Im not sure if this is right or not but i think it might be C.
Hope its right!!!!
Explanation:
The problem says that the hairless phenotype never breeds true. That means that it's not the result of a homozygous genotype (H₁H₁ or H₂H₂), so it is caused by the heterozygous genotype (H₁H₂).
The <u>expected </u>offspring from the cross between two Mexican hairless would be:
<h3>P
H₁H₂ x
H₁H₂</h3><h3>F1 1/4
H₁H₁, 2/4
H₁H₂ and 1/4
H₂H₂.</h3>
And the <u>expected</u> phenotypic ratio 3:1. However, the observed offspring shows a 2:1 ratio. What's happening?
If the observed phenotypic ratio in the offspring of a monohybrid cross (a single gene with two alleles) is 2:1, we can suspect that one of the genotypes is lethal in homozygosis and therefore does not appear in the progeny (the puppies are born dead).
If we proposed that the H₂ allele is lethal in homozygosis, then:
- The H₁H₁ genotype would cause normal puppies --> 1
- The H₁H₂ genotype would cause hairless puppies --> 2
- The H₂H₂ is lethal and causes the death of puppies --> 0
The phenotypic ratios change to 2:1, as observed in the experiment.
THE EXCRETORY SYSTEM
Answer: The attachment shows the nephron which is the functional unit of the kidney.
It does the work of urine formation through 3 distinct processes.
-Ultra filtration( Small molecules are forced out of the selectively permeable membrane of the glomerulus into the Bowman's capsule under regulated pressure.these molecules are from the blood in the glomerulus brought in by the afferent arteriole.
- Selective reabsorption ( Useful molecule and iron such as glucose and sodium are reabsorbed back into the blood as the filtrate flows through the tubule(nephron)
-Tubular Secretion. ( Movement of molecules not filtered by the glomerulus during the initial stage of filtration back into the filtrate through the renal capillaries.
Stella's urine sample shows the presence of large protein (ULTRAFILTRATION)
John's blood test report indicates a high toxin level ( ULTRAFILTRATION AND TUBULAR SECRETION)
Miguel's blood test shows an increase in metabolic waste( ULTRAFILTRATION AND TUBULAR SECRETION)
Janice's urine report shows the presence of vital materials ( SELECTIVE REABSORPTION).
Step 1: glucose is broken down into 2 molecules of pyruvate
Step 2: Completes breakdown of carbon dioxide, makes small amounts of ATP, provides electrons
Step 3: electron transport chain, chemiosmosis; energy from electrons-- produces 32 ATP
