body system to fight against infection.

This antibodies are produced by genes that are not mutated.

They provide the immediate defence against disease or infection and help in keeping the body fit.

They are useful in the regulation of body function or activities i.e homeostasis

Autoantibodies are useful as biological marker in targeting some disease that are present in the body as they could be specific to certain organs in the body.

) How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Answers

Answer:

5 * 10^(10)

Explanation:

The question is not complete. Remaining part of the question is as follows - Minimal growth medium for bacteria such as E. coli includes various salts with  characteristic concentrations in the mM range and a carbon source. The carbon source is  typically glucose and it is used at 0.5% (a concentration of 0.5 g/100 mL). For nitrogen,  minimal medium contains ammonium chloride (NH4Cl) with a concentration of 0.1g/100 mL

How many cells can be grown in a 5 mL culture using minimal medium before the medium exhausts the carbon?

Solution -

We will first find the mass concentration of 0.5 g/100 mL of solution.

(0.5)/(100) gram per ml of glucose

The chemical formula of glucose is C_6H_(12)O_6

The molecular weight of glucose molecule is 180 grams per mole

Now, we will find the number of moles of glucose in a 5 ml medium -

((0.5)/(100) * 5)/(180) \n1.39 * 10^(-4) mole

The number of carbon atom in each glucose molecule is equal to six, thus, number of minimal carbon mole is equal to

1.39 * 10^(-4) * 6\n= 8.34* 10^(-4)mole

Number of carbon atoms is equal to

8.34* 10^(-4) * 6.023 * 10^(23)\n= 5 * 10^(20)\n Carbons

One bacteria has 10^(10) carbon molecule.Thus, 5 ml medium will have 5 * 10^(10) bacteria