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leonid [27]
3 years ago
12

Albinism in humans is inherited as a simple recessive trait. Consider two families. In one family, two parents without albinism

have five children, four without albinism and one with albinism. In the other one, a male without albinism and a female with albinism have six children, all without albinism. Part A Which of Mendel's postulates do these families demonstrate?
Biology
1 answer:
alexandr402 [8]3 years ago
6 0

Mendel's law of Dominance is demonstrated by these families.

Explanation:

Mendel's Law of Dominance states that the recessive gene will not be expressed in front of Dominant gene. He found that in a heterozygous gene the allele for Dominant trait will mask the trait expressed by recessive trait, also one trait will overshadow the other trait both the traits will not appear in offspring.

Aa X Aa (In first family both parents are heterozygous)

AA X  aa ( In second family the father is dominant and mother is recessive and albino, but no child will be albino since all would be heterozygous for the trait.)

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The relationship between heredity and environment is
Lilit [14]
Not the same and they is not in a relationship
7 0
3 years ago
Three linked autosomal loci were studied in smurfs.
cupoosta [38]

Answer:

height -------- color --------- mood

           (13.2cM)      (14.5cM)

C=0.421

I = 0.579

Explanation:

We have the number of descendants of each phenotype product of the tri-hybrid cross.

Phenotype Number

  • pink, tall, happy            580
  • blue, dwarf, gloomy     601
  • pink, tall, gloomy         113
  • blue, dwarf, happy      107
  • blue, tall, happy              8
  • pink, dwarf, gloomy        6
  • blue, tall, gloomy          98
  • pink, dwarf, happy      101

Total number of individuals = 1614 = N

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the phenotypes of the parental with the ones of the double recombinants. We can recognize the parental in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

  • Pink, tall, happy            580 individuals
  • Blue, dwarf, gloomy      601 individuals

Simple recombinant)

  • Pink Tall Gloomy           113 individuals
  • Blue, Dwarf, Happy       107 individuals
  • Blue Tall Gloomy             98 individuals
  • Pink Dwarf Happy          101 individuals

Double Recombinant)  

  • Blue Tall Happy                 8 individuals
  • Pink  Dwarf Gloomy           6 individuals  

Comparing them we realize that parental and double recombinant individuals differ in the position of the gene codifying for <u>color</u><u>.</u> They only change in the position of Blue and Pink. This suggests that the position of the color gene is in the middle of the other two genes, height and mood, because in a double recombinant only the central gene changes position in the chromatid.  

So, the alphabetic order of the genes is:

---- height ---- color ----- mood ----

Now we will call Region I to the area between Height and Color, and Region II to the area between Color and Mood.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between Height and color genes, and P2 to the recombination frequency between color and mood.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region (the ones that have an intermediate phenotypic frequency), DR is the number of double recombinants in each region, and N is the total number of individuals.  So:

Region I

Tall------ Pink--------happy  (Parental) 580 individuals

Dwarf ---Pink------- Happy (Simple Recombinant) 101 individuals

Dwarf--- Pink-------Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Tall ------Blue------- Gloomy (Simple Recombinant)  98 individuals

Tall ----- Blue------- Happy   (Double Recombinant) 8 individuals  

Region II

Tall------ Pink--------happy (Parental) 580 individuals

Tall-------Pink------- Gloomy (Simple Recombinant) 113 individuals

Dwarf----Pink------- Gloomy (Double Recombinant) 6 individuals

Dwarf----Blue-------Gloomy (Parental) 601 individuals

Dwarf ----Blue-------Happy (Simple Recombinant) 107 individuals

Tall ----- Blue------- Happy   (Double Recombinant) 8 individuals

In each region, the highlighted traits are the ones that suffered recombination.

  • P1 = (R + DR) / N

P1 = (101+6+98+8)/1614

P1 = 213/1614

P1 = 0.132    

  • P2= = (R + DR) / N

P2 = (113+6+107+8)/1614

P1 = 234/1614

P1 = 0.145

Now, to calculate the recombination frequency between the two extreme genes, height and mood, we can just perform addition or a sum:

  • P1 + P2= Pt

0.132 + 0.145 = Pt

0.277=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).  

The map unit is the distance between the pair of genes for which every 100 meiotic products, one results in a recombinant product.  

Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.132 x 100 = 13.2 MU = 13.2 cM

GD2= P2 x 100 = 0.145 x 100 = 14.5 MU = 14.5 cM

GD3=Pt x 100 = 0.277 x 100 = 27.7 MU = 27.7 cM

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:  

-observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals

-expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

  • CC= ((6 + 8)/1614)/0.132x0.145

        CC=0.008/0.019

        CC=0.421

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.421

I = 0.579

8 0
3 years ago
A drug that inhibits the pumping of sodium and chloride ions out of the ascending limb of the loop of henle would result in
Elina [12.6K]
Hey there,

Question: <span>A drug that inhibits the pumping of sodium and chloride ions out of the ascending limb of the loop of henle would result in</span>

Answer: less water removed from the descending limb

Hope this helps :))

<em>~Top♥</em>
4 0
3 years ago
Viruses, bacteria, fungi, and parasites are all different types of infectious agents that may cause disease. What do these agent
valkas [14]

Answer:

Single celled? probably

3 0
3 years ago
Which of the following disorders results from the hypo-secretion of antidiuretic hormone (ADH)?
QveST [7]

Answer: I think the answer would be C

Explanation: Diabetes insipidus (DI) is a condition caused by hyposecretion of, or insensitivity to the effects of, antidiuretic hormone (ADH), also known as arginine vasopressin (AVP). ... Cranial DI: decreased secretion of ADH. Decreased secretion of ADH reduces the ability to concentrate urine and so causes polyuria and polydipsia.

Hope this helped

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8 0
3 years ago
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